Courses

Courses for Kids

Free study material

Store

Talk to our experts

1800-120-456-456

NCERT Solutions

Class 12

Maths

Chapter 5 Continuity And Differentiability

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

Q: 9. What is the use of continuity in real life?

Continuity is used in many real-world scenarios:Motion: Modeling the continuous movement of objects (e.g., a car accelerating)Signal transmission: Ensuring smooth flow of information in electrical circuits or communication networksEconomics: Analyzing trends in prices, market behavior, or economic growth over time

Complete Resource for NCERT Class 12 Maths Chapter 5 Solutions - Free PDF Download

The NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability cover all of the chapter's questions (All Exercises and Miscellaneous Exercise solutions). These NCERT Solutions for Class 12 Maths have been carefully compiled and created in accordance with the most recent CBSE Syllabus 2023-24 updates. Students can use these NCERT Solutions for Class 12 to reinforce their foundations. Subject experts at Vedantu has created the continuity and differentiability class 12 NCERT solutions to ensure they match the current curriculum and help students while solving or practicing problems.

Table of Content

Glance of NCERT Solutions for Class 12 Maths Chapter 5 | Vedantu

A function is considered continuous at a point if its output (y-value) changes smoothly as the input (x-value) approaches that point. There are no abrupt jumps or breaks.
The chapter explores different types of continuities - one-sided limits, two-sided limits, and continuity at a point.
Differentiability essentially means the function has a smooth, well-defined slope at each point. The slope is represented by the derivative of the function.
The chapter covers methods to find the derivative of various functions, including algebraic functions, exponential and logarithmic functions, trigonometric functions, inverse trigonometric functions, and functions defined parametrically.
This article contains chapter notes, exercises, links and important questions for Chapter 5 - Continuity and Differentiability which you can download as PDFs.
There are eight exercises and one miscellaneous exercise (144 fully solved questions) in class 12th maths chapter 5 Continuity and Differentiability.

Access Exercise Wise NCERT Solutions for Chapter 5 Maths Class 12

S.No.	Current Syllabus Exercises of Class 12 Maths Chapter 5
1	NCERT Solutions of Class 12 Maths Continuity and Differentiability Exercise 5.1
2	NCERT Solutions of Class 12 Maths Continuity and Differentiability Exercise 5.2
3	NCERT Solutions of Class 12 Maths Continuity and Differentiability Exercise 5.3
4	NCERT Solutions of Class 12 Maths Continuity and Differentiability Exercise 5.4
5	NCERT Solutions of Class 12 Maths Continuity and Differentiability Exercise 5.5
6	NCERT Solutions of Class 12 Maths Continuity and Differentiability Exercise 5.6
7	NCERT Solutions of Class 12 Maths Continuity and Differentiability Exercise 5.7
8	NCERT Solutions of Class 12 Maths Continuity and Differentiability Miscellaneous Exercise

Courses

More Free Study Material for Continuity and Differentiability

Revision notes

850.5k views • 10k downloads

Important questions

776.4k views • 12k downloads

Ncert books

787.8k views • 11k downloads

Exercises under NCERT Class 12 Maths Chapter 5 Continuity and Differentiability

Chapter 5 of NCERT Class 12 Maths textbook covers Continuity and Differentiability. The chapter is divided into nine exercises which are briefly summarized as follows:

Exercise 5.1: This exercise deals with the concept of continuity of a function, and how to check the continuity of a function using algebraic techniques.

Exercise 5.2: The Intermediate Value Theorem is introduced in this exercise, which is used to prove the existence of roots of equations.

Exercise 5.3: This exercise covers the concept of differentiability of a function, and the differentiability of algebraic and trigonometric functions are explored.

Exercise 5.4: The chain rule is introduced in this exercise, which is used to find the derivative of composite functions.

Exercise 5.5: This exercise covers the derivative of implicit functions and related rates problems.

Exercise 5.6: The derivatives of inverse trigonometric functions and logarithmic functions are explored in this exercise.

Exercise 5.7: This exercise covers the derivatives of functions expressed in parametric form.

Exercise 5.8: This exercise explores the concept of derivatives of functions expressed in polar form.

Miscellaneous Exercise: This exercise contains additional problems related to continuity and differentiability.

Overall, these exercises provide a comprehensive understanding of the concepts of continuity and differentiability, and how to apply them to solve various mathematical problems.

Access NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability

Exercise 5.1

1. Prove that the function $f (x) =5x-3$ is a continuous at $x=0$ , at $x=-3$ and at $x=5$ .

Ans: The given function is $f (x) =5x-3$ .

At $x=0$ , $f (0) =5 \times 0-3=-3$ .

Taking limit as $x \to 0$ both sides of the function give

$lim_{x \to 0} f (x) = lim_{x \to 0} (5x-3) =5 \times 0-3=-3$

$∴ lim_{x \to 0} f (x) =f (0)$ .

Thus, $f$ satisfies continuity at $x=0$ .

Again, at $x = -3, f (-3) =5 \times (-3) -3=-18$ .

Now, taking limit as $x \to 3$ both sides of the function give

$lim_{x \to 3} f (x) = lim_{x \to 3} f (5x-3) =5 \times (-3) -3=-18$

$∴ lim_{x \to 3} f (x) =f (-3)$ .

Therefore, $f$ satisfies continuity at $x=-3$

Also, at $x=5, f (x) =f (5) =5 \times 5-3=25-3=22$ .

Taking limit as $x \to 5$ both sides of the function give

$lim_{x \to 5} f (x) = lim_{x \to 5} (5x-3) =5 \times 5-3=22$

$∴ lim_{x \to 5} f (x) =f (5)$ .

Hence, $f$ satisfies continuity at $x=5$ .

2. Examine the continuity of the function $f (x) =2 x^{2} -1$ at $x=3$ .

Ans: The given function is $f (x) =2 x^{2} -1$ .

Now, at $x=3, f (3) =2 \times 3^{2} -1=17$ .

Taking limit as $x \to 3$ both sides of the function give

$lim_{x \to 3} f (x) = lim_{x \to 3} (2 x^{2} -1) =2 \times 3^{2} -1=17$

$∴ lim_{x \to 3} f (x) =f= (3)$ .

Hence, $f$ satisfies continuity at $x=3$ .

3. Examine the following functions for continuity.

(a) $f (x) =x-5$

Ans: The given function is $f (x) =x-5$ .

It is assured that for every real number $k$ , $f$ is defined and its value at $k$ is $k-5$ . Also, it can be noted that

$lim_{x \to k} f (x) = lim_{x \to k} f (x-5) =k=k-5=f (k)$ .

$∴ lim_{x \to k} f (x) =f (k)$

Hence, $f$ satisfies continuity at every real number and so, it is a continuous function.

(b) $f (x) = \frac{1}{x-5},x \neq 5$

Ans: The given function is

$f (x) = \frac{1}{x-5}$ .

Let $k \neq 5$ is any real number, then taking limit as $x \to k$ both sides of the function give

$lim_{x \to k} f (x) = lim_{x \to k} \frac{1}{x-5} = \frac{1}{k-5}$

Also, $f(k)= \frac{1}{k-5}$ , since $k \neq 5$

$∴ lim_{x \to k} f (x) =f (k)$

Therefore, $f$ satisfies continuity at every point in the domain of $f$ and so, it is a continuous function.

Ans: The given function is

$f (x) = \frac{x^{2} -25}{x+5}, x \neq 5$

Now let $c \neq -5$ be any real number, then taking limit as $x \to c$ on both sides of the function give

$lim_{x \to c} f (x) = lim_{x \to c} \frac{x^{2} -25}{x+5} = lim_{x \to c} \frac{(x+5) (x-5)}{x+5} = lim_{x \to c} (x-5) = (c-5)$

Again, $f (c) = \frac{(c+5) (c-5)}{c+5} = (c-5)$ , since $c \neq 5$ .

Hence, $f$ satisfies continuity at every point in the domain of $f$ and so it is a continuous function.

(d) $f (x) = | x-5 |$

Ans: The given function is $f (x) = | x-5 | = {\begin{aligned} 5-x, if x < 5 \\ x-5, if x \geq 5 \end{aligned}$ .

Note that, $f$ is defined at all points in the real line. So, let assume $c$ be a point on a real line.

Then, we have $c < 5$ or $c = 5$ or $c > 5$ .

Now, let's discuss these three cases one by one.

Case (i). $c < 5$

Then, the function becomes $f (c) =5-c$ .

Now, $lim_{x \to c} f (x) = lim_{x \to c} (5-x) =5-c$ .

$∴ lim_{x \to c} f (x) =f (c)$ .

Therefore, $f$ is continuous at all real numbers which are less than $5$ .

Case (ii). $c=5$

Then, $f (c) =f (5) = (5-5) =0$ .

Now,

$lim_{x \to 5^{-}} f (x) = lim_{x \to 5} (5-x) = (5-5) =0$ and

$lim_{x \to 5^{+}} f (x) = lim_{x \to 5} (x-5) =0$ .

Therefore, we have

$lim_{x \to c^{-}} f (x) = lim_{x \to c^{+}} f (x) =f (c)$ .

Thus, $f$ satisfies continuity at $x=5$ , and so $f$ is continuous at $x=5$ .

Case (iii). $c > 5$

Then we have, $f (c) =f (5) =c-5$ .

Now,

$lim_{x \to c} f (x) = lim_{x \to c} (x-5) =c-5$ .

Therefore,

$lim_{x \to c} f (x) =f (c)$ .

So, $f$ is continuous at all real numbers that are greater than $5$ .

Thus, $f$ satisfies continuity at every real number and hence, it is a continuous function.

4. Prove that the function $f (x) = x^{n}$ is continuous at $x=n$ , where $n$ is a positive integer.

Ans: The given function is $f (x) = x^{n}$ .

We noticed that the function $f$ is defined at all positive integers $n$ and also its value at $x=n$ is $n^{n}$ .

Therefore, $lim_{x \to n} f (n) = lim_{x \to n} f (x^{n}) = n^{n}$ .

So, $lim_{x \to n} f (x) =f (n)$ .

Thus, the function $f (x) = x^{n}$ is continuous at $x=n$ , where $n$ is a positive integer.

5. Is the function $f$ defined by

$f (x) = {\begin{aligned} x, if x \leq 1 \\ 5, if x > 1 \end{aligned}$

continuous at $x = 0$ , $x = 1$ ? At $x = 2$ ?

Ans: The given function is $f (x) = {\begin{aligned} x, if x \leq 1 \\ 5, if x > 1 \end{aligned}$

It is obvious that the function $f$ is defined at $x= 0$ and its value at $x= 0$ is $0$ .

Now, $lim_{x \to 0} f (x) = lim_{x \to 0} x=0$ .

So, $lim_{x \to 0} f (x) =f (0)$ .

Hence, the function $f$ satisfies continuity at $x=0$ .

It can be observed that $f$ is defined at $x=1$ and its value at this point is $1$ .

Now, the left-hand limit of the function $f$ at $x=1$ is

$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} x=1$ .

Also, the right-hand limit of the function $f$ at $x=1$ is

$lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} f (5)$

Therefore, $lim_{x \to 1^{-}} f (x) \neq lim_{x \to 1^{+}} f (x)$ .

Thus, $f$ is not continuous at $x=1$

It can be found that $f$ is defined at $x=2$ and its value at this point is $5$ .

That is, $lim_{x \to 2} f (x) = lim_{x \to 2} f (5) =5$ .

Therefore, $lim_{x \to 2} f (x) =f (2)$

Hence, $f$ satisfies continuity at $x=2$ .

Find all points of discontinuity of $f$ , where $f$ is defined by

6. $f (x) = {\begin{aligned} 2x+3, if x \leq 2 \\ 2x-3, if x > 2 \end{aligned}$ .

Ans: The given function is $f (x) = {\begin{aligned} 2x+3, if x \leq 2 \\ 2x-3, if x > 2 \end{aligned}$

It can be observed that the function $f$ is defined at all the points in the real line.

Let consider $c$ be a point on the real line. Then, three cases may arise.

I. $c < 2$

II. $c > 2$

III. $c = 2$

Case (i). When $c < 2$

Then, we have $lim_{x \to c} f (x) = lim_{x \to \infty} (2x+3) =2c+3$ .

Therefore,

$lim_{x \to c} f (x) =f (c)$ .

Hence, $f$ attains continuity at all points $x$ , where $x < 2$ .

Case (ii). When $c > 2$

Then, we have $f (c) =2c-3$ .

So,

$lim_{x \to c} f (x) = lim_{x \to \infty} (2x-3) =2c-3$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

Hence, $f$ satisfies continuity at all points $x$ , where $x > 2$ .

Case(iii). When $c=2$

Then, the left-hand limit of the function $f$ at $x=2$ is

$lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} (2x+3) =2 \times 2+3=7$ and

the right-hand limit of the function $f$ at $x=2$ is,

$lim_{x \to 2^{+}} f (x) = lim_{x \to 2 +} (2x+3) =2 \times 2-3=1$ .

Thus, at $x=2$ , $lim_{x \to 2^{-}} f (x) \neq lim_{x \to 2^{+}} f (x)$ .

So, the function $f$ does not satisfy continuity at $x=2$ .

Hence, $x=2$ is the only point of discontinuity of the function $f (x)$ .

7. $f (x) = {\begin{aligned} | x | + 3, if x \leq - 3 \\ - 2 x, if - 3 < x < 3 \\ 6 x + 2, if x \geq 3 \end{aligned}$

Ans: The given function is $f (x) = {\begin{aligned} | x | +3, if x \leq -3 \\ -2x, if -3 < x < 3 \\ 6x+2, if x \geq 3 \end{aligned}$

Observe that, $f$ is defined at all the points in the real line.

Now, let assume $c$ as a point on the real line.

Then five cases may arise. Either $c < -3$ , or $c = -3$ or $-3 < c < 3$ , or $c = 3$ , or $c > 3$ .

Let's discuss the five cases one by one.

Case I. When $c < -3$

Then, $f (c) =-c+3$ and

$lim_{x \to c} f (x) = lim_{x \to c} (-x+3) =-c+3$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

Hence, $f$ satisfies continuity at all points $x$ , where $x < -3$ .

Case II. When $c=-3$

Then, $f (-3) =- (-3) +3=6$ .

Also, the left-hand limit

$lim_{x \to - 3^{-}} f (x) = lim_{x \to - 3^{-}} (-x+3) =- (-3) +3=6$ .

and the right-hand limit

$lim_{x \to - 3^{+}} f (x) = lim_{x \to - 3^{+}} f (- 2 x) =-2 \times (-3) =6$ .

Therefore, $lim_{x \to - 3} f (x) =f (-3)$ .

Hence, $f$ satisfies continuity at $x=-3$ .

Case III. When $-3 < c < 3$ .

Then, $f (c) = - 2 c$ and also

$lim_{x \to c} f (x) = lim_{x \to c} (- 2 x) = - 2 c$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

Hence, $f$ satisfies continuity at $x$ , where $-3 < x < 3$ .

Case IV. When $c=3$

Then, the left-hand limit of the function $f$ at $x=3$ is

$lim_{x \to 3^{-}} f (x) = lim_{x \to 3^{-}} f (-2x) =-2 \times 3=-6$ and

the right-hand limit of the function $f$ at $x=3$ is

$lim_{x \to 3^{+}} f (x) = lim_{x \to 3^{+}} f (6x+2) =6 \times 3+2=20$ .

Thus, at $x=3$ , $lim_{x \to 3^{-}} f (x) \neq lim_{x \to 3^{+}} f (x)$ .

Hence, $f$ does not satisfy continuity at $x=3$ .

Case V. When $c > 3$ .

Then $f (c) =6c+2$ and also

$lim_{x \to c} f (x) = lim_{x \to c} (6x+2) =6c+2$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

So, $f$ satisfies continuity at all points $x$ , when $x > 3$ .

Thus, $x=3$ is the only point of discontinuity of the function $f$ .

8. $f (x) = {\begin{aligned} \frac{| x |}{x}, if x \neq 0 \\ 0, if x = 0 \end{aligned}$ .

Ans:

The given function is $f (x) = {\begin{aligned} \frac{| x |}{x}, if x \neq 0 \\ 0, if x = 0 \end{aligned}$ .

Now, $f (x)$ can be rewritten as

$f (x) = {\begin{aligned} \frac{| x |}{x} = \frac{- x}{x} = - 1 if x < 0 \\ 0, if x = 0 \\ \frac{| x |}{x} = \frac{x}{x} = 1 if x > 0 \end{aligned}$

It can be noted that the function $f$ is defined at all points of the real line.

Now, let assume $c$ as a point on the real line.

Then three cases may arise, either $c < 0$ , or $c = 0$ , or $c > 0$ .

Let's discuss three cases one by one.

Case I. When $c < 0$ .

Then, $f (c) =-1$ and

$lim_{x \to c} f (x) = lim_{x \to c} (-1) =-1$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

Hence, $f$ satisfies continuity at all the points $x$ where $x < 0$ .

Case II. When $c=0$ .

Then, the left-hand limit of the function $f$ at $x=0$ is

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (-1) =-1$ and

the right-hand limit of the function $f$ at $x=0$ is

$lim_{x \to 0 +} f (x) = lim_{x \to 0^{+}} (1) =1$ .

At $x=0$ , $lim_{x \to 0^{-}} f (x) \neq lim_{x \to 0^{+}} f (x)$ .

Hence, the function $f$ does not satisfy continuity at $x=0$ .

Case III. When $c > 0$ .

Then $f (c) =1$ and also

$lim_{x \to c} f (x) = lim_{x \to c} (1) =1$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

So, the function $f$ is continuous at all the points $x$ , for $x > 0$ .

Thus, $x=0$ is the only point of discontinuity for the function $f$ .

9. $f (x) = {\begin{aligned} \frac{x}{| x |}, if x < 0 \\ -1, if x \geq 0 \end{aligned}$

Ans: The given function is $f (x) = {\begin{aligned} \frac{x}{| x |}, if x < 0 \\ -1, if x \geq 0 \end{aligned}$

Now, we know that, if $x < 0$ , then $| x | =-x$ .

Therefore, the $f (x)$ can be written as

$f (x) = {\begin{aligned} \frac{x}{| x |} = \frac{x}{-x} = -1 if x < 0 \\ - 1, if x = 0 \\ - 1, if x > 0 \end{aligned}$

$\Rightarrow f (x) =-1$ for all positive real numbers.

Now, let's assume $c$ as any real number.

Then three cases may arise, either $c < 0$ , or $c = 0$ , or $c > 0$ .

Let's discuss three cases one by one.

Case I. When $c < 0$ .

Then, $f (c) =-1$ and

$lim_{x \to c} f (x) = lim_{x \to c} (-1) =-1$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

Hence, $f$ satisfies continuity at all the points $x$ where $x < 0$ .

Case II. When $c=0$ .

Then, the left-hand limit of the function $f$ at $x=0$ is

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (-1) =-1$ and

the right-hand limit of the function $f$ at $x=0$ is

$lim_{x \to 0 +} f (x) = lim_{x \to 0^{+}} (-1) =-1$ .

At $x=0$ , $lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{+}} f (x)$ .

Hence, the function $f$ satisfies continuity at $x=0$ .

Case III. When $c > 0$ .

Then $f (c) =-1$ and also

$lim_{x \to c} f (x) = lim_{x \to c} (-1) =-1$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

So, the function $f$ is continuous at all the points $x$ , for $x > 0$ .

Then, we have $lim_{x \to c} f(x) = lim_{x \to c} (- 1) = - 1$ and $f (c) =-1 = lim_{x \to e} f (x)$ .

Therefore, the function $f (x)$ is a continuous function.

Thus, there does not exist any point of discontinuity.

10. $f(x)= {\begin{aligned} x+1, if x \geq 1 \\ x^{2} +1, if x < 1 \end{aligned}$

Ans: The given function is $f(x)= {\begin{aligned} x+1, if x \geq 1 \\ x^{2} +1, if x < 1 \end{aligned}$ .

Note that, $f (x)$ is defined at all the points of the real line.

Now, let's assume $c$ as a point on the real line.

Then three cases may arise, either either $c < 1$ , or $c = 1$ , or $c > 1$ .

Let's discuss the three cases one by one.

Case I. When $c < 1$ .

Then, $f (c) = c^{2} +1$ and also

$lim_{x \to c} f (x) = lim_{x \to c} f (x^{2} +1) = c^{2} +1$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

Hence, $f$ satisfies continuity at all the points $x$ , where $x < 1$ .

Case II. When $c=1$ .

Then, we have $f (c) =f (1) =1+1=2$ .

Now, the left-hand limit of $f$ at $x=1$ is

$lim_{x \to 1^{-}} f(x)= lim_{x \to 1^{-}} (x^{2} +1)= 1^{2} +1=2$ and the right-hand limit of $f$ at $x=1$ is, $lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} (x+1) =1+1=2$ .

Therefore, $lim_{x \to 1} f (x) =f (c)$ .

Hence, $f$ satisfies continuity at $x=1$ .

Case III. When $c > 1$ .

Then, we have $f (c) =c+1$ and

$lim_{x \to c} f (x) = lim_{x \to c} (x+1) =c+1$ .

Therefore,

$lim_{x \to c} f (x) =f (c)$ .

So, $f$ satisfies continuity at all the points $x$ , where $x > 1$ .

Hence, there does not exist any discontinuity points.

11. $f (x) = {\begin{aligned} x^{3} - 3, if x \leq 2 \\ x^{2} + 1, if x > 2 \end{aligned}$

Ans: The given function is $f (x) = {\begin{aligned} x^{3} -3, if x \leq 2 \\ x^{2} +1, if x > 2 \end{aligned}$

Observe that, the function $f$ is defined at all points in the real line.

Now, let assume $c$ as a point on the real line.

Case I. When $c < 2$ .

Then, we have $f (c) = c^{3} -3$ and also $lim_{x \to c} f (x) = lim_{x \to c} (x^{3} -3) = c^{3} -3$ .

Therefore, the function $f$ attains continuity at all the points $x$ , where $x < 2$ .

Case II. When $c=2$ .

Then, we have $f (c) =f (2) = 2^{3} -3=5$ .

Now the left-hand limit of the function is

$lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} (x^{3} - 3) = 2^{3} -3=5$ and the right-hand limit is

$lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} (x^{2} +1) = 2^{2} +1=5$ .

Therefore, $lim_{x \to 2} f (x) =f (2)$ .

Hence, the function $f$ is continuous at $x=2$ .

Case III. When $c > 2$ .

Then, $f (c) = c^{2} + 1$ and

$lim_{x \to c} f (x) = lim_{x \to c} (x^{2} +1) = c^{2} +1$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

So, $f$ attains continuity at all the points $x$ , where $x > 2$ .

Thus, the function $f$ is continuous at all the points on the real line.

Hence, $f$ does not have any point of discontinuity.

12. $f (x) = {\begin{aligned} x^{10} -1, if x \leq 1 \\ x^{2}, if x > 1 \end{aligned}$ .

Ans: The given function is $f (x) = {\begin{aligned} x^{10} -1, if x \leq 1 \\ x^{2}, if x > 1 \end{aligned}$

Observe that, the function $f$ is defined at every point of the real line.

Now, let assume $c$ as a point on the real number line.

Case I. When $c < 1$ .

Then $f (c) = c^{10} -1$ .

Also, $lim_{x \to c} f (x) = lim_{x \to c} (x^{10} -1) = c^{10} -1$

Therefore, $lim_{x \to c} f (x) =f (c)$ .

Hence, the function $f$ attains continuity at every point $x$ , for $x < 1$ .

Case II. When $c=1$ .

Then the left-hand limit of the function $f (x)$ at $x=1$ is

$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (x^{10} - 1) = 1^{10} - 1 = 1 - 1 = 0$ and

the right-hand limit of the function $f$ at $x=1$ is

$lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} (x^{2}) = 1^{2} =1$ .

So, we can notice that, $lim_{x \to 1^{-}} f (x) \neq lim_{x \to 1^{+}} f (x)$ .

Hence, the function $f$ does not satisfy continuity at $x=1$ .

Case III. When $c > 1$ .

Then, $f (c) = c^{2}$ .

Also, $lim_{x \to c} f (x) = lim_{x \to c} (x^{2}) = c^{2}$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

Thus, the function $f$ attains continuity at every point $x$ , for $x > 1$ .

Hence, we can conclude that $x=1$ is the only point of discontinuity for the function $f$ .

13. Is the function defined by $f (x) = {\begin{aligned} x+5, if x \leq 1 \\ x-5, if x > 1 \end{aligned}$ a continuous function?

Ans: The given function is $f (x) = {\begin{aligned} x+5, if x \leq 1 \\ x-5, if x > 1 \end{aligned}$ .

It can be noted that the function $f$ is defined at every point on the real line.

Now, let assume $c$ as a point on the real line.

Case I. When $c < 1$ .

Then, $f (c) =c+5$ .

Also, $lim_{x \to c} f (x) = lim_{x \to c} (x+5) =c+5$ .

Hence, $f$ satisfies continuity at every point $x$ , for $x < 1$ .

Case II. When $c=1$ .

Then, $f (1) =1+5=6$ .

Now, the left-hand limit of the function $f$ at $x=1$ is

$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (x+5) =1+5=6$ and the right-hand limit of the function at $x=1$ is $lim_{x \to 1^{+}} f(x)= lim_{x \to 1^{+}} (x-5)=1-5=4$ .

Thus, it is seen that, $lim_{x \to 1^{-}} f (x) \neq lim_{x \to 1^{+}} f (x)$ .

Hence, $f$ does not attain continuity at $x=1$ .

Case III. When $c > 1$ .

Then $f (c) =c-5$ .

Also, $lim_{x \to c} f (x) = lim_{x \to c} (x-5) =c-5$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

Thus, the function $f$ is continuous at every point $x$ , for $x > 1$ .

Hence, we can conclude that $x=1$ is the only point of discontinuity for the function $f$ .

14. Discuss the continuity of the function $f$ , where $f$ is defined by

$f (x) = {\begin{aligned} 3, if 0 \leq x \leq 1 \\ 4, if 1 < x < 3 \\ 5, if 3 \leq x \leq 10 \end{aligned}$ .

Ans: The given function is $f (x) = {\begin{aligned} 3, if 0 \leq x \leq 1 \\ 4, if 1 < x < 3 \\ 5, if 3 \leq x \leq 10 \end{aligned}$ .

Therefore, $f$ is defined in the interval $[0,10]$ .

Now let's assume $c$ as a point in the interval $[0,10]$ .

Then there may arise five cases.

Case I. When $0 \leq c < 1$ .

Then $f (c) =3$ .

Also, $lim_{x \to c} f (x) = lim_{x \to c} (3) =3$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

Hence, the function $f$ attains continuity at the interval $[0,1]$ .

Case II. When $c=1$ .

Then $f (3) =3$ .

Also, the left-hand-limit of the function at $x=1$ is

$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (3) =3$ and the right-hand-limit of the function at $x=1$ is

$lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} (4) =4$ .

Thus, it is noticed that $lim_{x \to 1^{-}} f (x) \neq lim_{x \to 1^{+}} f (x)$ .

Hence, the function $f$ does not satisfy continuity at $x=1$ .

Case III. When $1 < c < 3$ .

Then $f (c) =4$ .

Also, $lim_{x \to c} f (x) = lim_{x \to c} (4) =4$ .

Thus, $lim_{x \to c} f (x) =f (c)$ .

Hence, the function $f$ attains continuity at every point in the interval $[1,3]$ .

Case IV. When $c=3$ .

Then $f (c) =5$ .

Now, the left-hand-limit of the function $f$ at $x=3$ is

$lim_{x \to 3^{-}} f (x) = lim_{x \to 3^{-}} (4) =4$ and the right-hand-limit of the function $f$ at $x=3$ is

$lim_{x \to 3^{+}} f (x) = lim_{x \to 3^{+}} (5) =5$ .

Therefore, it is noted that $lim_{x \to 3^{-}} f (x) \neq lim_{x \to 3^{+}} f (x)$ .

Hence, the function $f$ is not continuous at $x=3$ .

Case V. When $3 < c \leq 10$ .

Then $f (c) =5$ .

Also, $lim_{x \to c} f (x) = lim_{x \to c} (5) =5$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

So, the function $f$ attains continuity at every point in the interval $[3,10]$ .

Hence, the function $f$ is not continuous at $x=1$ and $x=3$ .

15. $f (x) = {\begin{aligned} 2x, if x < 0 \\ 0, if 0 \leq x \leq 1 \\ 4x, if x > 1 \end{aligned}$ .

Ans: The given function is

$f (x) = {\begin{aligned} 2x, if x < 0 \\ 0, if 0 \leq x \leq 1 \\ 4x, if x > 1 \end{aligned}$

Now, let consider $c$ be a point on the real number line.

Then, five cases may arrive.

Case I. When $c < 0$ .

Then, $f (c) =2c$ .

Also, $lim_{x \to c} f (x) = lim_{x \to c} (2x) =2c$ .

Therefore, $lim_{x \to c} f (x) =f (c)$ .

Hence, the function $f$ attains continuity at every point $x$ whenever $x < 0$ .

Case II. When $c = 0$ .

Then, $f (c) =f (0) =0$ .

Now, the left-hand-limit of the function $f$ at $x = 0$ is

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (2x) = 0$ and the right-hand limit of the function $f$ at $x = 0$ is,

$lim_{x \to 0^{+}} (x) = lim_{x \to 0^{+}} (0) =0$ .

Therefore, $lim_{x \to 0} f (x) =f (0)$ .

Thus, the function $f$ attains continuity at $x = 0$ .

Case III. When $0 < c < 1$

Then, $f(x)=0$ .

Also, $lim_{x \to c} f(x)= lim_{x \to c} (0)=0$ .

Therefore, $lim_{x \to c} f(x)=f(c)$ .

Hence, $f$ attains continuity at every point in the interval $(0,1)$ .

Case IV. When $c =1$ .

Then, $f(c)=f(1)=0$ .

Now, the left-hand-limit at $x = 1$ is

$lim_{x \to 1^{-}} f(x)= lim_{x \to 1^{-}} (0)=0$ and the right-hand-limit at $x = 1$ is

$lim_{x \to 1^{+}} f(x)= lim_{x \to 1^{+}} (4x)=4 \times 1=4$ .

Thus, it is noticed that, $lim_{x \to 1^{-}} f(x) \neq lim_{x \to 1^{+}} f(x)$ .

Hence, the function $f$ is not continuous at $x = 1$ .

Case V. When $c > 1$ .

Then, $f(c)=f(1)=0$ .

Also, $lim_{x \to c} f(x)= lim_{x \to c} (4x)=4c$

Therefore, $lim_{x \to c} f(x)=f(c)$ .

So, the function $f$ attains continuity at every point $x$ , for $c > 1$ .

Hence, the function $f$ is discontinuous only at $x = 1$ .

16. $f(x)= {\begin{aligned} -2, if x \leq -1 \\ 2x, if -1 < x \leq 1 \\ 2, if x > 1 \end{aligned}$ .

Ans: The given function is $f(x)= {\begin{aligned} -2, if x \leq -1 \\ 2x, if -1 < x \leq 1 \\ 2, if x > 1 \end{aligned}$

Note that, $f$ is defined at every point in the interval $[- 1, \infty)$ .

Now, let assume $c$ is a point on the real number line.

Case I. When $c < - 1$ .

Then, $f(c)=-2$ .

Also, $lim_{x \to c} f(x)= lim_{x \to c} (-2)=-2$ .

Therefore, $lim_{x \to c} f(x)=f(c)$ .

Hence, the function $f$ attains continuity at every point $x$ , for $x < - 1$ .

Case II. When $c=-1$ .

Then, $f(c)=f(-1)=-2$ .

Now, the left-hand-limit of the function at $x=-1$ is

$lim_{x \to 1^{-}} f(x)= lim_{x \to 1^{-}} (-2)=-2$ and the right-hand-limit at $x=-1$ is

$lim_{x \to 1^{+}} (x)= lim_{x \to 1^{+}} =2 \times (-1)=-2$ .

Therefore, $lim_{x \to - 1} f(x)=f(-1)$ .

Hence, the function $f$ satisfies continuity at $x=-1$ .

Case III. When $-1 < c < 1$ .

Then, $f(c)=2c$ and $lim_{x \to c} f(x)= lim_{x \to c} (2x)=2c$

Therefore, $lim_{x \to c} f(x)=f(c)$ .

Hence, the function $f$ attains continuity at every point in the interval $(-1,1)$ .

Case IV. When $c=1$ .

Then, $f(c)=f(1)=2 \times 1=2$

Now, the left-hand-limit of the function at $x = 1$ is

$lim_{x \to 1^{-}} f(x)= lim_{x \to 1^{-}} (2x)=2 \times 1=2$ and the right-hand-limit at $x = 1$ is

$lim_{x \to 1^{+}} f(x)= lim_{x \to 1^{+}} 2=2$ .

Therefore, $lim_{x \to 1} f(x)=f(c)$ .

Thus, the function $f$ attains continuity at $x=2$ .

Case V. When $c > 1$ .

Then $f(c)=2$ .

Also, $lim_{x \to 2} f(x)= lim_{x \to 2} (2)=2$ .

Therefore, $lim_{x \to c} f(x)=f(c)$ .

Hence, the function $f$ is continuous at every point $x$ , for $x > 1$ .

Thus, it can be concluded that the function $f$ is continuous for all the points.

17. Find the relationship between $a$ and $b$ so that the function $f$ defined by $f(x)= {\begin{aligned} ax+1, if x \leq 3 \\ bx+3, if x > 3 \end{aligned}$ is continuous at $x=3$ .

Ans: The given function is $f(x)= {\begin{aligned} ax+1, if x \leq 3 \\ bx+3, if x > 3 \end{aligned}$ .

The function $f$ will be continuous at $x = 3$ if

$lim_{x \to 3^{-}} f(x)= lim_{x \to 3^{-}} f(x)=f(3)$ , ..….. (1)

$lim_{x \to 3^{-}} f(x)= lim_{x \to 3^{-}} f(ax+1)=3a+1$ ,

$lim_{x \to 3^{+}} f(x)= lim_{x \to 3^{+}} f(bx+1)=3b+3$ , …… (2)

and

$f(3)=3a+1$ . …… (3)

Therefore, from the equation (1), (2), and (3) gives

$\Rightarrow 3a+1=3b+3$

$\Rightarrow 3a=3b+2$

$\Rightarrow a=b+ \frac{2}{3}$

Hence, the required relationship between $a$ and $b$ is given by $a=b+ \frac{2}{3}$ .

18. For what value of $λ$ is the function defined by $f (x) = {\begin{aligned} λ (x^{2} - 2 x), if x \leq 0 \\ 4 x + 1, if x > 0 \end{aligned}$ is continuous at $x = 0$ . What about continuity at $x = 1$ ?

Ans: The given function is $f(x)= {\begin{aligned} λ (x^{2} -2x), if x \leq 0 \\ 4x+1, if x > 0 \end{aligned}$

Now the function will be continuous at $x = 0$ if

$lim_{x \to 0^{-}} f(x)= lim_{x \to 0^{+}} f(x)=f(0)$ .

Also, the R.H.L and L.H.L are given by,

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} (4 x + 1) = 4 (0) + 1 = 1$ ,

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} λ (x^{2} -2x) = λ (0^{2} -2 \times 0)=0$ .

So, $lim_{x \to 0^{+}} f (x) \neq lim_{x \to 0^{-}} f (x)$ .

Thus, there does not exist any value of $λ$ for which $f$ is continuous at $x = 0$ .

Now, at $x = 1$ ,

$f(1)=4x+1=4 \times 1+1=5$ and

$lim_{x \to 1} (4x+1)=4 \times 1+1=5$ .

Therefore, $lim_{x \to 1} f(x)=f(1)$ .

Hence, the function $f$ is continuous at $x = 1$ , for all values of $λ$ .

19. Show that the function defined by $g(x)=x- [x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$ .

Ans: The given function is $g(x)=x- [x]$ .

Note that the function is defined at every integral point.

Now, let assume that $n$ is an integer.

Then, $g(n)=n- [n] =n-n=0$ .

Now taking left-hand-limit as $x \to n$ to the function $g$ gives

$lim_{x \to n^{-}} g(x)= lim_{x \to n^{-}} [x- [x]] = lim_{x \to n^{-}} (x) - lim_{x \to n-} [x] =n- (n-1) =1$ .

Again, the right-hand-limit on the function at $x=n$ is

$lim_{x \to n^{+}} g(x)= lim_{x \to n^{+}} [x- [x]] = lim_{x \to n^{+}} (x) - lim_{x \to n^{+}} [x] =n-n=0$ .

Note that, $lim_{x \to n^{-}} g(x) \neq lim_{x \to n^{+}} g(x)$ .

Thus, the function $f$ is cannot be continuous at $x=n,$

Hence, the function $g$ is not continuous at any integral point.

20. Is the function defined by $f(x)= x^{2} -sinx+5$ continuous at $x= π$ ?

Ans: The given function is $f(x)= x^{2} -sinx+5$ .

Now, at $x= π$ ,

$f(x) = f(π) = π^{2} -sin π + 5 = π^{2} -0+5 = π^{2} +5$

Taking limit as $x \to π$ on the function $f (x)$ gives

$lim_{x \to π} f(x)= lim_{x \to π} (x^{2} -sinx+5)$ .

Now substitute $x= π +h$ into the function $f (x)$ .

When $x \to π$ , then $h \to 0$ .

Therefore,

$lim_{x \to π} f(x)= lim_{x \to π} (x^{2} -sinx) +5$ .

$= lim_{h \to 0} [(π + h^{2}) -sin (π +h) +5]$

$= lim_{h \to 0} {(π +h)}^{2} - lim_{h \to 0} sin (π +h) + lim_{h \to 0} 5$

$= {(π +0)}^{2} - lim_{h \to 0} [sin π \cdot cosh+cos π \cdot sinh] +5$

$= π^{2} - lim_{h \to 0} (sin π \cdot cosh) - lim_{h \to 0} (cos π \cdot sinh) +5$

$= π^{2} - sin π \cdot cos0 - cos π \cdot sin0+5$

$= π^{2} - 0 \times 1 - (- 1) \times 0+5= π^{2} +5$ .

So, $lim_{x \to x} f(x)=f(π)$ .

Hence, it is concluded that the function $f$ is continuous at $x=n$ .

21. Discuss the continuity of the following functions:

(a) $f (x) = sinx + cosx$

Ans. Suppose that, ‘ $a$ ’ is any real number. Therefore,

$lim_{x \to a^{+}} f (x) = lim_{h \to 0} f (a + h)$

Now,

$\begin{aligned} lim_{h \to 0} f (a + h) = lim_{h \to 0} [\sin (a + h) + \cos (a + h)] \\ = lim_{h \to 0} [\sin a \cdot \cosh + \cos a \cdot \sinh + \cos a \cdot \cosh - \sin a \cdot \sinh] \\ = \sin a \cdot \cos 0 + \cos a \cdot \sin 0 + \cos a \cdot \cos 0 - \sin a \cdot \sin 0 \end{aligned}$

$= \sin a + \cos a$ ( since, $\cos 0 = 1$ and $\sin 0 = 0$ )

$= f (a)$ .

In a similar way, it can be shown that,

$lim_{x \to a^{-}} = f (a)$

Thus, $lim_{x \to a^{+}} f (x) = f (a) = lim_{x \to a^{-}} f (x)$ .

Hence, $f (x)$ is continuous at $x = a$ , $\forall a \in R$ , that is,

the function $f(x)=sinx+cosx$ is continuous.

(b) $f (x) = sinx - cosx$

Ans. Suppose that, ‘ $a$ ’ is any real number. Therefore,

$lim_{x \to a^{+}} f (x) = lim_{h \to 0} f (a + h)$

Now,

$\begin{aligned} lim_{h \to 0} f (a + h) = lim_{h \to 0} [\sin (a + h) - \cos (a - h)] \\ = lim_{h \to 0} [\sin a \cdot \cosh + \cos a \cdot \sinh - \cos a \cdot \cosh - \sin a \cdot \sinh] \\ = \sin a \cdot \cos 0 + \cos a \cdot \sin 0 - \cos a \cdot \cos 0 - \sin a \cdot \sin 0 \end{aligned}$

$= \sin a - \cos a$ [ since, $\cos 0 = 1$ and $\sin 0 = 0$ ]

$= f (a)$ .

In a similar way, it can be shown that,

$lim_{x \to a^{-}} = f (a)$

Thus, $lim_{x \to a^{+}} f (x) = f (a) = lim_{x \to a^{-}} f (x)$ .

Hence, $f (x)$ is continuous at $x = a$ , $\forall a \in R$ , that is,

the function $f(x)=sinx - cosx$ is continuous.

Ans. Suppose that, ‘ $a$ ’ is any real number. Therefore,

$lim_{x \to a^{+}} f (x) = lim_{h \to 0} f (a + h)$ .

Now,

$\begin{aligned} lim_{h \to 0} f (a + h) = lim_{h \to 0} [\sin (a + h) \cdot \cos (a + h)] \\ = lim_{h \to 0} (\sin a \cdot \cosh + \cos a \cdot \sinh) (\cos a \cdot \cosh - \sin a \cdot \sinh) \\ = (\sin a \cdot \cos 0 + \cos a \cdot \sin 0) (\cos a \cdot \cos a - \sin a \cdot \sin 0) \\ = (\sin a + 0) (\cos a - 0) \\ = \sin a \cdot \cos a \end{aligned}$

$= f (a)$ .

In a similar way, it can be shown that,

$lim_{x \to a^{-}} f (x) = f (a)$

Thus, $lim_{x \to a^{+}} f (x) = f (a) = lim_{x \to a^{-}} f (x)$ .

Hence, $f (x)$ is continuous at $x = a$ , $\forall a \in R$ , that is, the function $f(x)=sinx \cdot cosx$ is continuous.

22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Ans:

Sine function:

It can be observed that the function $g(x)=sinx$ is defined for all real numbers.

Now, let's consider $c$ to be a real number and substitute $x=c+h$ into the function $g$ .

When, $x \to c$ , then $h \to 0$ .

So, $g(c)=sinc$ and

$\begin{aligned} lim_{x \to c} g(x)= lim_{x \to c} sinx \\ = lim_{h \to 0} sin (c+h) \\ = lim_{h \to 0} [sinc \cdot cosh+cosc \cdot sinh] \\ = lim_{h \to 0} (sinc \cdot cosh) + lim_{h \to 0} (cosc \cdot sinh) \\ =sinc \cdot cos0+cosc \cdot sin0 \\ =sinc+0 \\ =sinc \end{aligned}$

Therefore, $lim_{x \to c} g(x)=g(c)$ .

Thus, the function $g (x) =sinx$ is continuous.

Cosine function:

let $h(x)=cosx$ .

It can be noted that $h(x)=cosx$ is defined for all real numbers.

Now, let's consider $c$ to be a real number and substitute $x=c+h$ into the function $h$ .

Then, $h(c)=cosc$ and

$\begin{aligned} lim_{x \to c} h(x)= lim_{x \to c} cosx \\ = lim_{h \to 0} cos (c+h) \\ = lim_{h \to 0} [cosc \cdot cosh-sinc \cdot sinh] \\ = lim_{h \to 0} (cosc \cdot cosh) - lim_{h \to 0} (sinc \cdot sinh) \\ =cosc \cdot cos0 - sinc \cdot sin0 \\ =cosc \times 1 - sinc \times 0 \\ =cosc \end{aligned}$

Therefore, $lim_{h \to 0} h(x)=h(c)$ .

Thus, the function $h(x)=cosx$ is continuous.

Cosecant function:

We know that if a function $g (x)$ is continuous, then

$\frac{1}{g(x)},g(x) \neq 0$ is continuous.

Now, let $g (x) = \sin x$ .

Then, $cosec x= \frac{1}{sinx},$ and $sinx \neq 0$ .

Therefore, $\cos e c x = \frac{1}{\sin x}$ , when $x \neq n π$ , $n \in N$ .

Since, the function $\sin x$ is continuous, so $f (x) = cosec x$ is continuous for all $x$ , except at $x = n π, n \in N$ .

Secant function:

We know that if a function $h (x)$ is continuous, then

$\frac{1}{h(x)},h(x) \neq 0$ is continuous.

Now, let $h (x) = \cos x$ .

Again, $secx = \frac{1}{cosx},$ and $cosx \neq 0$ .

Therefore, $\sec x = \frac{1}{\cos x}$ and $x \neq (2 n + 1) \frac{π}{2}$

Since, the function $\cos x$ is continuous, so the function $f (x) = \sec x$ is continuous for all $x$ , except at $x = (2n+1) \frac{π}{2}, n \in N$ .

Cotangent function:

We know that if two functions say $g$ and $h$ are continuous, then

$\frac{h(x)}{g(x)},g(x) \neq 0$ is continuous.

Now, let $h (x) = \cos x$ and $g (x) = \sin x$ .

Then, $cotx = \frac{cosx}{sinx}$ and $sinx \neq 0$ .

That is, $\cot x = \frac{\cos x}{\sin x}$ and $x \neq n π, n \in N$ .

Since, $\cos x$ and $\sin x$ are both continuous, so the function $f (x) = \cot x$ is continuous except at $x = n π, n \in N$ .

23.Find all points of discontinuity of $f$ , where $f(x)= {\begin{aligned} \frac{sinx}{x}, if x < 0 \\ x+1, if x \geq 0 \end{aligned}$ .

Ans: The given function is $f(x)= {\begin{aligned} \frac{sinx}{x}, if x < 0 \\ x+1, if x \geq 0 \end{aligned}$ .

Note that, the function $f$ is defined at every point on the real number line.

Now, let's consider $c$ be a real number.

Then there may arise three cases, either $c < 0$ , or $c > 0$ , or $c = 0$ .

Let us discuss one after another.

Case I. When $c < 0$ .

Then, $f(c)= \frac{sinc}{c}$ .

Also, $lim_{x \to c} f(x) (\frac{sinx}{x}) = \frac{sinc}{c}$ .

Therefore, $lim_{x \to c} f(x)=f(c)$ .

Hence, the function $f$ is continuous at every point $x$ , for $x0$ .

Case II. When $c > 0$ .

Then $f(c)=c+1$ .

Also, $lim_{x \to c} f(x)= lim_{x \to c} (x+1)=c+1$ .

Therefore, $lim_{x \to c} f(x)=f(c)$ .

Hence, the function $f$ is continuous at every point, where $x0$ .

Case III. When $c = 0$ .

Then $f(c)=f(0)=0+1=1$ .

Now, the left-hand-limit of the function $f$ at $x=0$ is

$lim_{x \to 0^{-}} f(x)= lim_{x \to 0^{-}} \frac{sinx}{x} =1$ and the right-hand-limit is

$lim_{x \to 0^{-}} f(x)= lim_{x \to 0^{-}} (x+1)=1$

Therefore, $lim_{x \to 0^{-}} f(x)= lim_{x \to 0^{-}} f(x)=f(0)$ .

So, the function $f$ is continuous at $x = 0$ .

Thus, the function $f$ is continuous at every real point.

Hence, the function $f$ does not have any point of discontinuity.

24. Determine if $f(x) = {\begin{aligned} x^{2} sin \frac{1}{x}, if x \neq 0 \\ 0, if x = 0 \end{aligned}$

is a continuous function?

Ans:

The given function is $f(x) = {\begin{aligned} x^{2} sin \frac{1}{x}, if x \neq 0 \\ 0, if x = 0 \end{aligned}$

We can observe that the function $f$ is defined at every point on the real number line.

Now, let's consider $c$ to be a real number.

Then, there may arise two cases, either $c \neq 0$ or $c=0$ .

Let us discuss the cases one after another.

Case I. When $c \neq 0$ .

Then $f(c)= c^{2} sin \frac{1}{c}$ .

Also,

$lim_{x \to c} f(x)= lim_{x \to c} (x^{2} sin \frac{1}{x}) = (lim_{x \to c} x^{2}) (lim_{x \to c} sin \frac{1}{x}) = c^{2} sin \frac{1}{c}$ .

Therefore, $lim_{x \to c} f(x)=f(c)$ .

Hence, the function $f$ is continuous at every point $x \neq 0$ .

Case II. When $c = 0$ .

Then $f(0)=0$ and also

$lim_{x \to 0^{-}} f(x)= lim_{x \to 0^{-}} (x^{2} sin \frac{1}{x}) = lim_{x \to 0} (x^{2} sin \frac{1}{x})$

Now, we know that,

$-1 \leq sin \frac{1}{x} \leq 1, x \neq 0$ .

$\begin{aligned} \Rightarrow - x^{2} \leq sin \frac{1}{x} \leq x^{2} \\ \Rightarrow lim_{x \to 0} (- x^{2}) \leq lim_{x \to 0} (x^{2} sin \frac{1}{x}) \leq 0 \\ \Rightarrow 0 \leq lim_{x \to 0} (x^{2} sin \frac{1}{x}) \leq 0 \\ \Rightarrow lim_{x \to 0} (x^{2} sin \frac{1}{x}) = 0 \end{aligned}$

Therefore, $lim_{x \to 0^{-}} f(x)=0$ .

Similarly, we have,

$lim_{x \to 0^{+}} f(x)= lim_{x \to 0^{+}} (x^{2} sin \frac{1}{x}) = lim_{x \to 0} (x^{2} sin \frac{1}{x}) =0$

Therefore, $lim_{x \to 0^{-}} f(x)=f(0)= lim_{x \to 0^{+}} f(x)$ .

Thus, the function $f$ is continuous at the point $x = 0$ .

So, the function $f$ is continuous at all real points.

Hence, the function $f$ is continuous.

25. Examine the continuity of $f$ , where $f$ is defined by $f(x)= {\begin{aligned} sinx-cosx, if x \neq 0 \\ 1 if x=0 \end{aligned}$ .

Ans: The given function is` $f(x)= {\begin{aligned} sinx-cosx, if x \neq 0 \\ 1 if x=0 \end{aligned}$ .

It can be observed that the function $f$ is defined at every point on the real number line.

Now, let's consider $c$ to be a real number.

Then, there may arise two cases, either $c \neq 0$ or $c=0$ .

Let us discuss the cases one after another.

Case I. When $c \neq 0$ .

Then, $f(c)=sinc-cosc$ .

Also, $lim_{x \to c} f(x)= lim_{x \to c} (sinx-cosx)=sinc-cosc$ .

Therefore, $lim_{x \to c} f(x)=f(c)$ .

Hence, the function $f$ is continuous at every point $x$ for $x \neq 0$ .

Case II. When $c =0$ .

Then, $f(0)=-1$ .

Now the left-hand-limit of the function $f$ at $x=0$ is

$lim_{x \to 0^{-}} f(x) = lim_{x \to 0^{-}} (sinx-cosx)=sin0-cos0=0-1=-1$ and the right-hand-limit is

$lim_{x \to 0^{+}} f(x)= lim_{x \to 0^{+}} (sinx-cosx)=sin0-cos0=0-1=-1$ .

Therefore, $lim_{x \to 0^{-}} f(x)= lim_{x \to 0^{+}} f(x)=f(0)$ .

So, the function $f$ is continuous at $x = 0$ .

Thus, the function $f$ is continuous at all real points.

Hence, the function $f$ is continuous.

Find the values of $k$ so that the function $f$ is continuous at the indicated point in Exercises 26 to 29.

26. $f(x) = {\begin{aligned} \frac{kcosx}{π -2x}, if x \neq \frac{π}{2} \\ 3, if x= \frac{π}{2} \end{aligned}$

at $x = \frac{π}{2}$ .

Ans: The given function is $f(x) = {\begin{aligned} \frac{kcosx}{π -2x}, if x \neq \frac{π}{2} \\ 3, if x= \frac{π}{2} \end{aligned}$ .

Observe that, $f$ is defined and continuous at $x= \frac{π}{2}$ , since the value of the $f$ at $x= \frac{π}{2}$ is equal with the limiting value of $f$ at $x= \frac{π}{2}$ .

Since, $f$ is defined at $x= \frac{π}{2}$ and $f (\frac{π}{2}) =3$ , so

$lim_{x \to \frac{π}{2}} f(x) = lim_{x \to \frac{π}{2}} \frac{kcosx}{π -2x}$ .

Substitute $x= \frac{π}{2} +h$ into the function $f (x)$ .

So, we have, $x \to \frac{π}{2} \Rightarrow h \to 0$ .

Then,

$lim_{x \to \frac{π}{2}} f(x)= lim_{x \to \frac{π}{2}} \frac{kcosx}{π -2x} = lim_{h \to 0} \frac{kcos (\frac{π}{2} +h)}{π -2 (\frac{π}{2} +h)}$ .

$\Rightarrow k lim_{h \to 0} \frac{-sinh}{-2h} = \frac{k}{2} lim_{h \to 0} \frac{sinh}{h} = \frac{k}{2} .1 = \frac{k}{2}$

Therefore, $lim_{x \to \frac{π}{2}} f(x)=f (\frac{π}{2})$

$\begin{aligned} \Rightarrow \frac{k}{2} =3 \\ \Rightarrow k=6 \end{aligned}$

Hence, the value of $k$ is $6$ for which the function $f$ is continuous.

27. $f(x)= {\begin{aligned} k x^{2}, if x \leq 2 \\ 3, if x > 2 \end{aligned}$ at $x = 2$ .

Ans: The given function is $f(x)= {\begin{aligned} k x^{2}, if x \leq 2 \\ 3, if x > 2 \end{aligned}$

It is known that, $f$ is continuous at $x = 2$ only if $f$ is defined at $x=2$ and if the value of $f$ at $x = 2$ is equal with the limiting value of $f$ at $x = 2$ .

So, at $x = 2$ ,

Now, the left-hand-limit and right-hand-limit of the function $f (x)$ at $x = 2$ respectively are,

$lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} (k x^{2}) = k {(2)}^{2} = 4k$ ,

and $lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} (3) = 3$ .

Since, the function is continuous at $x = 2$ , so

$\begin{aligned} lim_{x \to 2^{-}} f(x)= lim_{x \to 2^{+}} f(x)=f(2) \\ \Rightarrow 4k=3 \\ \Rightarrow k= \frac{3}{4} \end{aligned}$

Hence, the value of $k$ is $\frac{3}{4}$ for which the function $f$ is continuous.

28. $f(x)= {\begin{aligned} kx+1, if x \leq π \\ cosx, if x > π \end{aligned}$ at $x = π$ .

Ans: The given function is $f(x)= {\begin{aligned} kx+1, if x \leq π \\ cosx, if x > π \end{aligned}$ .

It is known that, $f$ is continuous at $x= π$ only if the value of $f$ at $x= π$ is equal with the limiting value of $f$ at $x= π$ .

It is provided that the function $f$ is defined at $x= π$ .

Also, $f(π)=k π +1$ .

Now, the left-hand-limit,

$lim_{x \to π^{-}} f (x) = lim_{x \to π^{-}} (kx+1)=k π +1$ .

Also, the right-hand-limit,

$lim_{x \to π^{+}} f (x) = lim_{x \to π^{+}} cosx=cos π =-1$ .

Since, the function $f$ is continuous, so

$lim_{x \to π^{-}} f (x) = lim_{x \to π^{+}} f (x) = f (π)$

$\begin{aligned} \Rightarrow k π +1=-1 \\ \Rightarrow k π =-2 \\ \Rightarrow k=- \frac{2}{π} \end{aligned}$

Hence, the value of $k$ is $- \frac{2}{π}$ for which the function $f$ is continuous at $x= π$ .

29. $f(x)= {\begin{aligned} kx+1, if x \leq 5 \\ 3x-5, if x > 5 \end{aligned}$ at $x = 5$

Ans: The given function is $f(x)= {\begin{aligned} kx+1, if x \leq 5 \\ 3x-5, if x > 5 \end{aligned}$ .

Recall that, the function $f$ is continuous at $x = 5$ only if the value of $f$ at $x = 5$ is equal to the limiting value of $f$ at $x = 5$ .

Note that, the function $f$ is defined at $x = 5$ .

Also, $f(5)=kx+1=5k+1$ .

Then, the left-hand-limit of the function,

$lim_{x \to 5^{-}} f(x)= lim_{x \to 5^{-}} (kx+1) =5k+1$ .

The right-hand-limit of the function,

$lim_{x \to 5^{+}} f (x) = lim_{x \to 5^{+}} (3x-5) =3 (5) - 5 = 15 - 5 = 10$ .

Since, the function $f$ is continuous, so

$\begin{aligned} lim_{x \to 5^{-}} f(x)= lim_{x \to 5^{+}} (3x-5)=5k+1 \\ \Rightarrow 5k+1=10 \\ \Rightarrow 5k=9 \\ \Rightarrow k= \frac{9}{5} \end{aligned}$

Hence, the value of $k$ is $\frac{9}{5}$ for which the function $f$ is continuous at $x=5$ .

30. Find the values of $a$ and $b$ such that the function defined by

$f$ such that $f(x)= {\begin{aligned} 5, if x \leq 2 \\ ax+b, if 2 < x < 10 \\ 21, if x \geq 10 \end{aligned}$ is a continuous function.

Ans: The given function is $f(x)= {\begin{aligned} 5, if x \leq 2 \\ ax+b, if 2 < x < 10 \\ 21, if x \geq 10 \end{aligned}$ .

Note that, $f$ is defined at every point on the real number line.

Now, realise that if the function $f$ is continuous then $f$ is continuous at every real number.

So, let $f$ satisfies continuity at $x=2$ and $x=10$ .

Then, since $f$ is continuous at $x=2$ , so

$\begin{aligned} lim_{x \to 2^{-}} f(x)= lim_{x \to 2^{+}} f(x)=f(2) \\ \Rightarrow lim_{x \to 2^{-}} (5)= lim_{x \to 2^{+}} (ax+b)=5 \\ \Rightarrow 5=2a+b \end{aligned}$

$\Rightarrow 2a+b=5$ …… (1)

Again, since $f$ attains continuity at $x=10$ , so

$\begin{aligned} lim_{x \to 1 0^{-}} f(x)= lim_{x \to 10^{+}} f(x)=f(10) \\ \Rightarrow lim_{x \to 1 0^{-}} (ax+b)= lim_{x \to 10^{+}} (21)=21 \end{aligned}$

$\Rightarrow 1 0a+b=21$ …… (2)

Subtracting the equation (1) from the equation (2), gives

$8a=16 \Rightarrow a=2$

Substituting $a=2$ in the equation (1), gives

$\begin{aligned} 2 \times 2+b=5 \\ \Rightarrow 4+b=5 \Rightarrow b=1 \end{aligned}$

Hence, the values of $a$ and $b$ are $2$ and $1$ respectively for which $f$ is a continuous function.

31. Show that the function defined by $f(x)=cos(x^{2})$ is a continuous function.

Ans: The given function is $f(x)=cos(x^{2})$ .

Note that, $f$ is defined for all real numbers and so $f$ can be expressed as the composition of two functions as, $f=g \circ h$ , where $g(x)=cosx$ and $h(x)= x^{2}$ .

$[∴ (goh)(x)=g(h(x))=g(x^{2})=cos(x^{2})=f(x)]$

Now, it is to be Proven that, the functions $g(x)=cosx$ and $h(x)= x^{2}$ are continuous.

Since the function $g$ is defined for all the real numbers, let 's consider $c$ be a real number.

Then, $g(c)=cosc$ .

Substitute $x=c+h$ into the function $g$ .

When, $x \to c$ , then $h \to 0$ .

Then we have,

$\begin{aligned} lim_{x \to c} g(x)= lim_{x \to c} cosx \\ = lim_{h \to 0} cos (c+h) \\ = lim_{h \to 0} [cosc \cdot cosh-sinc \cdot sinh] \\ = lim_{h \to 0} (cosc \cdot cosh) - lim_{h \to 0} (sinc \cdot sinh) \\ =cosc \cdot cos0 - sinc \cdot sin0 \\ =cosc \times 1 - sinc \times 0 \\ =cosc \end{aligned}$

Therefore, $lim_{x \to c} g(x)=g(c)$ .

Hence, the function $g(x)=cosx$ is continuous.

Again, $h(x)= x^{2}$ is defined for every real point.

So, let consider $k$ be a real number, then $h(k)= k^{2}$ and

$lim_{x \to k} h(x)= lim_{x \to k} x^{2} = k^{2}$ .

Therefore, $lim_{x \to k} h(x)=h(k)$ .

Hence, the function $h$ is continuous.

Now, remember that for real valued functions $g$ and $h$ , such that $(g \circ h)$ is defined at $c$ , if $g$ is continuous at $c$ and $f$ is continuous at $g(c)$ , then $(f \circ h)$ is continuous at $c$ .

Hence, the function $f(x)=(g \circ h)(x)=cos(x^{2})$ is continuous.

32. Show that the function defined by $f(x)= | cosx |$ is a continuous function.

Ans: The given function is $f(x)= | cosx |$ .

Note that, the function $f$ is defined for all real numbers. So, the function $f$ can be expressed as the composition of two functions as, $f=g \circ h$ , where $g(x)= | x |$ and $h(x)=cosx$ .

$[∵ (goh)(x)=g(h(x))=g(cosx)= | cosx | =f(x)]$

Now, it is to be proved that the functions $g(x)= | x |$ and $h(x)=cosx$ are continuous.

Remember that, $g(x)= | x |$ , can be written as

$g(x)= {\begin{aligned} -x, if x < 0 \\ x, if x \geq 0 \end{aligned}$ .

Now, since the function $g$ is defined for every real number, let consider $c$ be a real number.

Then there may arise three cases, either $c < 0$ , or $c > 0$ , or $c=0$ .

Let's discuss the cases one after another.

Case I. When $c < 0$ .

Then, $g(c)=-c$ .

Also, $lim_{x \to c} g(x)= lim_{x \to c} (-x)=-c$ .

Therefore, $lim_{x \to c} g(x)=g(c)$ .

Hence, the function $g$ is continuous at every point $x$ , for $x < 0$ .

Case II. When $c > 0$ .

Then, $g(c)=c$ .

Also, $lim_{x \to c} g(x)= lim_{x \to c} x=c$ .

Therefore, $lim_{x \to c} g(x)=g(c)$ .

Hence, the function $g$ is continuous at every point $x$ for $x > 0$ .

Case III. When $c=0$ .

Then, $g(c)=g(0)=0$ .

Now, the left-hand-limit of the function $g$ at $x=0$ is

$lim_{x \to 0^{-}} g(x)= lim_{x \to 0^{-}} (-x)=0$ and the right-hand-limit is

$lim_{x \to 0^{+}} g(x)= lim_{x \to 0^{+}} (x)=0$ .

Therefore, $lim_{x \to 0^{-}} g(x)= lim_{x \to 0^{+}} g(x)=g(0)$ .

Hence, the function $g$ is continuous at $x=0$ .

By observing the above three discussions, we can conclude that the function $g$ is continuous at every real point.

Now, since the function $h(x)=cosx$ is defined for all real numbers, let 's consider $c$ be a real number. Then, substitute $x=c+h$ into the function $h$ .

So, when $x \to c$ , then $h \to 0$ .

Then, we have

$h (c) = cosc$ and

$lim_{x \to c} h(x)= lim_{x \to c} cosx$

$= lim_{h \to 0} cos (c+h)$

$= lim_{h \to 0} [cosc \cdot cosh-sinc \cdot sinh]$

$= lim_{h \to 0} (cosc \cdot cosh) - lim_{h \to 0} (sinc \cdot sinh)$

$\begin{aligned} =cosc \cdot cos0 - sinc \cdot sin0 \\ =cosc \times 1-sinc \times 0 \\ =cosc \end{aligned}$

Therefore, $lim_{x \to c} h (x) = h (c)$ .

Hence, the function $h(x)=cosx$ is continuous.

Now remember that, for real valued functions $g$ and $h$ , such that $(g \circ h)$ is defined at $x=c$ only if $g$ is continuous at $c$ and $f$ is continuous at $g(c)$ , then the composition functions $(f \circ g)$ is continuous at $x=c$ .

Thus, the function $f(x)=(goh)(x)=g(h(x))=g(cosx)= | cosx |$ is continuous.

33. Examine that $\sin | x |$ is a continuous function.

Ans: First suppose that, $f (x) = \sin | x |$ .

Now, note that the function $f$ is defined for all real numbers and so $f$ can be expressed as the composition of functions as, $f=g \circ h,$ where $g(x)=sinx$ and $h(x)= | x |$ .

$[(g \circ h)(x)=g(h(x))=g(| x |)=sin | x | =f(x)]$

So, it is to be proved that the functions $g(x)=sinx$ and $h(x)= | x |$ are continuous.

Now, remember that, the function $h(x)= | x |$ can be written as

$h(x)= {\begin{aligned} -x, if x < 0 \\ x, if x \geq 0 \end{aligned}$

Note that, the function $h$ is defined for every real number, and so let consider $c$ be a real number.

Then, there may arise three cases, either $c < 0$ , or $c > 0$ , or $c=0$ .

Let us discuss the cases one after another.

Case I. When $c < 0$ .

Then $h(c)=-c$ .

Also, $lim_{x \to c} (-x)= lim_{x \to c} x=-c$ .

Therefore, $lim_{x \to c} h(x)=h(c)$ .

Hence, the function $h$ is continuous at every point $x$ for $x < 0$ .

Case II. When $c > 0$ .

Then, $h(c)=c$

Also, $lim_{x \to c} (-x)= lim_{x \to c} x=c$ .

Therefore, $lim_{x \to c} h(x)=h(c)$ .

Thus, the function $h$ is continuous at every point $x$ for $x > 0$ .

Case III. When $c = 0$ .

Then, $h(c)=h(0)=0$ .

Also, the left-hand-limit of the function $h$ at $x=0$ is

$lim_{x \to 0^{-}} h(x)= lim_{x \to 0^{-}} (-x)=0$ and the right-hand -limit is

$lim_{x \to 0^{+}} h(x)= lim_{x \to 0^{+}} (x)=0$ .

Therefore, $lim_{x \to 0^{-}} h(x)= lim_{x \to 0^{+}} (x)=h(0)$ .

Thus, the function $h$ is continuous at $x = 0$ .

By observing the above three discussions, we can conclude that the function $h$ is continuous at every point.

Again, since the function $g(x)=sinx$ is defined for all real numbers, so let consider $c$ be a real number and substitute $x=c+k$ into the function.

Now, when $x \to c$ then $k \to 0$ .

Then, we have

$g(c)=sinc$ .

Also,

$\begin{aligned} lim_{x \to c} g(x)= lim_{x \to c} sinx \\ = lim_{k \to 0} sin (c+k) \\ = lim_{k \to 0} [sinc \cdot cosk+cosc \cdot sink] \\ = lim_{k \to 0} (sinc \cdot cosk) + lim_{h \to 0} (cosc \cdot sink) \\ =sinc \cdot cos0+cosc \cdot sin0 \\ =sinc+0 \\ =sinc \end{aligned}$

Therefore, $lim_{x \to c} g(x)=g(c)$ .

Hence, the function $g$ is continuous.

Now, remember that, for any two real valued functions $g$ and $h$ , such that the composition of functions $g \circ h$ is defined at $c$ , if $g$ is continuous at $c$ and $f$ is continuous at $g(c)$ , then the composition function $g \circ h$ is continuous at $c$ .

Thus, the function $(g \circ h)(x)=g(h(x))=g(| x |)=sin | x | =f(x)$ is continuous.

34.Find all the points of discontinuity of $f$ defined by $f (x) = | x | - | x + 1 |$ .

Ans: The given function is $f(x)= | x | - | x+1 |$ .

Let consider two functions

$g(x)= | x |$ and $h(x)= | x+1 |$ .

Then we get, $f=g-h$ .

Now, the function $g(x)= | x |$ can be written as

$g(x)= {\begin{aligned} -x, if x < 0 \\ x, if x \geq 0 \end{aligned}$ .

Note that, the function $g$ is defined for every real number and so let consider $c$ be a real number.

Then there may arise three cases, either $c < 0$ , or $c > 0$ , or $c=0$ .

Let us discuss the cases one after another.

Case I. When $c < 0$ .

Then, $g(c)=g(0)=-c$ .

Also, $lim_{x \to c} g(x)= lim_{x \to c} (-x)=-c$ .

Therefore, $lim_{x \to c} g(x)=g(c)$ .

Hence, the function $g$ is continuous at every point $x$ for $x < 0$ .

Case II. When $c > 0$ .

Then $g(c)=c$ .

Also, $lim_{x \to c} g(x)= lim_{x \to c} x=c$ .

Therefore, $lim_{x \to c} g(x)=g(c)$ .

Hence, the function $g$ is continuous at every point $x$ , where $x > 0$ .

Case III. When $c = 0$ .

Then $g(c)=g(0)=0$ .

Also, the left-hand-limit of the function $g$ at $x=0$ is

$lim_{x \to 0^{-}} g(x)= lim_{x \to 0^{-}} (-x)=0$ and the right-hand-limit is

$lim_{x \to 0^{+}} g(x)= lim_{x \to 0^{+}} (x)=0$ .

Therefore, $lim_{x \to 0^{-}} g(x)= lim_{x \to 0^{+}} (x)=g(0)$ .

Hence, the function $g$ is continuous at $x = 0$ .

Thus, we can conclude by observing the above three discussions that $g$ is continuous at every real point.

Now, remember that, the function $h(x)= | x+1 |$ can be written as

$h(x)= {\begin{aligned} -x(x+1), if, x < -1 \\ x+1, if, x \geq -1 \end{aligned}$ .

Note that, the function $h$ is defined for all real numbers, and so let consider $c$ be a real number.

Case I. When $c < -1$ .

Then $h(c)=-(c+1)$ .

Also, $lim_{x \to c} [-(x+1)] =-(c+1)$ .

Therefore, $lim_{x \to c} h(x)=h(c)$ .

Hence, the function $h$ attains continuity at every real point $x$ , where $x < -1$ .

Case II. When $c > -1$ .

Then, $h(c)=c+1$ .

Also, $lim_{x \to c} h(x)= lim_{x \to c} (x+1)=(c+1)$ .

Therefore, $lim_{x \to c} h(x)=h(c)$ .

Hence, the function $h$ satisfies continuity at every real point $x$ for $x > -1$ .

Case III. When $c =-1$ .

Then, $h(c)=h(-1)=-1+1=0$ .

Also, the left-hand-limit of the function $h$ at $x=1$ is

$lim_{x \to 1^{-}} h(x)= lim_{x \to 1^{-}} [-(x+1)] =-(-1+1)=0$ and the right-hand-limit is

$lim_{x \to 1^{+}} h(x)= lim_{x \to 1^{+}} (x+1)=(-1+1)=0$ .

Therefore, $lim_{x \to 1^{-}} h= lim_{x \to 1^{+}} h(x)=h(-1)$ .

Thus, the function $h$ satisfies continuity at $x=-1$ .

Hence, by observing the above three discussions, we can conclude that the function $h$ is continuous for every real point.

Now, since the functions $g$ and $h$ are both continuous, so the function $f=g-h$ is also continuous.

Hence, the function $f$ does not have any discontinuity points.

Exercise 5.2

Differentiate the function with respect to $x$ in Exercises 1 to 8.

1. $f (x) =sin(x^{2} +5)$

Ans: Let $f(x)=sin(x^{2} +5), u(x)= x^{2} +5,$ and $v(t)=sint$

Then, $(v \circ u)(x)=v(u(x))=v(x^{2} +5)=sin(x^{2} +5)=f(x)$

Therefore, $f$ is a composition of two functions $u$ and $v$ .

Substitute $t=u(x)= x^{2} +5$ .

Then, it gives

$\begin{aligned} \frac{dv}{dt} = \frac{d}{dt} (sint)=cost=cos(x^{2} +5) \\ \frac{dt}{dx} = \frac{d}{dx} (x^{2} +5)= \frac{d}{dx} (x^{2})+ \frac{d}{dx} (5)=2x+0=2x \end{aligned}$

Applying the chain rule of derivatives gives

$\frac{df}{dx} = \frac{dv}{dt} . \frac{dt}{dx} =cos(x^{2} +5) \times 2x=2xcos(x^{2} +5)$

An alternate method.

$\begin{aligned} \frac{df}{dx} = \frac{dv}{dt} . \frac{dt}{dx} =cos(x^{2} +5) . \frac{d}{dx} (x^{2} +5) \\ =cos(x^{2} +5) . [\frac{d}{dx} (x^{2})+ \frac{d}{dx} (5)] \\ =cos(x^{2} +5) . [2x+0] \\ =2xcos(x^{2} +5) \end{aligned}$

Hence, the derivative of the function $f (x) = sin (x^{2} + 5)$ is $2xcos(x^{2} +5)$ .

2. $f (x) =cos(sinx)$

Ans: Let suppose that, $f(x)=cos(sinx), u(x)=sinx,$ and $v(t)=cost$

Then, $(v \circ u)(x)=v(u(x))=v(sinx)=cos(sinx)=f(x)$

Therefore, it is observed that $f$ is the composition of two functions $u$ and $v$ .

Now, substitute $t=u(x)=sinx$ .

Then,

$\frac{dv}{dt} = \frac{d}{dt} (cost)=-sint=-sin(sinx)$ and

$\frac{dt}{dx} = \frac{d}{dx} (sinx)=cosx$ .

Applying the chain rule of derivatives gives

$\frac{df}{dx} = \frac{dv}{dt} . \frac{dt}{dx} = -sin(sinx) .cosx=-cosxsin(sinx)$

An alternate method.

$\frac{d}{dx} [cos(sinx)] =-sin(sinx) . \frac{d}{dx} (sinx)=-sin(sinx)cosx=-cosxsin(sinx)$ .

Hence, the derivative of the function $f (x) = cos (sinx)$ is $-cosxsin(sinx)$ .

3. $f (x) =sin(ax+b)$ .

Ans: Let suppose that, $f(x)=sin(ax+b), u(x)=ax+b,$ , and $v(t)=sint$

Then we get, $(v \circ u)(x)=v(u(x))=v(ax+b)=sin(ax+b)=f(x)$ .

It is observed that the function $f$ is the composition of two functions $u$ and $v$ .

Now, substitute $t=u(x)=ax+b$ .

Therefore,

$\frac{dv}{dt} = \frac{d}{dt} (sint)=cost=cos(ax+b)$ and

$\frac{dt}{dx} = \frac{d}{dx} (ax+b)= \frac{dt}{dx} (ax)+ \frac{d}{dx} (b)=a+0=a$ .

Applying the chain rule derivatives, gives

$\frac{df}{dx} = \frac{dv}{dt} . \frac{dt}{dx} =cos(ax+b) .a=acos(ax+b)$ .

Alternate method.

$\begin{aligned} \frac{d}{dx} [sin(ax+b)] =cos(ax+b) . \frac{d}{dx} (ax+b) \\ =cos(ax+b) \times [\frac{d}{dx} (ax)+ \frac{d}{dt} (b)] \\ =cos(ax+b) \times (a+0) \\ =acos(ax+b) \end{aligned}$

Hence, the derivative of the function $f (x) = \sin (ax+b)$ is $acos(ax+b)$ .

4. $f (x) = \sec (\tan (\sqrt{x}))$ with respect to $x$ .

Ans: Let suppose that, $f(x)=sec(tan(\sqrt{x})), u(x)= \sqrt{x},v(t)=tant,$ and $w(s)=secs$

Then, we get, $(w \circ v \circ u)(x)=w [v(u(x))] =w [v(\sqrt{x})] =w(tan \sqrt{x})=f(x)$ .

It is observed that the function $g$ is the composition of three functions $u$ , $v$ and $w$ .

Now, substitute $s=v(t)$ and $t=u(x)= \sqrt{x}$ .

Then, we get

$\begin{aligned} \frac{dw}{ds} = \frac{d}{ds} (secs)=secs=sec(tant) \times tan(tant) [s=tant] \\ =sec(tan \sqrt{x}) \times tan(tan \sqrt{x}) [t= \sqrt{x}] \end{aligned}$

Thus, applying the chain rule of derivatives gives

$\begin{aligned} \frac{dt}{dx} = \frac{dw}{ds} . \frac{ds}{dt} \times \frac{dt}{dx} \\ =sec(tan(\sqrt{x}) \times (tan(\sqrt{x})xse c^{2} \sqrt{x} \times \frac{1}{2 \sqrt{x}} \\ = \frac{1}{2 \sqrt{x}} se c^{2} \sqrt{x} sec(tan \sqrt{x})tan(tan \sqrt{x}) \\ = \frac{se c^{2} \sqrt{x} sec(tan \sqrt{x})tan(tan \sqrt{x})}{2 \sqrt{x}} \end{aligned}$

An alternate method.

$\begin{aligned} \frac{d}{dx} [sec(tan(\sqrt{x}))] =sec(tan(\sqrt{x}) .(tan(\sqrt{x}) . \frac{d}{dx} (tan(\sqrt{x}) \\ =sec(tan \sqrt{x}) \times tan(tan \sqrt{x}) \times se c^{2} (\sqrt{x}) \times \frac{d}{dx} (\sqrt{x}) \\ =sec(tan \sqrt{x}) \times tan(tan \sqrt{x}) \times se c^{2} (\sqrt{x}) \times \frac{1}{2 \sqrt{x}} \\ = \frac{se c^{2} \sqrt{x} sec(tan \sqrt{x})tan(tan \sqrt{x})}{2 \sqrt{x}} \end{aligned}$

Hence, the derivative of the function $f (x) = sec (tan (\sqrt{x}))$ is $\frac{se c^{2} \sqrt{x} sec(tan \sqrt{x})tan(tan \sqrt{x})}{2 \sqrt{x}}$ .

5. $f (x) = \frac{sin(ax+b)}{cos(cx+d)}$ .

Ans: The given function is $f(x)= \frac{sin(ax+b)}{cos(cx+d)}$ .

Now, let $g(x)=sin(ax+b)$ and $h(x)=cos(cx+d)$ .

Here we will use the divide formula of derivatives ${f}' = \frac{g^{'} h-gh^{'}}{h^{2}}$ . …… (1)

First, consider the function $g(x)=sin(ax+b)$ .

Let's assume $u(x)=ax+b,$ and $v(t)=sint$ .

Then, we get $(v \circ u)(x)=v(u(x))=v(ax+b)=sin(ax+b)=g(x)$ .

Therefore, we observe that the function $g$ is the composition of two functions, $u$ and $v$ .

So, substitute $t=u(x)=ax+b$ .

Then,

$\frac{dv}{dt} = \frac{d}{dt} (sint)=cost=cos(ax+b)$ and

$\frac{dt}{dx} = \frac{d}{dx} (ax+b)= \frac{dt}{dx} (ax)+ \frac{d}{dx} (b)=a+0=a$ .

Therefore, applying the chain rule of derivatives gives

${g}'= \frac{dg}{dx} = \frac{dv}{dt} . \frac{dt}{dx} =cos(ax+b) .a=acos(ax+b)$ .

Now, consider the function $h(x)=cos(cx+d)$ .

Let's suppose $p(x)=cx+d,$ and $q(t)=cosy$ .

Then, we have $(q \circ p)(x)=q(p(x))=q(cx+d)=cos(cx+d)=h(x)$ .

Therefore, the function $h$ is the composition of two functions $p$ and $q$ .

Now, substitute $y=p(x)=cx+d$ .

Then we have,

$\frac{dq}{dy} = \frac{d}{dy} (cosy)=-siny=-sin(cx+d)$ and

$\frac{dy}{dx} = \frac{d}{dx} (cx+d)= \frac{d}{dx} (cx)+ \frac{d}{dx} (d)=c$ .

Therefore, applying the chain rule of derivatives gives

${h}'= \frac{dh}{dx} = \frac{dq}{dy} . \frac{dy}{dx} =-sin(cx+d) \times c=-csin(cx+d)$ .

Now, substituting all the obtained derivatives into the formula (1) gives

$\begin{aligned} {f}'= \frac{acos(ax+b) \times cos(cx+d)+sin(ax+b) \times csin(cx+d)}{[cos(cx+d]^{2}} \\ = \frac{acos(ax+b)}{cos(cx+d)} +csin(ax+b) \times \frac{sin(cx+d)}{cos(cx+d)} \times \frac{1}{cos(cx+d)} \\ =acos(ax+b)sec(cx+d)+csin(ax+b)tan(cx+d)sec(cx+d) \end{aligned}$

Hence, the derivative of the function $f(x)= \frac{sin(ax+b)}{cos(cx+d)}$ is $acos(ax+b)sec(cx+d)+c.sin(ax+b)tan(cx+d)sec(cx+d)$ .

6. $f (x) = \cos (x^{3}) \times si n^{2} (x^{5})$ .

Ans: The given function is $f (x) =cos (x^{3}) \times si n^{2} (x^{5})$ .

Then,

$\begin{aligned} \frac{d}{dx} [cos x^{3} \times si n^{2} (x^{5})] =si n^{2} (x^{5}) \times \frac{d}{dx} (cos x^{3})+cos x^{3} \times \frac{d}{dx} [si n^{2} (x^{5})] \\ =si n^{2} (x^{5}) \times (-sin x^{3}) \times \frac{d}{dx} (x^{3})+cos x^{3} +2sin(x^{5}) \times \frac{d}{dx} [sin x^{5}] \\ =-sin x^{3} si n^{2} (x^{5}) \times 3 x^{2} +2sin x^{5} cos x^{3} \times cos x^{5} \times \frac{d}{dx} (x^{5}) \\ =-3 x^{2} sin x^{3} \times si n^{2} (x^{5})+2sin x^{5} cos x^{5} cos x^{3} \times 5 x^{4} \\ =10 x^{4} sin x^{5} cos x^{5} cos x^{3} -3 x^{2} sin x^{3} si n^{2} (x^{5}) \end{aligned}$

Hence, the derivative of the function $f (x) =cos (x^{3}) \times si n^{2} (x^{5})$ is $10 x^{4} sin x^{5} cos x^{5} cos x^{3} -3 x^{2} sin x^{3} si n^{2} (x^{5})$ .

7. $f (x) = 2 \sqrt{\cot (x^{2})}$ .

Ans: The given function is $f (x) = 2 \sqrt{\cot (x^{2})}$ .

Then,

$\begin{aligned} \frac{d}{dx} [2 \sqrt{\cot (x^{2})}] \\ =2 \cdot \frac{1}{2 \sqrt{\cot (x^{2})}} \times \frac{d}{dx} [cot(x^{2})] \\ = - \frac{1}{\sqrt{\cot (x^{2})}} cose c^{2} (x^{2}) \times \frac{d}{dx} (x^{2}) \\ = - \frac{1}{\sqrt{\cot (x^{2})}} cose c^{2} (x^{2}) \times 2x \\ =-2xcose c^{2} (x^{2}) \sqrt{\tan (x^{2})} \end{aligned}$

Hence, the derivative of the function $f (x) = 2 \sqrt{\cot (x^{2})}$ is $-2xcose c^{2} (x^{2}) \sqrt{\tan (x^{2})}$ .

8. $f (x) =cos(\sqrt{x})$ .

Ans: The given function is $f(x)=cos(\sqrt{x})$

Now, let $u(x)= \sqrt{x}$ and $v(t)=cost$ .

Then, we have, $(v \circ u)(x)=v(u(x))=v(\sqrt{x})=cos \sqrt{x} =f(x)$ .

It is observed that the function $f$ is the composition of two functions $u$ and $v$ . So, let $t=u(x)= \sqrt{x}$ .

Then,

$\frac{dt}{dx} = \frac{d}{dx} (\sqrt{x})= \frac{d}{dx} (x^{\frac{1}{2}}) = \frac{1}{2} x^{\frac{1}{2}} = \frac{1}{2 \sqrt{x}}$ .

Also, $\frac{dv}{dt} = \frac{d}{dt} (cost)=-sint=sin(\sqrt{x})$ .

Now, by applying the chain rule of derivatives, gives

$\begin{aligned} \frac{dt}{dx} = \frac{dv}{dt} \times \frac{dt}{dx} \\ =-sin(\sqrt{x}) \times \frac{1}{2 \sqrt{x}} \\ =- \frac{1}{2 \sqrt{x}} sin(\sqrt{x}) \\ =- \frac{sin(\sqrt{x})}{2 \sqrt{x}} \end{aligned}$

An alternate method.

$\begin{aligned} \frac{d}{dx} [cos(\sqrt{x})] \\ =-sin(\sqrt{x}) . \frac{d}{dx} (\sqrt{x}) \\ =-sin(\sqrt{x}) \times \frac{d}{dx} (x^{\frac{1}{2}}) \\ =-sin \sqrt{x} \times \frac{1}{2} x^{\frac{1}{2}} \\ = \frac{-sin \sqrt{x}}{2 \sqrt{x}} \end{aligned}$

Hence, the derivative of the function $f(x)=cos(\sqrt{x})$ is $- \frac{sin(\sqrt{x})}{2 \sqrt{x}}$ .

9. Prove that the function $f$ given by $f(x)= | x-1 |, x \in R$ is not differentiable at $x = 1$ .

Ans: The given function is $f(x)= | x-1 |, x \in R$ .

We know that a function $f$ is called differentiable at a point $x=c$ in its domain if both the $lim_{k \to 0^{-}} \frac{f(c+h)-f(c)}{h}$ and $lim_{h \to 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

Now verify the differentiability for the function $f$ at the point $x =1$ .

First, the left-hand-derivative is

$lim_{h \to 0^{-}} \frac{f(1+h)-f(1)}{h} = lim_{h \to 0^{-}} \frac{f | 1+h-1 | | 1-1 |}{h}$

$lim_{h \to 0^{-}} \frac{f | h | -0}{h} = lim_{h \to 0^{+}} \frac{-h}{h} = 1$ , since $h0 \Rightarrow | h | =-h$ .

Now the right-hand-derivative is

$lim_{h \to 0^{+}} \frac{f(1+h)-f(1)}{h} = lim_{h \to 0^{+}} \frac{f | 1+h-1 | | 1-1 |}{h}$

$lim_{h \to 0^{+}} \frac{f | h | -0}{h} = lim_{h \to 0^{+}} \frac{-h}{h} = -1$ , since $h0 \Rightarrow | h | =h$ .

From the above, it is noted that $lim_{h \to 0^{-}} \frac{f(1+h)-f(1)}{h} \neq lim_{h \to 0^{+}} \frac{f(1+h)-f(1)}{h}$ .

Hence, the function $f(x)= | x-1 |, x \in R$ is not differentiable at the point $x=1$ .

10. Prove that the greatest integer function defined by $f(x)= [x], 0x3$ , is not differentiable at $x = 1$ and $x = 2$ .

Ans: The given function is $f(x)= [x], 0x3$ .

Remember that a function $f$ is called differentiable at a point $x=c$ in its domain if both the limits, $lim_{h \to 0^{-}} \frac{f(c+h)-f(c)}{h}$ and $lim_{h \to 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

First, take the left-hand-derivative of the function $f$ at $x=1$ such that

$lim_{h \to 0^{-}} \frac{f(1+h)-f(1)}{h} = lim_{h \to 0^{-}} \frac{[1+h] - [1]}{h} = lim_{h \to 0^{-}} \frac{(0-1)}{h} = lim_{h \to 0^{+}} \frac{-h}{h} = \infty$ .

Now, take the right-hand-derivative of the function $f$ at $x=1$ such that

$lim_{h \to 0^{+}} \frac{f(1+h)-f(1)}{h} = lim_{h \to 0^{+}} \frac{[1+h] [1]}{h} = lim_{h \to 0^{+}} \frac{1-1}{h} = lim_{h \to 0^{+}} 0=0$ .

Therefore, it is being noticed that, $lim_{h \to 0^{-}} \frac{f(1+h)-f(1)}{h} \neq lim_{h \to 0^{+}} \frac{f(1+h)-f(1)}{h}$ .

Thus, the function $f$ is not differentiable at $x=1$ .

Now, justify the differentiability of the function $f$ at $x=2$ .

First, take the left-hand-derivative of the function $f$ at $x=2$ , such that

$lim_{h \to 0^{-}} \frac{f(2+h)-f(2)}{h} = lim_{h \to 0^{-}} \frac{[2+h] - [2]}{h} = lim_{h \to 0^{-}} \frac{(1-2)}{h} = lim_{h \to 0^{+}} \frac{-1}{h} = \infty$

Now, take the right-hand-derivative of the function $f$ at $x=2$ , such that

$lim_{h \to 0^{+}} \frac{f(2+h)-f(2)}{h} = lim_{h \to 0^{+}} \frac{[2+h] [2]}{h} = lim_{h \to 0^{+}} \frac{1-2}{h} = lim_{h \to 0^{+}} 0=0$

It is observed from the above discussion that, $lim_{h \to 0^{-}} \frac{f(2+h)-f(2)}{h} \neq lim_{h \to 0^{+}} \frac{f(2+h)-f(2)}{h}$ .

Thus, the function $f$ is not differentiable at the point $x=2$ .

Exercise 5.3

Find $\frac{dy}{dx}$ in the following.

1. $2 x + 3 y = sinx$ .

Ans: The given equation is $2x+3y=sinx$ .

Differentiating both sides of the equation with respect to $x$ , gives

$\frac{d}{dy} (2x+3y)= \frac{d}{dx} (sinx)$

$\Rightarrow \frac{d}{dx} (2x)+ \frac{d}{dx} (3y)=cosx$ , applying the addition rule of derivatives.

$\Rightarrow 2+3 \frac{dy}{dx} =cosx$

$\Rightarrow 3 \frac{dy}{dx} =cosx-2$

Therefore, $\frac{dy}{dx} = \frac{cosx-2}{3}$ .

2. $2 x + 3 y = siny$ .

Ans: The given equation is $2 x + 3 y = siny$ .

Differentiating both sides of the equation with respect to $x$ , gives

$\frac{d}{dx} (2x)+ \frac{d}{dx} (3y)= \frac{d}{dx} (siny)$

$\Rightarrow 2+3 \frac{dy}{dx} =cosy \frac{dy}{dx}$ , applying the chain rule of derivatives.

$\Rightarrow 2 = (cosy-3) \frac{dy}{dx}$

Therefore, $\frac{dy}{dx} = \frac{2}{cosy-3}$ .

3. $ax + b y^{2} = cosy$ .

Ans: The given function is $ax+b y^{2} =cosy$ .

Differentiating both sides of the equation with respect to $x$ , gives $\frac{d}{dx} (ax)+ \frac{d}{dx} (b y^{2})= \frac{d}{dx} (cosy)$

$\Rightarrow a \times 1+b \frac{d}{dy} (y^{2}) \frac{dy}{dx} = \frac{d}{dy} (cosy) \frac{dy}{dx}$ , applying the chain rule of derivatives.

$\Rightarrow a+b \times 2y \frac{dy}{dx} =-siny \frac{dy}{dx}$

$\Rightarrow (2by+siny) \frac{dy}{dx} =-a$

Therefore, $\frac{dy}{dx} = \frac{-a}{2by+siny}$ .

4. $xy + y^{2} = tanx + y$ .

Ans: The given equation is $xy+ y^{2} =tanx+y$ .

Differentiating both sides of the equation with respect to $x$ , gives $\begin{aligned} \frac{d}{dx} (xy+ y^{2})= \frac{d}{dx} (tanx+y) \\ \Rightarrow \frac{d}{dx} (xy)+ \frac{dy}{dx} (y^{2})= \frac{d}{dx} (tanx)+ \frac{dy}{dx} \end{aligned}$

$\Rightarrow [y \times \frac{d}{dx} (x)+x \times \frac{dy}{dx}] +2y \frac{dy}{dx} =se c^{2} x+ \frac{dy}{dx}$ , applying chain rule of derivatives.

$\Rightarrow y \times 1+x \frac{dy}{dx} +2y \frac{dy}{dx} =se c^{2} x+ \frac{dy}{dx}$

Therefore, $\frac{dy}{dx} = \frac{se c^{2} x-y}{(x+2y-1)}$ .

5. $x^{2} + xy + y^{2} = 100$ .

Ans: The given equation is $x^{2} +xy+ y^{2} =100$ .

Differentiating both sides of the equation with respect to $x$ , gives $\begin{aligned} \frac{dy}{dx} (x^{2} +xy+ y^{2})= \frac{d}{dx} 100 \\ \Rightarrow \frac{dy}{dx} (x^{2})+ \frac{dy}{dx} (xy)+ \frac{dy}{dx} (y^{2})=0 \end{aligned}$

$\Rightarrow 2x+ [y \times \frac{d}{dx} (x)+x \times \frac{dy}{dx}] +2y \frac{dy}{dx} =0$ , applying the chain rule of derivatives.

$\begin{aligned} \Rightarrow 2x+y \times 1+x \times \frac{dy}{dx} +2y \frac{dy}{dx} =0 \\ \Rightarrow 2x+y+(x+2y) \frac{dy}{dx} =0 \end{aligned}$

Therefore, $\frac{dy}{dx} =- \frac{2x+y}{x+2y}$ .

6. $x^{3} + x^{2} y + x y^{2} + y^{3} = 81$ .

Ans: The given equation is $x^{3} + x^{2} y+x y^{2} + y^{3} =81$ .

Differentiating both sides of the equation with respect to $x$ , gives $\begin{aligned} \frac{dy}{dx} (x^{3} + x^{2} y+x y^{2} + y^{3})= \frac{d}{dx} (81) \\ \Rightarrow \frac{d}{dx} (x^{3})+ \frac{dy}{dx} (x^{2} y)+ \frac{dy}{dx} (x y^{2})+ \frac{dy}{dx} (y^{3})=0 \\ \Rightarrow 3x+ [y \frac{d}{dx} (x^{2})+ x^{2} \frac{dy}{dx}] + [y^{2} \frac{d}{dx} (x)+x \frac{d}{dx} (y^{2})] +3 y^{2} \frac{dy}{dx} =0 \end{aligned}$

$\Rightarrow 3x+ [y \times 2x+ x^{2} \frac{dy}{dx}] + [y^{2} \times 1+x \times 2y \times \frac{dy}{dx}] +3 y^{2} \frac{dy}{dx} =0$ , applying chain rule.

$\Rightarrow (x^{3} +2xy+3 y^{2}) \frac{dy}{dx} +(3x+2xy+ y^{2})=0$

Therefore, $\frac{dy}{dx} = \frac{-(2x+2xy+ y^{2})}{(x^{3} +2xy+3 y^{2})}$ .

7. $si n^{2} y+cosxy= k$ .

Ans: The given equation is $si n^{2} y+cosxy= k$ .

Differentiating both sides of the equation with respect to $x$ , gives $\begin{aligned} \frac{d}{dx} (si n^{2} y+cosxy)= \frac{d}{dx} (k) \\ \Rightarrow \frac{d}{dx} (si n^{2} y)+ \frac{d}{dx} (cosxy)=0 \end{aligned}$ …. (1)

Applying the chain rule of derivatives gives

$\frac{d}{dx} (si n^{2} y)=2siny \frac{d}{dx} (siny)=2sinycosy \frac{dy}{dx}$ ….. (2)

and

$\frac{d}{dx} (cosxy)=-sinxy \frac{d}{dx} (xy)=-sinxy [y \frac{d}{dx} (x)+x \frac{dy}{dx}] =-ysinxy-xsinxy \frac{dy}{dx}$ …. (3)

From (1), (2) and (3), we obtain

$\begin{aligned} 2sinycosy \frac{dy}{dx} -ysinxy-xsinxy \frac{dy}{dx} =0 \\ \Rightarrow (2sinycosy-xsinxy) \frac{dy}{dx} =ysinxy \\ \Rightarrow (sin2y-xsinxy) \frac{dy}{dx} =ysinxy \end{aligned}$

Therefore, $\frac{dy}{dx} = \frac{ysinxy}{sin2y-xsinxy}$ .

8. $si n^{2} x+co s^{2} y=1$ .

Ans: The given equation is $si n^{2} x+co s^{2} y=1$ .

Differentiating both sides of the equation with respect to $x$ , gives $\begin{aligned} \frac{dy}{dx} (si n^{2} x+co s^{2} y)= \frac{d}{dx} (1) \\ \Rightarrow \frac{d}{dx} (si n^{2} x)+ \frac{d}{dx} (co s^{2} y)=0 \\ \Rightarrow 2sinx \times \frac{d}{dx} (sinx)+2cosy \times \frac{d}{dx} (cosy)=0 \end{aligned}$

$\begin{aligned} \Rightarrow 2sinxcosx+2cosy(-siny) \times \frac{dy}{dx} =0 \\ \Rightarrow sin2x-sin2y \frac{dy}{dx} =0 \end{aligned}$

$∴ \frac{dy}{dx} = \frac{sin2x}{sin2y}$

9. $y = si n^{- 1} (\frac{2 x}{1 + x^{2}})$ .

Ans: The given equation is $y=si n^{-1} (\frac{2x}{1+ x^{2}})$ .

Now, $y=si n^{-1} (\frac{2x}{1+ x^{2}})$

$\Rightarrow siny= \frac{2x}{1+ x^{2}}$ .

Differentiating both sides of the equation with respect to $x$ , gives $\frac{d}{dx} (siny)= \frac{d}{dx} (\frac{2x}{1+ x^{2}})$

$\Rightarrow cosy \frac{dy}{dx} = \frac{d}{dx} (\frac{2x}{1+ x^{2}})$ …… (1)

Now, the function $\frac{2x}{1+ x^{2}}$ is of the form of $\frac{u}{v}$ .

Applying the quotient rule, gives

$\begin{aligned} \frac{d}{dx} (\frac{2x}{1+ x^{2}}) = \frac{(1+ x^{2}) \frac{d}{dx} (2x)-2x \times \frac{d}{dx} (1+ x^{2})}{{(1+ x^{2})}^{2}} \\ = \frac{(1+ x^{2}) \times 2-2x \times [0+2x]}{{(1+ x^{2})}^{2}} = \frac{2+2 x^{2} -4 x^{2}}{{(1+ x^{2})}^{2}} \end{aligned}$

Therefore, $\frac{d}{dx} (\frac{2x}{1+ x^{2}}) = \frac{2(1- x^{2})}{{(1+ x^{2})}^{2}}$ ……. (2)

It is given that, $siny= \frac{2x}{1+ x^{2}}$ $\Rightarrow cosy= \sqrt{1-si n^{2} y} = \sqrt{1- {(\frac{2x}{1+ x^{2}})}^{2}} = \sqrt{\frac{{(1+ x^{2})}^{2} -4 x^{2}}{{(1+ x^{2})}^{2}}}$

$\Rightarrow cosy= \sqrt{\frac{{(1- x^{2})}^{2}}{{(1- x^{2})}^{2}}} = \frac{1- x^{2}}{1+ x^{2}}$ ….. (3)

From the equation (1), (2) and (3), gives

$\frac{1- x^{2}}{1+ x^{2}} \frac{dy}{dx} = \frac{2(1- x^{2})}{{(1+ x^{2})}^{2}}$

Therefore, $\frac{dy}{dx} = \frac{2}{1+ x^{2}}$ .

10. $y=ta n^{-1} (\frac{3x- x^{3}}{1-3 x^{2}}), - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$ .

Ans: The given function is $y=ta n^{-1} (\frac{3x- x^{3}}{1-3 x^{2}})$ .

Now, $y=ta n^{-1} (\frac{3x- x^{3}}{1-3 x^{2}})$

$\Rightarrow tany= \frac{3x- x^{3}}{1-3 x^{2}}$ …… (1)

According to the trigonometric formulas,

$tany= \frac{3tan \frac{y}{3} -ta n^{3} \frac{y}{3}}{1-3ta n^{2} \frac{y}{3}}$ …... (2)

By comparing the equations (1) and (2), gives

$x=tan \frac{y}{3}$ . …… (3)

Differentiating both sides of the equation (3) with respect to $x$ , gives

$\begin{aligned} \frac{d}{dx} (x)= \frac{d}{dx} (tan \frac{y}{3}) \\ \Rightarrow 1=se c^{2} \frac{y}{3} \times \frac{d}{dx} (\frac{y}{3}) \\ \Rightarrow 1=se c^{2} \frac{y}{3} \times \frac{1}{3} \times \frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx} = \frac{3}{se c^{2} \frac{y}{3}} = \frac{3}{1+ta n^{2} \frac{y}{3}} \end{aligned}$

Therefore, $\frac{dy}{dx} = \frac{3}{1+ x^{2}}$ .

11. $y=co s^{-1} (\frac{1- x^{2}}{1+ x^{2}})$ , 0<x<1.

Ans: The given equation is $y=co s^{-1} (\frac{1- x^{2}}{1+ x^{2}})$

$\Rightarrow cosy= \frac{1- x^{2}}{1+ x^{2}}$

$\Rightarrow \frac{1-ta n^{2} \frac{y}{2}}{1+ta n^{2} \frac{y}{2}} = \frac{1- x^{2}}{1+ x^{2}}$ . …… (1)

By comparing both sides of the equation (1) give

$tan \frac{y}{2} =x$ …… (2)

Differentiating both sides of the equation (2) with respect to $x$ , gives $\begin{aligned} se c^{2} \frac{y}{2} \times \frac{d}{dx} (\frac{y}{2}) = \frac{d}{dx} (x) \\ \Rightarrow se c^{2} \frac{y}{2} \times \frac{1}{2} \frac{dy}{dx} =1 \end{aligned}$

$\begin{aligned} \Rightarrow \frac{dy}{dx} = \frac{2}{se c^{2} \frac{y}{2}} \\ \Rightarrow \frac{dy}{dx} = \frac{2}{1+ta n^{2} \frac{y}{2}} \end{aligned}$

Therefore, using the equation (2), gives

$\frac{dy}{dx} = \frac{2}{1+ x^{2}}$ .

12. $y=si n^{-1} (\frac{1- x^{2}}{1+ x^{2}})$ , 0<x<1.

Ans: The given equation is $y=si n^{-1} (\frac{1- x^{2}}{1+ x^{2}})$ .

Now, $y=si n^{-1} (\frac{1- x^{2}}{1+ x^{2}})$

$\Rightarrow siny= \frac{1- x^{2}}{1+ x^{2}}$ . …… (1)

Differentiating both sides of the equation with respect to $x$ , gives $\frac{d}{dx} (siny)= \frac{d}{dx} (\frac{1- x^{2}}{1+ x^{2}})$ …... (2)

Using chain rule, we get

$\frac{d}{dx} (siny)=cosy \times \frac{dy}{dx}$ …… (3)

$cosy= \sqrt{1-si n^{2} y} = \sqrt{1- {(\frac{1- x^{2}}{1+ x^{2}})}^{2}} = \sqrt{\frac{{(1+ x^{2})}^{2} -(1- x^{2})^{2}}{1+ x^{2}}} = \sqrt{\frac{4 x^{2}}{{(1+ x^{2})}^{2}}}$ , using the equation (1).

$\Rightarrow cosy= \frac{2x}{1+ x^{2}}$ …… (4)

Therefore, from the equation (3) and (4) gives

$\frac{d}{dx} (siny)= \frac{2x}{1+ x^{2}} \frac{dy}{dx}$ ….. (5)

Now,

$\frac{d}{dx} (\frac{1- x^{2}}{1+ x^{2}}) = \frac{(1+ x^{2})(1- x^{2})-(1- x^{2})(1+ x^{2})}{{(1+ x^{2})}^{2}}$ , applying the quotient rule.

$\begin{aligned} = \frac{(1+ x^{2})(-2x)-(1- x^{2})(2x)}{{(1+ x^{2})}^{2}} \\ = \frac{-2x-2 x^{3} -2x+2 x^{3}}{{(1- x^{2})}^{2}} \end{aligned}$

$\Rightarrow \frac{d}{dx} (\frac{1- x^{2}}{1+ x^{2}}) = - \frac{4}{{(1+ x^{2})}^{2}}$ …... (6)

Using the equations (2), (5), and (6), gives

$\frac{2x}{1+ x^{2}} \frac{dy}{dx} = \frac{-4x}{{(1+ x^{2})}^{2}}$

Therefore, $\frac{dy}{dx} = \frac{-2}{1+ x^{2}}$ .

An alternate method.

$\begin{aligned} y=si n^{-1} (\frac{1- x^{2}}{1+ x^{2}}) \\ \Rightarrow siny= \frac{1- x^{2}}{1+ x^{2}} \\ \Rightarrow (1+ x^{2})siny=1- x^{2} \\ \Rightarrow (1+siny) x^{2} =1-siny \\ \Rightarrow x^{2} = \frac{1-siny}{1+siny} \\ \Rightarrow x^{2} = \frac{{(cos \frac{y}{2} -sin \frac{y}{2})}^{2}}{{(cos \frac{y}{2} +sin \frac{y}{x})}^{2}} \\ \Rightarrow x= \frac{cos \frac{y}{2} -sin \frac{y}{2}}{cos \frac{y}{2} +sin \frac{y}{2}} \end{aligned}$

$\Rightarrow x=tan (\frac{π}{4} - \frac{π}{2})$

Differentiating both sides of the equation with respect to $x$ , gives $\begin{aligned} \frac{d}{dx} (x)= \frac{d}{dx} [tan (\frac{π}{4} - \frac{y}{2})] \\ \Rightarrow 1=se c^{2} (\frac{π}{4} - \frac{y}{2}) \times \frac{dy}{dx} (\frac{π}{4} - \frac{y}{2}) \\ \Rightarrow 1= [1+ta n^{2} (\frac{π}{4} - \frac{y}{2}) \times (- \frac{1}{2} \times \frac{dy}{dx})] \\ \Rightarrow 1=(1+ x^{2}) (- \frac{1}{2} \frac{dy}{dx}) \end{aligned}$

Therefore, $\frac{dx}{dy} = \frac{-2}{1+ x^{2}}$ .

13. $y=co s^{-1} (\frac{2x}{1+ x^{2}})$ , -1<x<1.

Ans: The given equation is $y=co s^{-1} (\frac{2x}{1+ x^{2}})$ .

Now, $y=co s^{-1} (\frac{2x}{1+ x^{2}})$

$\Rightarrow cosy= \frac{2x}{1+ x^{2}}$ . …… (1)

Differentiating both sides of the equation with respect to $x$ , gives $\frac{d}{dx} (cosy)= \frac{d}{dx} \times (\frac{2x}{1+ x^{2}})$

$\Rightarrow -siny \times \frac{dy}{dx} = \frac{(1- x^{2}) \times \frac{d}{dx} (2x)-2x \times \frac{d}{dx} (1+ x^{2})}{{(1+ x^{2})}^{2}}$ , applying the quotient rule.

$\Rightarrow - \sqrt{1-co s^{2} y} \frac{dy}{dx} = \frac{(1+ x^{2}) \times 2-2x \times 2x}{{(1+ x^{2})}^{2}}$

$\Rightarrow [\sqrt{1- {(\frac{2x}{1+ x^{2}})}^{2}}] \frac{dx}{dy} =- [\frac{2(1-x)^{2}}{{(1+ x^{2})}^{2}}]$ , using the equation (1).

$\begin{aligned} \Rightarrow \sqrt{\frac{{(1- x^{2})}^{2} -4 x^{2}}{{(1+ x^{2})}^{2}}} = \frac{dy}{dx} = \frac{-2(1-x)^{2}}{(1+ x^{2})} \\ \Rightarrow \sqrt{\frac{{(1- x^{2})}^{2}}{{(1+ x^{2})}^{2}}} \frac{dy}{dx} = \frac{-2(1-x)^{2}}{(1+ x^{2})} \\ \Rightarrow \frac{1- x^{2}}{1+ x^{2}} \times \frac{dy}{dx} = \frac{-2(1-x)^{2}}{(1+ x^{2})} \end{aligned}$

Therefore, $\frac{dy}{dx} = \frac{-2}{(1+ x^{2})}$ .

14. $y = si n^{- 1} (2 x \sqrt{1 - x^{2}}), - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$ .

Ans: The given equation is $y=si n^{-1} (2x \sqrt{1- x^{2}})$ .

Now, $y=si n^{-1} (2x \sqrt{1- x^{2}})$

$\Rightarrow siny=2x \sqrt{1- x^{2}}$ . …… (1)

Differentiating both sides of the equation with respect to $x$ , gives $cosy \frac{dy}{dx} =2 [x \frac{d}{dx} (\sqrt{1- x^{2}}) + \sqrt{1- x^{2}} \frac{dx}{dx}]$

$\Rightarrow \sqrt{1-si n^{2} y} \frac{dy}{dx} =2 [\frac{x}{2} \times \frac{-2}{\sqrt{1- x^{2}}} + \sqrt{1- x^{2}}]$

$\Rightarrow \sqrt{1-(2x \sqrt{1- x^{2}})^{2}} \frac{dy}{dx} =2 [\frac{- x^{2} +1- x^{2}}{\sqrt{1- x^{2}}}]$ , using the equation (1).

$\begin{aligned} \Rightarrow \sqrt{1-4 x^{2} (1- x^{2})} \frac{dy}{dx} =2 [\frac{1-2 x^{2}}{\sqrt{1- x^{2}}}] \\ \Rightarrow \sqrt{{(1-2x)}^{2}} \frac{dy}{dx} =2 [\frac{1-2 x^{2}}{\sqrt{1- x^{2}}}] \\ \Rightarrow (1-2 x^{2}) \frac{dy}{dx} =2 [\frac{1-2 x^{2}}{\sqrt{1- x^{2}}}] \end{aligned}$

Therefore, $\frac{dy}{dx} = \frac{2}{\sqrt{1- x^{2}}}$ .

15. $y = se c^{- 1} (\frac{1}{2 x^{2} - 1}), 0 < x < \frac{1}{\sqrt{2}}$ .

Ans: The given equation is $y=se c^{-1} (\frac{1}{2 x^{2} -1})$ .

Now,

$\begin{aligned} y=se c^{-1} (\frac{1}{2 x^{2} -1}) \\ \Rightarrow secy= \frac{1}{2 x^{2} -1} \\ \Rightarrow cosy=2 x^{2} -1 \\ \Rightarrow 2 x^{2} =1+cosy \\ \Rightarrow 2 x^{2} =2co s^{2} \frac{y}{2} \end{aligned}$

$\Rightarrow x=cos \frac{y}{2}$ ……. (1)

Differentiating both sides of the equation with respect to $x$ , gives

$\begin{aligned} \frac{d}{dx} (x)= \frac{d}{dx} (cos \frac{y}{2}) \\ \Rightarrow 1=sin \frac{y}{2} \times \frac{d}{dx} (\frac{y}{2}) \\ \Rightarrow \frac{-1}{sin \frac{y}{2}} = \frac{1}{2} \frac{dy}{dx} \end{aligned}$

$\Rightarrow \frac{d y}{d x} = \frac{- 2}{s i n \frac{y}{2}} = \frac{- 2}{\sqrt{1 - c o s^{2}} \frac{y}{2}}$

$= \frac{- 2}{\sqrt{1 - x^{2}}}$ , using the equation (1).

Therefore, $\frac{dy}{dx} = \frac{-2}{\sqrt{1 - x^{2}}}$ .

Exercise 5.4

Differentiate the following w.r.t. $x$ .

1. $y= \frac{e^{x}}{sinx}$ .

Ans: The given function is $y= \frac{e^{x}}{sinx}$ .

Then, we have

$\frac{dy}{dx} = \frac{sinx \frac{d}{dx} (e^{x})- e^{x} \frac{d}{dx} (sinx)}{si n^{2} x}$ , by applying the quotient rule of derivatives.

$= \frac{sinx \times (e^{x})- e^{x} \times (cosx)}{si n^{2} x}$

Therefore, the derivative of the function $y$ is

$\frac{dy}{dx} = \frac{e^{x} (sinx-cosx)}{si n^{2} x}, x \neq n π, n \in Z$ .

2. $y= e^{si n^{-1} x}$ .

Ans: The given function is $y= e^{si n^{-1} x}$ .

Then, we have

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} (e^{si n^{-1} x}) \\ = e^{si n^{-1} x} \times \frac{d}{dx} (si n^{-1} x) \\ = e^{si n^{-1} x} \times \frac{1}{\sqrt{1- x^{2}}} \\ = \frac{e^{si n^{-1} x}}{\sqrt{1- x^{2}}} \end{aligned}$

Therefore, the derivative of the function $y$ is

$\frac{dy}{dx} = \frac{e^{si n^{-1} x}}{\sqrt{1- x^{2}}}, x \in (-1,1)$ .

3. $y = e^{x^{3}}$ .

Ans: The given function is $y= e^{x^{3}}$ .

Then by applying the chain rule of derivatives we have,

$\frac{dy}{dx} = \frac{d}{dx} (e^{x^{3}})= e^{x^{3}} \times \frac{d}{dx} (x^{3})= e^{x^{3}} \times 3 x^{2}$ .

Therefore, the derivative of the function $y$ is

$\frac{dy}{dx} =3 x^{2} e^{x^{3}}$ .

4. $y = \sin (ta n^{- 1} e^{- x})$ .

Ans: The given function is $y=sin(ta n^{-1} e^{-x})$ .

Now, applying the chain rule of derivatives, give

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} [sin(ta n^{-1} e^{-x})] \\ =cos(ta n^{-1} e^{-x}) \times \frac{d}{dx} (ta n^{-1} e^{-x}) \\ =cos(ta n^{-1} e^{-x}) \times \frac{1}{1+(e^{-x})^{2}} \times \frac{d}{dx} (e^{-x}) \\ = - \frac{cos(ta n^{-1} e^{-x})}{1+ e^{-2x}} \times e^{-x} \end{aligned}$

Therefore, the derivative of the function $y$ is

$\frac{dy}{dx} = \frac{- e^{-x} cos(ta n^{-1} e^{-x})}{1+ e^{-2x}}$ .

5. $y = \log (\cos (e^{x}))$

Ans: Let $y=log (cos (e^{x}))$

Now, by applying the chain rule of derivatives give

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} [log (cos (e^{x}))] \\ = \frac{1}{cos e^{x}} \times \frac{d}{dx} (cos (e^{x})) \\ = \frac{1}{cos e^{x}} \times (-sin (e^{x})) \times \frac{d}{dx} (e^{x}) \\ = \frac{-sin e^{x}}{cos e^{x}} \times e^{x} \end{aligned}$

Therefore, the derivative of the function $y$ is

$\frac{dy}{dx} =- e^{x} tan (e^{x}), x \neq (2n+1) \frac{π}{2}, n \in N$ .

6. $y= e^{x} + e^{x^{2}} + . . . + e^{x^{5}}$ .

Ans: The given function is $y= e^{x} + e^{x^{2}} + . . . + e^{x^{5}}$ .

Then, differentiating with respect to $x$ both sides, give

$\frac{dy}{dx} = \frac{d}{dx} (e^{x} + e^{x^{2}} + . . . + e^{x^{5}})$

$= \frac{d}{dx} (e^{x})+ \frac{d}{dx} ({e^{x}}^{^{2}})+ \frac{d}{dx} ({e^{x}}^{^{4}})+ \frac{d}{dx} ({e^{x}}^{^{5}})$ , applying the sum rule of derivatives.

$\begin{aligned} = e^{x} + [e^{x^{2}} \times \frac{d}{dx} (x^{2})] + [e^{x^{3}} \times \frac{d}{dx} (x^{3})] + [e^{x^{4}} \times \frac{d}{dx} (x^{4})] + [e^{x^{5}} \times \frac{d}{dx} (x^{5})] \\ = e^{x} +(e^{x^{2}} \times 2x)+(e^{x^{3}} \times 3 x^{2})+(e^{x^{4}} \times 4 x^{3})+(e^{x^{5}} \times 5 x^{4}) \end{aligned}$

Therefore, the derivative of the function $y$ is

$\frac{dy}{dx} = e^{x} +2x e^{x^{2}} +3 x^{2} e^{x^{3}} +4 x^{3} e^{x^{4}} +5 x^{4} e^{x^{5}}$ .

7. $y = \sqrt{e^{\sqrt{x}}}, x > 0$ .

Ans: The given function is $y= \sqrt{e^{\sqrt{x}}}$ .

Then squaring both sides both sides of the equation give

$y^{2} = e^{\sqrt{x}}$

Now, differentiating both sides with respect to $x$ gives

$\frac{d}{dx} (y^{2}) = \frac{d}{dx} (e^{\sqrt{x}})$

$\begin{aligned} \Rightarrow 2 y \frac{dy}{dx} = e^{\sqrt{x}} \frac{d}{dx} (\sqrt{x}) \\ \Rightarrow 2 y \frac{dy}{dx} = e^{\sqrt{x}} \frac{1}{2} \times \frac{1}{\sqrt{x}} \\ \Rightarrow \frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4y \sqrt{x}} \end{aligned}$

$\Rightarrow \frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4 \sqrt{e^{\sqrt{x}}} \sqrt{x}}$ , substituting the value of $y$ .

Therefore,

$\frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}, x0$ .

8. $y = \log (logx), x > 1$ .

Ans: The given function is $y=log(logx)$ .

Now, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{d}{dx} [log(logx)]$

$= \frac{1}{logx} \times \frac{d}{dx} (logx)$ , by applying the chain rule of derivatives.

$= \frac{1}{logx} \times \frac{1}{x}$

Therefore, $\frac{dy}{dx} = \frac{1}{xlog (x)}, x>1$ .

9. $y = \frac{cosx}{logx}, x > 0$ .

Ans: The given function is $y= \frac{cosx}{logx}$ .

Differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{\frac{d}{dx} (cos) \times logx-cosx \times \frac{d}{dx} (logx)}{{(logx)}^{2}}$ , by applying the quotient rule.

$= \frac{-sinxlogx-cosx \times \frac{1}{x}}{{(logx)}^{2}}$

Therefore,

$\frac{dy}{dx} = \frac{- [xlogxsinx+cosx]}{x(logx)^{2}}, x>0$ .

10. $y = \cos (logx + e^{x}), x > 0$ .

Ans: The given function is $y=cos(logx+ e^{x})$ .

Then differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{d}{dx} [cos (logx+ e^{x})]$ .

$\Rightarrow \frac{dy}{dx} =-sin [logx+ e^{x}] \times \frac{d}{dx} (logx+ e^{x})$ , by applying the chain rule of derivatives.

$\begin{aligned} =-sin(logx+ e^{x}) \times [\frac{d}{dx} (logx)+ \frac{d}{dx} (e^{x})] \\ =-sin(logx+ e^{x}) \times (\frac{1}{x} + e^{x}) \end{aligned}$

Therefore, $\frac{dy}{dx} = - (\frac{1}{x} + e^{x}) sin(logx+ e^{x})$ .

Exercise 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t $x$ .

1. $y=cosx \times cos2x \times cos3x$ .

Ans: The given function is $y=cosx \times cos2x \times cos3x$ .

First, taking logarithm both sides of the equation give,

$logy=log(cosx \times cos2x \times cos3x)$

$\Rightarrow logy=log(cosx)+log(cos2x)+log(cos3x)$ , by the property of logarithm.

Now, differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{1}{y} \frac{dy}{dx} = \frac{1}{cosx} \times \frac{d}{dx} (cosx)+ \frac{1}{cos2x} \times \frac{d}{dx} (cos2x)+ \frac{1}{cos3x} \times \frac{d}{dx} (cos3x) \\ \Rightarrow \frac{dy}{dx} =y [- \frac{sinx}{cosx} - \frac{sin2x}{cos2x} \times \frac{d}{dx} (2x)- \frac{sin3x}{cos3x} \times \frac{d}{dx} (3x)] \end{aligned}$

Therefore,

$\frac{dy}{dx} =-cos x \times cos2x \times cos3x [tanx+2tan2x+3tan3x]$ .

2. $y = \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}}$ .

Ans: The given function is $y= \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$ .

First taking logarithm both sides of the equation give

$\begin{aligned} logy=log \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \\ \Rightarrow logy= \frac{1}{2} log [\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}] \\ \Rightarrow logy= \frac{1}{2} [log {(x-1)(x-2)} -log {(x-3)(x-4)(x-5)}] \\ \Rightarrow logy= \frac{1}{2} [log(x-1)+log(x-2)-log(x-3)-log(x-4)-log(x-5)] \end{aligned}$

Now, differentiating both sides of the equation with respect to $x$ give

$\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} [log(x-1)+log(x-2)-log(x-3)-log(x-4)-log(x-5)]$ .

$\begin{aligned} \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} [\frac{1}{x-1} \times \frac{d}{dx} (x-1)+ \frac{1}{x-2} \times \frac{d}{dx} (x-2)- \frac{1}{x-3} \times \frac{d}{dx} (x-3)- \frac{1}{x-4} \times \frac{d}{dx} (x-4) \\ - \frac{1}{x-5} \times \frac{d}{dx} (x-5)] \end{aligned}$

$\Rightarrow \frac{dy}{dx} = \frac{y}{2} (\frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5})$

Therefore,

$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} [\frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5}]$ .

3. Differentiate the function $y=(logx)^{cosx}$ with respect to $x$ .

Ans: The given function is $y=(logx)^{cosx}$ .

First, taking logarithm both sides of the equation give

$logy=cosx .log(logx)$ .

Now, differentiating both sides of the equation with respect to $x$ give

$\frac{1}{y} \times \frac{dy}{dx} = \frac{d}{dx} (cosx) \times log(logx)+cosx \times \frac{d}{dx} [log(logx)]$

$\Rightarrow \frac{1}{y} \times \frac{dy}{dx} =-sinxlog(logx)+cosx \times \frac{1}{logx} \times \frac{d}{dx} (logx)$ , by applying the chain rule.

$\Rightarrow \frac{dy}{dx} =y [-sinxlog(logx)+ \frac{cosx}{logx} \times \frac{1}{x}]$

Therefore,

$\frac{dy}{dx} =(logx)^{cosx} [\frac{cosx}{xlogx} -sinx \times log(logx)]$ .

4. $y= x^{x} - 2^{sinx}$ .

Ans: The given function is $y= x^{x} - 2^{sinx}$ .

Now, let $x^{x} =u$ …… (1)

and $2^{sinx} =v$ . …… (2)

Therefore, $y=u-v$ . …… (3)

Then differentiating the equation (3) with respect to $x$ gives

$\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$ …… (4)

Now, taking logarithm both sides of the equation (1) give

$\begin{aligned} log (u) = \log (x^{x}) \\ \Rightarrow \log u=xlogx \end{aligned}$

Differentiating both sides of the equation with respect to $x$ gives $\begin{aligned} \frac{1}{u} \frac{du}{dx} = [\frac{d}{dx} (x) \times logx+x \times \frac{d}{dx} (logx)] \\ \Rightarrow \frac{du}{dx} =u [1 \times logx+x \times \frac{1}{x}] \\ \Rightarrow \frac{du}{dx} = x^{x} (logx+1) \end{aligned}$

$\Rightarrow \frac{du}{dx} = x^{x} (1+logx)$ …… (5)

Now, taking logarithm both sides of the equation (2) give

$log (2^{sinx}) =logv$

$\Rightarrow logv=sinx \times log2$ .

Differentiating both sides of the equation with respect to $x$ , give

$\begin{aligned} \frac{1}{v} \times \frac{dv}{dx} =log2 \times \frac{d}{dx} (sinx) \\ \Rightarrow \frac{dv}{dx} =vlog2cosx \end{aligned}$

$\Rightarrow \frac{dv}{dx} = 2^{sinx} cosxlog2$ …… (6)

Therefore, from the equation (4), (5) and (6) give

$\frac{dy}{dx} = x^{x} (1+logx)- 2^{sinx} cosxlog 2$ .

5. $y = (x + 3)^{2} {(x + 4)}^{3} {(x + 5)}^{4}$ .

Ans: The given function is $y=(x+3)^{2} {(x+4)}^{3} {(x+5)}^{4}$ .

First, taking logarithm both sides of the equation give

$logy=log [{(x+3)}^{2} {(x+4)}^{3} {(x+5)}^{4}]$

$\Rightarrow logy=2log(x+3)+3log(x+4)+4log(x+5)$

Now, differentiating both sides of the equation with respect to $x$ , give

$\begin{aligned} \frac{1}{y} \times \frac{dy}{dy} =2 \times \frac{1}{x-3} \times \frac{d}{dz} (x+3)+3 \times \frac{1}{x+4} \times \frac{d}{dx} (x+4)+4 \times \frac{1}{x+5} \times \frac{d}{dx} (x+5) \\ \Rightarrow \frac{dy}{dx} =y [\frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}] \\ \Rightarrow \frac{dy}{dx} =(x+3)^{2} {(x+4)}^{3} {(x+5)}^{4} \times [\frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}] \\ \Rightarrow \frac{dy}{dx} =(x+3)^{2} {(x+4)}^{3} {(x+5)}^{4} \times [\frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}] \\ \Rightarrow \frac{dy}{dx} =(x+3)^{2} {(x+4)}^{2} {(x+5)}^{2} - [2(x^{2} +9x+20)+3(x^{2} +9x+15)+4(x^{2} +7x+12)] \end{aligned}$

Therefore,

$\frac{dy}{dx} =(x+3)(x+4)^{2} {(x+5)}^{3} (9 x^{2} +70x+133)$ .

6. $y = {(x + \frac{1}{x})}^{x} + x^{(1 + \frac{1}{x})}$ .

Ans: The given function is $y= {(x+ \frac{1}{x})}^{x} + x^{(1+ \frac{1}{x})}$ .

First, let $u= {(x+ \frac{1}{x})}^{x}$ and $v= x^{(1+ \frac{1}{x})}$

Therefore, $y=u+v$ . …… (1)

Differentiating the equation (1) both sides with respect to $x$ give

$\Rightarrow \frac{dy}{dx} = \frac{du}{dx} = \frac{dv}{dx}$ …... (2)

Now, $u= {(x+ \frac{1}{x})}^{x}$

$\Rightarrow logu=log {(x+ \frac{1}{x})}^{x}$

$\Rightarrow logu=xlog (x+ \frac{1}{x})$

Differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (x) \times log (x+ \frac{1}{x}) +x \times \frac{d}{dx} [log (x+ \frac{1}{x})] \\ \Rightarrow \frac{1}{u} \frac{du}{dx} =1 \times log (x+ \frac{1}{x}) +x \times \frac{1}{(x+ \frac{1}{x})} \times \frac{d}{dx} (x+ \frac{1}{x}) \\ \Rightarrow \frac{du}{dx} =u [log (x+ \frac{1}{x}) + \frac{x}{(x+ \frac{1}{x})} \times (x+ \frac{1}{x^{2}})] \\ \Rightarrow \frac{du}{dx} = {(x+ \frac{1}{x})}^{x} [log (x+ \frac{1}{x}) + \frac{(x- \frac{1}{x})}{(x+ \frac{1}{x})}] \\ \Rightarrow \frac{du}{dx} = {(x+ \frac{1}{x})}^{x} [log (x+ \frac{1}{x}) + \frac{x^{2} +1}{x^{2} -1}] \end{aligned}$

$\Rightarrow \frac{du}{dx} = {(x+ \frac{1}{x})}^{2} [\frac{x^{2} +1}{x^{2} -1} +log (x+ \frac{1}{x})]$ …… (3)

Also, $v= x^{(x+ \frac{1}{x})}$

$\begin{aligned} \Rightarrow logv=log [x^{x^{(x+ \frac{1}{x})}}] \\ \Rightarrow logv= (x+ \frac{1}{x}) logx \end{aligned}$

Differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{1}{v} \times \frac{dv}{dx} = [\frac{d}{dx} (1+ \frac{1}{x})] \times logx+ (1+ \frac{1}{x}) \times \frac{d}{dx} logx \\ \Rightarrow \frac{1}{v} \frac{dv}{dx} =- \frac{logx}{x^{2}} + \frac{1}{x} + \frac{1}{x^{2}} \end{aligned}$

$\Rightarrow \frac{dv}{dx} =v [\frac{-logx+x+1}{x^{2}}]$ ……. (4)

Hence, from the equations (2), (3) and (4), give

$\frac{dy}{dx} = {(x+ \frac{1}{x})}^{x} [\frac{x^{2} -1}{x^{2} +1} +log (x+ \frac{1}{x})] + x^{(x+ \frac{1}{x})} (\frac{x+1-logx}{x^{2}})$ .

7. $y = (logx)^{x} + x^{logx}$ .

Ans: The given function is $y=(logx)^{x} + x^{logx}$ .

Then, let $u=(logx)^{x}$ and $v= x^{logx}$ .

Therefore, $y=u+v$ .

Differentiating both sides of the equation with respect to $x$ gives

$\Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ ……. (1)

Now, $u=(logx)^{x}$

$\begin{aligned} \Rightarrow logu=log [{(logx)}^{x}] \\ \Rightarrow logu=xlog(logx) \end{aligned}$

Differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (x) \times log(logx)+x \times \frac{d}{dx} [log(logx)] \\ \Rightarrow \frac{du}{dx} =u [1 \times log(logx)+x \times \frac{1}{logx} \times \frac{d}{dx} (logx)] \\ \Rightarrow \frac{du}{dx} =(logx)^{x} [log(logx)+ \frac{x}{logx} \times \frac{1}{x}] \end{aligned}$

$\begin{aligned} \Rightarrow \frac{du}{dx} =(logx)^{x} [log(logx)+ \frac{1}{logx}] \\ \Rightarrow \frac{du}{dx} =(logx)^{x} = [\frac{log(logx) \times logx+1}{logx}] \end{aligned}$

$\frac{du}{dx} =(logx)^{x-1} [1+logx \times log(logx)]$ ……. (2)

Again, $v= x^{logx}$

$\begin{aligned} \Rightarrow \log v=log (x^{logx}) \\ \Rightarrow \log v=logxlogx= {(\log x)}^{2} \end{aligned}$

Differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{1}{v} \times \frac{dx}{dx} = \frac{d}{dx} [{(logx)}^{2}] \\ \Rightarrow \frac{1}{v} \times \frac{dx}{dx} =2(logx) \times \frac{d}{dx} (logx) \\ \Rightarrow \frac{dv}{dx} =2 x^{logx} \frac{logx}{x} \end{aligned}$

$\Rightarrow \frac{dv}{dx} =2 x^{logx - 1} \times logx$ ….…. (3)

Hence, from the equations (1), (2), and (3), gives

$\frac{dy}{dx} =(logx)^{x-1} [1+logx \times log(logx)] +2 x^{logx-1} \times logx$ .

8. $y = (sinx)^{x} + si n^{- 1} \sqrt{x}$ .

Ans: The given function is $y=(sinx)^{x} +si n^{-1} \sqrt{x}$ .

Now, let $u=(sinx)^{x}$ and $v=si n^{-1} \sqrt{x}$ .

Therefore, $y=u+v$ .

Then, differentiating both sides of the equation with respect to $x$ gives

$\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$ …….. (1)

Now, $u=(sinx)^{x}$

$\begin{aligned} \Rightarrow logu=xlog(sinx)^{x} \\ \Rightarrow logu=xlog(sinx) \end{aligned}$

Differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (x) \times log(sinx)+x \times \frac{d}{dx} [log(sinx)] \\ \Rightarrow \frac{du}{dx} =u [1 \times log(sinx)+x \times \frac{1}{sinx} \times \frac{d}{dx} (sinx)] \\ \Rightarrow \frac{du}{dx} =(sinx)^{x} [log(sinx)+ \frac{x}{sinx} \times cosx] \end{aligned}$

$\Rightarrow \frac{du}{dx} =(sinx)^{x} (xcotx+logsinx)$ ….… (2)

Again, $v=si n^{-1} \sqrt{x}$ .

Differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{dv}{dx} = \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \times \frac{d}{dx} (\sqrt{x}) \\ \Rightarrow \frac{dv}{dx} = \frac{1}{\sqrt{1-x}} \times \frac{1}{2 \sqrt{x}} \end{aligned}$

$\Rightarrow \frac{dv}{dx} = \frac{1}{2 \sqrt{x- x^{2}}}$

Hence, from the equations (1), (2) and (3), gives

$\frac{dv}{dx} =(sinx)^{x} (xcotx+logsinx)+ \frac{1}{2 \sqrt{x- x^{2}}}$ .

9. $y = x^{sinx} + (sinx)^{cosx}$ .

Ans: The given function is $y= x^{sinx} +(sinx)^{cosx}$ .

Then, let $u= x^{sinx}$ and $v=(sinx)^{cosx}$ .

Therefore, $y=u+v$ .

Differentiating both sides of the equation with respect to $x$ gives

$\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$ …… (1)

Now, $u= x^{sinx}$

$\begin{aligned} \Rightarrow logu=xlog(x^{sinx}) \\ \Rightarrow logu=sinxlogx \end{aligned}$

Differentiating both sides of the equation with respect to $x$ gives
$\begin{aligned} \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (sinx) \times logx+sinx \times \frac{d}{dx} (logx) \\ \Rightarrow \frac{du}{dx} =u= [cosxlogx+sinx \times \frac{1}{x}] \end{aligned}$

$\Rightarrow \frac{du}{dx} = x^{sinx} [cosxlogx+ \frac{sinx}{x}]$ …....(2)

Again, $v=(sinx)^{cosx}$

$\begin{aligned} \Rightarrow logv=log(sinx)^{cosx} \\ \Rightarrow logv=cosxlog(sinx) \end{aligned}$

Then, differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx} (cosx) \times log(sinx)+cosx \times \frac{d}{dx} [log(sinx)] \\ \Rightarrow \frac{dv}{dx} =v [-sinx \times log(sinx)+cosx \times \frac{1}{sinx} \times \frac{d}{dx} (sinx)] \\ \Rightarrow \frac{du}{dx} =(sinx)^{cosx} [-sinxlogsinx+cotxcosx] \end{aligned}$ $\Rightarrow \frac{dv}{dx} =(sinx)^{cosx} [cosxcotx-sinxlogsinx]$ …… (3)

Hence, from the equations (1), (2) and (3), gives

$\frac{du}{dx} = x^{sinx} (cosxlogx+ \frac{sinx}{x}) +(sinx)^{cosx} [cosxcotx-sinxlogsinx]$ .

10. $y = x^{xcosx} + \frac{x^{2} + 1}{x^{2} - 1}$ .

Ans: The given function is $y= x^{xcosx} + \frac{x^{2} +1}{x^{2} -1}$ .

First, let $u= x^{xcosx}$ and $v= \frac{x^{2} +1}{x^{2} -1}$ .

Therefore, $y=u+v$ .

Differentiating both sides of the equation with respect to $x$ gives

$\Rightarrow \frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$ ……. (1)

Now, $u= x^{xcosx}$ .

Then, differentiating both sides of the equation with respect to $x$ gives $\begin{aligned} \frac{1}{u} \frac{d u}{d x} = \frac{d}{d x} (x) \times \cos x \log x + x \times \frac{d}{d x} (\cos x) \times \log x + x \cos x \times \frac{d}{d x} (\log x) \\ \Rightarrow \frac{d u}{d x} = u [1 \times \cos x \times \log x + x \times (- \sin x) \log x + x \cos x \times \frac{1}{x}] \end{aligned}$

$\Rightarrow \frac{du}{dx} = x^{xcosx} (cosxlogx-xsinxlogx+cosx)$ …… (2)

Again, $v= \frac{x^{2} +1}{x^{2} -1}$

$\Rightarrow logv=log(x^{2} +1)-log(x^{2} -1)$

Differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{1}{v} = \frac{dv}{dx} = \frac{2x}{x^{2} +1} - \frac{2x}{x^{2} -1} \\ \Rightarrow \frac{dv}{dx} =v [\frac{2x(x^{2} -1)-2x(x^{2} +1)}{(x^{2} +1)(x^{2} -1)}] \\ \Rightarrow \frac{du}{dx} = \frac{x^{2} +1}{x^{2} -1} \times [\frac{-4x}{(x^{2} +1)(x^{2} -1)}] \end{aligned}$

$\Rightarrow \frac{dv}{dx} = \frac{-4x}{{(x^{2} -1)}^{2}}$ …….. (3)

Hence, from the equations (1), (2) and (3), give

$\frac{dv}{dx} = x^{xcosx} [cosx(1+logx)-xsinxlogx] - \frac{4x}{{(x^{2} -1)}^{2}}$ .

11. $y = (xcosx)^{x} + (xsinx)^{\frac{1}{x}}$ .

Ans: The given function is $y=(xcosx)^{x} +(xsinx)^{\frac{1}{x}}$ .

Then, let $u=(xcosx)^{x}$ and $v=(xsinx)^{\frac{1}{x}}$ .

Therefore, $y=u+v$ .

$\Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ ……. (1)

Again, $u=(cosx)^{x}$

$\begin{aligned} \Rightarrow logu=log(xcosx)^{x} \\ \Rightarrow logu=xlog(xcosx) \\ \Rightarrow logu=x [logx+logcosx] \\ \Rightarrow logu=xlogx+xlogcosx \end{aligned}$

Differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (xlogx+xlogcosx) \\ \Rightarrow \frac{du}{dx} =u [{logx \times \frac{d}{dx} (x)+x \times \frac{d}{dx} (logx)} + {logcosx \times \frac{d}{dx} (x)+x \times \frac{d}{dx} (logcosx)}] \\ \Rightarrow \frac{du}{dx} =(xcosx)^{x} [{logx \times 1+x \times \frac{1}{x}} + {logcosx-1+x \times \frac{1}{cosx} \times \frac{d}{dx} (cosx)}] \end{aligned}$

$\begin{aligned} \Rightarrow \frac{du}{dx} =(xcosx)^{x} [{logx+1} + {logcosx-1+ \frac{x}{cosx} \times (-sinx)}] \\ \Rightarrow \frac{du}{dx} =(xcosx)^{x} [(logx+1)+(logcosx-xtanx)] \end{aligned}$

$\Rightarrow \frac{du}{dx} =(xcosx)^{x} [1-xtanx+(logx+logcosx)]$

Therefore,

$\frac{du}{dx} =(xcosx)^{x} [1-xtanx+(logx(xcosx)]$ …….. (2)

Again, $v=(xsinx)^{\frac{1}{x}}$

$\begin{aligned} \Rightarrow logv=log(xsinx)^{\frac{1}{x}} \\ \Rightarrow logv= \frac{1}{x} log(xsinx) \\ \Rightarrow logv= \frac{1}{x} (logx+logsinx) \\ \Rightarrow logv= \frac{1}{x} logx+ \frac{1}{x} logsinx \end{aligned}$

Differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx} (\frac{1}{x} logx) + \frac{d}{dx} [\frac{1}{x} log(sinx)] \\ \Rightarrow \frac{1}{v} \frac{dv}{dx} = [\frac{1}{x} logx \times \frac{d}{dx} (\frac{1}{x}) + \frac{1}{x} \times \frac{d}{dx} (logx)] + [log(sinx) \times \frac{d}{dx} (\frac{1}{x}) + \frac{1}{x} \times \frac{d}{dx} {(logsinx)}] \\ \Rightarrow \frac{1}{v} \frac{dv}{dx} = [\frac{1}{x} logx \times (- \frac{1}{x^{2}}) + \frac{1}{x} \times \frac{1}{x}] + [log(sinx) \times (- \frac{1}{x^{2}}) + \frac{1}{x} \times \frac{1}{sinx} \times \frac{d}{dx} (sinx)] \\ \Rightarrow \frac{1}{v} \frac{dv}{dx} = \frac{1}{x^{2}} (1-logx)+ [\frac{1-logx}{x^{2}} + \frac{1}{xsinx} \times cosx] \\ \Rightarrow \frac{1}{v} \frac{dv}{dx} = \frac{1}{x^{2}} {(xsinx)}^{\frac{1}{x}} + [\frac{1-logx}{x^{2}} + \frac{-log(sinx)+xcotx}{x^{2}}] \\ \Rightarrow \frac{dv}{dx} =(xsinx)^{\frac{1}{x}} [\frac{1 - \log x - \log (\sin x) + x \cot x}{x^{2}}] \end{aligned}$

Therefore,

$\frac{dv}{dx} =(xsinx)^{\frac{1}{x}} [\frac{1-log(xsinx)+xcotx}{x^{2}}]$ ……. (3)

Hence, from the equations (1), (2) and (3), gives

$\Rightarrow \frac{dy}{dx} =(xcosx)^{x} [1-xtanx+log(xcosx)] +(xsinx)^{\frac{1}{x}} [\frac{1-log(xsinx)+xcotx}{x^{2}}]$ .

Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15.

12. $x^{y} + y^{x} = 1$ .

Ans: The given function is $x^{y} + y^{x} =1$ .

Then, let $x^{y} =u$ and $y^{x} =v$ .

Therefore, $u+v=1$ .

Differentiating both sides of the equation with respect to $x$ gives

$\frac{du}{dx} + \frac{dv}{dy} =0$

Now, $u= x^{y}$ ……. (1)

$\begin{aligned} \Rightarrow logu=log(x^{y}) \\ \Rightarrow logu=ylogx \end{aligned}$

Differentiating both sides of the equation with respect to $x$ gives

$\frac{1}{u} \frac{du}{dx} =logx \frac{dy}{dx} +y \times \frac{d}{dx} (logx)$

$\Rightarrow \frac{du}{dx} =u [logx \frac{dy}{dx} +y \times \frac{1}{x}]$

Therefore, $\frac{du}{dx} = x^{y} [logx \frac{dy}{dx} + \frac{y}{x}]$ ………… (2)

Also, $v= y^{x}$

Taking logarithm both sides of the equation give

$\begin{aligned} \Rightarrow logv=log(y^{3}) \\ \Rightarrow logv=xlogy \end{aligned}$

Differentiating both sides of the equation with respect to $x$ gives

$\frac{1}{v} \times \frac{dv}{dx} =logy \times \frac{d}{dx} (x)+x \times \frac{d}{dx} (logy)$

$\Rightarrow \frac{dv}{dx} =v (logy \times 1+x \times \frac{1}{y} \times \frac{dy}{dx})$

Therefore, $\frac{dv}{dx} = y^{x} (logy+ \frac{x}{y} \frac{dy}{dx})$ ……... (3)

So, from the equation (1), (2) and (3), gives

$\begin{aligned} x^{y} (logx \frac{dy}{dx} + \frac{y}{x}) + y^{x} (logy+ \frac{x}{y} \frac{dy}{dx}) =0 \\ \Rightarrow (x^{2} +logx+x y^{y-1}) \frac{dy}{dx} =- (y x^{y-1} + y^{x} logy) \end{aligned}$

Hence, $\frac{dy}{dx} = \frac{-(y x^{y-1} + y^{x} logy)}{x^{y} logx+x y^{x-1}}$ .

13. $y^{x} = x^{y}$ .

Ans: The given equation is $y^{x} = x^{y}$ .

Then, taking logarithm both sides of the equation give

$xlogy=ylogx$ .

Differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} logy \times \frac{d}{dx} (x)+x \times \frac{d}{dx} (logy)=logx \times \frac{d}{dx} (y)+y \times \frac{d}{dx} (logx) \\ \Rightarrow logy \times 1+x \times \frac{1}{y} \times \frac{dy}{dx} =logx \times \frac{dy}{dx} +y \times \frac{1}{x} \\ \Rightarrow logy+ \frac{x}{y} \frac{dy}{dx} =logx \frac{dy}{dx} + \frac{y}{x} \\ \Rightarrow (\frac{x}{y} -logx) \frac{dy}{dx} = \frac{y}{x} -logy \end{aligned}$

$\begin{aligned} \Rightarrow (\frac{x-ylogx}{y}) \frac{dy}{dx} = \frac{y-xlogy}{x} \\ \Rightarrow (\frac{x-ylogx}{y}) \frac{dy}{dx} = \frac{y-xlogy}{x} \end{aligned}$

Therefore, $\frac{dy}{dx} = \frac{y}{x} (\frac{y-xlogy}{x-ylogx})$ .

14. ${(cosx)}^{y} =(cosy)^{x}$ .

Ans: The given equation is ${(cosx)}^{y} =(cosy)^{x}$ .

Then, taking logarithm both sides of the equation give

$ylogcosx=xlogcosy$ .

Now, differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} logcosx \times \frac{dy}{dx} +y \times \frac{d}{dx} (logcosx)=logcosy \times \frac{d}{dx} (x)+x \times \frac{d}{dx} (logcosy) \\ \Rightarrow logcosx \frac{dy}{dx} + \frac{y}{cosx} \times (-sinx)=logcosy+ \frac{x}{cosy} (-siny) \times \frac{dy}{dx} \\ \Rightarrow logcosx \frac{dy}{dx} -ytanx=logcosy-xtany \frac{dy}{dx} \\ \Rightarrow (logcosx+xtany) \frac{dy}{dx} =ytanx+logcosy \end{aligned}$

Therefore, $\frac{dy}{dx} = \frac{ytanx+logcosy}{xtany+logcosx}$ .

15. $xy= e^{(x-y)}$ .

Ans: The given equation is $xy= e^{(x-y)}$ .

Then, taking logarithm both sides of the equation give

$\begin{aligned} log(xy)=log(e^{x-y}) \\ \Rightarrow logx+logy=(x-y)loge \\ \Rightarrow logx+logy=(x-y) \times 1 \\ \Rightarrow logx+logy=x-y \end{aligned}$

Now, differentiating both sides of the equation with respect to $x$ gives $\begin{aligned} \frac{d}{dx} (logx)+ \frac{d}{dx} (logy)= \frac{d}{dx} (x)- \frac{dy}{dx} \\ \Rightarrow \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} =1- \frac{1}{x} \\ \Rightarrow (1+ \frac{1}{y}) \frac{dy}{dx} = \frac{x-1}{x} \end{aligned}$

Therefore, $\frac{dy}{dx} = \frac{y(x-1)}{x(y+1)}$ .

16. Find the derivative of the function given by

$f(x)=(1+x)(1+ x^{2})(1+ x^{4})(1+ x^{8})$ and hence find $f^{'} (1)$ .

Ans: The given function is $f(x)=(1+x)(1+ x^{2})(1+ x^{4})(1+ x^{8})$ .

By taking logarithm both sides of the equation give

$logf(x)=log(1+x)+log(1+ x^{2})+log(1+ x^{4})+log(1+ x^{8})$

Now, differentiating both sides of the equation with respect to $x$ gives $\begin{aligned} \frac{1}{f(x)} \times \frac{d}{dx} [f(x)] = \frac{d}{dx} log(1+x)+ \frac{d}{dx} log(1+ x^{2})+ \frac{d}{dx} log(1+ x^{4})+ \frac{d}{dx} log(1+ x^{8}) \\ \Rightarrow \frac{1}{f(x)} \times f^{'} (x)= \frac{1}{1+x} \times \frac{1}{dx} (1+x)+ \frac{1}{1+ x^{2}} \times \frac{d}{dx} log(1+ x^{2})+ \frac{1}{1+ x^{4}} \times \frac{d}{dx} log(1+ x^{4}) \\ + \frac{1}{1+ x^{8}} \times \frac{d}{dx} log(1+ x^{8}) \\ \Rightarrow f^{'} (x)=f(x) [\frac{1}{1+x} + \frac{1}{1+ x^{2}} \times 2x+ \frac{1}{1+ x^{4}} \times 4 x^{3} + \frac{1}{1+ x^{8}} \times 8 x^{7}] \end{aligned}$

Therefore,

$f^{'} (x)=(1+x)(1+ x^{2})(1+ x^{4})(1+ x^{8}) [\frac{1}{1+x} + \frac{2x}{1+ x^{2}} + \frac{4 x^{3}}{1+ x^{4}} + \frac{8 x^{7}}{1+ x^{8}}]$

So,
$\begin{aligned} f^{'} (1)=(1+1)(1+ 1^{2})(1+ 1^{4})(1+ 1^{8}) [\frac{1}{1+1} + \frac{2 \times 1}{1+ 1^{2}} + \frac{4 \times 1^{3}}{1+ 1^{4}} + \frac{8 \times 1^{7}}{1+ 1^{8}}] \\ =2 \times 2 \times 2 \times 2 [\frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2}] \\ =16 \times (\frac{1+2+4+8}{2}) \\ =16 \times \frac{15}{2} =120 \end{aligned}$

Hence, ${f}' (1) =120$ .

17. Differentiate $y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$ in three ways as mentioned below. Do they all give the same answer?

(a) by using product rule.

Ans: The given function is $y= (x^{2} -5x+8) (x^{3} +7x+9)$ .

Now, let consider $u=(x^{2} -5x+8)$ and $v=(x^{3} +7x+9)$

Therefore, $y=uv$ .

$\begin{aligned} \Rightarrow \frac{dy}{dx} = \frac{du}{dv} .v+u . \frac{du}{dx} \\ \Rightarrow \frac{dy}{dx} = \frac{d}{dx} (x^{2} -5x+8) .(x^{3} +7x+9)+(x^{2} -5x+8) . \frac{d}{dx} (x^{3} +7x+9) \\ \Rightarrow \frac{dy}{dx} =(2x-5)(x^{3} +7x+9) .(x^{2} -5x+8)(3 x^{2} +7) \\ \Rightarrow \frac{dy}{dx} =2x(x^{3} +7x+9)-5(x^{2} -5x+8)+ x^{2} (3 x^{2} +7)-5x(3 x^{2} +7)-8(3 x^{2} +7) \\ \Rightarrow \frac{dy}{dx} =(2 x^{4} +14 x^{2} +18x)-5 x^{3} -35x-45+(3 x^{4} +7 x^{2})-15 x^{3} -35x+24 x^{2} +56 \end{aligned}$

Hence, $\frac{dy}{dx} =5 x^{4} -20 x^{3} +45 x^{2} +52x+11$ .

(b) by expanding the product to obtain a single polynomial.

Ans: The given function is

$y=(x^{2} -5x+8)(x^{3} +7x+9)$ .

Then, calculating the product, gives

$\begin{aligned} y= x^{2} (x^{3} +7x+9)-5 x^{4} (x^{3} +7x+9)+8(x^{3} +7x+9) \\ \Rightarrow y= x^{5} +7 x^{3} +9 x^{2} -5 x^{3} -26 x^{2} +11x+72 \end{aligned}$

Now, differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} (x^{5} +7 x^{3} +9 x^{2} -5 x^{3} -26 x^{2} +11x+72) \\ = \frac{d}{dx} (x^{5})-5 \frac{d}{dx} (x^{4})+15 \frac{d}{dx} (x^{3})-26 \frac{d}{dx} (x^{3})+11 \frac{d}{dx} (x)+ \frac{d}{dx} (72) \\ =5 x^{4} -5 \times 4 x^{3} +15 \times 3 x^{2} -26 \times 2x+11 \times 1+0 \end{aligned}$

Hence, $\frac{dy}{dx} =5 x^{4} -20 x^{3} +45 x^{2} -52x+11$ .

Ans: The given function is

$y=(x^{2} -5x+8)(x^{3} +7x+9$ .

Now, taking logarithm both sides of the function give

$logy=log(x^{2} -5x+8)+log(x^{3} +7x+9)$

Differentiating both sides of the equation with respect to $x$ gives $\begin{aligned} \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} log(x^{2} -5x+8)+ \frac{d}{dx} log(x^{3} +7x+9) \\ \Rightarrow \frac{1}{y} \frac{dy}{dx} = \frac{1}{x^{2} -5x+8} . \frac{d}{dx} (x^{2} -5x+8)+ \frac{1}{x^{3} +7x+9} . \frac{d}{dx} (x^{3} +7x+9) \\ \Rightarrow \frac{dy}{dx} =y [\frac{1}{x^{2} -5x+8} \times (2x-5)+ \frac{1}{x^{3} +7x+9} \times (3 x^{2} +7)] \\ \Rightarrow \frac{dy}{dx} =(x^{2} -5x+8)(x^{3} +7x+9) [\frac{2x-5}{x^{3} -5x+8} + \frac{3 x^{2} +7}{x^{3} +7x+9}] \\ \Rightarrow \frac{dy}{dx} =(x^{2} -5x+8)(x^{3} +7x+9) [\frac{(2x-5)(x^{3} +7x+9)+(3 x^{2} +7)(x^{2} -5x+8)}{(x^{3} -5x+8)+(x^{3} +7x+9)}] \\ \Rightarrow \frac{dy}{dx} =2x(x^{3} +7x+9 x^{2})-5(x^{3} +7x+9)+3 x^{2} (x^{2} -5x+8)+7(x^{3} +7x+9) \\ \Rightarrow \frac{dy}{dx} =(2 x^{4} +14 x^{2} +18x)+(5 x^{3} -35x+45)+(3 x^{4} -15 x^{3} +24 x^{2})+(7 x^{2} +35x+56) \end{aligned}$

Therefore, $\frac{dy}{dx} =5 x^{2} -20 x^{3} +45 x^{2} -52x+11$ .

Hence, comparing the above three results, it is concluded that the derivative $\frac{dy}{dx}$ are the same for all methods.

18. If $u$ , $v$ , and $w$ are functions of $x$ , then show that $\frac{d}{dx} (u .v .w)= \frac{du}{dx} v .w+u \frac{du}{dx} .w+u .v \frac{dw}{d x}$ in two ways - first by using repeated application of product rule, second by logarithmic differentiation.

Ans: Let the function $y=u .v .w=u .(v .w)$ .

Then applying the product rule of derivatives, give

$\frac{dy}{dx} = \frac{du}{dx} .(v .w)+u . \frac{d}{dx} (v .w)$

$\Rightarrow \frac{dy}{dx} = \frac{du}{dx} v .w+u [\frac{dv}{dx} .w+v . \frac{dv}{dx}]$ (Using the product rule again)

Thus,

$\frac{dy}{dx} = \frac{du}{dx} v .w+u . \frac{dv}{dx} .w+u .v \frac{dw}{dx}$ .

Now, take logarithm both sides of the function $y=u .v .w$ .

Then, we have $logy=logu+logv+logw$ .

Differentiating both sides of the equation with respect to $x$ gives $\begin{aligned} \frac{1}{y} . \frac{dy}{dx} = \frac{d}{dx} (logu)+ \frac{d}{dx} (logv)+ \frac{d}{dx} (logw) \\ \Rightarrow \frac{1}{y} . \frac{dy}{dx} = \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \\ \Rightarrow \frac{dy}{dx} =y (\frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}) \\ \Rightarrow \frac{dy}{dx} =u .v .w (\frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}) \end{aligned}$

Hence, $\frac{dy}{dx} = \frac{du}{dx} v .w+u \frac{dv}{dx} .w+u .v \frac{dw}{dx}$ .

Exercise 5.6

If $x, y$ are connected parametrically by the equation given below, without eliminating the parameter, Find $\frac{dy}{dx}$ .

1. $x=2a t^{2}, y=a t^{4}$ .

Ans: The given equations are

$x=2a t^{2}$ …… (1)

and $y=a t^{4}$ …… (2)

Then, differentiating both sides of the equation (1) with respect to $t$ gives

$\frac{dx}{dt} = \frac{d}{dt} (2a t^{2})=2a \times \frac{d}{dt} (t^{2})=2a \times 2t=4at$ . …… (3)

Also, differentiating both sides of the equation (2) with respect to $t$ gives

$\frac{dy}{dt} = \frac{d}{dt} (a t^{4})=a \times \frac{d}{dt} (t^{4})=a \times 4 \times t^{3} =4a t^{3}$ …… (4)

Now, dividing the equations (4) by (3) gives

$\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} = \frac{4a t^{3}}{4at} = t^{2}$ .

Hence, $\frac{dy}{dx} = t^{2}$ .

2. $x = acos θ, y = bcos θ$ .

Ans: The given equations are

$x=acos θ$ …… (1)

and $y=bcos θ$ …… (2)

Then, differentiating both sides of the equation (1) with respect to $θ$ gives

$\frac{dx}{d θ} = \frac{d}{d θ} (acos θ)=a(-sin θ)=-asin θ$ . …… (3)

Also, differentiating both sides of the equation (1) with respect to $θ$ gives

$\frac{dy}{d θ} = \frac{d}{d θ} (bcos θ)=b(-sin θ)=-bsin θ$ …… (4)

Therefore, dividing the equation (4) by (3) gives

$\frac{dy}{dx} = \frac{(\frac{dy}{d θ})}{(\frac{dx}{d θ})} = \frac{-bsin θ}{-asin θ} = \frac{b}{a}$ .

Hence, $\frac{dy}{dx} = \frac{b}{a}$ .

3. $x = sint, y = \cos 2 t$ .

Ans:

The given equations are

$x=sint$ …… (1)

and $y=cos2t$ …… (2)

Then, differentiating both sides of the equation (1) with respect to $t$ gives

$\frac{dx}{dt} = \frac{d}{dt} (sint)=cost$ . …… (3)

Also, differentiating both sides of the equation (2) with respect to $t$ gives

$\frac{dy}{dt} = \frac{d}{dt} (cos2t)=sin2t \times \frac{d}{dt} (2t)=-2sin2t$ …… (4)

Therefore, by dividing the equation (4) by (3) gives

$\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} = \frac{-2sin2t}{cost} = \frac{-2 \times 2sintcost}{cost} =-4sint$

Hence, $\frac{dy}{dx} =-4sint$ .

4. $x = 4 t, y = \frac{4}{t}$ .

Ans: The given equations are

$x=4t$ …… (1)

and $y= \frac{4}{t}$ …… (2)

Now, differentiating both sides of the equation (1) with respect to $t$ gives

$\frac{dx}{dt} = \frac{d}{dt} (4t)=4$ . …… (3)

Also, differentiating both sides of the equation (2) with respect to $t$ gives

$\frac{dy}{dt} = \frac{d}{dt} (\frac{4}{t}) =4 \times \frac{d}{dt} (\frac{1}{t}) =4 \times (\frac{-1}{t^{2}}) = \frac{-4}{t^{2}}$ …… (4)

Therefore, dividing the equation (4) by (3) gives

$\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} = \frac{(\frac{-4}{t^{2}})}{4} = \frac{-1}{t^{2}}$ .

Hence, $\frac{dy}{dx} =- \frac{1}{t^{2}}$ .

5. $x = \cos θ - \cos 2 θ, y = \sin θ - \sin 2 θ$ .

Ans: The given equations are

$x = cos θ - 2cos θ$ …… (1)

and $y=sin θ -sin2 θ$ …… (2)

Then, differentiating both sides of the equation (1) with respect to $θ$ gives

$\frac{dx}{d θ} = \frac{d}{d θ} (cos θ -cos2 θ)= \frac{d}{d θ} (cos θ)- \frac{d}{d θ} (cos2 θ)=-sin θ (-2sin2 θ)=2sin2 θ -sin θ$ … (3)

Also, differentiating both sides of the equation (2) with respect to $θ$ gives

$\frac{dy}{d θ} = \frac{d}{d θ} (sin θ -sin2 θ)= \frac{d}{d θ} (sin θ)- \frac{d}{d θ} (sin2 θ)=cos θ -2cos2 θ$ ….. (4)

Therefore, dividing the equation (4) by (3) gives

$\frac{dy}{dx} = \frac{(\frac{dy}{d θ})}{(\frac{dx}{d θ})} = \frac{cos θ -2cos2 θ}{2sin2 θ -sin θ}$ .

Hence, $\frac{dy}{dx} = \frac{cos θ -2cos 2 θ}{2sin2 θ -sin θ}$ .

6. $x = a (θ - \sin θ), y = a (1 + \cos θ)$ .

Ans: The given equations are

$x=a(θ -sin θ)$ …… (1)

and $y=a(1+cos θ)$ …… (2)

Then, differentiating on both sides of the equation (1) with respect to $θ$ gives

$\frac{dx}{d θ} =a [\frac{d}{d θ} (θ)- \frac{d}{d θ} (sin θ)] =a(1-cos θ)$ …… (3)

Also, differentiating both sides of the equation (2) with respect to $θ$ gives

$\frac{dy}{d θ} =a [\frac{d}{d θ} (1)+ \frac{d}{d θ} (cos θ)] =a [0+(-sin θ)] =-asin θ$ …… (4)

Therefore, by dividing the equation (4) by (3) gives

$\frac{dy}{dx} = \frac{(\frac{dy}{d θ})}{(\frac{dx}{d θ})} = \frac{-asin θ}{a(1-cos θ)} = \frac{-2sin \frac{θ}{2} cos \frac{θ}{2}}{2si n^{2} \frac{θ}{2}} = \frac{-cos \frac{θ}{2}}{sin \frac{θ}{2}} =-cot \frac{θ}{2}$ .

Hence, $\frac{dy}{dx} =-cot \frac{θ}{2}$ .

7. $x = - \frac{si n^{3} t}{\sqrt{\cos 2 t}}, y = \frac{co s^{3} t}{\sqrt{\cos 2 t}}$ .

Ans: The given equations are,

$x=- \frac{si n^{3} t}{\sqrt{cos2t}}$ …… (1)

and $y= \frac{co s^{3} t}{\sqrt{cos2t}}$ …… (2)

Then, differentiating both sides of the equation (1) with respect to $t$ gives

$\frac{dx}{dt} = \frac{d}{dt} [\frac{si n^{3} t}{\sqrt{cos2t}}]$

$\begin{aligned} = \frac{\sqrt{cos2t} \frac{d}{dt} (si n^{3} t)-si n^{3} t \times \frac{d}{dt} \sqrt{cos2t}}{cos2t} \\ = \frac{\sqrt{cos2t} \times 3si n^{2} t \times \frac{d}{dt} (sint)-si n^{3} t \times \frac{1}{2 \sqrt{cos2t}} \times \frac{d}{dt} (cos2t)}{cos2t} \\ = \frac{3cos2t \times si n^{2} tcost- \frac{si n^{3} t}{2} \times (-2sin2t)}{cos2t \sqrt{cos2t}} \end{aligned}$

Also, differentiating both sides of the equation (2) with respect to $t$ gives

$\frac{dx}{dt} = \frac{3cos2tsi n^{2} tcost+si n^{3} tsin2t}{cos2t \sqrt{cos2t}}$ . …… (3)

$\begin{aligned} \frac{dy}{dt} = \frac{d}{dt} [\frac{co s^{3} t}{\sqrt{cos2t}}] \\ = \frac{\sqrt{cos2t} \times \frac{d}{dt} (co s^{3} t)-co s^{3} t \times \frac{d}{dt} (\sqrt{cos2t})}{cos2t} \\ = \frac{3 \sqrt{cos2t} co s^{2} t(-sint)-co s^{3} t \times \frac{1}{2(\sqrt{cos2t})} \times \frac{d}{dt} (cos2t)}{cos2t} \end{aligned}$

$\frac{dy}{dt} = \frac{-3cos2t \times co s^{2} t \times sint+co s^{3} tsin2t}{cos2t \times \sqrt{cos2t}}$ …… (4)

Thus, dividing the equation (4) by the equation (3) gives

$\frac{dy}{dx} = \frac{(\frac{dx}{dt})}{(\frac{dx}{dt})} = \frac{-3cos2t \times co s^{2} t \times sint+co s^{3} tsin2t}{3cos2tcostsi n^{2} t+si n^{3} tsin2t}$

$= \frac{sintcost [-3cos2t \times cost+2co s^{3} t]}{sintcost [3cos2tsint+2si n^{3} t]}$

$= \frac{[-3(2co s^{2} t-1)cost+2co s^{3} t]}{[3(1-2si n^{3} t)sint+2si n^{3} t]}$

$[\begin{aligned} cos2t=(2co s^{2} t-1) \\ cos2t=(1-2si n^{2} t) \end{aligned}]$

$\begin{aligned} = \frac{-4co s^{3} t+3cost}{3sint-4si n^{3} t} \\ = \frac{-cos3t}{sin3t} \end{aligned}$
$[\begin{aligned} cos3t=4co s^{3} t-3cost \\ sin3t=3sint-4si n^{2} t \end{aligned}]$

Hence, $\frac{dy}{dx} =-cot3t$ .

8. $x = a (cost + logtan \frac{t}{2}), y = asint$ .

Ans: The given equations are

$x=a (cost+logtan \frac{t}{2})$ …… (1)

and $y=asint$ …… (2)

Then, differentiating both sides of the equation (1) with respect to $t$ gives

$\frac{dx}{dt} =a \times [\frac{d}{d θ} (cost)+ \frac{d}{d θ} (logtan \frac{t}{2})]$

$\begin{aligned} =a [-sint+ \frac{1}{tan \frac{t}{2}} \times \frac{d}{dt} (tan \frac{t}{2})] \\ =a [-sint+cot \frac{t}{2} \times se c^{2} \frac{t}{2} \times \frac{d}{dt} (\frac{t}{2})] \end{aligned}$

$=a (-sint+ \frac{cos \frac{t}{2}}{sin \frac{t}{2}} \times \frac{1}{co s^{2} \frac{t}{2}} \times \frac{1}{2})$

$\begin{aligned} =a (-sint+ \frac{1}{2sin \frac{t}{2} cos \frac{t}{2}}) \\ =a (-sint+ \frac{1}{sint}) \\ =a (\frac{-si n^{2} t+1}{sint}) \end{aligned}$

Therefore, $\frac{dx}{dt} =a \frac{co s^{2} t}{sint}$ …… (3)

Also, differentiating both sides of the equation (2) with respect to $t$ gives

$\frac{dy}{dt} =a \frac{d}{dt} (sint)=acost$ …… (4)

Thus, dividing the equation (4) by the equation (3) gives

$\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} = \frac{acost}{(a \frac{co s^{2} t}{sint})} = \frac{sint}{cost} =tant$ .

Hence, $\frac{dy}{dx} =tant$ .

9. $x = asec θ, y = btan θ$ .

Ans: The given equations are

$x=asec$ …… (1)

and $y=btan θ$ …… (2)

Then, differentiating both sides of the equation (1) with respect to $θ$ gives

$\frac{dx}{d θ} =a \times \frac{d}{d θ} (sec θ)=asec θ tan θ$ …… (3)

Also, differentiating both sides of the equation (2) with respect to $θ$ gives

$\frac{dy}{d θ} =b \times \frac{d}{d θ} (tan θ)=bse c^{2} θ$ …… (4)

Thus, dividing the equation (4) by the equation (3) gives

$\frac{dy}{dx} = \frac{(\frac{dy}{d θ})}{(\frac{dx}{d θ})} = \frac{bse c^{2} θ}{asec θ tan θ} = \frac{b}{a} sec θ tan θ =- \frac{bcos θ}{acos θ sin θ} = \frac{b}{a} \times \frac{1}{sin θ} = \frac{b}{a} cosec θ$

Hence, $\frac{dy}{dx} = \frac{b}{a} cosec θ$ .

10. $x=a(cos θ + θ sin θ), y=a(sin θ - θ cos θ)$ .

Ans: The given equations are

$x = a(cos θ + θ sin θ)$ …… (1)

and $y = a(sin θ - θ cos θ)$ …… (2)

Then, differentiating both sides of the equation (1) with respect to $θ$ gives

$\begin{aligned} \frac{dx}{d θ} = a [\frac{d}{d θ} cos θ + \frac{d}{d θ} (θ sin θ)] \\ = a [-sin θ + θ \frac{d}{d θ} (sin θ)+sin θ \frac{d}{d θ} (θ)] \end{aligned}$

$=a [-sin θ + θ cos θ +sin θ]$ .

Therefore, $\frac{dx}{d θ} =a θ cos θ$ …… (3)

Also, differentiating both sides of the equation (2) with respect to $θ$ gives

$\begin{aligned} \frac{dy}{d θ} = a [\frac{d}{d θ} (sin θ)- \frac{d}{d θ} (θ cos θ)] \\ = a [cos θ - {θ \frac{d}{d θ} (cos θ)+cos θ \times \frac{d}{d θ} (θ)}] \end{aligned}$

$\Rightarrow \frac{dy}{d θ} = a [cos θ + θ sin θ -cos θ]$

Therefore, $\frac{dy}{d θ} = a θ sin θ$ …… (4)

Thus, dividing the equation (4) by the equation (3) gives

$\begin{aligned} \frac{dy}{dx} = \frac{(\frac{dy}{d θ})}{(\frac{dx}{d θ})} \\ = \frac{a θ sin θ}{a θ sin θ} \\ =tan θ \end{aligned}$ .

Hence, $\frac{dy}{dx} =tan θ$ .

11.If $x = \sqrt{a^{si n^{- 1} t}}, y = \sqrt{a^{co s^{- 1} t}}$ , show that, $\frac{dy}{dx} = - \frac{y}{x}$ .

Ans: The given parametric equations are $x= \sqrt{a^{si n^{-1} t}}$ and $y= \sqrt{a^{co s^{-1} t}}$ .

Now, $x= \sqrt{a^{si n^{-1} t}}$ and $y= \sqrt{a^{co s^{-1} t}}$

$\Rightarrow x= (a^{si n^{-1} t})$ and $y= {(a^{co s^{-1} t})}^{\frac{1}{2}}$

$\Rightarrow x= a^{\frac{1}{2} si n^{-1t}}$ and

$y= a^{\frac{1}{2} co s^{-1} t}$

Therefore, first consider $x= a^{\frac{1}{2} si n^{-1} t}$ .

Take logarithms on both sides of the equation.

Then, we have

$logx= \frac{1}{2} si n^{-1} tloga$ .

Then, differentiating both sides of the equation with respect to $t$ gives

$\begin{aligned} \frac{1}{x} \times \frac{dx}{dt} = \frac{1}{2} loga \times \frac{d}{dt} (si n^{-1} t) \\ \Rightarrow \frac{dx}{dt} = \frac{x}{2} loga \times \frac{1}{\sqrt{1- t^{2}}} \end{aligned}$

Therefore, $\frac{dx}{dt} = \frac{xloga}{2 \sqrt{1- t^{2}}}$ . …… (1)

Again, consider the equation $y= a^{\frac{1}{2} co s^{-1} t}$ .

Take logarithm both sides of the equation.

Then, we have

$logy= \frac{1}{2} co s^{-1} tloga$

Differentiating both sides of the equation with respect to $t$ gives

$\begin{aligned} \frac{1}{y} \times \frac{dx}{dt} = \frac{1}{2} loga \times \frac{d}{dt} (co s^{-1} t) \\ \Rightarrow \frac{dx}{dt} = \frac{yloga}{2} \times (\frac{1}{\sqrt{1- t^{2}}}) \end{aligned}$

Therefore, $\frac{dx}{dt} = \frac{-yloga}{2 \sqrt{1- t^{2}}}$ . …… (2)

Thus, dividing the equation (2) by the equation (1) gives

$\begin{aligned} \frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} \\ = \frac{(\frac{-yloga}{2 \sqrt{1- t^{2}}})}{(\frac{xloga}{2 \sqrt{1- t^{2}}})} \\ = \frac{y}{x} \end{aligned}$ .

Hence, $\frac{dy}{dx} = - \frac{y}{x}$ .

Exercise 5.7

Find the second order derivatives of the functions given in Exercises 1 to 10.

1. $y= x^{2} +3x+2$ .

Ans: The given function is $y= x^{2} +3x+2$ .

Then, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{d}{dx} (x^{2})+ \frac{d}{dx} (3x)+ \frac{d}{dx} (2)=2x+3+0=2x+3$

That is,

$\frac{dy}{dx} = 2 x+3$ .

Again, differentiating both sides with respect to $x$ gives

$\frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (2x+3)= \frac{d}{dx} (2x)+ \frac{d}{dx} (3)=2+0=2$

Hence, $\frac{d^{2} y}{d x^{2}} =2$ .

2. $y = x^{20}$ .

Ans: The given function is $y= x^{20}$ .

Then, differentiating both sides with respect to $x$ gives $\frac{dy}{dx} = \frac{d}{dx} (x^{20})=20 x^{19}$

Again, differentiating both sides with respect to $x$ gives

$\frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (20 x^{19})=20 \frac{d}{dx} (x^{19})=20 (19) x^{18} =380 x^{18}$ .

Hence, $\frac{d^{2} y}{d x^{2}} =380 x^{18}$ .

3. $y = x \cdot cosx$ .

Ans: The given function is $y=x .cosx$ .

Then, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{d}{dx} (x .cosx)=cosx . \frac{d}{dx} (x)+x \frac{d}{dx} (cosx)=cosx .1+x(-sinx)=cosx-xsinx$

That is, $\frac{dy}{dx} =cosx-xsinx$ .

Again, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (cosx-xsinx)= \frac{d}{dx} (cosx)- \frac{d}{dx} (xsinx) \\ =-sinx- [sinx \cdot \frac{d}{dx} (x)+x \cdot \frac{d}{dx} (sinx)] \\ =-sinx-(sinx+xcosx) \end{aligned}$

Hence, $\frac{d^{2} y}{d x^{2}} =-(xcosx+2sinx)$ .

4. $y = logx$ .

Ans: The given function is $y=logx$ .

Then, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{d}{dx} (logx)= \frac{1}{x}$

Again, differentiating both sides with respect to $x$ gives

$\frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (\frac{1}{x}) = \frac{-1}{x^{2}}$

Hence, $\frac{d^{2} y}{d x^{2}} =- \frac{1}{x^{2}}$ .

5. $y= x^{3} logx$ .

Ans: The given function is $y= x^{3} logx$ .

Then, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{d}{dx} [x^{3} logx] =logx . \frac{d}{dx} (x^{3})+ x^{3} \frac{d}{dx} (logx)=logx .3 x^{2} + x^{3} . \frac{1}{x} =logx .3 x^{2} + x^{2}$

That is, $\frac{dy}{dx} = x^{2} (1+3logx)$ .

Again, differentiating both sides with respect to $x$ gives
$\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (x^{2} (1+3logx)) \\ =(1+3logx) . \frac{d}{dx} (x^{2})+ x^{2} \frac{d}{dx} (1+3logx) \\ =(1+3logx) .2x+ x^{3} . \frac{3}{x} \\ =2x+6logx+3x \\ =5x+6xlogx \end{aligned}$

Hence, $\frac{d^{2} y}{d x^{2}} =x(5+6logx)$ .

6. $y = e^{x} \sin 5 x$

Ans: The given function is $y= e^{x} sin5x$ .

Then, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} [e^{x} sin5x] =sinx \frac{d}{dx} (e^{x})+ e^{x} \frac{d}{dx} (sin5x) \\ \Rightarrow \frac{dy}{dx} =sin5x . e^{x} + e^{x} .cos5x . \frac{d}{dx} (5x) \end{aligned}$

That is, $\frac{dy}{dx} = e^{x} (sin5x+5cos5x)$ .

Again, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} [e^{x} (sin5x+5cos5x)] \\ =(sin5x+5cos5x) . \frac{d}{dx} (e^{x})+ e^{x} . \frac{d}{dx} (sin5x+5cos5x) \\ =(sin5x+5cos5x)(e^{x})+ e^{x} [cos5x . \frac{d}{dx} (5x)+5(-sin5x) . \frac{d}{dx} (5x)] \\ = e^{x} (sin5x+5cos5x)+ e^{x} (5cos5x-25sin5x) \end{aligned}$

$= e^{x} (10cos5x-24sin5x)$ .

Hence, $\frac{d^{2} y}{d x^{2}} =2 e^{x} (5cox5x-12sin5x)$ .

7. $y= e^{6x} cos3x$ .

Ans: The given function is $y= e^{6x} cos3x$ .

Then, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} (e^{6x} cos3x)=cos3x \times \frac{d}{dx} (e^{6x})+ e^{6x} \times \frac{d}{dx} (cos3x) \\ \Rightarrow \frac{dy}{dx} = cos3x \times e^{6x} \times \frac{d}{dx} (6x)+ e^{6x} \times (-sin3x) \times \frac{d}{dx} (3x) \end{aligned}$

Therefore,

$\frac{dy}{dx} =6 e^{6x} cos3x-3 e^{6x} sin3x$ …… (1)

Again, differentiating both sides with respect to $x$ gives

$\frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (6 e^{6x} cos3x-3 e^{6x} sin3x)=6 \times \frac{d}{dx} (e^{6x} cos3x)-3 \times \frac{d}{dx} (e^{6x} sin3x)$

$=6 \times [6 e^{6x} cos3x-3 e^{6x} sin3x] -3 \times [sin3x \times \frac{d}{dx} (e^{6x})+ e^{6x} \times \frac{d}{dx} (sin3x)]$ [using (1)]

$\begin{aligned} =36 e^{6x} cos3x-18 e^{6x} sin3x-3 [sin3x \times e^{6x} \times 6+ e^{6x} \times cos3x-3] \\ =36 e^{6x} cos3x-18 e^{6x} sin3x-18 e^{6x} sin3x-9 e^{6x} cos3x \end{aligned}$

Hence, $\frac{d^{2} y}{d x^{2}} =9 e^{6x} (3cos3x-4sin3x)$ .

8. $y = ta n^{- 1} x$ .

Ans: The given function is $y=ta n^{-1} x$ .

Then, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{d}{dx} ta n^{-1} x= \frac{1}{1- x^{2}}$

Again, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (\frac{1}{1+ x^{2}}) = \frac{d}{dx} {(1+ x^{2})}^{-1} =(-1) \times (1+ x^{2})^{- 2} \times \frac{d}{dx} (1+ x^{2}) \\ =- \frac{1}{{(1+ x^{2})}^{2}} \times 2x \end{aligned}$

Hence,

$\frac{d^{2} y}{d x^{2}} = \frac{-2x}{{(1+ x^{2})}^{2}}$ .

9. $y=log(logx)$ .

Ans: The given function is $y=log(logx)$ .

Now, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} [log(logx)] \\ = \frac{1}{logx} \times \frac{d}{dx} (logx) \\ \Rightarrow \frac{dy}{dx} =(xlogx)^{-1} \end{aligned}$

Again, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} [{(xlogx)}^{-1}] =(-1) \times (xlogx)^{-2} \frac{d}{dx} (xlogx) \\ = \frac{-1}{{(xlogx)}^{2}} \times [logx \times \frac{d}{dx} (x)+x \times \frac{d}{dx} (logx)] \\ = \frac{-1}{{(xlogx)}^{2}} \times [logx \times 1+x \times \frac{1}{x}] \end{aligned}$

Hence, $\frac{d^{2} y}{d x^{2}} = \frac{-(1+logx)}{{(xlogx)}^{2}}$ .

10. $y=sin(logx)$ .

Ans: The given function is $y=sin(logx)$ .

Now, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{d}{dx} [sin(logx)] =cos(logx) \times \frac{d}{dx} (logx) = \frac{cos(logx)}{x}$

Again, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} [\frac{cos(logx)}{x}] \\ = \frac{x [cos(logx)] -cos(logx) \times \frac{d}{dx} (x)}{x^{2}} \\ = \frac{x [-sin(logx) \times \frac{d}{dx} (logx)] -cos(logx) \times 1}{x^{2}} \\ = \frac{-xsin(logx) \times \frac{1}{x} -cos(logx)}{x^{2}} \end{aligned}$

Hence, $\frac{d^{2} y}{d x^{2}} = \frac{[-sin(logx)-cos(logx)]}{x^{2}}$ .

11. If $y=5cosx-3sinx$ , prove that $\frac{d^{2} y}{d x^{2}} +y=0$ .

Ans: The given equation is $y=5cosx-3sinx$ .

Then, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{d}{dx} (5cosx)- \frac{d}{dx} (3sinx)=5 \frac{d}{dx} (cosx)-3 \frac{d}{dx} (sinx)=5(-sinx)-3cosx$

Therefore, $\frac{dy}{dx} =-(5sinx+3cosx)$ .

Again, differentiating both sides with respect to $x$ gives

$\frac{d^{2} y}{d x^{2}} = \frac{d}{dx} [-(5sinx+3cosx)]$

$\begin{aligned} =- [5 \times \frac{d}{dx} (sinx)+3 \times \frac{d}{dx} (cosx)] \\ = [5cosx+3(-sinx)] \\ =-y \end{aligned}$

That is, $\frac{d^{2} y}{d x^{2}} =-y$ .

Hence, $\frac{d^{2} y}{d x^{2}} +y=0$ .

12. If $y = co s^{- 1} x$ , Find $\frac{d^{2} y}{d x^{2}}$ in the terms of $y$ alone.

Ans: The given function is $y=co s^{-1} x$ .

Now, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{d}{dx} (co s^{-1} x)= \frac{-1}{\sqrt{1- x^{2}}} =-(1- x^{2})^{\frac{-1}{2}}$

Again, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} [-(1- x^{2})^{\frac{-1}{2}}] \\ = (\frac{-1}{2}) \times (1- x^{2})^{\frac{-3}{2}} \times \frac{d}{dx} (1- x^{2}) \\ = \frac{1}{\sqrt[2]{{(1- x^{2})}^{3}}} \times (-2x) \end{aligned}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{-x}{\sqrt{{(1- x^{2})}^{3}}}$ …… (1)

Now, $y=co s^{-1} x \Rightarrow x=cosy$ .

Therefore, substituting $x=cosy$ into equation (1), gives

$\begin{aligned} \frac{d^{2} x}{d y^{2}} = \frac{-cosy}{\sqrt{{(1-co s^{2} y)}^{3}}} \\ = \frac{-cosy}{si n^{3} y} \\ = \frac{-cosy}{siny} \times \frac{1}{si n^{2} y} \end{aligned}$

Hence, $\frac{d^{2} y}{d x^{2}} = -coty \times cose c^{2} y$ .

13. If $y = 3 \cos (logx) + 4 \sin (logx)$ , show that $x^{2} y_{2} + x y_{1} + y = 0$ .

Ans: The given equations are $y=3cos(logx)+4sin(logx)$ …… (1)

and $x^{2} y_{2} +x y_{1} +y=0$ …… (2)

Then, differentiating both sides of the equation (1) with respect to $x$ gives

$\begin{aligned} y_{1} =3 \times \frac{d}{dx} [cos(logx)] +4 \times \frac{d}{dx} [sin(logx)] \\ =3 \times [-sin(logx) \times \frac{d}{dx} (logx)] +4 \times [cos(logx) \times \frac{d}{dx} (logx)] \end{aligned}$

$y_{1} = \frac{-3sin(logx)}{x} + \frac{4cos(logx)}{x} = \frac{4cos(logx)-3sin(logx)}{x}$

Again, differentiating both sides with respect to $x$ gives

$y_{2} = \frac{d}{dx} (\frac{4cos(logx)-3sin(logx)}{x})$

$\begin{aligned} = \frac{x \cdot \frac{d}{d x} [4 {cos(logx)} - {3sin(logx)}] - {4cos(logx)-3sin(logx)} \times 1}{x^{2}} \\ = \frac{x [-4sin(logx) \frac{d}{d x} (logx)-3cos(logx) \frac{d}{d x} (logx)] -4cos(logx)+3sin(logx)}{x^{2}} \end{aligned}$

$= \frac{x [-4sin(logx) \cdot \frac{1}{x} -3cos(logx) \cdot \frac{1}{x}] -4cos(logx)+3sin(logx)}{x^{2}}$

$= \frac{-4sin(logx)-3cos(logx)-4cos(logx)+3sin(logx)}{x^{2}}$

Therefore,

$y_{2} = \frac{-sin(logx)-7cos(logx)}{x^{2}}$ .

Now, substituting the derivatives $y_{1}$ , $y_{2}$ and $y$ into the LHS of the equation (2) gives

$\begin{aligned} x^{2} y_{2} +x y_{1} +y \\ = x^{2} (\frac{-sin(logx)-7cos(logx)}{x^{2}}) +x (\frac{4cos(logx)-3sin(logx)}{x^{2}}) +3cos(logx)+4sin(logx) \\ =-sin(logx)-7cos(logx)+4cos(logx)-3sin(logx)+4sin(logx) \\ =0 \end{aligned}$

Hence, it has been proved that $x^{2} y_{2} +x y_{1} +y = 0$ .

14. If $y = A e^{mx} + B e^{nx}$ , show that $\frac{d^{2} y}{d x^{2}} - (m + n) \frac{dy}{dx} + mny = 0$ .

Ans: The given equations are $y=A e^{mx} +B e^{nx}$ …… (1)

and $\frac{d^{2} y}{d x^{2}} -(m+n) \frac{dy}{dx} +mny=0$ ……. (2)

Then, differentiating both sides of the equation (1) with respect to $x$ gives

$\frac{dy}{dx} =A . \frac{d}{dx} (e^{mx})+B . \frac{d}{dx} (e^{mx})=A . e^{mx} . \frac{d}{dx} (mx)+B . e^{nx} . \frac{d}{dx} (nx)=Am e^{mx} +Bn e^{nx}$

Again, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (Am e^{mx} +Bn e^{nx})=Am . \frac{d}{dx} (e^{mx})+Bn . \frac{d}{dx} (e^{nx}) \\ =Am . e^{mx} . \frac{d}{dx} (mx)+Bn . e^{nx} . \frac{d}{dx} (nx) \end{aligned}$

Therefore, $\frac{d^{2} y}{d x^{2}} =A m^{2} e^{mx} +B n^{2} e^{nx}$ .

Thus, substituting the derivatives $y_{1}$ , $y_{2}$ and $y$ into the LHS of the equation (2) gives

$\begin{aligned} \frac{d^{2} y}{d x^{2}} -(m+n) \frac{dy}{dx} +mny \\ =A m^{2} e x^{mx} +B n^{2} e^{nx} -(m+n) .(Am e^{mx} +Bn e^{nx})+mn(A e^{mx} +B e^{nx}) \\ =A m^{2} e x^{mx} +B n^{2} e^{nx} -Ame x^{mx} +Bmn e^{nx} +Amn e^{mx} +B n^{2} e^{nx} +Amn e^{mx} +Bmn e^{nx} \\ =0 \end{aligned}$

Thus, it has been proved that $\frac{d^{2} y}{d x^{2}} -(m+n) \frac{dy}{dx} +mny=0$ .

15. If $y = 500 e^{7 x} + 600 e^{- 7 x}$ , show that $\frac{d^{2} y}{d x^{2}} = 49 y$ .

Ans: The given equation is $y=500 e^{7x} +600 e^{-7x}$ . …… (1)

Then, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{dy}{dx} =500 \times (e^{7x})+600 \times \frac{d}{dx} (-7x) \\ =500 \times e^{7x} \times \frac{d}{dx} (7x)+600 \times e^{-7x} \times \frac{d}{dx} (-7x) \\ =3500 e^{7x} -4200 e^{-7x} \end{aligned}$

Again, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{d^{2} y}{d x^{2}} =3500 \times \frac{d}{dx} (e^{7x})-4200 \times \frac{d}{dx} (e^{-7x}) \\ =3500 \times e^{7x} \times \frac{d}{dx} (7x)-4200 \times e^{-7x} \times \frac{d}{dx} (-7x) \\ =7 \times 3500 \times e^{7x} +7 \times 4200 \times e^{-7x} \\ =49 \times 500 e^{7x} +49 \times 600 e^{-7x} \\ =49(500 e^{7x} +600 e^{-7x}) \end{aligned}$

$=49y$ , using the equation (1).

Thus, it has been proved that $\frac{d^{2} y}{d x^{2}} =49y$ .

16. If $e^{y} (x + 1) = 1$ , show that $\frac{d^{2} y}{d x^{2}} = {(\frac{dy}{dx})}^{2}$ .

Ans: The given equation is $e^{y} (x+1)=1$ .

Now, $e^{y} (x+1)=1 \Rightarrow e^{y} = \frac{1}{x+1}$ .

So, taking logarithm bth sides of the equation gives

$y=log \frac{1}{(x+1)}$

Therefore, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} =(x+1) \frac{d}{dx} (\frac{1}{x+1}) =(x+1) \times \frac{-1}{{(x+1)}^{2}} = \frac{-1}{x+1}$

That is,

$\frac{dy}{dx} = \frac{-1}{x+1}$ …… (1)

Again, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} = (\frac{1}{x+1}) =- (\frac{-1}{{(x+1)}^{2}}) = \frac{1}{{(x+1)}^{2}} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = {(\frac{-1}{x+1})}^{2} \end{aligned}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = {(\frac{dy}{dx})}^{2}$ , using the equation (1).

Thus, it is proved that $\frac{d^{2} y}{d x^{2}} = {(\frac{dy}{dx})}^{2}$ .

17.If $y = (ta n^{- 1} x)^{2}$ , show that ${(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2$ .

Ans: The given equations are $y=(ta n^{-1} x)^{2}$ .

Then, differentiating both sides with respect to $x$ gives

$\begin{aligned} y_{1} =2ta n^{-1} x \frac{d}{dx} (ta n^{-1} x) \\ \Rightarrow y_{1} =2ta n^{-1} x \times \frac{1}{1+ x^{2}} \\ \Rightarrow (1+ x^{2}) y_{1} =2ta n^{-1} x \end{aligned}$

Again, differentiating both sides with respect to $x$ gives

$\begin{aligned} (1+ x^{2}) y_{2} +2x y_{1} =2 (\frac{1}{1+ x^{2}}) \\ \Rightarrow (1+ x^{2}) y_{2} +2x(1+ x^{2}) y_{1} =2 \end{aligned}$

Thus, it has been proved that $(1+ x^{2}) y_{2} +2x(1+ x^{2}) y_{1} =2$ .

Miscellaneous Exercise

Differentiate w.r.t. to $x$ , the following function.

1. $y = (3 x^{2} - 9 x + 5)^{9}$ .

Ans: The given function is

$y=(3 x^{2} -9x+5)^{9}$ .

Differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} {(3 x^{2} -9x+5)}^{9} \\ =9(3 x^{2} -9x+5)^{8} \times \frac{d}{dx} (3 x^{2} -9x+5) \\ =9(3 x^{2} -9x+5)^{8} \times (6x-9x) \\ =9(3 x^{2} -9x+5)^{8} \times 3(2x-3) \\ =27(3 x^{2} -9x+5)^{8} (2x-3) \end{aligned}$

2. $y = si n^{3} x + co s^{6} x$ .

Ans: The given function is

$y=si n^{3} x+co s^{6} x$ .

Differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} =(si n^{3} x)+ \frac{d}{dx} (co s^{6} x) \\ =3si n^{2} x \times \frac{d}{dx} (sinx)+6co s^{5} x \frac{d}{dx} (cosx) \\ =3si n^{2} x \times cosx+6co s^{5} x(-sinx) \\ {=3sin}^{2} xcosx(sinx-2co s^{4} x) \end{aligned}$

3. $y = (5 x)^{3 \cos 2 x}$ .

Ans: The given function is $y=(5x)^{3cos2x}$ .

First, take the logarithm of both sides of the function.

$\log y = 3 \cos 2 x \log 5 x$ .

Then, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{1}{y} \frac{dy}{dx} =3 [log5 . \frac{d}{dx} (cos2x)+cos2x . \frac{d}{dx} (log5x)] \\ \Rightarrow \frac{dy}{dx} =3y [log5x(-sin2x) . \frac{d}{dx} (2x)+cos2x . \frac{1}{5x} . \frac{d}{dx} (5x)] \\ \Rightarrow \frac{dy}{dx} =3y [-2sin2xlog5x+ \frac{cos2x}{x}] \\ \Rightarrow \frac{dy}{dx} =3y [\frac{3cos2x}{x} -6sin2xlog5x] \end{aligned}$

Hence, $\frac{dy}{dx} =(5x)^{3cos2x} [\frac{3cos2x}{x} -6sin2xlog5x]$ .

4. $y=si n^{-1} (x \sqrt{x}), 0 \leq x \leq 1$ .

Ans: The given function is $y=si n^{-1} (x \sqrt{x})$ .

Then, differentiating both sides with respect to $x$ by using the chain rule gives

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} si n^{-1} (x \sqrt{x}) \\ = \frac{1}{\sqrt{1- {(x \sqrt{x})}^{2}}} \times \frac{d}{dx} (x \sqrt{x}) \\ = \frac{1}{\sqrt{1- x^{3}}} . \frac{d}{dx} (x^{\frac{3}{2}}) \\ = \frac{1}{\sqrt{1- x^{3}}} \times \frac{3}{2} . x^{\frac{1}{2}} \\ = \frac{3 \sqrt{x}}{2 \sqrt{1- x^{3}}} \end{aligned}$

Hence,

$\frac{dy}{dx} = \frac{3}{2} \sqrt{\frac{x}{1- x^{3}}}$ .

5. $y= \frac{co s^{-1} \frac{x}{2}}{\sqrt{2x+7}}, -2$ < $x$ < $2$ .

Ans: The given function is

$y= \frac{co s^{-1} \frac{x}{2}}{\sqrt{2x+7}}$ .

Then, differentiating both sides with respect to $x$ using the quotient rule gives

$\begin{aligned} \frac{dy}{dx} = \frac{\sqrt{2x+7} \frac{d}{dx} (co s^{-1} \frac{x}{2}) - (co s^{-1} \frac{x}{2}) \frac{d}{dx} (\sqrt{2x+7})}{{(\sqrt{2x+7})}^{2}} \\ = \frac{\sqrt{2x+7} [\frac{-1}{\sqrt{1- {(\frac{x}{2})}^{2}}} . \frac{d}{dx} (\frac{x}{2})] - (co s^{-1} \frac{x}{2}) \frac{1}{2 \sqrt{2x+7}} . \frac{d}{dx} (2x+7)}{2x+7} \\ = \frac{\sqrt{2x+7} \frac{-1}{\sqrt{4- x^{2}}} - (co s^{-1} \frac{x}{2}) \frac{2}{2 \sqrt{2x+7}}}{2x+7} \\ = \frac{- \sqrt{2x+7}}{\sqrt{4- x^{2}} \times (2x+7)} - \frac{co s^{-1} \frac{x}{2}}{(\sqrt{2x+7}) (2x+7)} \end{aligned}$

Hence, $\frac{dy}{dx} =- [\frac{1}{\sqrt{4 - x^{2}} \sqrt{2 x + 7}} + \frac{co s^{-1} \frac{x}{2}}{{(2x+7)}^{\frac{3}{2}}}]$ .

6. $y = co t^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}], 0 < x < 2$ .

Ans: The given function is $y=co t^{-1} [\frac{\sqrt{1+sinx} + \sqrt{1-sinx}}{\sqrt{1+sinx} - \sqrt{1-sinx}}]$ ……. (1)

Now,

$\frac{\sqrt{1+sinx} + \sqrt{1-sinx}}{\sqrt{1+sinx} - \sqrt{1-sinx}}$

$\begin{aligned} = \frac{(\sqrt{1+sinx} + \sqrt{1-sinx})}{(\sqrt{1+sinx} - \sqrt{1-sinx}) \sqrt{1+sinx} + \sqrt{1-sinx}} \\ = \frac{(1+sinx)+(1-sinx)+2 \sqrt{(1+sinx)-(1-sinx)}}{(1+sinx)-(1-sinx)} \\ = \frac{2+2 \sqrt{1-si n^{2} x}}{2sinx} \\ = \frac{1+cosx}{sinx} \\ = \frac{2co s^{2} \frac{x}{2}}{2sinx \frac{x}{2} cos \frac{x}{2}} \end{aligned}$

Therefore,

$\frac{\sqrt{1+sinx} + \sqrt{1-sinx}}{\sqrt{1+sinx} - \sqrt{1-sinx}} =cot \frac{x}{2}$ . …… (2)

So, from the equations (1) and (2) we obtain,

$\begin{aligned} y=co t^{-1} (co t^{\frac{x}{2}}) \\ \Rightarrow y= \frac{x}{2} \end{aligned}$

Now, differentiating both sides with respect to $x$ gives $\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (x)$ Hence, $\frac{dy}{dx} = \frac{1}{2}$ .

7. $y=(logx)^{logx}, x > 1$ .

Ans: The given function is $y=(logx)^{logx}$ .

First take logarithm both sides of the function.

$logy=logx \times log(logx)$ .

Now, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} [logx \times log(logx)] \\ \Rightarrow \frac{1}{y} \frac{dy}{dx} =log(logx) \times \frac{d}{dx} (logx)+ \frac{d}{dx} [log(logx)] \\ \Rightarrow \frac{dy}{dx} =y [log(logx) \times \frac{1}{x} +logx \times \frac{1}{logx} \times \frac{d}{dx} (logx)] \\ \Rightarrow \frac{dy}{dx} =y [\frac{1}{x} log(logx)+ \frac{1}{x}] \end{aligned}$

Hence, $\frac{dy}{dx} =(logx)^{logx} [\frac{1}{x} + \frac{log(logx)}{x}]$ .

8. $y = \cos (acosx + bsinx)$ , for some constants $a$ and $b$ .

Ans: The given function is $y=cos(acosx+bsinx)$ .

Now, differentiating both sides with respect to $x$ by using the chain rule of derivatives gives

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} cos(acosx+bsinx) \\ \Rightarrow \frac{dy}{dx} =-sin(acosx+bsinx) \times \frac{d}{dx} (acosx+bsinx) \\ = -sin(acosx+bsinx) \times [a(-sinx)+bcosx] \end{aligned}$

Hence, $\frac{dy}{dx} = (asinx-bcosx) \times sin(acosx+bsinx)$ .

9. $y = (sinx - cosx)^{(sinx - cosx)}, \frac{π}{4} < x < \frac{3 π}{4}$ .

Ans: The given function is $y=(sinx-cosx)^{(sinx-cosx)}$ .

First take logarithm both sides of the function.

$\begin{aligned} logy=log [{(sinx-cosx)}^{(sinx-cosx)}] \\ \Rightarrow logy=(sinx-cosx) \times log(sinx-cosx) \end{aligned}$

Now, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} [(sinx-cosx) \times log(sinx-cosx)] \\ \Rightarrow \frac{1}{y} \frac{dy}{dx} =log(sinx-cosx) \times \frac{d}{dx} (sinx-cosx)+(sinx-cosx) \times \frac{d}{dx} log(sinx-cosx) \\ \Rightarrow \frac{1}{y} \frac{dy}{dx} =log(sinx-cosx) \times (cosx+sinx)+(sinx-cosx) \times \frac{1}{(sinx-cosx)} \times \frac{d}{dx} (sinx-cosx) \\ \Rightarrow \frac{dy}{dx} =(sinx-cosx)^{(sinx-cosx)} [(cosx+sinx) \times log(sinx-cosx)+(cosx+sinx)] \end{aligned}$

Hence, the required derivative is

$\frac{dy}{dx} =(sinx-cosx)^{(sinx-cosx)} (cosx+sinx) [1+log(sinx-cosx)]$ .

10. $y = x^{x} + x^{a} + a^{x} + a^{a}$ , for some fixed $a > 0$ and $x > 0$ .

Ans: The given function is

$y= x^{x} + x^{a} + a^{x} + a^{a}$ .

Now, assume that $x^{x} =u$ ,

$x^{a} =v$ , $a^{x} =w$ and $a^{a} =s$

Therefore, we have $y=u+v+w+s$ .

So, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} + \frac{ds}{dx}$ …… (1)

Also, $u= x^{x}$

$\begin{aligned} \Rightarrow logu=log x^{x} \\ \Rightarrow logu=xlogx \end{aligned}$

Then, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{1}{u} \frac{du}{dx} =logx . \frac{d}{dx} (x)+x . \frac{d}{dx} (logx) \\ \Rightarrow \frac{du}{dx} =u [logx .1+x . \frac{1}{x}] \end{aligned}$

Thus, $\frac{du}{dx} = x^{x} [logx+1] = x^{x} (1+logx)$ ……. (2)

Again, $v= x^{a}$

Then, differentiating both sides with respect to $x$ gives

$\frac{du}{dx} = \frac{d}{dx} (x^{a})$

$\Rightarrow \frac{dv}{dx} =a x^{a-1}$ …… (3)

Also, $w= a^{x}$

$\begin{aligned} \Rightarrow logw=log a^{x} \\ \Rightarrow logw=xloga \end{aligned}$

So, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{1}{w} . \frac{dw}{dx} =loga . \frac{d}{dx} (x) \\ \Rightarrow \frac{dw}{dx} =wloga \end{aligned}$

$\Rightarrow \frac{dw}{dx} = a^{x} loga$ ……… (4)

and

$s= a^{a}$

Then differentiating both sides with respect to $x$ gives

$\frac{ds}{dx} =0$ , ……(5)

as $a$ is constant, and so $a^{a}$ is also a constant.

Now, from the equations (1), (2), (3), (4), and (5) we have

$\frac{dy}{dx} = x^{x} (1+logx)+a x^{a-1} + a^{x} loga+0$

Hence, $\frac{dy}{dx} = x^{x} (1+logx)+a x^{a-1} + a^{x} loga$ .

11. $y = x^{x^{2} - 3} + (x - 3)^{x^{2}}$ , for $x > 3$ .

Ans: The given function is

$y= x^{x^{2} -3} +(x-3)^{x^{2}}$ .

Now suppose that $u= x^{x^{2} -3}$ and $v=(x-3)^{x^{2}}$

Therefore, $y=u+v$ .

Now, differentiating both sides with respect to $x$ gives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ ……. (1)

Also, $u= x^{x^{2} -3}$ .

Take logarithm both sides of the equation.

$\begin{aligned} \Rightarrow logu=log(x^{x^{2} -3}) \\ \Rightarrow logu=(x^{2} -3)logx \end{aligned}$

Differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{1}{u} \frac{du}{dx} =logx . \frac{d}{dx} (x^{2} -3)+(x^{2} -3) . \frac{d}{dx} (logx) \\ \Rightarrow \frac{1}{u} \frac{du}{dx} =logx .2x+(x^{2} -3) . \frac{1}{x} \end{aligned}$

Hence, $\frac{du}{dx} = x^{x^{2} -3} . [\frac{x^{2} -3}{x} +2 \times logx]$ . …… (2)

Again, $v=(x-3)^{x^{2}}$ .

Take logarithm both sides of the equation.

$\begin{aligned} \Rightarrow logv=log(x-3)^{x^{2}} \\ \Rightarrow logv= x^{2} log(x-3) \end{aligned}$

Now, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{1}{u} . \frac{dv}{dx} =log(x-3) . \frac{d}{dx} (x^{2})+ x^{2} . \frac{d}{dx} [log(x-3)] \\ \Rightarrow \frac{1}{u} . \frac{dv}{dx} =log(x-3) .2x+ x^{2} . \frac{1}{x-3} . \frac{d}{dx} (x-3) \\ \Rightarrow \frac{dv}{dx} =v . [2xlog(x-3)+ \frac{x^{2}}{x-3} .1] \end{aligned}$

Hence, $\frac{dv}{dx} =(x-3)^{x^{2}} [\frac{x^{2}}{x-3} +2xlog(x-3)]$ …… (3)

Thus, from the equations (1), (2) and (3) we obtain

$\frac{dy}{dx} = x^{x^{2} -3} [\frac{x^{2} -3}{x} +2xlogx] +(x-3)^{x^{2}} [\frac{x^{2}}{x-3} +2xlog(x-3)]$ .

12. Find $\frac{dy}{dx}$ if $y = 12 (1 - cost), x = 10 (t - sint), - \frac{π}{2} < t < \frac{π}{2}$ .

Ans: The given equations are $y=12(1-cost),$ …… (1)

and $x=10(t-sint)$ …… (2)

Then differentiating the equations (1) and (2) with respect to $x$ gives

$\begin{aligned} \frac{dx}{dt} = \frac{d}{dt} [10(t-sint)] =10 \times \frac{d}{dt} (t-sint)=10(1-cost) \\ \frac{dy}{dt} = \frac{d}{dt} [12(1-cost)] =12 \times \frac{d}{dt} (1-cost)=12 \times [0-(-sint)] =12sint \end{aligned}$

Therefore, by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$ we have,

$\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} = \frac{12sint}{10(1-cost)} = \frac{12 \times 2sin \frac{t}{2} \times cos \frac{t}{2}}{10 \times 2si n^{2} \frac{t}{2}}$

Hence, $\frac{dy}{dx} = \frac{6}{5} cot \frac{t}{2}$ .

13. Find $\frac{dy}{dx}$ , if $y=si n^{-1} x+si n^{-1} \sqrt{1- x^{2}}, 0 < x < 1$ .

Ans: The given equation is $y=si n^{-1} x+si n^{-1} \sqrt{1- x^{2}}$ .

Differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} [si n^{-1} x+si n^{-1} \sqrt{1- x^{2}}] \\ \Rightarrow \frac{dy}{dx} = \frac{d}{dx} (si n^{-1} x)+ \frac{d}{dx} (si n^{-1} \sqrt{1- x^{2}}) \end{aligned}$

$\begin{aligned} \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1- x^{2}}} + \frac{1}{\sqrt{1- {(\sqrt{1- x^{2}})}^{2}}} \times \frac{d}{dx} (\sqrt{1- x^{2}}) \\ \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1- x^{2}}} - \frac{1}{\sqrt{1 - 1 + x^{2}}} \frac{1}{2 \times \sqrt{1- x^{2}}} \times 2 x \\ \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1- x^{2}}} - \frac{1}{\sqrt{1- x^{2}}} \end{aligned}$

Hence, $\frac{dy}{dx} =0$ .

14. If $x \sqrt{1 + y} + y \sqrt{1 + x} = 0$ , for $- 1$ < $x$ < $1$ , prove that

$\frac{dy}{dx} = - \frac{1}{{(1 + x)}^{2}}$ .

Ans: The given equation is

$\begin{aligned} x \sqrt{1+y} +y \sqrt{1+x} =0 \\ \Rightarrow x \sqrt{1+y} =y \sqrt{1+x} \end{aligned}$

Now, squaring both sides of the equation, gives

$\begin{aligned} x^{2} (1+y)= y^{2} (1+x) \\ \Rightarrow x^{2} + x^{2} y= y^{2} +x y^{2} \\ \Rightarrow x^{2} - y^{2} =x y^{2} - x^{2} y \\ \Rightarrow x^{2} - y^{2} =xy(y-x) \\ \Rightarrow (x+y)(x-y)=xy(y-x) \\ ∴ x+y=-xy \\ \Rightarrow (1+x)y=-x \\ \Rightarrow y= \frac{-x}{(1+x)} \end{aligned}$

Now, differentiating both sides of the equation with respect to $x$ gives

$\frac{dy}{dx} =- \frac{(1+x) \frac{d}{dx} (x)-x \frac{d}{dx} (1+x)}{{(1+x)}^{2}} =- \frac{(1+x)-x}{{(1+x)}^{2}}$

Hence, $\frac{dy}{dx} =- \frac{1}{{(1+x)}^{2}}$ .

15. If ${(x - a)}^{2} + (y - b)^{2} = c^{2}$ , for some constant $c > 0$ , prove that $\frac{{[1 + {(\frac{dy}{dx})}^{2}]}^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}$ is a constant independent of $a$ and $b$ .

Ans: The given equation is ${(x-a)}^{2} +(y-b)^{2} = c^{2}$ .

Differentiating both sides of the equation with respect to $x$ gives $\begin{aligned} \frac{d}{dx} = [(x-a)^{2}] + \frac{d}{dx} [(y-b)^{2}] = \frac{d}{dx} (c^{2}) \\ \Rightarrow 2(x-a) . \frac{d}{dx} (x-a)+2(y-b) . \frac{d}{dx} (y-b)=0 \\ \Rightarrow 2(x-a) .1+2(y-b) . \frac{dy}{dx} =0 \end{aligned}$

Hence, $\frac{dy}{dx} = \frac{-(x-a)}{y-b}$ ..…... (1)

Again, differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} [\frac{-(x-a)}{y-b}] \\ =- \frac{[(y-b) . \frac{d}{dx} (x-a)-(x-a) . \frac{d}{dx} (y-b)]}{{(y-b)}^{2}} \\ =- [\frac{(y-b)-(x-a) . \frac{dy}{dx}}{{(y-b)}^{2}}] \\ =- [\frac{(y-b)-(x-a) . {\frac{-(x-a)}{y-b}}}{{(y-b)}^{2}}] \\ =- [\frac{{(y-b)}^{2} +(x+a)^{2}}{{(y-b)}^{2}}] \end{aligned}$

Therefore,

${[\frac{1+ {(\frac{dy}{dx})}^{2}}{\frac{d^{2} y}{d x^{2}}}]}^{\frac{3}{2}} = \frac{{[(1+ \frac{{(x-a)}^{2}}{{(y-b)}^{2}})]}^{\frac{3}{2}}}{- [\frac{{(y-a)}^{2} +(x-a)^{2}}{{(y-a)}^{3}}]} = \frac{{[\frac{{(y-b)}^{2} +(x-a)^{2}}{{(y-b)}^{2}}]}^{\frac{3}{2}}}{- [\frac{{(y-a)}^{2} +(x-a)^{2}}{{(y-a)}^{3}}]} = \frac{{[\frac{c^{2}}{{(y-b)}^{2}}]}^{\frac{3}{2}}}{- \frac{c^{2}}{{(y-b)}^{3}}}$

$\Rightarrow {[\frac{1+ {(\frac{dy}{dx})}^{2}}{\frac{d^{2} y}{d x^{2}}}]}^{\frac{3}{2}} = \frac{\frac{c^{2}}{{(y-b)}^{3}}}{\frac{c^{2}}{{(y-b)}^{3}}} = -c$ , is a constant, and is independent of $a$ and $b$ .

16. If $cosy=xcos(a+y)$ , with $cosa \neq \pm 1$ , prove that $\frac{dy}{dx} = \frac{co s^{2} (a + y)}{sina}$ .

Ans: The given equation is $cosy=xcos(a+y)$ .

Then, differentiating both sides of the equation with respect to $x$ gives

$\begin{aligned} \frac{d}{dx} [cosy] = \frac{d}{dx} [xcos(a+y)] \\ \Rightarrow -siny \frac{dy}{dx} =cos(a+y) . \frac{d}{dx} (x)+x . \frac{d}{dx} [cos(a+y)] \\ \Rightarrow -siny \frac{dy}{dx} =cos(a+y)+x . [-sin(a+y)] \frac{dy}{dx} \end{aligned}$

$\Rightarrow [xsin(a+y)-siny] \frac{dy}{dx} =cos(a+y)$ …….. (1)

Since $cosy=xcos(a+y) \Rightarrow x= \frac{cosy}{cos(a+y)}$ , so from the equation (1) gives

$\begin{aligned} [\frac{cosy}{cos(a+y)} .sin(a+y)-siny] \frac{dy}{dx} =cos(a+y) \\ \Rightarrow [cosy .sin(a+y)-siny .cos(a+y)] . \frac{dy}{dx} =co s^{2} (a+y) \\ \Rightarrow sin(a+y-y) \frac{dy}{dx} =co s^{2} (a+y) \end{aligned}$

Hence, it has been proved that $\frac{dy}{dx} = \frac{co s^{2} (a+y)}{sina}$ .

17. If $x = a (cost + tsint)$ and $y = a (sint - tcost)$ , find $\frac{d^{2} y}{d x^{2}}$ .

Ans: The given equations are

$x=a(cost+tsint)$ …… (1)

and $y=a(sint-tcost)$ …… (2)

Then, differentiating both sides of the equation (1) with respect to $x$ gives

$\begin{aligned} \frac{dx}{dt} =a [-sint+sint . \frac{d}{dx} (t)+t . \frac{d}{dt} (sint)] \\ =a [-sint+sint+cost] =atcost \end{aligned}$

Again, differentiating both sides of the equation (2) with respect to $x$ gives

$\begin{aligned} \frac{dy}{dt} =a . \frac{d}{dt} (sint-tcost) \\ a [cost- {cost . \frac{d}{dt} (t)+t . \frac{d}{dt} (cost)}] \\ a [cost- {cost-tsint}] =atsint \end{aligned}$

Therefore,

$\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dx})} = \frac{atsint}{atcost} =tant$

Now, differentiating both sides with respect to $x$ gives

$\frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (\frac{dy}{dx}) = \frac{d}{dx} (tant)=se c^{2} t . \frac{dt}{dx} =se c^{2} t . \frac{1}{atcost}$

Hence, $\frac{d^{2} y}{d x^{2}} = \frac{se c^{3} (t)}{at}$ .

18. If $f (x) = {| x |}^{3}$ , show that ${f^{'}}^{'} (x)$ exists for all real $x$ and find it.

Ans: Remember that, $| x | = {\begin{aligned} x,if x \geq 0 \\ -x,if x0 \end{aligned}}$

Therefore, if $x \geq 0,$ then $f(x)= {| x |}^{3} = x^{3}$ .

Then, ${f}' (x)=3 x^{2}$ .

Differentiating both sides with respect to $x$ gives

${{f}'}' (x)=6x$ .

Now, if $x0,$ then $f(x)= {| x |}^{3} =(- x^{3})= x^{3}$ .

So, ${f}' (x)=3 x^{2}$ .

Therefore, differentiating both sides with respect to $x$ gives

${{f}'}' (x)=6x$ .

Hence, for $f(x)= {| x |}^{3},$ ${{f}'}' (x)$ exists for all real values of $x$ and is provided as

${{f}'}'(x)= {\begin{aligned} 6x,if x \geq 0 \\ -6x,if x0 \end{aligned}}$

19. Using the fact that $\sin (A + B) = sinAcosB + cosAsinB$ and the differentiation, obtain the sum formula for cosines.

Ans: The given sum formula is $sin(A+B)=sinAcosB+cosAsinB$ .

Now, differentiating both sides with respect to $x$ gives

$\frac{d}{dx} [sin(A+B)] = \frac{d}{dx} (sinAcosB)+ \frac{d}{dx} (cosAsinB)$

$\begin{aligned} \Rightarrow cos(A+B) \times \frac{d}{dx} (A+B)=cosB \times \frac{d}{dx} (sinA)+sinA \times \frac{d}{dx} (cosB)+sinB \times \frac{d}{dx} (cosA) \\ +cosA \times \frac{d}{dx} (sinB) \\ \Rightarrow cos(A+B) \times \frac{d}{dx} (A+B)=cosB \times cosA \frac{d}{dx} +sinA(-sinB) \frac{dB}{dx} +sinB(-sinA) \times \frac{dA}{dx} \\ +cosAcosB \frac{d B}{d x} \\ \Rightarrow cos(A+B) [\frac{dA}{dx} + \frac{dB}{dx}] =(cosAcosB-sinAsinB) \times [\frac{dA}{dx} + \frac{dB}{dx}] \end{aligned}$

Hence the required sum formula for cosines is $cos(A+B)=cosAcosB-sinAsinB$ .

20. Does there exist a function which is continuous everywhere but not differentiable at exactly two points?

Ans: Let take the function $f (x) = | x | + | x - 1 |$

Observe that, the function $f$ is continuous everywhere, but not differentiable at $x = 0$ and $x = 1$ .

21.If $y = | \begin{matrix} f (x) & g (x) & h (x) \\ l & m & n \\ a & b & c \end{matrix} |$ , prove that $\frac{dy}{dx} = | \begin{matrix} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ l & m & n \\ a & b & c \end{matrix} |$ .

Ans: The given function is $y= | \begin{aligned} f(x) g(x) h(x) \\ l m n \\ a b c \end{aligned} |$

Evaluate the determinant.

$y=(mc-nb)f(x)-(lc-na)g(x)+(lb-ma)h(x)$ .

Now, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{dy}{dx} = \frac{d}{dx} [(mc-nb)f(x)] - \frac{d}{dx} [(lc-na)g(x)] + \frac{d}{dx} [(lb-ma)h(x)] \\ =(mc-nb)f(x)-(lc-na)g(x)+(lb-ma)h(x) \\ = | \begin{array}{c} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ l & m & n \\ a & b & c \end{array} | \end{aligned}$

Hence, $\frac{dy}{dx} = | \begin{matrix} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ l & m & n \\ a & b & c \end{matrix} |$

22.If $y = e^{aco s^{- 1} x}, -1 \leq x \leq 1$ , show that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{dy}{dx} - a^{2} y = 0$ .

Ans: The given equation is

$y= e^{aco s^{-1} x}$ .

Then take logarithm both sides of the equation.

$\begin{aligned} logy=aco s^{-1} xloge \\ \Rightarrow logy=aco s^{-1} x \end{aligned}$

Now, differentiating both sides with respect to $x$ gives

$\begin{aligned} \frac{1}{y} \frac{dy}{dx} =ax \frac{1}{\sqrt{1- x^{2}}} \\ \frac{dy}{dx} = \frac{-ax}{\sqrt{1- x^{2}}} \end{aligned}$

Therefore, squaring both the sides of the equation, gives

$\begin{aligned} {(\frac{dy}{dx})}^{2} = \frac{a^{2} y^{2}}{1- x^{2}} \\ \Rightarrow (1- x^{2}) {(\frac{dy}{dx})}^{2} = a^{2} y^{2} \\ \Rightarrow (1- x^{2}) {(\frac{dy}{dx})}^{2} = a^{2} y^{2} \end{aligned}$

Again, differentiating both sides with respect to $x$ gives $\begin{aligned} {(\frac{dy}{dx})}^{2} \frac{d}{dx} (1- x^{2})+(1- x^{2}) \times \frac{d}{dx} [{(\frac{dy}{dx})}^{2}] = a^{2} \frac{d}{dx} (y^{2}) \\ \Rightarrow {(\frac{dy}{dx})}^{2} (-2x)+(1- x^{2}) \times 2 \frac{dy}{dx} \times \frac{d^{2} y}{d x^{2}} = a^{2} \times 2y \times \frac{dy}{dx} \\ \Rightarrow x \frac{dy}{dx} +(1- x^{2}) \frac{d^{2} y}{d x^{2}} = a^{2} \times y \end{aligned}$

Hence, it is proved that $(1- x^{2}) \frac{d^{2} y}{d x^{2}} -x \frac{dy}{dx} - a^{2} y=0$

NCERT Solutions for Class 12 Maths Chapter 5

5.1 Introduction

In continuity and differentiability class 12 NCERT solutions, you will be introduced to Continuity and differentiability and its advanced theorem and logarithms. You will learn the advanced form of what you have studied in your previous class. You will explore important concepts of Continuity, differentiability and relations between them. Moreover, you will learn the differentiation of inverse trigonometric functions. Study a new concept called exponential and logarithmic functions and powerful techniques of logarithms by geometrically distinct conditioning through differential calculus.

5.2 Continuity

In this class 12 continuity and differentiability NCERT solutions, you will learn the rigorous formulation of the intuitive concept of function that varies with no abrupt breaks or jumps. A function is a relationship in which every value of an independent variable is associated with a dependent variable. You will engage with the topic of the algebra of continuous functions. You will study some analogous algebra of uninterrupted functions since the continuity of a function at a point is reasonable to expect results similar to the case of limits. You will learn essential definitions and theorems and how to prove the theorems.

5.3 Differentiability

In this continuity and differentiability class 12 solutions PDF, you will recall some facts you have already learnt in the previous class, such as the derivative of a real function. You will learn about the list of derivatives of the specific standard function. You will discover three theorems and how to prove them. You will also learn that the result of every differentiable function is continuous. You will learn derivatives of composite functions, derivatives of implicit functions and inverse trigonometric functions. You will learn about the Chain Rule theorem and its functionality.

5.4 Exponential and Logarithmic Functions

In this continuity and differentiability class 12 solutions, you will explore aspects of different classes of functions like polynomial functions, rational functions and trigonometric functions. You can learn about a new class of related functions and logarithmic functions. This section will emphasize on statements made that are motivational and precise proofs of theorems. You will learn new definitions of exponential function and logarithms. You will find several examples in this section, which will help you grasp and understand exponential functions and logarithmic functions.

5.5 Logarithmic Differentiation

In this continuity and differentiability class 12 NCERT solution, you will learn to differentiate the of functions given in specific form mentioned in the textbook. This section will provide more examples than theory since the topic demands a practical approach and less of a theoretical approach. This section can give you an example of bringing out a particular class of functions provided in certain forms.

5.6 Derivatives of Functions in Parametric Forms

In class 12 maths chapter 5, you will learn the relation between two variables is neither explicit nor implicit. You will learn how to link a third variable with each of two variables separately by establishing a relation between the first two variables. You will learn why in such a situation; the third variable is called a parameter. You will be given several examples to understand the topic better and grasp the concepts mentioned in this section.

5.7 Second Order Derivative

In this NCERT Solutions for, class 12 maths chapter 5 solutions you will further learn how to work with the second-order derivative. Again, this section will provide you with more examples than theory as it requires more of a practical approach than a theoretical approach. These examples will help you grasp and hold on to the theorems and concepts you have learnt in previous sections.

Key Learnings from NCERT Solutions for Class 12 Maths chapter 5

A real-valued function is continuous at a point in its domain if the limit of the function at that point equals the value of the function at that point. A function is continuous if it is continuous on the whole of its domain.
Continuous Function: A function is called a continuous function on a given interval if it is continuous at every point of the interval.
Arithmetic of Continuous Functions: Arithmetic operations (sum, difference, product, and quotient of continuous functions) performed on continuous functions is continuous. For example, if g and f are continuous functions, then

(f + g) (x) = f(x) + g(x) is continuous.

(f - g) (x) = f(x) - g(x) is continuous.

(f . g) (x) = f(x) . g (x) is continuous.

(f / g ) (x) = f(x ) / g(x) (wherever g(x) ≠ 0) is continuous.

Differentiable: A function is called to be differentiable if the derivative of the function exists at each point in its domain. If a function f(x) is differentiable at x = a, then f′(a) exists in the domain. Some of the derivative rules to find out the derivatives of a given function are listed below:

i) (f + g)′ = f′ + g′

ii) (f - g)′ = f′ - g′

iii) (fg)′ = f′g + fg′

iv) (f/g)′ = (f′g−fg′)/g2

Logarithmic Differentiation is a powerful technique to differentiate functions of the form $f (x) - [11 (x)]^{rix}$ . Here both $f (x)$ and $u (x)$ need to be positive for this technique to make sense
Logarithmic Functions: Logarithmic functions are the inverses of exponential functions. The logarithmic function y = logax is said to be equivalent to the exponential equation x = ay.

Rolle’s Theorem: If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b) such that f(a) = f(b), then there exists c in (a, b) such that f ′(c) = 0.
Following are some of the standard derivatives (appropriate domains):

$\frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}} \frac{d}{d x} (\cos^{- 1} x) = - \frac{1}{\sqrt{1 - x^{2}}} \frac{d}{d x} (\tan^{- 1} x) = \frac{1}{1 + x^{2}} \frac{d}{d x} (\cot^{- 1} x) = \frac{- 1}{1 + x^{2}} \frac{d}{d x} (\sec^{- 1} x) = \frac{1}{x \sqrt{1 - x^{2}}} \frac{d}{d x} ({cosec}^{- 1} x) = \frac{- 1}{x \sqrt{1 - x^{2}}} \frac{d}{d x} (e^{x}) - e^{x} \frac{d}{d x} (\log^{x} x) = \frac{1}{x}$

Overview of Deleted Syllabus for CBSE Class 12 Maths Chapter 5

Chapter	Dropped Topics
Continuity and Differentiability	Examples 22 and 23, and 27
	5.8 Mean Value Theorem, Exercise 5.8
	Miscellaneous Example 44 (ii)
	Questions 19 (Miscellaneous Exercise)
	Summary points 5 (derivatives of $\cot^{- 1} x, \sec^{- 1} x, {cosec}^{- 1} x$ ), 7 and 8

Class 12 Maths Chapter 4: Exercises Breakdown

Exercise	Number of Questions
Exercise 5.1 Solutions	34 Questions Solutions (10 Short Answers, 24 Long Answers)
Exercise 5.2 Solutions	10 Questions Solutions (2 Short Answers, 8 Long Answers)
Exercise 5.3 Solutions	15 Questions Solutions (9 Short Answers, 6 Long Answers)
Exercise 5.4 Solutions	10 Questions Solutions (5 Short Answers, 5 Long Answers)
Exercise 5.5 Solutions	18 Questions Solutions (4 Short Answers, 14 Long Answers)
Exercise 5.6 Solutions	11 Questions Solutions (7 Short Answers, 4 Long Answers)
Exercise 5.7 Solutions	17 Questions Solutions (10 Short Answers, 7 Long Answers)
Miscellaneous Exercise Solutions	22 Questions Solutions

Conclusion

NCERT Solutions for Class 12 Maths Chapter 5 on Continuity and Differentiability, provided by Vedantu, offer crucial insights into fundamental mathematical concepts. It's important to grasp the idea of continuity, where a function has no sudden jumps or breaks, and differentiability, which deals with the smoothness of a function's graph. Focus on understanding how to identify and analyze points of discontinuity, and how to calculate derivatives using various rules and techniques. In previous year question papers, around 3-5 questions were asked from this chapter. Mastering these concepts lays a strong foundation for advanced mathematical studies and real-world applications. Keep practicing and seeking clarity on challenging topics to excel in this subject.

Other Study Materials of CBSE Class 12 Maths Chapter 5

S.No.	Important Links for Chapter 5 Continuity and Differentiability
1	Class 12 Continuity and Differentiability Important Questions
2	Class 12 Continuity and Differentiability Revision Notes
3	Class 12 Continuity and Differentiability Formulas
4	Class 12 Continuity and Differentiability NCERT Exemplar Solutions
5	Class 12 Continuity and Differentiability RS Aggarwal Solutions

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

S.No.	NCERT Solutions Class 12 Maths Chapter-wise List
1	Chapter 1 - Relations and Functions Solutions
2	Chapter 2 - Inverse Trigonometric Functions Solutions
3	Chapter 3 - Matrices Solutions
4	Chapter 4 - Determinants Solutions
5	Chapter 6 - Application of Derivatives Solutions
6	Chapter 7 - Integrals Solutions
7	Chapter 8 - Application of Integrals Solutions
8	Chapter 9 - Differential Equations Solutions
9	Chapter 10 - Vector Algebra Solutions
10	Chapter 11 - Three Dimensional Geometry Solutions
11	Chapter 12 - Linear Programming Solutions
12	Chapter 13 - Probability Solutions

Important Related Links for NCERT Class 12 Maths

S.No	Important Resources Links for Class 12 Maths
1	CBSE Class 12 Maths Revision Notes
2	CBSE Syllabus for Class 12 Maths
3	CBSE Class 12 Maths Important Questions
4	CBSE Class 12 Maths Previous Year Question Paper
5	CBSE Class 12 Maths Sample Paper
6	NCERT Books for Class 12 Maths
7	NCERT Exemplar for Class 12 Maths
8	CBSE Class 12 Maths Formulas
9	RS Aggarwal Class 12 Maths Solutions
10	RD Sharma Class 12 Maths Solutions

FAQs on NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

1. How many topics are there in Chapter 3 of Class 12 Mathematics NCERT textbook?

Class 12 Mathematics Chapter 3 deals with Continuity and Differentiability. It is one of most of the important chapters in Class 12 Mathematics. Before studying the chapter, every student should go through all the topics and subtopics first. That is why we have given the topics and sub-topics of the chapter here - 5.1 Introduction, 5.2 Continuity, 5.2.1 Algebra of continuous functions, 5.3. Differentiability, 5.3.1 Derivatives of composite functions, 5.3.2 Derivatives of implicit functions, 5.3.3 Derivatives of inverse trigonometric functions, 5.4 Exponential and Logarithmic Functions, 5.5. Logarithmic Differentiation, 5.6 Derivatives of Functions in Parametric Forms, 5.7 Second Order Derivative.

2. Give me an overview of all the topics of Class 12 Mathematics Chapter 3.

A. Before you dig into the Class 12 Maths Chapter 3, you must learn a little about all the topics and sub-topics.

5.1 – Introduction - You will study about the limit, differential, and continuity, their properties, methods to find derivatives by limits.

5.2 – Continuity- You will study about the continuity of a function at a given point i.e. a function is continuous if it is continuous on the whole of its domain, properties of a continuous function, types of discontinuity, intermediate value theorem, Cauchy’s definition, Heine’s definition.

5.3 – Differentiability- You will study about the concept of differentiability – it is the change of quantity with respect to another quantity, standard differentiation, fundamental rules, types of derivatives, the relationship between continuity and differentiability.

5.4 – Exponential and Logarithmic Functions- You will study the concept of exponential and logarithmic functions i.e. differentiation of the exponential function and differentiation of the logarithmic functions.

5.5 – Logarithmic Differentiation- You will study about logarithmic differentiation i.e. differentiation of the function in the form of a logarithm.

5.6 – Derivatives of Functions in Parametric Forms- You will study about derivatives of functions in parametric form.

5.7 – Second Order Derivative- You will study and learn to differentiate the functions of second and third-order.

3. How many exercises are there in Class 12 Maths Chapter 3 NCERT book?

A. Class 1 Maths Chapter 3 has a total of eight exercises and each exercise consists of a bunch of questions of all types.

EXERCISE 5.1 - 34 Questions
EXERCISE 5.2 - 10 Questions
EXERCISE 5.3 - 15 Questions
EXERCISE 5.4 - 10 Questions
EXERCISE 5.5 - 18 Questions
EXERCISE 5.6 - 11 Questions
EXERCISE 5.7 - 17 Questions
Miscellaneous Exercise - 6 Questions

4. Why should I opt for Chapter 3 of Class 12 Mathematics NCERT Solutions?

There are several benefits for opting Chapter 3 of Class 12 Mathematics NCERT Solutions. Class 12 Mathematics Chapter 3 of NCERT Solutions are designed and created by the best subject matter experts from the relevant industry. All the topics covered in all the answers and each answer come with an in-depth explanation of every concept to make them understandable. These NCERT Solutions for Class 12 Chapter 3 Mathematics plays a crucial role in your preparation for various competitive exams apart from the board exams as well such as JEE Main, Olympiad etc. These solutions will be very helpful when you will study at home as per your own convenience.

5. What is the relationship between continuity and differentiability of a function?

Differentiability implies continuity: If a function is differentiable at a specific point, it must also be continuous at that same point. fundamentally, a well-defined slope (derivative) at a point suggests the function's output changes smoothly there (continuity).
Continuity does not necessarily imply differentiability: A function can be continuous at a point but not differentiable. Imagine a sharp turn or corner in the function's graph. The output changes continuously as you approach the point, but the slope becomes undefined at that exact point.

6. What is the use of continuity and differentiability?

Continuity: Ensures functions behave predictably, with no "jumps" or sudden changes. This is crucial in various fields like physics (modeling motion) or economics (analyzing market trends).
Differentiability: Allows us to calculate rates of change (slopes) and analyze functions in more detail. It's fundamental in optimization problems (finding minimum/maximum values) and many areas of engineering and science.

7. What is the important formula for continuity?

There's no single formula for continuity. The formal definition involves proving that for any small change in the input (ε), there's a corresponding small enough change in the output (δ) to ensure the function's output stays close. However, understanding the concept of "no jumps" is often sufficient for most purposes.

8. What is the chain rule for continuity and differentiability?

The chain rule is a formula used in calculus to find the derivative of a composite function (a function within another function). It doesn't directly relate to continuity, but ensures that if the individual functions are continuous/differentiable, the composite function most likely inherits those properties as well.

9. What is the use of continuity in real life?

Continuity is used in many real-world scenarios:

Motion: Modeling the continuous movement of objects (e.g., a car accelerating)
Signal transmission: Ensuring smooth flow of information in electrical circuits or communication networks
Economics: Analyzing trends in prices, market behavior, or economic growth over time

S.No.	Related NCERT Solutions for Class 12 Maths
1	NCERT Book for Class 12 Maths in Hindi
2	NCERT Solutions for Class 12 Maths in hindi

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

Complete Resource for NCERT Class 12 Maths Chapter 5 Solutions - Free PDF Download

Glance of NCERT Solutions for Class 12 Maths Chapter 5 | Vedantu

Access Exercise Wise NCERT Solutions for Chapter 5 Maths Class 12

Exercises under NCERT Class 12 Maths Chapter 5 Continuity and Differentiability

Access NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability

Miscellaneous Exercise

NCERT Solutions for Class 12 Maths Chapter 5

5.1 Introduction

5.2 Continuity

5.3 Differentiability

5.4 Exponential and Logarithmic Functions

5.5 Logarithmic Differentiation

5.6 Derivatives of Functions in Parametric Forms

5.7 Second Order Derivative

Key Learnings from NCERT Solutions for Class 12 Maths chapter 5

Overview of Deleted Syllabus for CBSE Class 12 Maths Chapter 5

Class 12 Maths Chapter 4: Exercises Breakdown

Conclusion

Other Study Materials of CBSE Class 12 Maths Chapter 5

NCERT Solutions for Class 12 Maths | Chapter-wise List

Related Links for NCERT Class 12 Maths in Hindi

Important Related Links for NCERT Class 12 Maths

FAQs on NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability