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Differentiation and Integration Formula

Formula

Differentiation and Integration Formula

Reviewed by:

Rama Sharma

Introduction

Differentiation and Integration are both quite crucial concepts in calculus which are typically used to learn the change. Calculus is not only restricted to mathematics but has a huge array of applications in various domains of science as well as the economy. Also, we may be able to spot calculus in establishing an analysis in finance as well as in the stock market. In this chapter, we will study some differentiation and integration formula with examples besides the interesting concept!

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Find below some of the basic formulas of differentiation and integration.

Differentiation and Integration Formulas

Differentiation Formulas	Integration Formulas
1.\[\frac{d}{dx}(x) = 1\]	1.\[∫1dx = x + C\]
2.\[\frac{d}{dx}(ax) = a\]	2.\[∫adx = ax + C\]
3.\[\frac{d}{dx}(x^n)=nx^{n−1}\]	3.\[∫x^ndx=\frac{x^{n+1}}{n+1} + C, n ≠ -1\]
4.\[\frac{d}{dx}(cosx) = -sinx\]	4.\[∫sinxdx = -cosx + C\]
5.\[\frac{d}{dx}(sinx) = cosx\]	5.\[∫cosxdx = sinx + C\]
6.\[\frac{d}{dx}(tanx)=sec^{2}x\]	6.\[∫sec^{2}xdx = tanx + C\]
7.\[\frac{d}{dx}(cotx)=−csc^{2}\]	7.\[∫csc^{2}xdx = - cotx + C\]
8.\[\frac{d}{dx}(secx) = secx tanx\]	8.\[∫secx(tanx)dx=secx + C \]
9.\[\frac{d}{dx}(cscx) = -cscx(cotx) \]	9.\[∫cscx(cotx)dx=−cscx + C \]
10.\[\frac{d}{dx}(lnx)=\frac{1}{x}\]	10.\[∫\frac{1}{x} dx = ln \|x\| + C \]
11.\[\frac{d}{dx}(ex)=e^{x}\]	11.\[∫e^{x}dx=e^x + C \]
12.\[\frac{d}{dx}(a^x)=(lna)a^x\]	12.\[∫a^xdx= \frac{a^x}{lna} + C, a > 0, a≠ 1\]
13.\[\frac{d}{dx}(sin^{−1}x)=\frac{1}{\sqrt{1-x^2}}\]	13. \[\int \frac{1}{\sqrt{1-x^{2}}} dx = sin^{-1}x + C \]
14.\[\frac{d}{dx}(tan^{−1}x)=\frac{1}{\sqrt{1+x^2}}\]	14.\[∫\frac{1}{1+x^2}dx=tan^{−1}x + C\]
15.\[\frac{d}{dx}(sec^{−1}x)=\frac{1}{\|x\|\sqrt{x^2−1}}\]	15.\[∫\frac{1}{\|x\|\sqrt{x^2−1}}dx=sec^{−1}x + C\]

What is Differentiation in Maths?

The mathematical concept of Differentiation facilitates us to find rates of change. For example, it enables us to detect the rate of change of velocity with respect to time (called the acceleration). In addition, It also enables us to find the rate of change of x with reference to y, which graphically at y against x is the gradient of the curve. There are a set of simple rules which can be used in order to differentiate a number of functions easily.

If y = some function of x, then the derivative of y (in respect to x) is written as dy/dx.

For Example: Find out the gradient of the curve y = 2x⁴ at the point (4, 56)?

Using the formula, dy/dx = 8 x²

When x = 4, dy/dx = 8 × 7 = 56.

What is Integration in Mathematics?

Integration is a mathematical technique to find a function g(x) the derivative of which, Dg(x), is equivalent to a provided function f(x). This is denoted by the integral sign “∫,” or ∫f(x), generally termed the indefinite integral of the function. The sign dx denotes a displacement of an infinitesimal along x; therefore ∫f(x) dx becomes the aggregate sum of the product of f(x) and dx. The definite integral is written as: - “∫a^b

With ‘a’ and ‘b’ referred to the limits of integration, is equivalent (=) to g(b) − g(a), where Dg(x) = f(x).

Usefulness of Integration

Integration is a mathematical approach of adding slices to find the whole. Applying the theory of Integration, we are able to easily find areas, volumes, median points of many useful figures. That said, it is still simplest to start with finding the area under the curve of a function like one below:

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Differentiation Under the Integral Sign

Differentiation under the integral sign is an algebraic operation in calculus that is performed in order to assess certain integrals. Under a reasonably loose situation on the function being integrated, this operation enables us to swap the order of integration and differentiation. In its simplified version, called the Leibniz integral rule, differentiation under the integral sign models the ensuing equation legitimate under the formula on:

\[\frac{d}{dt} \int^{b}_{a} f(x,t)dx = \int^{b}_{a} \partial_{t} f(x,t)dx\]

The Usefulness of Differentiation under the Integral Sign

Many integrals that would otherwise be impracticable or need a substantially more complex approach can be solved by this method.

Differentiation under the integral sign rule can be used to assess the certain unusual definite integrals such as given below:

\[I’(b) \frac{d}{db} \int^{1}_{0} \frac{x^{b}-1}{In{x}}dx \]

\[\frac{d}{db} \int^{1}_{0} \frac{x^b-1}{In(x)} = = \int^{1}_{0} \frac{\partial}{\partial{b}}\frac{x^b -1}{In(x)} \]

\[I’(b) = \int^1_0 \frac{In(x)x^b}{In(x)} dx\]

Solved Examples For You

Example1: Differentiate the given functions

\[Y = \sqrt[3]{x^2}{(2x-x^2)}\]

Solution1:

In the given function we would not be able to just differentiate the 1st term, and then differentiate the 2nd term and multiply the two together. This would just NOT work win the case given. However, it is still possible to perform the derivative of this function. All we are supposed to do is convert the radical into fractional exponents and multiply this across the parenthesis.

\[Y = x^{\frac{2}{3}}(2x-x^2) = 2x^{\frac{5}{3}} -x^{\frac{8}{3}}\]

Here, we can differentiate the function

\[Y^{1} = \frac{10}{3} x^{\frac{2}{3}} - \frac{8}{3}x^{\frac{5}{3}}\]

Example2: Find out if the given equation

\[f(x) = 2x^3+ \frac{300}{x^{-3}}+4 \]rising, reducing or not changing at x= −2?

Solution2:

We are already aware that the rate of change of a function is provided by the derivative of the function. Thus all we are supposed to do is to rewrite the function. Rewriting the function is required in order to deal with the 2nd term and then take the derivative.

\[f(x) = 2x^3+ \frac{300}{x^{-3}}+4\] → \[f^{1} (x) = 6x^{2} - 900^{x^{-4}} = 6x^{2} - \frac{900}{x^{4}}\]

Remember that had rewritten the last term in the derivative back as a fraction.

This is not something we’ve performed up to this step and is only being done here to help with the assessment in the next step. Note that it’s often simpler to do the assessment with positive exponents.

So, upon assessing the derivative we now obtain,

\[f^{1} (-2) = 6(4) -\frac{900}{16} = -\frac{129}{4} = -32.25\]

Thus, at x = −2, the derivative turns out to be negative and, hence, the function is decreasing at x = −2

FAQs on Differentiation and Integration Formula

1. What is Meant by Integration by Parts?

Integration by parts is a unique mathematical technique of integration which is most commonly used when two functions are multiplied together but is also quite useful in other ways. It is actually a technique of using incorporating the product rule in reverse. Below is the rule of Integration by Parts:

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Where, u is the function, denoted by u(x)

v is the function, denoted by v(x)

u' is the derivative of the function, denoted by u(x)

Moreover,

The formula for Integration by Parts is as given:-

∫u dv = u. v −∫v du

2. How do we Find the Derivative?

You might find it really difficult, but it isn't. The trick is to first simplify the given expression: perform the division (divide each term on the numerator by 3x^½. Use the law of indices in what equation we get. Now, differentiate term by term. Note that there are different ways of writing the derivative. They are although all typically the same. In other words:

(1) If y = x³,

dy/dx = 2x

This implies that if y = x³ the derivative of y, in respect to x is 3x

(2) d/dx (x³) = 2x

This implies that the derivative of x³ in respect to x is 2x.

3. What is the use of setting limits tending to zero in differentiation?

Setting limits to zero while dealing with differentiation makes sure the interval for which the differentiation operation is being pursued is extremely small and hence the slope of the curve for the particular moment can be calculated. For ex- The slope of a circle at a certain point is considered as the tangent because it is at a right angle to the radius vector. It is also helpful in finding out the instantaneous velocity of a moving object.

4. What is the difference between definite and indefinite integration?

Indefinite integration, the function is integrated using the integration operation and then the values are substituted to give a definite value of the integration operation. While in indefinite integration, a constant C is added after the integration as the interval is not defined. It is done because integration is considered as the reverse function of differentiation and when the differentiation of a function is performed, the differentiation of a constant is always 0.

5. What is integration by parts?

When two functions are asked to be integrated simultaneously, which contain the same variable, that possesses a tough task as the integration of 1 function can be done simply but the integration of two functions is confusing because, at the very core, integration is about calculating the area under a curve. For this, a technique called integration by parts is used where the first and second functions are separated and the respective operation is performed over both of them.

6. What is the use of differentiation in three-dimensional coordinate geometry while determining slopes of a function?

To calculate the slope of a function, the function is differentiated with respect to the other variable. This is fairly simple in two-dimensional geometry as there are only two variables and the slope can be easily calculated as dy/dx. But in 3D geometry, there are 3 variables so the process of finding slopes becomes tedious. For this, we use a method called partial differentiation where a function is differentiated along with one variable while keeping the other variable constant.

7. I am facing trouble while dealing with the exercises of differentiation and integration, what shall I do?

Differentiation and Integration are advanced-level topics and every student has some or other doubts while dealing with the exercises of this chapter. However, this is also the bedrock of higher-level mathematics and hence a good command of this subject is essential to excel in the entrance examinations. These topics are also very heavily used in the Physics curriculum. So, if you are having issues, ask your peers to help with your doubts and you can access the Vedantu platform to get answers and solutions to exercises.