Question

Find the derivative of ${\cos }^{2}x$ by using the first principle of derivatives.

Answer 1

Hint: The first principle of derivatives: Given a function $y=f\left(x\right)$ , its first derivative, the rate of change of y with respect to the change in x, is defined by: ${\displaystyle \frac{dy}{dx}}=\underset{h\to 0}{lim}\left[{\displaystyle \frac{f(x+h)-f\left(x\right)}{(x+h)-\left(x\right)}}\right]$ .
Finding the derivative of a function by computing this limit is known as differentiation from first principles.
Use the identity $\sin (A+B)\sin (A-B)={\sin }^{2}A-{\sin }^{2}B={\cos }^{2}B-{\cos }^{2}A$ .
We know that $\underset{x\to 0}{lim}{\displaystyle \frac{\sin x}{x}}=1$ .

Complete step by step answer:
Let's say that the given function is $y=f(x)={\cos }^{2}x$ .
For a change from $x$ to $x+h$ , the value of y changes from $f(x)$ to $f(x+h)$ .
The rate of change of y with respect to the change in x, will be given by:
 ${\displaystyle \frac{\text{Change in the value of y}}{\text{Change in the value of x}}}={\displaystyle \frac{f(x+h)-f(x)}{(x+h)-(x)}}$
This rate for very small values of the change in x, is called the derivative of the function and is represented by ${\displaystyle \frac{dy}{dx}}$ .
∴ ${\displaystyle \frac{dy}{dx}}=\underset{h\to 0}{lim}\left[{\displaystyle \frac{f(x+h)-f\left(x\right)}{(x+h)-\left(x\right)}}\right]$
⇒ ${\displaystyle \frac{d}{dx}}\left({\cos }^{2}x\right)=\underset{h\to 0}{lim}\left[{\displaystyle \frac{{\cos }^2(x+h)-{\cos }^2\left(x\right)}{(x+h)-\left(x\right)}}\right]$
Using the identity ${\cos }^{2}B-{\cos }^{2}A=\sin (A+B)\sin (A-B)$ , we get:
= $\underset{h\to 0}{lim}\left[{\displaystyle \frac{\sin [(x+h)+\left(x\right)]\sin [(x+h)-\left(x\right)]}{h}}\right]$
= $\underset{h\to 0}{lim}\left[{\displaystyle \frac{\sin (2x+h)\sin \left(h\right)}{h}}\right]$
Which can be written as:
= $\underset{h\to 0}{lim}\sin (2x+h)\times \underset{h\to 0}{lim}{\displaystyle \frac{\sin \left(h\right)}{h}}$
Applying the limit and using $\underset{x\to 0}{lim}{\displaystyle \frac{\sin x}{x}}=1$ , we get:
= $\sin (2x+0)\times 1$
= $\sin 2x$
Therefore, the derivative of ${\cos }^{2}x$ is $\sin 2x$ .

Note: Differentiability of a Function: A function $f(x)$ is differentiable at $x=a$ in its domain, if its derivative is continuous at $a$ .
This means that $f^{\prime }(a)$ must exist, or equivalently: $\underset{x\to a^{+}}{lim}{f}^{\prime }(x)=\underset{x\to a^{-}}{lim}{f}^{\prime }(x)=\underset{x\to a}{lim}{f}^{\prime }(x)=f^{\prime }(a)$ .
A continuous function is always differentiable but a differentiable function needs not be continuous.
Indeterminate Forms: Any expression whose value cannot be defined, like ${\displaystyle \frac{0}{0}},\pm {\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }}},0^0,{\mathrm{\infty }}^0$ etc.
L'Hospital's Rule: For the differentiable functions f(x) and g(x), the $\underset{x\to c}{lim}{\displaystyle \frac{f(x)}{g(x)}}$ , if $f(x)$ and $g(x)$ are both 0 or $\pm \mathrm{\infty }$ (i.e. an Indeterminate Form) is equal to the $\underset{x\to c}{lim}{\displaystyle \frac{f^{\prime }(x)}{g^{\prime }(x)}}$ , if it exists.