Question

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

Answer 1

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: $\dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x+h \right)-f\left( x \right)}{\left( x+h \right)-\left( x \right)} \right]$ .
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 $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ .

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:
$\dfrac{\text{Change in the value of y}}{\text{Change in the value of x}}=\dfrac{f(x+h)-f(x)}{(x+h)-(x)}$
This rate for very small values of the change in x, is called the derivative of the function and is represented by $\dfrac{dy}{dx}$ .
∴ $\dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x+h \right)-f\left( x \right)}{\left( x+h \right)-\left( x \right)} \right]$
⇒ $\dfrac{d}{dx}\left( {{\cos }^{2}}x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{{{\cos }^{2}}\left( x+h \right)-{{\cos }^{2}}\left( x \right)}{\left( x+h \right)-\left( x \right)} \right]$
Using the identity ${{\cos }^{2}}B-{{\cos }^{2}}A=\sin (A+B)\sin (A-B)$ , we get:
= $\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ \left( x+h \right)+\left( x \right) \right]\sin \left[ \left( x+h \right)-\left( x \right) \right]}{h} \right]$
= $\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{\sin \left( 2x+h \right)\sin \left( h \right)}{h} \right]$
Which can be written as:
= $\underset{h\to 0}{\mathop{\lim }}\,\sin \left( 2x+h \right)\times \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( h \right)}{h}$
Applying the limit and using $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ , we get:
= $\sin \left( 2x+0 \right)\times 1$
= $\sin 2x$
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 ${f}'(a)$ must exist, or equivalently: $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,{f}'(x)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,{f}'(x)=\underset{x\to a}{\mathop{\lim }}\,{f}'(x)={f}'(a)$ .
A continuous function is always differentiable but a differentiable function needs not be continuous.
Indeterminate Forms: Any expression whose value cannot be defined, like $\dfrac{0}{0},\pm \dfrac{\infty }{\infty },{{0}^{0}},{{\infty }^{0}}$ etc.
L'Hospital's Rule: For the differentiable functions f(x) and g(x), the $\underset{x\to c}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}$ , if $f(x)$ and $g(x)$ are both 0 or $\pm \infty$ (i.e. an Indeterminate Form) is equal to the $\underset{x\to c}{\mathop{\lim }}\,\dfrac{{f}'(x)}{{g}'(x)}$ , if it exists.