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: $ \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.