Question

Find the derivative of $\tan x$ with respect to $x$ from first principle.

Answer 1

From the first principle

$\begin{array}{c}{f}^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ ={lim}_{h\to 0}\frac{\tan \left(x+h\right)-\tan x}{h}\\ ={lim}_{h\to 0}\frac{\frac{\sin \left(x+h\right)}{\cos \left(x+h\right)}-\frac{\sin x}{\cos x}}{h}\end{array}$

$=\underset{h\to 0}{lim}\frac{\sin \left(x+h\right)\cos x-\sin x\cos x\left(x+h\right)}{h\cos \left(x+h\right)\cos x}$

$=\underset{h\to 0}{lim}\frac{\frac{\sin \left(2x+h\right)+\sinh }{2}-\frac{\sin \left(2x+h\right)-\sinh }{2}}{h\cos \left(x+h\right)\cos x}$

$=\underset{h\to 0}{lim}\frac{\sinh }{h\cos \left(x+h\right)\cos x}$

$=\underset{h\to 0}{lim}\frac{\sinh }{h}\times \underset{h\to 0}{lim}\frac{1}{\cos \left(x+h\right)\cos x}$

$=1\times \frac{1}{\cos x\times \cos x}$

$=1\times \frac{1}{{\cos }^{2}x}$

$={\sec }^{2}x$