Question

Differentiate $\sec x$ by first principle.

Answer 1

Let $f\left(x\right)=\sec x$

$\therefore f(x+h)\sec (x+h)$

$\frac{d}{dx}f\left(x\right)=\underset{h\to 0}{lim}\left[\frac{f(x+h)-f\left(x\right)}{h}\right]$

$\frac{d}{dx}\left(\sec x\right)=\underset{h\to 0}{lim}\left[\frac{\sec (x+h)-\sec x}{h}\right]$

$\frac{d}{dx}\left(\sec x\right)=\underset{h\to 0}{lim}\left[\frac{1}{h}\left(\frac{1}{\cos (x+h)}-\frac{1}{\cos x}\right)\right]$

$\frac{d}{dx}\left(\sec x\right)=\underset{h\to 0}{lim}\left[\frac{1}{h}\left(\frac{\cos \left(x\right)-\cos (x+h)}{\cos (x+h)\cos x}\right)\right]$

$\frac{d}{dx}\left(\sec x\right)=\underset{h\to 0}{lim}\left[\frac{1}{h}\frac{2\sin \left(\frac{2x+h}{2}\right)\sin \left(\frac{x+h-x}{2}\right)}{h.\cos (x+h)\cos x}\right]$

and $\frac{d}{dx}\left(\sec x\right)=\underset{h\to 0}{lim}\left[\frac{\sin \left(x+\frac{h}{2}\right)}{\cos (x+h)\cos x}\right]$ $\cdot \underset{h\to 0}{lim}\left[\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right]$

$\frac{d}{dx}\left(\sec x\right)=\frac{\sin x}{\cos x\cdot \cos x}\times 1$

$\frac{d}{dx}\left(\sec x\right)=\sec x\cdot \tan x$.