Question

Find the derivative of the following functions:$2\tan x-7\sec x$

Answer 1

Let $f\left(x\right)=2\tan x-7\sec x$
Thus using first principle,
$f^{{}^{\prime }}=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
= $\underset{h\to 0}{lim}\frac{1}{h}[2\tan (x+h)-7\sec (x+h)-2\tan x+7\sec x]$
= $\underset{h\to 0}{lim}\frac{1}{h}[2\{\tan (x+h)-\tan x\}-7\{\sec (x+h)-\sec x\}]$
= $2\underset{h\to 0}{lim}\frac{1}{h}[\tan (x+h)-\tan x]-7\underset{h\to 0}{lim}\frac{1}{h}[\sec (x+h)-\sec x]$
= $2\underset{h\to 0}{lim}\frac{1}{h}[\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}]-7\underset{h\to 0}{lim}\frac{1}{h}[\frac{1}{\cos (x+h)}-\frac{1}{\cos x}]$
= $2\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{\sin (x+h)\cos x-\sin x\cos (x+h)}{\cos x\cos (x+h)}\right]$$-7\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{\cos x-\cos (x+h)}{\cos x\cos (x+h)}\right]$
= $2\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{\sin (x+h-x)}{\cos x\cos (x+h)}\right]-7\underset{h\to 0}{lim}\frac{1}{h}$$\left[\frac{-2\left(\frac{x+x+h}{2}\right)\sin \left(\frac{x-x-h}{2}\right)}{\cos x\cos (x+h)}\right]$
= $2\underset{h\to 0}{lim}\left[\left(\frac{\sin h}{h}\right)\frac{1}{\cos x\cos (x+h)}\right]-7\underset{h\to 0}{lim}\frac{1}{h}$$\left[\frac{-2\sin \left(\frac{2x+h}{2}\right)\sin (-\frac{h}{2})}{\cos x\cos (x+h)}\right]$
= $2\left(\underset{h\to 0}{lim}\frac{\sin h}{h}\right)\left(\underset{h\to 0}{lim}\frac{1}{\cos x\cos (x+h)}\right)-7\left(\underset{\frac{h}{2}\to 0}{lim}\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)$$\left(\underset{h\to 0}{lim}\frac{\sin \left(\frac{2x+h}{2}\right)}{\cos x\cos (x+h)}\right)$
= 2.1. $\frac{1}{\cos x\cos x}-7\cdot 1\left(\frac{\sin x}{\cos x\cos x}\right)$
= $2{\sec }^{2}x-7\sec x\tan x$