Question

Find the derivative of the following functions:
$5\sec x+4\cos x$

Answer 1

Let $f\left(x\right)=5\sec x+4\cos x$
Thus using first principle,
$f^{{}^{\prime }}\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
= $\underset{h\to 0}{lim}\frac{5\sec (x+h)+4\cos (x+h)-[5\sec x+4\cos x]}{h}$
= $5\underset{h\to 0}{lim}\frac{[\sec (x+h)-\sec x]}{h}+4\underset{h\to 0}{lim}\frac{[\cos (x+h)-\cos x]}{h}$
= $5\underset{h\to 0}{lim}\frac{1}{h}[\frac{1}{\cos (x+h)}-\frac{1}{\cos x}]+4\underset{h\to 0}{lim}\frac{1}{h}[\cos (x+h)-\cos x]$
= $5\underset{h\to 0}{lim}\left[\frac{\cos x-\cos (x+h)}{\cos x\cos (x+h)}\right]+4\underset{h\to 0}{lim}\frac{1}{h}[\cos x\cos h-\sin x\sin h-\cos x]$
= $\frac{5}{\cos x}\cdot \underset{h\to 0}{lim}\frac{1}{h}\left[\frac{-2\sin \left(\frac{2x+h}{2}\right)\sin (-\frac{h}{2})}{\cos (x+h)}\right]+4[-\cos x\underset{h\to 0}{lim}\frac{(1-\cos h)}{h}-\sin x\underset{h\to 0}{lim}\frac{\sin h}{h}]$
= $\frac{5}{\cos x}\cdot \underset{h\to 0}{lim}\left[\frac{\sin \left(\frac{2x+h}{2}\right)\cdot \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}}{\cos (x+h)}\right]+4[(-\cos x)\cdot \left(0\right)-\left(\sin x\right)\cdot 1]$
= $\frac{5}{\cos x}\cdot [\underset{h\to 0}{lim}\frac{\sin \left(\frac{2x+h}{2}\right)}{\cos (x+h)}\cdot \underset{h\to 0}{lim}\frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}]-4\sin x$
= $\frac{5}{\cos x}\cdot \frac{\sin x}{\cos x}\cdot 1-4\sin x$
= $5\sec x\tan x\cdot -4\sin x$