Question

Find the derivative of the following functions: $\cos ecx$

Answer 1

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