Question

Find the derivative of the following functions from first principle:
$\cos (x-\frac{\pi }{8})$

Answer 1

Let $f\left(x\right)=\cos (x-\frac{\pi }{8})$
Thus using first principle
$f^{\prime }\left(x\right)=\underset{x\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
=$\underset{x\to 0}{lim}\frac{1}{h}[\cos (x+h-\frac{\pi }{8})-\cos (x-\frac{\pi }{8})]$
=$\underset{x\to 0}{lim}\frac{1}{h}[-2\sin \frac{(x+h-\frac{\pi }{8}+x-\frac{\pi }{8})}{2}\sin \left(\frac{x+h-\frac{\pi }{8}-x+\frac{\pi }{8}}{2}\right)]$
$=\underset{x\to 0}{lim}\frac{1}{h}[-2\sin \left(\frac{2x+h-\frac{\pi }{4}}{2}\right)\sin \frac{h}{2}]$
$=\underset{x\to 0}{lim}[-\sin \left(\frac{2x+h-\frac{\pi }{4}}{2}\right)\frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}]$
$=-\sin \left(\frac{2x+0-\frac{\pi }{4}}{2}\right)\cdot 1=-\sin (x-\frac{\pi }{8})$