Question

Find the derivative of the following functions from the first principals w.r.t to $x$.
$\sin 2x$

Answer 1

We know the first principle is,

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

Here, $f\left(x\right)=\sin 2x$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\sin \left(2.\right(x+h\left)\right)-\sin 2x}{h}$

$=\underset{h\to 0}{lim}\frac{\sin (2x+2h)-\sin 2x}{h}$

$=\underset{h\to 0}{lim}\frac{2.\cos \left(\frac{2x+2h+2x}{2}\right).\sin \left(\frac{2x+2h-2x}{2}\right)}{h}$

$=\underset{h\to 0}{lim}\frac{2.\cos (2x+h).\sin h}{h}$

$=\underset{h\to 0}{lim}2.\cos (2x+h).\frac{\sin h}{h}$

$=\underset{h\to 0}{lim}2.\cos (2x+h).\underset{h\to 0}{lim}\left(\frac{\sin h}{h}\right)$

We know, $\underset{h\to 0}{lim}\left(\frac{\sin h}{h}\right)=1$ and substituting $h=0$

$=2.\cos 2x$

$\therefore$ $f^{\prime }\left(x\right)=2.\cos 2x$