Question

Find the differential coefficient of $\sin x$ by first principle.

Answer 1

Find the differential coefficient of $\sin x$ by first principle. Suppose that
$f\left(x\right)=\sin x$
$\therefore f(x+h)=\sin (x+h)$
$\therefore f(x+h)-f\left(x\right)=\sin (x+h)-\sin x$
$\Rightarrow \frac{f(x+h)-f\left(x\right)}{h}=\frac{\sin (x+h)-\sin x}{h}$
$\Rightarrow \underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{\sin (x+h)-\sin x}{h}$
$\Rightarrow \frac{d}{dx}f\left(x\right)=\underset{h\to 0}{lim}\left[\frac{2\cos (x+\frac{h}{2}).\sin \frac{h}{2}}{h}\right]$
$\Rightarrow \frac{d}{dx}\sin x=\underset{h\to 0}{lim}\cos (x+\frac{h}{2}).\underset{h\to 0}{lim}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)$
$=\cos (x+0).\underset{\frac{h}{2}\to 0}{lim}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)\because h\to 0$
$\therefore \frac{h}{2}\to 0$
$=\cos x.1$
$\Rightarrow \frac{d}{dx}\sin x=\cos x$.