Find the derivative of the following functions

Let $y=\sin ^{n} x$.

Accordingly, for $n=1, y=\sin x$.

$\therefore \frac{d y}{d x}=\cos x$, i.e., $\frac{d}{d x} \sin x=\cos x$

For $n=2, y=\sin ^{2} x$

$\therefore \frac{d y}{d x}=\frac{d}{d x}(\sin x \sin x)$

$=(\sin x)^{\prime} \sin x+\sin x(\sin x)^{\prime}$ [By Leibnitz product rule]

$=\cos x \sin x+\sin x \cos x$

$=2 \sin x \cos x$ $\ldots(1)$

For $n=3, y=\sin ^{3} x$

$\therefore \frac{d y}{d x}=\frac{d}{d x}\left(\sin x \sin ^{2} x\right)$

$=(\sin x)^{\prime} \sin ^{2} x+\sin x\left(\sin ^{2} x\right)^{\prime}$ [By Leibnitz product rule] 

$=\cos x \sin ^{2} x+\sin x(2 \sin x \cos x)$ $[$ Using (1) $]$

$=\cos x \sin ^{2} x+2 \sin ^{2} x \cos x$

$=3 \sin ^{2} x \cos x$

We assert that $\frac{d}{d x}\left(\sin ^{n} x\right)=n \sin ^{(n-1)} x \cos x$

Let our assertion be true for n = k.

i.e., $\frac{d}{d x}\left(\sin ^{k} x\right)=k \sin ^{(k-1)} x \cos x$ $\ldots(2)$

Consider

$\frac{d}{d x}\left(\sin ^{h+1} x\right)=\frac{d}{d x}\left(\sin x \sin ^{k} x\right)$

$=(\sin x)^{\prime} \sin ^{k} x+\sin x\left(\sin ^{k} x\right)^{\prime}$ [By Leibnitz product rule]

$=\cos x \sin ^{k} x+\sin x\left(k \sin ^{(k-1)} x \cos x\right)$ $[$ Using $(2)]$

$=\cos x \sin ^{k} x+k \sin ^{2} x \cos x$

$=(k+1) \sin ^{k} x \cos x$

Thus, our assertion is true for n = k + 1.

Hence, by mathematical induction, $\frac{d}{d x}\left(\sin ^{n} x\right)=n \sin ^{(n-1)} x \cos x$