Let $y = \sin^nx$

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

$ \therefore \large\frac{dy}{dx}$$= \cos x$, i.e.,$\large\frac{d}{dx}$$ \sin x = \cos x$

For $n=2,y=\sin^2 x$.

$ \therefore \large\frac{dy}{dx} = \large\frac{d}{dx} $$(\sin x \sin x)$

$ = (\sin x)' \sin x+ \sin x (\sin x)'$$\qquad \qquad$ [By Leibnitz product rule ]

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

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

For $n=3,y=\sin^3 x$.

$ \therefore \large\frac{dy}{dx} = \large\frac{d}{dx} $$(\sin x \sin^2 x)$

$ = (\sin^2 x)' \sin x+ \sin x (\sin^2 x)'$$\qquad \qquad$ [By Leibnitz product rule ]

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

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

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

We assert that $ \large\frac{d}{dx}$$(\sin^nx)=n\sin^{(n-1)}x\cos x$

Let our assertion be true for $n=k$

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

Consider

$ \large\frac{d}{dx}$$(\sin^{k+1}x)= \large\frac{d}{dx}$$(\sin x \sin^k x)$

$ = ( \sin x)' \sin^k x+ \sin x ( \sin^kx)'$$\qquad$ [ By Leibnitz product rule]

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

$ = \cos x \sin^kx+k \sin^k x \cos x $

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

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

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