# Find the derivative of the following functions ( it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers ) $\sin^n x$

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$