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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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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 ) $\sin^n x$

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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 $
answered Apr 14, 2014 by thanvigandhi_1
 

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