Question

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed nonzero constants and m and n are integers): sin n x

Answer 1

Let y = sinn x.
Accordingly, for n = 1, y = sin x.
$∴\frac{dy}{dx}$ = cosx, i.e., $\frac{d}{dx }$ sin x = cosx
For n = 2, y = sin2 x.
$∴\frac{dy}{dx} =\frac{d}{dx}$ (sin x sin x)
= (sin x)' sin x + sin x (sin x)' [By Leibnitz product rule]
= cos x sin x + sin x cos x
= 2 sin x cos x ..(1)
For n = 3, y = sin3 x.
$∴\frac{dy}{dx}$ =$\frac{d}{dx}$ (sin x sin2 x)
= (sin x)' sin2 x + sin x (sin2 x)' [By Leibnitz product rule]
=cosxsin2 x + sin x(2 sin x cos x) [Using (1)]
= cos xsin2x+2 sin2 x cos x
=3sin2 x cos.

We assert that $\frac{d}{dx}$ (sin n x) n sin(k-1) x cos x
Let our assertion be true for n = k.
i.e., $\frac{d}{dx}$ (sink) = k sin(k-1) x cos x ...(2)
Consider
d/dx(sink+1 x) d/dx (sin x sin k x)
=(sin x)' sink x+ sin x (sink x)' [By Leibnitz product rule]
=cos x sink x + sin x (k sin(k-1) x cos x) [Using (2)]
=(k+1) sink x cos x

Thus, our assertion is true for n = k + 1.
Hence, by mathematical induction, d/dx(sin n x) = n sin(n-1) x cos x