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Question

Mathematics Question on Differentiability

Differentiate sin⁡  (sin⁡  2x).\sin \,\,(\sin \,\,2x).

A

2 cos⁡ 2x. cos⁡ 2x2\,\cos \,2x.\,\cos \,2x

B

2 cos⁡ 2x. cos⁡ (sin⁡ 2x)2\,\cos \,2x.\,\cos \,(\sin \,2x)

C

2 cos⁡ 2x.sin⁡2x2\,\cos \,2x.\sin 2x

D

cos⁡ 2x.cos⁡ (sin⁡2x)\cos \,2x.\cos \,(\sin 2x)

Answer

2 cos⁡ 2x. cos⁡ (sin⁡ 2x)2\,\cos \,2x.\,\cos \,(\sin \,2x)

Explanation

Solution

Let y=sin⁡(sin⁡ 2x)y=\sin (\sin \,2x) Then, dydx=ddx  [sin⁡ (sin⁡2x)]\frac{dy}{dx}=\frac{d}{dx}\,\,[\sin \,(\sin 2x)]
=cos⁡(sin⁡2x).cos⁡ 2x.2=\cos (\sin 2x).\cos \,2x.2 dydx=2cos⁡ 2x.cos⁡(sin⁡2x)\frac{dy}{dx}=2\cos \,2x.\cos (\sin 2x)