A) \[2\,\cos \,2x.\,\cos \,2x\]
B) \[2\,\cos \,2x.\,\cos \,(\sin \,2x)\]
C) \[2\,\cos \,2x.\sin 2x\]
D) \[\cos \,2x.\cos \,(\sin 2x)\]
Correct Answer: B
Solution :
Let \[y=\sin (\sin \,2x)\] Then, \[\frac{dy}{dx}=\frac{d}{dx}\,\,[\sin \,(\sin 2x)]\] \[=\cos (\sin 2x).\cos \,2x.2\] \[\frac{dy}{dx}=2\cos \,2x.\cos (\sin 2x)\]
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