Differentiate in the form of wrtx? sin^3xsin3x

1 Answer
Jun 7, 2018

(sin^3x\cdot\sin(3x))'=3\sin^2x(\cos x\sin(3x)+\sin x \cos(3x))

Explanation:

We need to use the Product Rule , and the Chain Rule to differentiate \sin^3x=(\sin x)^3 and \sin(3x).
Thus
(sin^3x\cdot\sin(3x))'=(\sin^3x)'\cdot \sin(3x)+(\sin^3 x)\cdot (\sin(3x))'=3\sin^2 x \cos x\sin(3x)+\sin^3x\cdot (3\cos(3x))=3\sin^2x(\cos x\sin(3x)+\sin x \cos(3x))