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How do you find the second derivative of #sin(2x)#?

1 Answer

Apr 13, 2017

#f''(x)=-4sin 2x#

Explanation:

Let #f(x)=sin 2x#.

By #(sin x)'=cos x# and Chain Rule,

#f'(x)=cos 2x cdot(2x)'=2cos 2x#

By #(cos x)'=-sin x# and Chain Rule,

#f''(x)=-2sin 2x cdot (2x)'=-4sin 2x#

I hope that this was clear.

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