Question

What is the derivative of `f(x)=-3sin(6x)*cos^2(3x)`?

Answer 1

`f'(x)=18cos(3x)(sin(3x)sin(6x)-cos(3x)cos(6x))`

Explanation:

Use product rule.

`f'(x)=cos^2(3x)d/dx(-3sin(6x))-3sin(6x)d/dx(cos^2(3x))`

Find each derivative separately, using chain rule each time.

`d/dx(-3sin(6x))=-3cos(6x)*6=-18cos(6x)`

This one requires you to deal with the exponent first, then the cosine function.

`d/dx(cos^2(3x))=2cos(3x)d/dx(cos(3x))`

`=2cos(3x) * -sin(3x) * 3=-6sin(3x)cos(3x)`

Plug these back in to find `f'(x)`.

`f'(x)=cos^2(3x) * -18cos(6x)-3sin(6x) * -6sin(3x)cos(3x)`

`=18cos(3x)(sin(3x)sin(6x)-cos(3x)cos(6x))`