What is the derivative of $f (x) = - 3 sin (6 x) \cdot {cos}^{2} (3 x)$ $f(x)=-3sin(6x)*cos^2(3x)$ ?

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Answer 1 · 2016-01-02T04:54:47

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

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

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 $f'(x)$ .

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

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

What is the derivative of f(x)=-3sin(6x)*cos^2(3x)f(x)=−3sin(6x)⋅cos2(3x)f(x)=-3sin(6x)*cos^2(3x)?