How do you differentiate #f(x)=cos(3x)*(2/3tanx)# using the product rule?

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Answer 1 · 2015-12-08T23:52:31

#f'(x)=(2sec^2xcos3x-6tanxsin3x)/3#

According to the product rule:

#f'(x)=2/3tanxd/dx[cos3x]+cos3xd/dx[2/3tanx]#

Find each derivative separately. The first will require the chain rule:

#d/dx[cos3x]=-sin3xd/dx[3x]=-3sin3x#

#d/dx[2/3tanx]=2/3d/dx[tanx]=2/3sec^2x#

Plug these back in:

#f'(x)=2/3tanx(-3sin3x)+cos3x(2/3sec^2x)#

#f'(x)=2/3sec^2xcos3x-2tanxsin3x#

#f'(x)=(2sec^2xcos3x-6tanxsin3x)/3#