If #f(x) =xe^x# and #g(x) = sin3x #, what is #f'(g(x)) #?

1 Answer
Bill K.
Jan 11, 2016

I think you meant to write this as: what is #(f(g(x)))'#? (or #d/dx(f(g(x)))#?...since it's supposed to be a Chain Rule problem). In that case, the answer is

#3e^{sin(3x)}cos(3x)+3e^{sin(3x)}cos(3x)sin(3x)#.

Explanation:

Let #h(x)=f(g(x))=e^{sin(3x)}sin(3x)#. We want to find #h'(x)=(f(g(x)))'=d/dx(f(g(x)))#.

The Chain Rule says that #h'(x)=f'(g(x)) * g'(x)#.

In this case, the Product Rule gives
#f'(x)=e^[x}+xe^[x}# and we also know (by the Chain Rule again), that #g'(x)=cos(3x) * d/dx(3x)=3cos(3x)#.

Putting these things together leads to the final answer:

#h'(x)=3e^{sin(3x)}cos(3x)+3e^{sin(3x)}cos(3x)sin(3x)#.

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