If $f (x) = x - x e^{2 x + 4}$ $f(x) =x-xe^(2x+4)$ and $g (x) = cos 9 x$ $g(x) = cos9x$ , what is $f'(g(x))$ ?

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Answer 1 · 2018-01-02T08:03:19

$9sin9x(2cos9xe^(2cos9x+4)+e^(2cos9x+4)-1)$

Since $color(red)g(x)=color(red)(cos9x)$ ,

then

$f(color(red)g(x))=f(color(red)(cos9x))=color(red)(cos9x)-color(red)(cos9x)*e^(2color(red)(cos9x)+4)$

Let's differentiate it as a chained function and get:

$f'(g(x))=-sin9x*9-(-sin9x*9*e^(2cos9x+4)+cos9x*e^(2cos9x+4)* 2* (-sin9x*9))$

$=-9sin9x+9sin9xe^(2cos9x+4)+18sin9xcos9xe^(2cos9x+4)$

$=9sin9x(2cos9xe^(2cos9x+4)+e^(2cos9x+4)-1)$

If f(x) =x-xe^(2x+4) f(x)=x−xe2x+4f(x) =x-xe^(2x+4) and g(x) = cos9x g(x)=cos9xg(x) = cos9x , what is f'(g(x)) f'(g(x)) ?