If $f(x) =-e^(2x-1)$ and $g(x) = 2cos^2x$ , what is $f'(g(x))$ ?

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Answer 1 · 2017-02-02T15:31:30

WOW...this is steep!

Ok, first let us find $f(g(x))$ .
We get:
$f(g(x))=-e^((2*color(red)(2cos^2x))-1)=-e^(4cos^2x-1)$

Now we derive using the Chain Rule:

$f'(g(x))=-e^(4cos^2x-1)*[8cos(x)(-sin(x)]=8cos(x)sin(x)*e^(4cos^2x-1)$

So:
$f'(g(x))=8cos(x)sin(x)*e^(4cos^2x-1)$

If f(x) =-e^(2x-1) f(x) =-e^(2x-1) and g(x) = 2cos^2x g(x) = 2cos^2x , what is f'(g(x)) f'(g(x)) ?