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

1 Answer

Feb 2, 2017

WOW...this is steep!

Explanation:

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)#

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