If #f(x) =-e^(2x+4) # and #g(x) = tan3x #, what is #f'(g(x)) #?

1 Answer
Oct 26, 2017

Answer:

#f'(g(x)) = -2e^(2tan(3x)+4)#

Explanation:

#f'(g(x))# indicates that we need to take the derivative of #f#, and then plug in #g(x)# for #x#.

First, let's find #f'(x)#:

We know that #d/dxe^u = e^u d/dx(u)#

Therefore:

#d/dx(-e^(2x+4)) = -e^(2x+4) * d/dx(2x+4)#

#= -2e^(2x+4)#

So #f'(x) = -2e^(2x+4)#.

We also know that #g(x) = tan(3x)#

Therefore, #f'(g(x)) = -2e^(2tan(3x)+4)#

Final Answer