If $f (x) = - e^{2 x + 4}$ $f(x) =-e^(2x+4)$ and $g (x) = tan 3 x$ $g(x) = tan3x$ , what is $f'(g(x))$ ?

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Answer 1 · 2017-10-26T13:30:49

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

$f'(g(x))$ 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

If f(x) =-e^(2x+4) f(x)=−e2x+4f(x) =-e^(2x+4) and g(x) = tan3x g(x)=tan3xg(x) = tan3x , what is f'(g(x)) f'(g(x)) ?