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

Oct 26, 2017

$f ' \left(g \left(x\right)\right) = - 2 {e}^{2 \tan \left(3 x\right) + 4}$

#### Explanation:

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

First, let's find $f ' \left(x\right)$:

We know that $\frac{d}{\mathrm{dx}} {e}^{u} = {e}^{u} \frac{d}{\mathrm{dx}} \left(u\right)$

Therefore:

$\frac{d}{\mathrm{dx}} \left(- {e}^{2 x + 4}\right) = - {e}^{2 x + 4} \cdot \frac{d}{\mathrm{dx}} \left(2 x + 4\right)$

$= - 2 {e}^{2 x + 4}$

So $f ' \left(x\right) = - 2 {e}^{2 x + 4}$.

We also know that $g \left(x\right) = \tan \left(3 x\right)$

Therefore, $f ' \left(g \left(x\right)\right) = - 2 {e}^{2 \tan \left(3 x\right) + 4}$