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

Jul 19, 2016

$f ' \left[g \left(x\right)\right] = - 2 {e}^{x} \left({e}^{x} + 1\right)$

#### Explanation:

$f \left(x\right) = - {x}^{2} - 2 x$
$g \left(x\right) = {e}^{x}$

Replacing $x$ by $g \left(x\right)$ in $f \left(x\right)$

$f \left[g \left(x\right)\right] = - {\left({e}^{x}\right)}^{2} - 2 \cdot {e}^{x}$

$f ' \left[g \left(x\right)\right] = \frac{d}{\mathrm{dx}} \left(- {\left({e}^{x}\right)}^{2} - 2 \cdot {e}^{x}\right) = = \frac{d}{\mathrm{dx}} \left(- {e}^{2 x} - 2 \cdot {e}^{x}\right)$

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

$= - 2 {e}^{x} \left({e}^{x} + 1\right)$