# Question #e365f

Jun 2, 2016

Same as you would if the exponent were positive - with a slight adjustment for the chain rule.

#### Explanation:

Consider differentiating ${e}^{x}$:
$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

It's a special function because its derivative is itself. Now, try ${e}^{-} x$:
$\frac{d}{\mathrm{dx}} {e}^{-} x = - {e}^{-} x$

We see that the derivative is the same as itself again - except this time, it's negative. Why?

The answer has to do with the chain rule, which states that the derivative of a composite function (a function within a function) is the derivative of the inner function times the derivative of the function in general. This is best demonstrated with an example:
$\frac{d}{\mathrm{dx}} {e}^{{x}^{2}}$

We have one function, ${x}^{2}$, nestled within another function, ${e}^{x}$. By the chain rule, the derivative is the derivative of ${x}^{2}$ times the derivative of ${e}^{{x}^{2}}$:
$\frac{d}{\mathrm{dx}} {e}^{{x}^{2}} = \left({x}^{2}\right) ' \cdot \left({e}^{{x}^{2}}\right) ' = \textcolor{red}{2 x} \textcolor{b l u e}{{e}^{{x}^{2}}}$
Because $\frac{d}{\mathrm{dx}} {x}^{2} = \textcolor{red}{2 x}$ and $\frac{d}{\mathrm{dx}} {e}^{{x}^{2}} = \textcolor{b l u e}{{e}^{{x}^{2}}}$

The same logic holds for functions with negative exponents, like ${e}^{-} x$. In this case, we have $- x$ nestled within ${e}^{x}$; the derivative is $\left(- x\right) ' \cdot \left({e}^{-} x\right) '$:
$\frac{d}{\mathrm{dx}} {e}^{-} x = - 1 \cdot {e}^{-} x = - {e}^{-} x$

This makes intuitive sense too. We know that ${e}^{-} x$ is constantly getting smaller, which means the slope is always negative. The derivative, $- {e}^{-} x$ should reflect that, and it does: $- {e}^{-} x$, because of the negative sign put there by the chain rule, is always negative.

So for $f \left(x\right) = {x}^{3} {e}^{-} x$, you would use the product rule:
$\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$
In this case $u = {x}^{3} \to u ' = 3 {x}^{2}$ and $v = {e}^{-} x \to v ' = - {e}^{-} x$
$\frac{d}{\mathrm{dx}} \left({x}^{3} {e}^{-} x\right) = \left(3 {x}^{2}\right) \left({e}^{-} x\right) + \left({x}^{3}\right) \left(- {e}^{-} x\right)$
$= 3 {x}^{2} {e}^{-} x - {x}^{3} {e}^{-} x$