# What is the derivative of (3x^3)/e^x?

Sep 18, 2017

$9 {x}^{2} {e}^{-} x - 3 {x}^{3} {e}^{-} x = \frac{3 {x}^{2} \left(3 - x\right)}{e} ^ x$

#### Explanation:

Note that $\frac{1}{e} ^ x = {e}^{-} x \to f \left(x\right) = 3 {x}^{3} {e}^{z} x$.

Now we can use the product rule. $f \left(x\right) = g \left(x\right) h \left(x\right) \to f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + f \left(x\right) g ' \left(x\right)$

The power rule and definition of the derivative of ${e}^{u}$ give us $\frac{d}{\mathrm{dx}} \left(3 {x}^{3}\right) = 9 {x}^{2} , \frac{d}{\mathrm{dx}} \left({e}^{-} x\right) = - {e}^{-} x$

Thus...

$f ' \left(x\right) = 9 {x}^{2} {e}^{-} x - 3 {x}^{3} {e}^{-} x = \frac{3 {x}^{2} \left(3 - x\right)}{e} ^ x$