# How do you find the derivative of f(x) = 2x^3 e^(9x)?

##### 1 Answer
May 17, 2015

The answer is : $f ' \left(x\right) = 6 {e}^{9 x} {x}^{2} \left(1 + 3 x\right)$

$f \left(x\right) = 2 {x}^{3} {e}^{9 x} = h \left(x\right) g \left(x\right)$, where $h \left(x\right) = 2 {x}^{3}$ and $g \left(x\right) = {e}^{9 x}$.

We will find the derivative with the product rule :

$f ' \left(x\right) = h ' \left(x\right) g \left(x\right) + h \left(x\right) g ' \left(x\right) = \left(2 {x}^{3}\right) ' {e}^{9 x} + 2 {x}^{3} \left({e}^{9 x}\right) '$

$f ' \left(x\right) = 2 \cdot 3 {x}^{2} \cdot {e}^{9 x} + 2 {x}^{3} \cdot 9 {e}^{9 x} = 6 {x}^{2} {e}^{9 x} + 18 {x}^{3} {e}^{9 x}$

$f ' \left(x\right) = 6 {e}^{9 x} \left({x}^{2} + 3 {x}^{3}\right) = 6 {e}^{9 x} {x}^{2} \left(1 + 3 x\right)$