# How do you differentiate g(x) = x^3e^(2x) using the product rule?

Jan 2, 2016

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

#### Explanation:

According to the product rule,

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(x\right)\right] = f ' \left(x\right) g \left(x\right) + g ' \left(x\right) f \left(x\right)$

Thus,

$g ' \left(x\right) = {e}^{2 x} \frac{d}{\mathrm{dx}} \left({x}^{3}\right) + {x}^{3} \frac{d}{\mathrm{dx}} \left({e}^{2 x}\right)$

Find each derivative.

$\frac{d}{\mathrm{dx}} \left({x}^{3}\right) = 3 {x}^{2}$

This will require the chain rule:

$\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = u ' {e}^{u}$, so

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

Plug these back in to find $g ' \left(x\right)$.

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

Optionally factored:

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