# How do you differentiate g(x) = (2e^(x^2) + 4e^x) ( 2x + 2x^2) using the product rule?

Dec 24, 2015

The product rule states that for a function $y = f \left(x\right) g \left(x\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$.

#### Explanation:

We'll also need chain rule, which indicates that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \left(4 x {e}^{x} + 4 {e}^{x}\right) \left(2 x + 2 {x}^{2}\right) + \left(2 {e}^{{x}^{2}} + 4 {e}^{x}\right) \left(2 + 4 x\right)$

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \left(4 {e}^{x} \left(x + 1\right)\right) \left(2 x \left(x + 1\right)\right) + \left(2 {e}^{x} \left({e}^{x} + 2\right)\right) \left(2 + 4 x\right)$

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \left(x + 1\right) \left(4 {e}^{x} + 2 x\right) + \left(2 {e}^{x} \left({e}^{x} + 2\right)\right) \left(2 + 4 x\right)$