# What is the derivative of? : (x^2+2)e^(4x)

## (Question Restore: portions of this question have been edited or deleted!)

Jan 15, 2018

(D) is an intermediate step.

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

#### Explanation:

We will apply the Product Rule for Differentiation:

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$, or, $\left(u v\right) ' = \left(\mathrm{du}\right) v + u \left(\mathrm{dv}\right)$

So with $y = \left({x}^{2} + 2\right) {e}^{4 x}$;

$\left\{\begin{matrix}\text{Let" & u = x^2+2 & => (du)/dx = 2x \\ "And} & v = {e}^{4 x} & \implies \frac{\mathrm{dv}}{\mathrm{dx}} = 4 {e}^{4 x}\end{matrix}\right.$

Then:

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$

Giving:

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