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

Nov 6, 2015

Apply the product rule to the functions ${f}_{1} \left(x\right) = 2 {x}^{2} - 4 x + 4$ and ${f}_{2} \left(x\right) = {e}^{x}$ to obtain $\frac{d}{\mathrm{dx}} {f}_{1} \left(x\right) {f}_{2} \left(x\right) = 2 {x}^{2} {e}^{x}$

#### Explanation:

The product rule states that for any functions $f \left(x\right)$ and $g \left(x\right)$
$\frac{d}{\mathrm{dx}} f \left(x\right) g \left(x\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

Let ${f}_{1} \left(x\right) = 2 {x}^{2} - 4 x + 4$ and ${f}_{2} \left(x\right) = {e}^{x}$

Then ${f}_{1} ' \left(x\right) = 4 x - 4$ and ${f}_{2} ' \left(x\right) = {e}^{x}$

Note that

$\frac{d}{\mathrm{dx}} \left(2 {x}^{2} - 4 x + 4\right) {e}^{x} = \frac{d}{\mathrm{dx}} {f}_{1} \left(x\right) {f}_{2} \left(x\right)$

So, applying the product rule gives us

$\frac{d}{\mathrm{dx}} \left(2 {x}^{2} - 4 x + 4\right) {e}^{x} = {f}_{1} ' \left(x\right) {f}_{2} \left(x\right) + {f}_{1} \left(x\right) {f}_{2} ' \left(x\right)$
$\implies \frac{d}{\mathrm{dx}} \left(2 {x}^{2} - 4 x + 4\right) {e}^{x} = \left(4 x - 4\right) {e}^{x} + \left(2 {x}^{2} - 4 x + 4\right) {e}^{x}$

Simplifying the result gives us the final answer of
$\frac{d}{\mathrm{dx}} \left(2 {x}^{2} - 4 x + 4\right) {e}^{x} = 2 {x}^{2} {e}^{x}$