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

Mar 9, 2016

Use expansion, apply linearity and product rule to get:
$g ' \left(x\right) = 4 \left(2 \left(x + 2\right) {e}^{x} + x \left(3 x + 2\right)\right)$

Explanation:

The approach that I am going to use to differentiate $g \left(x\right) = \left(2 {x}^{2} + 4 {e}^{x}\right) \left(2 x + 2\right)$ is expansion:
$g \left(x\right) = \left(2 {x}^{3} + 8 x {e}^{x}\right) + \left(4 {x}^{2} + 8 {e}^{x}\right)$
Now we can apply linearity and write:
$g ' \left(x\right) = 2 \frac{d \left({x}^{3}\right)}{\mathrm{dx}} + \frac{d \left(\textcolor{b l u e}{8 x {e}^{x}}\right)}{\mathrm{dx}} + 4 \frac{d \left({x}^{2}\right)}{\mathrm{dx}} + 8 \frac{d \left({e}^{x}\right)}{\mathrm{dx}}$ Now you can differentiate each term in straight forward manner except for $\textcolor{b l u e}{b l u e}$. For $\textcolor{b l u e}{b l u e}$ you can use product rule: $\textcolor{b l u e}{b l u e ' = 8 {e}^{x} + 8 x {e}^{x}}$
Putting it all together you have:
$g ' \left(x\right) = 4 \left(2 \left(x + 2\right) {e}^{x} + x \left(3 x + 2\right)\right)$