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

Sep 8, 2016

$g ' \left(x\right) = \frac{2 x + 1}{x}$

#### Explanation:

$g \left(x\right) = \ln \left(x {e}^{2 x}\right)$

$g ' \left(x\right) = \frac{1}{x {e}^{2 x}} \cdot \frac{d}{\mathrm{dx}} \left(x {e}^{2 x}\right)$ Standerd differential and chain rule

$= \frac{1}{x {e}^{2 x}} \cdot \left(x \cdot {e}^{2 x} \cdot 2 + {e}^{2 x} \cdot 1\right)$ Product rule, chain rule and standard differential

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

$= \frac{\cancel{{e}^{2 x}} \left(2 x + 1\right)}{x \cancel{{e}^{2 x}}}$

$= \frac{2 x + 1}{x}$