# How do you differentiate g(x) = e^(2x)ln3x using the product rule?

Dec 19, 2015

I found: ${e}^{2 x} \left[2 \ln \left(3 x\right) + \frac{1}{x}\right]$

#### Explanation:

The Product Rule tells you that if you have:

$g \left(x\right) = f \left(x\right) \cdot h \left(x\right)$
then
$g ' \left(x\right) = f ' \left(x\right) h \left(x\right) + f \left(x\right) h ' \left(x\right)$

In your case for each function you will need, when derived, to use also the Chain Rule (in red) to get:

$g ' \left(x\right) = \textcolor{red}{2 {e}^{2 x}} \ln \left(3 x\right) + {e}^{2 x} \textcolor{red}{\frac{1}{3 x} \cdot 3} =$

$= {e}^{2 x} \left[2 \ln \left(3 x\right) + \frac{1}{x}\right]$