# How do you differentiate  y =( ln(3x + 2))^2 using the chain rule?

Jan 31, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 \ln \left(3 x + 2\right)}{3 x + 2}$

#### Explanation:

Differentiate using the $\textcolor{b l u e}{\text{ chain rule}}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln \left(3 x + 2\right) \frac{d}{\mathrm{dx}} \left(\ln \left(3 x + 2\right)\right)$

$= 2 \ln \left(3 x + 2\right) . \frac{1}{3 x + 2} \frac{d}{\mathrm{dx}} \left(3 x + 2\right)$

$= 2 \ln \left(3 x + 2\right) . \frac{1}{3 x + 2} . 3$

$= \frac{6 \ln \left(3 x + 2\right)}{3 x + 2}$