# How do you differentiate y = ln root3(6x + 7)?

Jan 7, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{6 x + 7}$

#### Explanation:

Rewrite using laws of logarithms:

$y = \ln {\left(6 x + 7\right)}^{\frac{1}{3}}$

$y = \frac{1}{3} \ln \left(6 x + 7\right)$

$3 y = \ln \left(6 x + 7\right)$

Now, let $y = \ln u$ and $u = 6 x + 7$. Then $\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 6$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \cdot 6$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{6 x + 7} \cdot 6$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6}{6 x + 7}$

Differentiate the left hand side now implicitly.

$\frac{d}{\mathrm{dx}} \left(3 y\right) = 3 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Put this together:

$3 \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6}{6 x + 7}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6}{3 \left(6 x + 7\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{6 x + 7}$

Hopefully this helps!