# How do I find the derivative of ln*root3((x-1)/(x+1))?

##### 1 Answer
Nov 9, 2016

$\frac{d}{\mathrm{dx}} \ln \sqrt[3]{\frac{x - 1}{x + 1}} = \frac{2}{3 {x}^{2} - 3}$

#### Explanation:

$y = \ln \sqrt[3]{\frac{x - 1}{x + 1}}$
$\therefore y = \ln {\left(\frac{x - 1}{x + 1}\right)}^{\frac{1}{3}}$
$\therefore y = \frac{1}{3} \ln \left(\frac{x - 1}{x + 1}\right)$
$\therefore y = \frac{1}{3} \left\{\ln \left(x - 1\right) - \ln \left(x + 1\right)\right\}$ (law of logs)

Differentiating (using the chain rule) gives:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \left\{\frac{1}{x - 1} - \frac{1}{x + 1}\right\}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \left\{\frac{\left(x - 1\right) - \left(x + 1\right)}{\left(x - 1\right) \left(x + 1\right)}\right\}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \left\{\frac{- 2}{1 - {x}^{2}}\right\}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3} \frac{1}{{x}^{2} - 1}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3 {x}^{2} - 3}$