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

May 5, 2016

$- \frac{2}{{x}^{2} - 1}$, x > 1.

#### Explanation:

under the condition x > 1 for log(x-1).
$f \left(x\right) = \ln \left(\frac{x + 1}{x - 1}\right) = \ln \left(x + 1\right) - \ln \left(x - 1\right)$.

Applying function of function rule,

$f ' \left(x\right) = \frac{1}{x + 1} \left(x + 1\right) ' - \frac{1}{x - 1} \left(x - 1\right) '$

$= \frac{1}{x + 1} \left(1\right) - \frac{1}{x - 1} \left(1\right)$

$= \frac{\left(x - 1\right) - \left(x + 1\right)}{\left(x + 1\right) \left(x - 1\right)}$

$= - \frac{2}{{x}^{2} - 1}$