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

Jul 3, 2016

Simplify using natural log properties, take the derivative, and add some fractions to get $\frac{d}{\mathrm{dx}} \ln \left(\frac{x + 1}{x - 1}\right) = - \frac{2}{{x}^{2} - 1}$

#### Explanation:

It helps to use natural log properties to simplify $\ln \left(\frac{x + 1}{x - 1}\right)$ into something a little less complicated. We can use the property $\ln \left(\frac{a}{b}\right) = \ln a - \ln b$ to change this expression to:
$\ln \left(x + 1\right) - \ln \left(x - 1\right)$

Taking the derivative of this will be a lot easier now. The sum rule says we can break this up into two parts:
$\frac{d}{\mathrm{dx}} \ln \left(x + 1\right) - \frac{d}{\mathrm{dx}} \ln \left(x - 1\right)$

We know the derivative of $\ln x = \frac{1}{x}$, so the derivative of $\ln \left(x + 1\right) = \frac{1}{x + 1}$ and the derivative of $\ln \left(x - 1\right) = \frac{1}{x - 1}$:
$\frac{d}{\mathrm{dx}} \ln \left(x + 1\right) - \frac{d}{\mathrm{dx}} \ln \left(x - 1\right) = \frac{1}{x + 1} - \frac{1}{x - 1}$

Subtracting the fractions yields:
$\frac{x - 1}{\left(x + 1\right) \left(x - 1\right)} - \frac{x + 1}{\left(x - 1\right) \left(x + 1\right)}$
$= \frac{\left(x - 1\right) - \left(x + 1\right)}{{x}^{2} - 1}$
$= \frac{x - 1 - x - 1}{{x}^{2} - 1}$
$= - \frac{2}{{x}^{2} - 1}$