# Whats the derivative of y=1/2 ln[(1+x)/(1-x)]?

Sep 3, 2017

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

#### Explanation:

$y = \frac{1}{2} \ln \left[\frac{1 + x}{1 - x}\right] .$

Using the Familiar Rules of $\log .$ fun., we get,

$y = \frac{1}{2} \left[\ln \left(1 + x\right) - \ln \left(1 - x\right)\right] .$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left[\frac{d}{\mathrm{dx}} \left\{\ln \left(1 + x\right)\right\} - \frac{d}{\mathrm{dx}} \left\{\ln \left(1 - x\right)\right\}\right] ,$

$= \frac{1}{2} \left[\frac{1}{1 + x} \cdot \frac{d}{\mathrm{dx}} \left(1 + x\right) - \frac{1}{1 - x} \cdot \frac{d}{\mathrm{dx}} \left(1 - x\right)\right] \ldots \left[\because , \text{ the Chain Rule}\right] ,$

$= \frac{1}{2} \left[\frac{1}{1 + x} \left(1\right) - \frac{1}{1 - x} \left(- 1\right)\right] ,$

$= \frac{1}{2} \left[\frac{1}{1 + x} + \frac{1}{1 - x}\right] ,$

$= \frac{1}{2} \left[\frac{\left(1 + x\right) + \left(1 - x\right)}{\left(1 + x\right) \left(1 - x\right)}\right] ,$

$= \frac{1}{\cancel{2}} \left[\frac{\cancel{2}}{1 - {x}^{2}}\right] .$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 - {x}^{2}} ,$ is the desired Derivative.

Enjoy Maths.!