# Whats the derivative of ln((1+x)/(1-x))?

Nov 6, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{1 - {x}^{2}}$

#### Explanation:

Let $y = \ln u$ and $u = \frac{1 + x}{1 - x}$.

We will use the chain rule to differentiate this problem. However, we must first find the derivative of each function.

$y ' = \frac{1}{u}$

By the quotient rule:

$u ' = \frac{1 \left(1 - x\right) - \left(- 1 \left(1 + x\right)\right)}{1 - x} ^ 2$

$u ' = \frac{1 - x - \left(- 1 - x\right)}{1 - x} ^ 2$

$u ' = \frac{1 - x + 1 + x}{1 - x} ^ 2$

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

Now, by the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \times \frac{2}{1 - x} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\frac{1 + x}{1 - x}} \times \frac{2}{1 - x} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - x}{1 + x} \times \frac{2}{1 - x} ^ 2$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{1 - {x}^{2}}$

Hopefully this helps!