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

Dec 17, 2016

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

#### Explanation:

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

Then $\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$ and $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{{e}^{x} \left(1 - {e}^{x}\right) - \left(- {e}^{x} \left(1 + {e}^{x}\right)\right)}{1 - {e}^{x}} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

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

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

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

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

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

Hopefully this helps!