# What is the derivative of [e^x / (1 - e^x)]?

Jul 31, 2015

${y}^{'} = {e}^{x} / {\left(1 - {e}^{x}\right)}^{2}$

#### Explanation:

The quotient rule will work just fine for this function because you can write it as

$y = {e}^{x} / \left(1 - {e}^{x}\right) = f \frac{x}{g} \left(x\right)$

In such cases, the derivative of the function can be found by

color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2, with $g \left(x\right) \ne 0$.

Apart from this, all you really need to know is that

$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$

So, the derivative of $y$ will be

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left({e}^{x}\right)\right] \cdot \left(1 - {e}^{x}\right) - {e}^{x} \cdot \frac{d}{\mathrm{dx}} \left(1 - {e}^{x}\right)}{1 - {e}^{x}} ^ 2$

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

${y}^{'} = \frac{{e}^{x} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{2 x}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{2 x}}}}}{1 - {e}^{x}} ^ 2$

${y}^{'} = \textcolor{g r e e n}{{e}^{x} / {\left(1 - {e}^{x}\right)}^{2}}$