# How do you differentiate f(x) = (e^x-1)/(x-e^x) using the quotient rule?

Feb 9, 2017

The answer is $f ' \left(x\right) = \frac{x {e}^{x} - 2 {e}^{x} + 1}{x - {e}^{x}} ^ 2$

#### Explanation:

We use

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

Here,

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

$u = {e}^{x} - 1$, $\implies$, $u ' = {e}^{x}$

$v = x - {e}^{x}$, $\implies$, $v ' = 1 - {e}^{x}$

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

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

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