# How do you differentiate f(x) = (e^x-4x)/(2x-e^x) using the quotient rule?

Jan 23, 2017

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

#### Explanation:

The quotient rule is

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

Here,

$u = {e}^{x} - 4 x$, $\implies$, $u ' = {e}^{x} - 4$

$v = 2 x - {e}^{x}$, $\implies$, $v ' = 2 - {e}^{x}$

Therefore,

$f ' \left(x\right) = \frac{\left({e}^{x} - 4\right) \left(2 x - {e}^{x}\right) - \left({e}^{x} - 4 x\right) \left(2 - {e}^{x}\right)}{2 x - {e}^{x}} ^ 2$

$= \frac{2 x {e}^{x} - {\cancel{e}}^{2 x} - \cancel{8} x + 4 {e}^{x} - 2 {e}^{x} + {\cancel{e}}^{2 x} + \cancel{8} x - 4 x {e}^{x}}{2 x - {e}^{x}} ^ 2$

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

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