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

Apr 9, 2018

$\frac{{\mathrm{de}}^{x} / \left(x {e}^{x} - 4\right)}{\mathrm{dx}} = - {e}^{x} \frac{{e}^{x} + 4}{x {e}^{x} - 4} ^ 2$.

#### Explanation:

The quotient rule says

$d \frac{g \frac{x}{h \left(x\right)}}{\mathrm{dx}} = \frac{g ' \left(x\right) \cdot h \left(x\right) - g \left(x\right) \cdot h ' \left(x\right)}{h \left(x\right)} ^ 2$

Here

$g \left(x\right) = {e}^{x}$,

$g ' \left(x\right) = {e}^{x}$,

$h \left(x\right) = x {e}^{x} - 4$, and

$h ' \left(x\right) = {e}^{x} + x {e}^{x}$ by the product rule

so

$\frac{{\mathrm{de}}^{x} / \left(x {e}^{x} - 4\right)}{\mathrm{dx}} = \frac{{e}^{x} \left(x {e}^{x} - 4\right) - {e}^{x} \left({e}^{x} + x {e}^{x}\right)}{x {e}^{x} - 4} ^ 2$

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