# How do you differentiate f(x) = (e^(4-x))/(e^(1-x)-1) using the quotient rule?

Apr 20, 2017

$f ' \left(x\right) = {e}^{4 - x} / {\left({e}^{1 - x} - 1\right)}^{2}$

#### Explanation:

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

So, we need $u '$ and $v '$:

$u ' = \left({e}^{4 - x}\right) ' = {e}^{4 - x} \cdot \left(4 - x\right) ' = - {e}^{4 - x}$

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

$f ' \left(x\right) = \frac{- {e}^{4 - x} \cdot \left({e}^{1 - x} - 1\right) + {e}^{1 - x} \cdot {e}^{4 - x}}{{e}^{1 - x} - 1} ^ 2$

$f ' \left(x\right) = \frac{- {e}^{5 - 2 x} + {e}^{4 - x} + {e}^{5 - 2 x}}{{e}^{1 - x} - 1} ^ 2$

$f ' \left(x\right) = {e}^{4 - x} / {\left({e}^{1 - x} - 1\right)}^{2}$