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

May 22, 2017

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

We apply the Quotient Rule for Differentiation:

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$, or less formally, $\text{ } \left(\frac{u}{v}\right) ' = \frac{v \left(\mathrm{du}\right) - u \left(\mathrm{dv}\right)}{v} ^ 2$

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with $f \left(x\right) = {e}^{x} / \left({e}^{x - 2} + 2 x\right)$ :

$\left\{\begin{matrix}\text{Let & "u=e^x & => & (du)/dx=e^xx \\ "And & } v = {e}^{x - 2} + 2 x & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = {e}^{x - 2} + 2\end{matrix}\right.$

Then:

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