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

Jul 15, 2017

d/(dx) [(e^x)/(e^(x-2)-4)] = color(blue)((-4e^x)/((e^(x-2)-4)^2)

#### Explanation:

We're asked to find the derivative

$\frac{d}{\mathrm{dx}} \left[\frac{{e}^{x}}{{e}^{x - 2} - 4}\right]$

Using the quotient rule, which is

$\frac{d}{\mathrm{dx}} \left[\frac{u}{v}\right] = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{{v}^{2}}$

where

• $u = {e}^{x}$

• $v = {e}^{x - 2} - 4$:

$= \frac{\left({e}^{x - 2} - 4\right) \left(\frac{d}{\mathrm{dx}} \left[{e}^{x}\right]\right) - \left({e}^{x}\right) \left(\frac{d}{\mathrm{dx}} \left[{e}^{x - 2} - 4\right]\right)}{{\left({e}^{x - 2} - 4\right)}^{2}}$

The derivative of ${e}^{x}$ is ${e}^{x}$ (fundamental!):

$= \frac{\left({e}^{x - 2} - 4\right) \left({e}^{x}\right) - \left({e}^{x}\right) \left(\frac{d}{\mathrm{dx}} \left[{e}^{x - 2} - 4\right]\right)}{{\left({e}^{x - 2} - 4\right)}^{2}}$

The derivative of ${e}^{x - 2}$ is thus also ${e}^{x - 2}$:

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

= color(blue)(((e^(x-2)-4)(e^x) - e^(2x-2))/((e^(x-2)-4)^2)

You can simplify this to

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

= color(blue)((-4e^x)/((e^(x-2)-4)^2)