# How do you find the derivative of (2x+8)/(x-8)?

Oct 21, 2016

$\frac{d}{\mathrm{dx}} \left(\frac{2 x + 8}{x - 8}\right) = \frac{- 24}{x - 8} ^ 2$

#### Explanation:

You need to use the quotient rule; $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

$\frac{d}{\mathrm{dx}} \left(\frac{2 x + 8}{x - 8}\right) = \frac{\left(x - 8\right) \frac{d}{\mathrm{dx}} \left(2 x + 8\right) - \left(2 x + 8\right) \frac{d}{\mathrm{dx}} \left(x - 8\right)}{x - 8} ^ 2$
$\therefore \frac{d}{\mathrm{dx}} \left(\frac{2 x + 8}{x - 8}\right) = \frac{\left(x - 8\right) \left(2\right) - \left(2 x + 8\right) \left(1\right)}{x - 8} ^ 2$
$\therefore \frac{d}{\mathrm{dx}} \left(\frac{2 x + 8}{x - 8}\right) = \frac{2 x - 16 - 2 x - 8}{x - 8} ^ 2$
$\therefore \frac{d}{\mathrm{dx}} \left(\frac{2 x + 8}{x - 8}\right) = \frac{- 24}{x - 8} ^ 2$