# How do you differentiate f(x)= -1 / (-2x-4 ) using the quotient rule?

Oct 31, 2016

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

#### Explanation:

I would first remove the -ve sign as $- 1$ can be factored out from the denominator.

So $f \left(x\right) = - \frac{1}{- 2 x - 4}$
$\therefore f \left(x\right) = \frac{1}{2 x + 4}$

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

$f ' \left(x\right) = \left\{\frac{\left(2 x + 4\right) \left(\frac{d}{\mathrm{dx}} 1\right) - \left(1\right) \left(\frac{d}{\mathrm{dx}} \left(2 x + 4\right)\right)}{2 x + 4} ^ 2\right\}$
$\therefore f ' \left(x\right) = \frac{0 - \left(1\right) \left(2\right)}{2 x + 4} ^ 2$
$\therefore f ' \left(x\right) = \frac{- 2}{2 x + 4} ^ 2$