# How do you differentiate f(x)= -1 / (2x-7 ) using the quotient rule?

Jun 8, 2016

$\frac{d}{\mathrm{dx}} \left(- \frac{1}{2 x - 7}\right) = \frac{2}{{\left(2 x - 7\right)}^{2}}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(- \frac{1}{2 x - 7}\right)$

Taking the constant out,

${\left(a \cdot f\right)}^{'} = a \cdot {f}^{'}$

$= - \frac{d}{\mathrm{dx}} \left(\frac{1}{2 x - 7}\right)$

$= - \frac{d}{\mathrm{dx}} \left({\left(2 x - 7\right)}^{- 1}\right)$

Applying chain rule,

$\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$L e t , 2 x - 7 = u$

$= - \frac{d}{\mathrm{du}} \left({u}^{- 1}\right) \frac{d}{\mathrm{dx}} \left(2 x - 7\right)$

WE know,

$\frac{d}{\mathrm{du}} \left({u}^{- 1}\right) = - \frac{1}{{u}^{2}}$

$\frac{d}{\mathrm{dx}} \left(2 x - 7\right) = 2$

$= - \left(- \frac{1}{{u}^{2}}\right) 2$

Substitute back $= - \left(- \frac{1}{{u}^{2}}\right)$
$= - \left(- \frac{1}{{\left(2 x - 7\right)}^{2}}\right) 2$

$= - \left(- \frac{1}{{\left(2 x - 7\right)}^{2}}\right) 2$

Simplify,
$\frac{2}{{\left(2 x - 7\right)}^{2}}$