# How do you differentiate f(x)= (2 x^2 - x - 6 )/ (x- 5 ) using the quotient rule?

Jan 17, 2017

The answer is $= \frac{2 {x}^{2} - 20 x + 11}{x - 5} ^ 2$

#### Explanation:

We need

$\left({x}^{n}\right) ' = n {x}^{n - 1}$

We use the formula

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

Here,

$u = 2 {x}^{2} - x - 6$, $\implies$, $u ' = 4 x - 1$

$v = x - 5$, $\implies$, $v ' = 1$

Therefore,

$f ' \left(x\right) = \frac{\left(4 x - 1\right) \left(x - 5\right) - \left(2 {x}^{2} - x - 6\right) \cdot 1}{x - 5} ^ 5$

$= \frac{4 {x}^{2} - 21 x + 5 - 2 {x}^{2} + x + 6}{x - 5} ^ 2$

$= \frac{2 {x}^{2} - 20 x + 11}{x - 5} ^ 2$