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

Apr 9, 2016

$f ' \left(x\right) = 1 - \frac{44}{x - 7} ^ 2$

#### Explanation:

The quotient rule states that
$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{u ' v - u v '}{v} ^ 2$

In our case, we have
$u = {x}^{2} - x + 2 \to u ' = 2 x - 1$
$v = x - 7 \to v ' = 1$

Applying the power rule and doing some algebra,
$f ' \left(x\right) = \frac{\left({x}^{2} - x + 2\right) ' \left(x - 7\right) - \left({x}^{2} - x + 2\right) \left(x - 7\right) '}{x - 7} ^ 2$
$\textcolor{w h i t e}{X X} = \frac{\left(2 x - 1\right) \left(x - 7\right) - \left({x}^{2} - x + 2\right)}{x - 7} ^ 2$
$\textcolor{w h i t e}{X X} = \frac{2 {x}^{2} - 15 x + 7 - {x}^{2} + x - 2}{x - 7} ^ 2$
$\textcolor{w h i t e}{X X} = \frac{{x}^{2} - 14 x + 5}{x - 7} ^ 2$
$\textcolor{w h i t e}{X X} = \frac{{x}^{2} - 14 x + 49 - 44}{x - 7} ^ 2$
$\textcolor{w h i t e}{X X} = \frac{{\left(x - 7\right)}^{2} - 44}{x - 7} ^ 2$
$\textcolor{w h i t e}{X X} = {\left(x - 7\right)}^{2} / {\left(x - 7\right)}^{2} - \frac{44}{x - 7} ^ 2$
$\textcolor{w h i t e}{X X} = 1 - \frac{44}{x - 7} ^ 2$