# How do you differentiate f(x)= (7 x^2 + 2x - 6 )/ (x- 1 ) using the quotient rule?

Jan 11, 2016

$\frac{{x}^{2} - 2 x + 4}{x - 1} ^ 2$

#### Explanation:

for a rational function let f(x) =$\frac{u}{v}$

where u and v are both functions of x.Then :

$f ' \left(x\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

differentiating using this and the chain rule gives :

$f ' \left(x\right) = \frac{\left(x - 1\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 2 x - 1\right) - \left({x}^{2} + 2 x - 6\right) \frac{d}{\mathrm{dx}} \left(x - 1\right)}{x - 1} ^ 2$

$= \frac{\left(x - 1\right) \left(2 x + 2\right) - \left({x}^{2} + 2 x - 6\right) \left(1\right)}{x - 1} ^ 2$

'tidying up ' the numerator :

$f ' \left(x\right) \frac{2 {x}^{2} - 2 x + 2 x - 2 - {x}^{2} - 2 x + 6}{x - 1} ^ 2$

$\Rightarrow f ' \left(x\right) = \frac{{x}^{2} - 2 x + 4}{x - 1} ^ 2$