# How do you differentiate f(x)= (4 x^2 + 5x -3 )/ (x- 1 ) using the quotient rule?

$f ' \left(x\right) = \frac{4 {x}^{2} - 8 x - 2}{x - 1} ^ 2$

#### Explanation:

From the given $f \left(x\right) = \frac{4 {x}^{2} + 5 x - 3}{x - 1}$

Use $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{{v}^{2}}$

let $u = 4 {x}^{2} + 5 x - 3$

let $v = x - 1$

Jan 15, 2016

I found $f ' \left(x\right) = \frac{4 {x}^{2} - 8 x - 2}{x - 1} ^ 2$

#### Explanation:

Consider the Quotient Rule:

We can apply this to your function to get:
$f ' \left(x\right) = \frac{\left(8 x + 5\right) \left(x - 1\right) - 1 \cdot \left(4 {x}^{2} + 5 x - 3\right)}{x - 1} ^ 2 =$
$= \frac{8 {x}^{2} - 8 x \cancel{+ 5 x} - 5 - 4 {x}^{2} \cancel{- 5 x} + 3}{x - 1} ^ 2 =$
$= \frac{4 {x}^{2} - 8 x - 2}{x - 1} ^ 2$