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

Mar 7, 2018

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

Explanation:

This is the quotient rule.

$f \left(x\right) = \frac{\textcolor{red}{u}}{\textcolor{b l u e}{v}}$

$f ' \left(x\right) = \frac{\textcolor{b l u e}{v} \textcolor{red}{u '} - \textcolor{b l u e}{v '} \textcolor{red}{u}}{\textcolor{b l u e}{{v}^{2}}}$

Hence, to differentiate the given $f \left(x\right)$, you do the following

$f ' \left(x\right) = \frac{\left(\textcolor{b l u e}{x - 1}\right) \left(\textcolor{red}{8 x + 2}\right) - \left(\textcolor{b l u e}{1}\right) \left(\textcolor{red}{4 {x}^{2} + 2 x - 3}\right)}{\textcolor{b l u e}{{\left(x - 1\right)}^{2}}}$

$f ' \left(x\right) = \frac{8 {x}^{2} + 2 x - 8 x - 2 - 4 {x}^{2} - 2 x + 3}{x - 1} ^ 2$

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

There is no need to simplify any further after this point