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

Dec 22, 2016

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

#### Explanation:

$\text{Given " f(x)=(g(x))/(h(x))" then }$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{here } g \left(x\right) = 2 {x}^{2} - x + 1 \Rightarrow g ' \left(x\right) = 4 x - 1$

$\text{and } h \left(x\right) = 2 x - 1 \Rightarrow h ' \left(x\right) = 2$

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

simplifying the numerator.

$f ' \left(x\right) = \frac{8 {x}^{2} - 6 x + 1 - 4 {x}^{2} + 2 x - 2}{2 x - 1} ^ 2$

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