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

${f}^{'} \left(x\right) = \frac{4 {x}^{2} - 4 x - 1}{{\left(2 x - 1\right)}^{2}}$
Quotient rule states that if $h$ and $k$ are two differentiable functions on ]a,b[, then forall x in ]a,b[ such that $k \left(x\right) \ne 0$ the following equality holds:
${\left[\frac{h \left(x\right)}{k \left(x\right)}\right]}^{'} = \frac{{h}^{'} \left(x\right) k \left(x\right) - h \left(x\right) {k}^{'} \left(x\right)}{{k}^{2} \left(x\right)}$
In our case $h \left(x\right) = 2 {x}^{2} - x + 1$ and $k \left(x\right) = 2 x - 1$. The two derivatives are ${h}^{'} \left(x\right) = 4 x - 1$ and ${k}^{'} \left(x\right) = 2$.
${f}^{'} \left(x\right) = \frac{\left(4 x - 1\right) \left(2 x - 1\right) - \left(2 {x}^{2} - x + 1\right) 2}{{\left(2 x - 1\right)}^{2}} = \frac{8 {x}^{2} - 6 x + 1 - 4 {x}^{2} + 2 x - 2}{{\left(2 x - 1\right)}^{2}} = \frac{4 {x}^{2} - 4 x - 1}{{\left(2 x - 1\right)}^{2}}$