Answer
$\displaystyle \frac{dy}{dx}=\frac{-18x^{2}+2x+1}{(2x-1)^{6}}$
Work Step by Step
$y=\displaystyle \frac{3x^{2}-x}{(2x-1)^{5}}= \frac{u(x)}{v(x)}$
Use the quotient rule
$\displaystyle \frac{dy}{dx}=\frac{v(t)\cdot u^{\prime}(t)-u(t)\cdot v^{\prime}(t)}{[v(t)]^{2}}$
For $v^{\prime}(x)$, we need the chain rule:
$v^{\prime}(x)=[(2x-1)^{5}]^{\prime} =5(2x-1)^{4}\cdot(2x-1)^{\prime}=5(2x-1)^{4}\cdot 2]$
$\displaystyle \frac{dy}{dx}=\frac{(2x-1)^{5}(6x-1)-(3x^{2}-x)[5(2x-1)^{4}\cdot 2]}{[(2x-1)^{5}]^{2}}$
$=\displaystyle \frac{(2x-1)^{5}(6x-1)-10(3x^{2}-x)(2x-1)^{4}}{(2x-1)^{10}}$
... factor out $(2x-1)^{4}$ in the numerator...
$=\displaystyle \frac{(2x-1)^{4}[(2x-1)(6x-1)-10(3x^{2}-x)]}{(2x-1)^{10}}$
... reduce the ratio with $(2x-1)^{4}$, simplify
$=\displaystyle \frac{12x^{2}-2x-6x+1-30x^{2}+10x}{(2x-1)^{6}}$
$\displaystyle \frac{dy}{dx}=\frac{-18x^{2}+2x+1}{(2x-1)^{6}}$
