Answer
$\dfrac{dy}{dx}=\dfrac{2x+1}{4x-2} (\dfrac{2}{2x+1}-\dfrac{4}{4x-2})$
Work Step by Step
We have: $y=\dfrac{2x+1}{4x-2}$
We differentiate both sides with respect to $x$.
$\ln y=\ln (2x+1) -\ln (4x-2) \\ \dfrac{1}{y} \dfrac{dy}{dx}=\dfrac{2}{2x+1}-\dfrac{4}{4x-2}\\ \dfrac{dy}{dx}=y (\dfrac{2}{2x+1}-\dfrac{4}{4x-2})\\ \dfrac{dy}{dx}=\dfrac{2x+1}{4x-2} (\dfrac{2}{2x+1}-\dfrac{4}{4x-2})$
