Answer
$\dfrac{(3x+1)^2}{4x (2x-1)^3} (\dfrac{6}{3x+1}-\dfrac{1}{x}-\dfrac{6}{2x-1}) $
Work Step by Step
We have: $y= \dfrac{(3x+1)^2}{4x (2x-1)^3}$
We differentiate both sides with respect to $x$.
$\ln y=\ln (3x+1)^2-\ln (4x) -\ln (2x-1)^3 \\ \dfrac{1}{y} \dfrac{dy}{dx}=\dfrac{6}{3x+1}-\dfrac{1}{x}-\dfrac{6}{2x-1}\\ \dfrac{dy}{dx}=y (\dfrac{6}{3x+1}-\dfrac{1}{x}-\dfrac{6}{2x-1}) \\ \dfrac{dy}{dx}= \dfrac{(3x+1)^2}{4x (2x-1)^3} (\dfrac{6}{3x+1}-\dfrac{1}{x}-\dfrac{6}{2x-1}) $
