Answer
$\dfrac{dy}{dx}=\dfrac{(2x+1)^4(3x+4)}{(x+1)(3x-1)^3}[\dfrac{8}{2x+1}+\dfrac{3}{3x+4}-\dfrac{1}{x+1}-\dfrac{9}{3x-1}]$
Work Step by Step
We have: $y=\dfrac{(2x+1)^4(3x+4)}{(x+1)(3x-1)^3}$
or, $\ln y=\ln [\dfrac{(2x+1)^4(3x+4)}{(x+1)(3x-1)^3}]$
We differentiate both sides with respect to $x$.
$\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{8}{2x+1}+\dfrac{3}{3x+4}-\dfrac{1}{x+1}-\dfrac{9}{3x-1}$
This implies that $\dfrac{dy}{dx}=y[\dfrac{8}{2x+1}+\dfrac{3}{3x+4}-\dfrac{1}{x+1}-\dfrac{9}{3x-1}]$
Therefore, $\dfrac{dy}{dx}=\dfrac{(2x+1)^4(3x+4)}{(x+1)(3x-1)^3}[\dfrac{8}{2x+1}+\dfrac{3}{3x+4}-\dfrac{1}{x+1}-\dfrac{9}{3x-1}]$
