Answer
$\displaystyle \frac{dy}{dx}=\frac{3x^{3}+15x^{2}-8x}{2(x+1)^{3}\sqrt{3x-2}}$
Work Step by Step
$y=\displaystyle \frac{x^{2}\sqrt{3x-2}}{(x+1)^{2}}$
$...$apply ln( ) to both sides
... on the RHS, apply$ \displaystyle \ln\frac{M}{N}=\ln M-\ln M$
$\ln y=\ln(x^{2}(3x-2)^{1/2})-(x+1)^{2}$
... now, apply $\ln(M\cdot N)=\ln M+\ln N$ and $\ln M^{n}=n\ln M$
$\displaystyle \ln y=2\ln x+\frac{1}{2}\ln(3x-2)-2\ln(x+1)$
$\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{2}{x}+\frac{3}{2(3x-2)}-\frac{2}{x+1}$
$\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{2\cdot 2(3x-2)(x+1)+3\cdot x(x+1)-2\cdot 2x(3x-2)}{2x(3x-2)(x+1)}$
$\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{4(3x^{2} +x-2)+3x^{2}+3x-12x^{2}+8x}{2x(3x-2)(x+1)}\qquad.../\times y$
$\displaystyle \frac{dy}{dx}=y[\frac{3x^{2}+15x-8}{2x(3x-2)(x+1)}]\quad$ ... insert expression for y
$\displaystyle \frac{dy}{dx}=\frac{x^{2}(3x-2)^{1/2}}{(x+1)^{2}}\cdot\frac{3x^{2}+15x-8}{2x(3x-2)(x+1)}$
$\displaystyle \frac{dy}{dx}=\frac{3x^{3}+15x^{2}-8x}{2(x+1)^{3}(3x-2)^{1/2}}$
$\displaystyle \frac{dy}{dx}=\frac{3x^{3}+15x^{2}-8x}{2(x+1)^{3}\sqrt{3x-2}}$
