Answer
$$y'=\frac{(x+1)^3}{(4x-2)^2}\left(\frac{3}{x+1}-\frac{4}{2x-1}\right)
.$$
Work Step by Step
Since $ y=\frac{(x+1)^3}{(4x-2)^2}$, then by taking the $\ln $ for both sides, we have
$$\ln y=\ln\left( \frac{(x+1)^3}{(4x-2)^2}\right)=\ln(x+1)^3-\ln (4x-2)^2\\
=3\ln(x+1)-2\ln (4x-2).$$
Now, taking the derivative, we get
$$\frac{y'}{y}=\frac{3}{x+1}-\frac{8}{4x-2}$$
Hence $$ y'=y\left(\frac{3}{x+1}-\frac{8}{4x-2}\right)\\
y'=\frac{(x+1)^3}{(4x-2)^2}\left(\frac{3}{x+1}-\frac{4}{2x-1}\right)
.$$
