## Calculus (3rd Edition)

$$y'=\frac{(x+1)^3}{(4x-2)^2}\left(\frac{3}{x+1}-\frac{4}{2x-1}\right) .$$
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 $$y'=y\left(\frac{3}{x+1}-\frac{8}{4x-2}\right)\\ y'=\frac{(x+1)^3}{(4x-2)^2}\left(\frac{3}{x+1}-\frac{4}{2x-1}\right) .$$