Answer
$y'=$ $\frac{2x^{2}+2x-1}{(x^{3}-x)}$ $\left(\frac{x(x-1)^{3/2}}{\sqrt{(x+1)}}\right)$
Work Step by Step
$y=\frac{x(x-1)^{3/2}}{\sqrt(x+1)}$ $, $ $x\gt1$
$Apply $ $\ln() $ $on$ $both $ $sides $
$lny=$ $\ln[\frac{x(x-1)^{3/2}}{\sqrt(x+1)}]$
$lny=ln(x)+\frac{3}{2}ln(x-1)-\frac{1}{2}ln(x+1)$
$\frac{y'}{y}=$ $\frac{1}{x}+$ $\frac{3}{2}(\frac{1}{x-1})-$ $\frac{1}{2}(\frac{1}{x+1})$
$\frac{y'}{y}=$ $\frac{1}{x}+$ $\frac{3}{2(x-1)}-$ $\frac{1}{2(x+1)}$
$\frac{y'}{y}=$ $\frac{1}{x}+$ $\frac{6(x+1)-2(x-1)}{4 (x^{2}-1)}$
$\frac{y'}{y}=$ $\frac{1}{x}+$ $\frac{6x+6-2x+2}{4 (x^{2}-1)}$
$\frac{y'}{y}=$ $\frac{1}{x}+$ $\frac{4x+8}{4 (x^{2}-1)}$
$\frac{y'}{y}=$ $\frac{1}{x}+$ $\frac{4(x+2)}{4 (x^{2}-1)}$
$\frac{y'}{y}=$ $\frac{1}{x}+$ $\frac{x+2}{x^{2}-1}$
$\frac{y'}{y}=$ $\frac{x^{2}-1+x^{2}+2x}{x(x^{2}-1)}$
$\frac{y'}{y}=$ $\frac{2x^{2}+2x-1}{(x^{3}-x)}$
$y'=$ $\frac{2x^{2}+2x-1}{(x^{3}-x)}$ $(y) $
$y'=$ $\frac{2x^{2}+2x-1}{(x^{3}-x)}$ $\left(\frac{x(x-1)^{3/2}}{\sqrt{(x+1)}}\right)$
