Answer
$$y' =\left(\frac{(x^2+1)^4}{(2x+1)^3(3x-1)^5}\right)\left(\frac{8x}{x^2+1}- \frac{6}{2x+1}- \frac{15}{3x-1}\right)$$
Work Step by Step
Given $$y=\frac{(x^2+1)^4}{(2x+1)^3(3x-1)^5}$$
Take $\ln $ for both sides
\begin{align*}
\ln y&=\ln (x^2+1)^4-\ln (2x+1)^3(3x-1)^5\\
&=4\ln (x^2+1)-3\ln (2x+1)-5\ln (3x-1)
\end{align*}
Then
\begin{align*}
\frac{y'}{y} &=4\frac{2x}{x^2+1}-3\frac{2}{2x+1}-5\frac{3}{3x-1}\\
&= \frac{8x}{x^2+1}- \frac{6}{2x+1}- \frac{15}{3x-1}\\
y'&=\left(\frac{(x^2+1)^4}{(2x+1)^3(3x-1)^5}\right)\left(\frac{8x}{x^2+1}- \frac{6}{2x+1}- \frac{15}{3x-1}\right)
\end{align*}
