Answer
$G'(y)=\frac{8y^2+10-y}{(2y+1)(y^2+1)}$
Work Step by Step
$G(y)=\ln \frac{(2y+1)^5}{\sqrt{y^2+1}}\\
G(y)=5\ln(2y+1)-\frac{1}{2}\ln(y^2+1)\\
G'(y)=\frac{5\times2}{2y+1}-\frac{1\times2y}{2\times(y^2+1)}\\
G'(y)=\frac{10}{2y+1}-\frac{y}{(y^2+1)}\\
G'(y)=\frac{10y^2+10-2y^2-y}{(2y+1)(y^2+1)}\\
G'(y)=\frac{8y^2+10-y}{(2y+1)(y^2+1)}$