Answer
$\frac{10}{2y+1}-\frac{y}{y^2+1}$
Work Step by Step
$\frac{d}{dy}\ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$
In this case, it is much easier to simplify the expression using the logarithm laws (Theorem 3 on page 422) before taking the derivative.
$=\frac{d}{dy}(\ln(2y+1)^5-\ln\sqrt{y^2+1})$
$=\frac{d}{dy}(5\ln(2y+1)-\frac{1}{2}\ln(y^2+1))$
$=5*\frac{1}{2y+1}\frac{d}{dy}(2y+1)-\frac{1}{2}*\frac{1}{y^2+1}*\frac{d}{dy}(y^2+1)$
$=5*\frac{1}{2y+1}*2-\frac{1}{2}*\frac{1}{y^2+1}*2y$
$=\frac{10}{2y+1}-\frac{y}{y^2+1}$
