Calculus 8th Edition

$\frac{10}{2y+1}-\frac{y}{y^2+1}$
$\frac{d}{dy}\ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$ In this case, it is much easier to simplify the expression using the logarithm laws (near the bottom of page 439) before taking the derivative. $=\frac{d}{dy}(\ln(2y+1)^5-\ln\sqrt{y^2+1})$ $=\frac{d}{dy}(5\ln(2y+1)-\frac{1}{2}\ln(y^2+1))$ $=5*\frac{1}{2y+1}\frac{d}{dy}(2y+1)-\frac{1}{2}*\frac{1}{y^2+1}*\frac{d}{dy}(y^2+1)$ $=5*\frac{1}{2y+1}*2-\frac{1}{2}*\frac{1}{y^2+1}*2y$ $=\frac{10}{2y+1}-\frac{y}{y^2+1}$