Differentiate the Function: $g(y)=ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$ $g(y)=ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$
$g(y)= ln(2y+1)^5-ln\sqrt{y^2-1}$
g'(y)=$\frac{5(2y+1)^4\cdot (2)}{(2y+1)^5}-\frac{2(y^2+1)(2y)}{(y^2+1)^2}$
At this point I can cancel the (2y+1) from the numerator and denominator of the first rational equation and from the second rational equation $(y^2+1)$. However, how would I get $\frac{y}{y^2+1}$ my answer is 
$g'(y)=\frac{10}{(2y+1)}-\frac{4y}{y^2+1)^2}$

HINT:
Write
$$\frac{d}{dy}\left(\log \sqrt{y^2+1}\right)=\frac{d}{dy}\left(\frac12\log (y^2+1)\right)=\frac12\left(\frac{2y}{y^2+1}\right)$$
To address the OP's comment, here $g(y)=y^2+1$ and $g'(y)=2y$.
