<s>[INST] 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 
$g'(y)=\frac{10}{(2y+1)}-\frac{4y}{y^2+1)^2}$
 [/INST] Hint
$$g=\log\frac{(2y+1)^5}{\sqrt{y^2+1}}=5\log(2y+1)-\frac 12 \log(y^2+1)$$ $$g'=5 \frac 2{2y+1}-\frac 12 \frac {2y}{y^2+1}=\frac {10}{2y+1}-\frac {y}{y^2+1}=\frac{8 y^2-y+10}{(2y+1)(y^2+1)}$$
</s>