Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 30

Determine the derivative of the function $\displaystyle G(y) = \left(\frac{y^2}{y+1}\right)^5$


$
\begin{equation}
\begin{aligned}
G'(y) &= \frac{d}{dy} \left(\frac{y^2}{y+1}\right)^5\\
\\
G'(y) &= 5 \left( \frac{y^2}{y+1} \right)^4 \frac{d}{dy} \left( \frac{y^2}{y+1}\right)\\
\\
G'(y) &= 5\left(\frac{y^2}{y+1}\right)^4 \left[ \frac{(y+1)\frac{d}{dy}(y^2) - (y^2) \frac{d}{dy} (y+1)}{(y+1)^2}\right]\\
\\
G'(y) &= 5\left(\frac{y^2}{y+1}\right)^4 \left[ \frac{(y+1)(2y)-(y^2)(1)}{(y+1)^2}\right]\\
\\
G'(y) &= 5\left(\frac{y^2}{y+1}\right)^4 \left[\frac{2y^2+2y-y^2}{(y+1)^2} \right]\\
\\
G'(y) &= 5\left(\frac{y^2}{y+1}\right)^4 \left[ \frac{y^2 + 2y }{(y+1)^2}\right]\\
\\
G'(y) &= \frac{5(y^2)^4(y^2+2y)}{(y+1)^4(y+1)^2}\\
\\
G'(y) &= \frac{5(y^8)(y^2+2y)}{(y+1)^6}\\
\\
G'(y) &= \frac{5y^{10}+ 10y^9}{(y+1)^6}

\end{aligned}
\end{equation}
$