Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 26

Determine the derivative of the function $\displaystyle G(y) = \frac{(y-1)^4}{(y^2+2y)^5}$

$
\begin{equation}
\begin{aligned}
G'(y) &= \frac{(y-1)^4}{(y^2+2y)^5}\\
\\
G'(y) &= \frac{d}{dy} \left[ \frac{(y-1)^4}{(y^2+2y)^5}\right]\\
\\
G'(y) &= \frac{\left[ (y^2+2y)^5 \cdot \frac{d}{dy} (y-1)^4\right] - \left[ (y-1)^4 \cdot \frac{d}{dy} (y^2+2y)^5 \right]}{\left[ (y^2+2y)^5\right]^2}\\
\\
G'(y) &= \frac{\left[ (y^2+2y)^5 \cdot 4(y-1)^3 \frac{d}{dy} (y-1)\right] - \left[ (y-1)^4 \cdot 5(y^2+2y)^4 \frac{d}{dy} (y^2+2y) \right]}{(y^2+2y)^{10}}\\
\\
G'(y) &= \frac{\left[ (y^2+2y)^5 \cdot 4(y-1)^3 (1)\right] - \left[ (y-1)^4 \cdot 5 (y^2 +2y)^4 (2y+2)\right]}{(y^2+2y)^{10}}\\
\\
G'(y) &= \frac{\left[ 4(y^2+2y)^5(y-1)^3\right]-\left[ 5(y-1)^4 (y^2+2y)^4(2y+2)\right]}{(y^2+2y)^{10}}\\
\\
G'(y) &= \frac{(y-1)^3(y^2+2y)^4\left[ 4(y^2+2y)-5(y-1)(2y+2) \right]}{(y^2+2y)^{10}}\\
\\
G'(y) &= \frac{(y-1)^3 \left[ 4y^2 + 8y - 5(2y^2+ \cancel{2y} - \cancel{2y} - 2)\right]}{(y^2+2y)^6}\\
\\
G'(y) &= \frac{(y-1)^3(4y^2+8y-10y^2+10)}{(y^2+2y)^6}\\
\\
G'(y) &= \frac{(y-1)^3(-6y^2+8y+10)}{(y^2+2y)^6}
\end{aligned}
\end{equation}
$