Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 21

Determine the derivative of the function $\displaystyle y = \left(\frac{x^2+1}{x^2-1}\right)^3$

$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx}\left( \frac{x^2+1}{x^2-1}\right)^3\\
\\
y' &= 3 \left(\frac{x^2+1}{x^2-1} \right)^2 \frac{d}{dx} \left( \frac{x^2+1}{x^2-1}\right)\\
\\
y' &= 3 \left(\frac{x^2+1}{x^2-1} \right)^2 \left[ \frac{(x^2-1) \frac{d}{dx} (x^2+1) - (x^2+1)\frac{d}{dx}(x^2-1)}{(x^2-1)^2}\right]\\
\\
y' &= 3 \left(\frac{x^2+1}{x^2-1} \right)^2 \left[ \frac{(x^2-1)(2x)-(x^2+1)(2x)}{(x^2-1)^2}\right]\\
\\
y' &= 3 \left(\frac{x^2+1}{x^2-1} \right)^2 \left[ \frac{\cancel{2x^3}-2x-\cancel{2x^3}-2x}{(x^2-1)^2}\right]\\
\\
y' &= 3 \left(\frac{x^2+1}{x^2-1} \right)^2 \left[ \frac{-4x}{(x^2-1)^2}\right]\\
\\
y' &= \frac{-12x(x^2+1)^2}{(x^2-1)^4}
\end{aligned}
\end{equation}
$