Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 35

Determine the derivative of the function $\displaystyle y = \left( \frac{1-\cos 2x}{1 + \cos 2x}\right)^4$


$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} \left( \frac{1-\cos 2x}{1 + \cos 2x}\right)^4\\
\\
y' &= 4 \left( \frac{1-\cos 2x}{1 + \cos 2x}\right)^3 \cdot \frac{d}{dx} \left( \frac{1-\cos 2x}{1+\cos 2x}\right)\\
\\
y' &= 4 \left( \frac{1-\cos 2x}{1 + \cos 2x}\right)^3 \left[ \frac{(1+\cos 2x) \frac{d}{dx}(1-\cos 2x) - (1-\cos 2x) \frac{d}{dx} (1+\cos 2x) }{(1+\cos2x)^2}\right]\\
\\
y' &= 4 \left( \frac{1-\cos 2x}{1 + \cos 2x}\right)^3 \left[ \frac{(1+\cos 2x)(-(-\sin2x)) \frac{d}{dx} (2x) - (1-\cos 2x)(-\sin 2x) \frac{d}{dx} (2x) }{(1+\cos 2x)^2}\right]\\
\\
y' &= 4 \left( \frac{1-\cos 2x}{1 + \cos 2x}\right)^3 \left[ \frac{(1+\cos2x)(\sin2x)(2)-(1-\cos2x)(-\sin2x)(2)}{(1+\cos2x)^2}\right]\\
\\
y' &= 4 \left( \frac{1-\cos 2x}{1 + \cos 2x}\right)^3 \left[ \frac{(1+\cos2x)(2\sin 2x)+(1-\cos2x)(2\sin2x)}{(1+\cos2x)^2}\right]\\
\\
y' &= 4 \left( \frac{1-\cos 2x}{1 + \cos 2x}\right)^3 \left[ \frac{2 \sin 2x + \cancel{2\cos2x\sin2x}+2\sin 2x - \cancel{2\cos2x\sin2x}}{(1+\cos2x)^2}\right]\\
\\
y' &= 4 \left( \frac{1-\cos 2x}{1 + \cos 2x}\right)^3 \left[ \frac{4\sin 2x}{(1+\cos2x)^2} \right]\\
\\
y' &= \frac{(1-\cos2x)^3(16\sin2x)}{(1+\cos2x)^3(1+\cos2x)^2}\\
\\
y' &= \frac{(1-\cos2x)^3(16\sin2x)}{(1+\cos2x)^5}
\end{aligned}
\end{equation}
$