Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 30

Determine the derivative of the function $y = \arctan \sqrt{\frac{1-x}{1+x}} $ and simplify if possible.
If $y = \arctan \sqrt{\frac{1-x}{1+x}}$, then


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