Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 34

Determine the derivative of the function $\displaystyle y = \tan^{-1} \left( \frac{x}{a} \right) + \ln \sqrt{\frac{x-a}{x+a}}$ and simplify if possible.
If $\displaystyle y = \tan^{-1}\left( \frac{x}{a} \right) + \ln \sqrt{\frac{x-a}{x+a}}$, then...

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