Answer
$\frac{{dy}}{{dx}} = \frac{{2{a}{x^2}}}{{{x^4} - {a^4}}}$
Work Step by Step
$$\eqalign{
& y = {\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + \ln \sqrt {\frac{{x - a}}{{x + a}}} \cr
& {\text{Rewrite the function}} \cr
& y = {\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + \ln {\left( {\frac{{x - a}}{{x + a}}} \right)^{1/2}} \cr
& y = {\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + \frac{1}{2}\ln \left( {\frac{{x - a}}{{x + a}}} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right) + \frac{1}{2}\ln \left( {\frac{{x - a}}{{x + a}}} \right)} \right] \cr
& {\text{Use sum rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right] + \frac{1}{2}\left[ {\ln \left( {\frac{{x - a}}{{x + a}}} \right)} \right] \cr
& {\text{Use }}\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}},{\text{ and }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {x/a} \right)}^2}}}\frac{d}{{dx}}\left[ {\frac{x}{a}} \right] + \frac{1}{2}\left( {\frac{{x + a}}{{x - a}}} \right)\underbrace {\frac{d}{{dx}}\left[ {\frac{{x - a}}{{x + a}}} \right]}_{{\text{Use quotient rule}}} \cr
& {\text{Computing the derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{{a^2}}}{{{a^2} + {x^2}}}\left( {\frac{1}{a}} \right) + \frac{1}{2}\left( {\frac{{x + a}}{{x - a}}} \right)\left( {\frac{{x + a - x + a}}{{{{\left( {x + a} \right)}^2}}}} \right) \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{a}{{{a^2} + {x^2}}} + \frac{1}{2}\left( {\frac{1}{{x - a}}} \right)\left( {\frac{{2a}}{{x + a}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{a}{{{a^2} + {x^2}}} + \frac{a}{{{x^2} - {a^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{a{x^2} - {a^3} + {a^3} + {a^2}{x^2}}}{{{x^4} - {a^4}}} \cr
& \frac{{dy}}{{dx}} = \frac{{2{a}{x^2}}}{{{x^4} - {a^4}}} \cr} $$
