Finding the Derivative of $\arctan \frac{x}{a - \sqrt{a^2 - x^2}}$ 
I am trying to simplify the derivative of $\arctan \frac{x}{a - \sqrt{a^2 - x^2}}$.

My work:
Everybody knows that $\frac{d}{dx} (\arctan \space u) = \frac{1}{1 + u^2} \frac{du}{dx}$
We let $u =  \frac{x}{a - \sqrt{a^2 - x^2}} .$ To get the $du,$ I remember that $\frac{d}{dx} \left( \frac{u}{v} \right)= \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}.$
So:
 $$\frac{d}{dx} \left ( \frac{x}{a - \sqrt{a^2 - x^2}}  \right) = \frac{a-\sqrt{a^2 - x^2} \frac{d}{dx} (x) - x \frac{d}{dx} (a-\sqrt{a^2 - x^2)}}{(a-\sqrt{a^2 - x^2)^2}} $$
$$ = \frac{a - \sqrt{a^2 - x^2} -x \left(\frac{x}{(\sqrt{a^2 - x^2})}\right)}{a^2 - 2\sqrt{a^2 - x^2} +a^2 - x^2}$$
$$ = \frac{ \frac{a - \sqrt{a^2 - x^2}}{1} + \frac{-x^2}{ \sqrt{a^2 - x^2} }}{2a^2 - 2\sqrt{a^2 - x^2} - x^2}$$
$$ = \frac{\frac{(a^2 - x^2)^{\frac{1}{2}} (a - (a^2 - x^2)^\frac{1}{2}) - x^2}{(a^2 - x^2)^{\frac{1}{2}}}}{2a^2 - x^2 - 2\sqrt{a^2 - x^2} }$$
$$ \frac{d}{dx} \left( \frac{x}{a - \sqrt{a^2 - x^2}}\right) = \frac{a(a^2- x^2)^\frac{1}{2} - (a^2- x^2) - x^2}{2a^2 - x^2 - 2\sqrt{a^2 - x^2} }$$
Then:
$$\frac{d}{dx} \left( \arctan \frac{x}{a - \sqrt{a^2 - x^2}} \right) = \frac{1}{1 + \left( \frac{x}{a - \sqrt{a^2 - x^2}} \right)^2}  \left( \frac{a(a^2- x^2)^{\frac{1}{2}} - (a^2- x^2) - x^2}{2a^2 - x^2 - 2\sqrt{a^2 - x^2} } \right) $$
At this point, simplifying it is difficult. How do you get the derivative of $\arctan \frac{x}{a - \sqrt{a^2 - x^2}}?$

Using $$u = \frac{x}{a-\sqrt{a^2-x^2}} = \frac{a+\sqrt{a^2-x^2}}{x}$$  results in $$\frac{du}{dx} = -\frac{a^2 + a \sqrt{a^2-x^2}}{x^2\sqrt{a^2-x^2}},$$ and $$\frac{d}{dx}\tan^{-1}(u) = -\frac{1}{2\sqrt{a^2-x^2}}.$$
