Derivative of $\tan^{-1}(f(x))$ What is derivative of $$\tan^{-1}\left(\frac{{\sqrt{4+x}+\sqrt{4-x}}}{\sqrt{4+x}-\sqrt{4-x}}\right).$$  So I tried to write it as $\tan(\tan^{-1}(...))$ to get the $f(x)=\frac{\pi}{4}+\tan^{-1}\left(\sqrt{\frac{4+x}{4-x}}\right)$ but still it's not better. Thanks help appreciated

Easy way : 
Put $$x=4\cos 2\theta$$
Then $$y=\tan ^{-1} \left(\frac{\sqrt{4+4 \cos 2\theta}+\sqrt{4-4 \cos 2\theta}}{\sqrt{4+4 \cos 2\theta}-\sqrt{4-4 \cos 2\theta}} \right)$$
$$y=\tan ^{-1} \left(\frac{2\sqrt{1+ \cos 2\theta}+2\sqrt{1- \cos 2\theta}}{2\sqrt{1+ \cos 2\theta}-2\sqrt{1- \cos 2\theta}} \right)$$
$$y=\tan ^{-1} \left(\frac{\sqrt{1+ \cos 2\theta}+\sqrt{1- \cos 2\theta}}{\sqrt{1+ \cos 2\theta}-\sqrt{1- \cos 2\theta}} \right)$$
Since $1+\cos 2\theta =2\cos^2 \theta $ and $1-\cos 2\theta =2\sin^2 \theta$
$$y=\tan ^{-1} \left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta} \right)$$
Now divide both denominator and numerator by $\cos \theta$ ,
$$y=\tan ^{-1} \left(\frac{1+\tan \theta}{1-\tan \theta} \right)$$
$$y=\tan ^{-1} \left(\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4}\cdot\tan \theta} \right)$$
$$y=\tan ^{-1} \left(\tan\left( \frac{\pi}{4}-\theta \right)\right)$$
$$y=\frac{\pi}{4}-\theta$$
So $$\frac{dy}{dx}=0+\frac{d\theta}{dx}$$
Since $x=4 \cos 2\theta$
$$\frac{dx}{d \theta} = -8 \sin 2 \theta $$
So $$\frac{dy}{dx}=\frac{d\theta}{dx}= -\frac{1}{8\sin 2\theta}=-\frac{1}{8 \sqrt{1-\frac{x^2}{16}}}$$
