<s>[INST] Find the $n^{th}$ derivative of $y=\tan^{-1} \left(\frac {1+x}{1-x}\right)$ Find the $n^{th}$ derivative of $y=\tan^{-1} \left(\dfrac {1+x}{1-x}\right)$
My Attempt:
$$y=\tan^{-1} \left(\dfrac {1+x}{1-x}\right)$$
Put $x=\cos (\theta)$
$$y=\tan^{-1} \left(\dfrac {1+\cos (\theta)}{1-\cos (\theta)}\right)$$
$$y=\tan^{-1} \left(\dfrac {2\cos^2 \left(\dfrac {\theta}{2}\right)}{2\sin^2\left(\dfrac {\theta}{2}\right)}\right)$$
$$y=\tan^{-1} \left(\cot^2 \left(\dfrac {\theta}{2}\right)\right)$$
 [/INST] $$D^{n}y= D^{n-1} Dy=D^{n-1}\frac{1}{1+x^2}= D^{n-1} \frac{1}{2i} \left( \frac{1}{x-i}- \frac{1}{x+i} \right)$$
The nth derivative of $(x+b)^{-1}$ is givwn as
$$D^n (x+b)^{-1}=(-1)^n n!(x+b)^{-n-1}$$ 
Using this
$$D^n y= (-1)^{(n-1)} (n-1)! ~ \Im (x-i)^{-n-2}=(-1)^{n-1} (n-1)!~ (x^2+1)^{(-n-2)/2} \sin[(n+2) \tan^{-1}(1/x)].$$
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