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 $x=\cos (\theta)$
$$y=\tan^{-1} \left(\dfrac {1+\cos (\theta)}{1-\cos (\theta)}\right)$$
$$y=\tan^{-1} \left(\dfrac {2\cos^2 \left(\dfrac {\theta}{2}\right)}{2\sin^2\left(\dfrac {\theta}{2}\right)}\right)$$
$$y=\tan^{-1} \left(\cot^2 \left(\dfrac {\theta}{2}\right)\right)$$
 A: Once you find the first derivative, which is
$$
f'(x)=\frac{1}{x^2+1}
$$
your question reduces to the following one:
What is the $n^\text{th}$ derivative of $f(x)=\frac{1}{1+x^2}$
or more generally, this one:
How find this $\left(\frac{1}{x^2+a^2}\right)^{(n)}$

To find the first derivative, you can simply apply the chain rule. 
A: In this question the derivatives given are incorrect starting with the third (your fourth). Also the accepted answer could be worked forward a little, so we provide another answer. First off,
$$\tan^{-1}\left(\frac{1+x}{1-x}\right)=\tan^{-1}\left(\tan\left(\tan^{-1}1+\tan^{-1}x\right)\right)=\tan^{-1}1+\tan^{-1}x+[-\pi]$$
The $[-\pi]$ happens when $x>1$. So for $x\ne1$,
$$\frac d{dx}\tan^{-1}\left(\frac{1+x}{1-x}\right)=\frac d{dx}\tan^{-1}x=\frac1{1+x^2}=\frac{-\frac12i}{x-i}+\frac{\frac12i}{x+i}$$
Then for $x\ne1$,
$$\begin{align}\frac{d^n}{dx^n}\tan^{-1}\left(\frac{1+x}{1-x}\right)&=\frac{-\frac12i(-1)^{n-1}(n-1)!}{(x-i)^n}+\frac{\frac12i(-1)^{n-1}(n-1)!}{(x+i)^n}\\
&=\frac{\frac12i(-1)^n(n-1)!\left[(x+i)^n-(x-i)^n\right]}{(x^2+1)^n}\\
&=\frac{\frac12i(-1)^n(n-1)!}{(x^2+1)^n}\sum_{k=0}^n{n\choose k}x^{n-k}\left[i^k-(-i)^k\right]\\
&=\frac{\frac12i(-1)^n(n-1)!}{(x^2+1)^n}\sum_{k=0}^n{n\choose k}x^{n-k}i^k\left[1-(-1)^k\right]\\
&=\frac{\frac12i(-1)^n(n-1)!}{(x^2+1)^n}\sum_{k=0}^{\lfloor\frac{n-1}2\rfloor}{n\choose {2k+1}}x^{n-2k-1}i(-1)^k(2)\\
&=\frac{(-1)^{n+1}(n-1)!\sum_{k=0}^{\lfloor\frac{n-1}2\rfloor}(-1)^k{n\choose {2k+1}}x^{n-2k-1}}{(x^2+1)^n}\end{align}$$
Written out,
$$\begin{array}{rl}f^{\prime}(x)&=\frac1{x^2+1}\\
f^{\prime\prime}(x)&=\frac{-(2x)}{(x^2+1)^2}\\
f^{\prime\prime\prime}(x)&=\frac{2(3x^2-1)}{(x^2+1)^3}\\
f^{(4)}(x)&=\frac{-6(4x^3-4x)}{(x^2+1)^4}\\
f^{(5)}(x)&=\frac{24(5x^4-10x^2+1)}{(x^2+1)^5}\\
f^{(6)}(x)&=\frac{-120(6x^5-20x^3+6x)}{(x^2+1)^6}\end{array}$$
A: Let $\arctan x=z\implies x=\tan z$
$$\tan y=\tan\left(\dfrac\pi4+z\right)$$
$$y=n\pi+\dfrac\pi4+\arctan x$$ where $n$ is an arbitrary integer constant so that  $-\dfrac\pi2<y<\dfrac\pi2$
$$y_1=\dfrac{dy}{dx}=\dfrac1{1+x^2}$$
$$y_1(1+x^2)=1$$
Taking $m(\ge2)$th derivative using General Leibniz rule
$$y_{m+1}(1+x^2)+\binom m1y_m(2x)+\binom m2 y_{m-1}(2)=0$$
A: $$D^{n}y= D^{n-1} Dy=D^{n-1}\frac{1}{1+x^2}= D^{n-1} \frac{1}{2i} \left( \frac{1}{x-i}- \frac{1}{x+i} \right)$$
The nth derivative of $(x+b)^{-1}$ is givwn as
$$D^n (x+b)^{-1}=(-1)^n n!(x+b)^{-n-1}$$ 
Using this
$$D^n y= (-1)^{(n-1)} (n-1)! ~ \Im (x-i)^{-n-2}=(-1)^{n-1} (n-1)!~ (x^2+1)^{(-n-2)/2} \sin[(n+2) \tan^{-1}(1/x)].$$
