Closed form of the $n^{th}$-derivative of $f(x) = \arctan x$ My problem is about characterization of the $n^{th}$-derivative of 
$$f(x) = \arctan x$$
We were asked to solve the following questions 


*

*Show that $$ f^{(n)} (x) = \frac{P_n(x)}{(1+x^2)^{n}}~~~n\ge1$$
Where $P_n$ is polynomial of degree to be specify.

*Find a recurrent relationships Between the $P_n's.$

*Then, Give the expression of $P_n(x) . $


My attempt I was able to prove just comparison with derivative that $$ P_{n+1}(x) = (1+x^2)P'_n(x) -2nx P_n(x)$$
This therefore gives a solution to questions 1. and 2. 
Note that from this, since $P_1 = 1$ we have $$\deg P_n = n-1$$

I do not know how to solve the last Question.

 A: We know that $\frac{d}{dx}\arctan(x)=\frac{1}{x^2+1} = \frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)$, hence
$$\begin{eqnarray*} \frac{d^{n+1}}{dx^{n+1}}\arctan(x)&=&\frac{1}{2i}\,\frac{d^n}{dx^n}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)\\&=&\frac{(-1)^n n!}{2i}\left(\frac{1}{(x-i)^{n+1}}-\frac{1}{(x+i)^{n+1}}\right)\\&=&\frac{(-1)^n n!}{\color{blue}{2i}}\cdot\frac{\color{blue}{(x+i)^{n+1}-(x-i)^{n+1}}}{(x^2+1)^{n+1}}\\&=&\frac{(-1)^n n!}{(x^2+1)^{n+1}}\sum_{k\leq (n+1)/2}\binom{n+1}{2k+1}x^{n-2k}(-1)^k\tag{1,3}\end{eqnarray*}$$
and of course, 
$$ q_n(x)=\frac{(x+i)^n-(x-i)^n}{2i} $$
is a Fibonacci-like sequence of polynomials, fulfilling
$$ \color{purple}{q_{n+2}(x) = 2x\,q_{n+1}(x)-(x^2+1)\,q_n(x)},\qquad q_0(x)=0, q_1(x)=1.$$
Additionally, $f^{(n)}(x)=\frac{P_n(x)}{(x^2+1)^n}$ implies $f^{(n+1)}(x)=\frac{P_{n+1}(x)}{(x^2+1)^{n+1}}=\frac{d}{dx}\left(\frac{P_n(x)}{(x^2+1)^n}\right) $ and $$P_{n+1}(x)=(x^2+1)P_n'(x)-2n x\,P_n(x)$$
as already remarked. On the other hand, by differentiating $q_n(x)$ we get that $\frac{q_n'(x)}{n}$, for any $n\geq 1$, fulfills a recurrence relation of the $\color{purple}{\text{purple}}$ form, too ($2$).
A: Not a full answer, but an ansatz: 
You know the degree $d(n)$ of $P_n$.
Write 
$$P_n(x) =\sum^{d(n)}_{k=0}a_k^{(n)}x^k$$
and then write down explicitly the relation between the $P_n$ and their derivates (the formula you wrote down in your question) using this formula. Then note that $x^k$ and $x^l$ are linearly independent if $k\neq l$ and write down the expression you get from this for the coefficients of each power. This will lead to relations between the $a_k^{(n)}$ and $a_l^{(n-1)}$. (Look for the highest order term first). 
