I was trying to derive a recurrence relantionship for computing the $k$-th derivative of the function (hoping no error during copy)
$$ f(x) = \frac{\arctan(x)}{x} $$
Using maple I've seen the following derivatives $$ \begin{array}{l} f^{(0)}(x) = \frac{\arctan(x)}{x} \\ f^{(1)}(x) = \frac{1}{(x^2 + 1)x} - \frac{f(x)}{x} \\ f^{(2)}(x) = \frac{-2}{(x^2+1)^2} + \frac{-2}{(x^2 + 1)x^2} + \frac{2f(x)}{x^2} \\ f^{(3)}(x) = \frac{8}{(x^2+1)^3x^{-1}} + \frac{4}{(x^2+1)^2x} + \frac{6}{(x^2+1)x^3} - \frac{6f(x)}{x^3} \\ f^{(4)}(x) = \frac{-48}{(x^2+1)^4x^{-2}} + \frac{-8}{(x^2+1)^3} + \frac{-16}{(x^2+1)^2 x^2} + \frac{-24}{(x^2+1)x^4} + \frac{24 f(x)}{x^4} \\ f^{(5)}(x) = \frac{384}{(x^2+1)^5 x^{-3}} + \frac{-48}{(x^2+1)^4x^{-1}} + \frac{64}{(x^2+1)^3x} + \frac{80}{(x^2+1)^2x^3} + \frac{120}{(x^2+1)x^5} - \frac{120f(x)}{x^5} \end{array} $$
This lead me to the following expression, as summation, for the $k$-th derivative
$$ f^{(k)}(x) = \sum_{j=1}^k \frac{c_{j,k}}{(x^2+1)^jx^{k-2j+2}} + (-1)^k \frac{k! f(x)}{x^k} $$
Firstly do you agree with me that the coefficients $c_{j,k}$ fully describes the $k$-th derivative of $f$? If not can you point out the mistake I made?
The next step I would writing $f^{(k+1)}$ in terms of the coefficients $c_{j,k}$ from there I should be able to derive a recurrence relationship for the coefficients (like a triangular table that would allow me to derive the coefficients $c_{j,k}$ for given $k$ and $j = 1 \ldots k$).
Do you think there's a smarter way to do achieve the same result? maybe easier? the computations involved here are quite messy.
Update : here is my attempt
$$ f^{(k+1)}(x) = \sum_{j=1}^k c_{j,k}\left[ \frac{-2jx}{(x^2+1)^{j+1}x^{k-2j+2}} + \frac{k-2j+2}{(x^2+1)^jx^{k-2j+3}}\right] + \frac{(-1)^k {k!}}{(x^2+1)x^{k+1}} + (-1)^{k+1} {(k+1)!} \frac{f(x)}{x^{k+1}} $$
I got stuck now...
$n\in\mathbb{N}$
It's $$\frac{d^n}{dx^n }\arctan x=\frac{d^n}{dx^n}(xf(x))=n\frac{d^{n-1}}{dx^{n-1} }f(x)+ x\frac{d^n}{dx^n }f(x)$$ and for $x>0$ (and proofed by induction) one gets $$\frac{d^n}{dx^n }\arctan x=(-1)^{n-1}(n-1)!\frac{\sin(n \arctan\frac{1}{x})}{\sqrt{1+x^2}^n}$$
It follows that the recursion for the derivatives of $\displaystyle f(x):=\frac{\arctan x}{x}$ is
$$\frac{d^n}{dx^n }f(x)=(-1)^{n-1}(n-1)!\frac{\sin(n \arctan\frac{1}{x})}{x\sqrt{1+x^2}^n}-\frac{n}{x} \frac{d^{n-1}}{dx^{n-1}}f(x)$$
Note:
Instead of $\arctan\frac{1}{x}$ I should write $\,$arccot(x)$\,$ (so that we can also use $x<0$) but it's not declared here.
Hint:
A proof can be found e.g. in http://www.math.nthu.edu.tw/~amen/2010/090408-2.pdf with $\arcsin\frac{1}{\sqrt{1+x^2}}$.