Let $r\in \lbrack 0,\infty)$. Define $f(x) = 1/(1+x^2)^r$. It would seem to be the case that $$|f^{(k)}(x)|\leq \frac{2r \cdot (2r+1) \dotsb (2r + k-1)}{(1+x^2)^{r + k/2}}$$ for all even $k\geq 0$. What would be a clean, short proof?


Note that
$$\left(\frac{1}{(1+x^2)^r}\right)^{(k)} = \frac{P\_k(x)}{(1+x^2)^{r+k}},$$
where $P\_k\in \mathbb{Z}\lbrack x\rbrack$ is given by the recurrence formula
$$P\_k(x)=1,\;\;\;\;\;\;P\_{k+1}(x) = P\_k'(x) (1+ x^2) + 2 (r+k)\cdot x P\_k(x).$$
It should not be too hard to work out all the coefficents of $P\_k(x)$; the leading terms are $$(-1)^{k+1} \cdot 2r \cdot (2r+1) \dotsb (2r + k-1)\cdot \left(x^k - \binom{k}{2} \frac{x^{k-2}}{2 r + 1} + \dotsc\right).$$ The desired bound should then follow after some work -- but I am hoping for a better proof.



Bounding higher derivatives of $f(x) = 1/(1+x^2)^r$
Well,
$$
(a^2+x^2)^{-r}=(x+ai)^{-r}\cdot (x-ai)^{-r}.
$$
Differentiate this $k$ times, we get
$$
\left((a^2+x^2)^{-r}\right)^{(k)}=k!\sum\_{s=0}^k {-r\choose s}{-r\choose k-s}
(x+ai)^{-r-s}(x-ai)^{-r-(k-s)}.
$$
Estimate this by the triangle inequality, you get some constant times $(x^2+a^2)^{-r-k/2}$. But for $a=0$ we have equality (because the signs of all summands are equal to $(-1)^k$), thus the constant is
$$
(-1)^kx^{2r+k}\left(x^{-2r}\right)^{(k)}=2r(2r+1)\ldots(2r+k-1).
$$


