Recently I discovered the differential identity


$$ \frac{d^{k+1}}{dx^{k+1}} (1+x^2)^{k/2} = \frac{(1 \times 3 \times \dots \times k)^2}{(1+x^2)^{(k+2)/2}}$$


valid for any odd natural number $k$; for instance $\frac{d^6}{dx^6} (1+x^2)^{5/2} = \frac{225}{(1+x^2)^{7/2}}$. This identity was surprising at first, since usually the repeated application of the product rule and chain rule leads to far messier expressions than this, but there are now several proofs that adequately explain this identity (collected at [this blog post of mine](https://terrytao.wordpress.com/2015/05/30/a-differentiation-identity/)). There is also the more general identity


$$ |\frac{d}{dx}|^{2s-1} (1+x^2)^{s-1} = \frac{2^{2s-1}\Gamma(s)}{\Gamma(1-s)} (1+x^2)^{-s}$$


valid for any complex $s$ (if everything is interpreted distributionally), which is related to the isomorphisms between principal series representations of $PGL\_2({\bf R})$.


The purpose of my question here is **not** to ask for more proofs of this identity (but you are welcome to visit the above-mentioned blog post to contribute another proof, if you wish). Instead, I am asking as to whether this identity (or something close to it) already appears in the literature - I find it hard to believe that such a simple identity has been missed for centuries, given that it could easily have been discovered and proven by (say) Euler. The closest match that I know of so far are the [Rodrigues formulae](http://en.wikipedia.org/wiki/Rodrigues%27_formula) for the classical orthogonal polynomials, but I was not quite able to place the above identity as a special case of these formulae (the exponents don't quite match up).



Is this differential identity known?
It's the Rodrigues formula for [Gegenbauer polynomials](https://en.wikipedia.org/wiki/Gegenbauer_polynomials) $C\_n^{(\alpha)}(x)$, in the special case $\alpha=-n/2$ when the polynomial is just unity.


The general formula reads


$$C\_n^{(\alpha)}(x)=\frac{(-2)^n}{n!}\frac{\Gamma(n+\alpha)\Gamma(n+2\alpha)}{\Gamma(\alpha)\Gamma(2n+2\alpha)}(1-x^2)^{-\alpha+1/2}\frac{d^n}{dx^n}\left[(1-x^2)^{n+\alpha-1/2}\right].$$


Substitution of $\alpha=-n/2$, with $C\_n^{(-n/2)}(x)=1$ for $n$ even, gives


$$1=\frac{(-2)^n(n/2)!}{n!(n-1)!}\lim\_{\alpha\rightarrow-n/2}\frac{\Gamma(n+2\alpha)}{\Gamma(\alpha)}(1-x^2)^{(n+1)/2}\frac{d^n}{dx^n}\left[(1-x^2)^{(n-1)/2}\right]$$


$$\qquad = (-1)^{n/2}\left(\frac{1}{(n-1)!!}\right)^2(1-x^2)^{(n+1)/2}\frac{d^n}{dx^n}\left[(1-x^2)^{(n-1)/2}\right],$$


or with $x\to ix$,


$$\frac{d^n}{dx^n}\left[(1+x^2)^{(n-1)/2}\right]=[(n-1)!!]^2(1+x^2)^{-(n+1)/2},$$


which is the desired expression.


