Is this differential identity known? 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).  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 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).
 A: It's the Rodrigues formula for 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.
A: This can be considered as a case of the Lagrange reversion formula. 
Consider the equation $v = x + y \sqrt{1+v^2}$ (where $v$ is a function of $x$ and $y$).   It has an explicit solution
$$ v = 
\dfrac{\sqrt{1+x^2-y^2}\; y - x}{y^2-1}
$$
Lagrange reversion (with $f(x) = (1+x^2)^{1/2}$ and $g = \arctan$) can be written as
$$ \eqalign{\arctan(v) - \arctan(x) &= \sum_{j=0}^\infty \dfrac{y^{j+1}}{(j+1)!} \left(\dfrac{\partial}{\partial x}\right)^{j} (f(x)^{j+1} g'(x))\cr
&= \sum_{j=0}^\infty \dfrac{y^{j+1}}{(j+1)!} \left(\dfrac{\partial}{\partial x}\right)^{j}(1+x^2)^{(j-1)/2} } \tag{1} $$
Take the partial derivative  with respect to $y$; it turns out that
$$ \eqalign{\dfrac{\partial }{\partial y} \arctan(v) &= \dfrac{1}{1+v^2} \dfrac{\partial v}{\partial y} = \dfrac{-1}{\sqrt{1+x^2-y^2}}\cr
&= -\sum _{k=0}^{\infty } (1+x^2)^{-k-1/2}{\frac { 
 \left( 2\,k \right) !}{ 4^k \;\left( k! \right) ^{2}}} y^{2k}
}$$
Thus equating the coefficients of $y^{2k}$ here and in the partial derivative of the right side of (1), you get your formula.
A: Here is a direct induction argument. Let $h_k(x):=(1+x^2)^{k/2}$, $g(x):=x$,
$p_k:=(1\times3\times\dots\times k)^2$. The identity in question is 
$$D^{k+1}h_k=p_k/h_{k+2}$$
for positive odd $k$, where $D$ is the differentiation operator. It is easy to check it for $k=1$. 
Then for odd $k\ge3$ 
$$D^{k+1}h_k=k D^k(gh_{k-2})=kg D^k h_{k-2}+k^2 D^{k-1}h_{k-2}$$
$$=kg D(p_{k-2}/h_k)+k^2 p_{k-2}/h_k=p_k/h_{k+2},$$
as desired; for the third equality in the above display we use the induction, and for the second we use the Leibniz rule. 
