Here’s a cute identity I discovered by accident recently. Observe that

$\displaystyle \frac{d}{dx} (1+x^2)^{0/2} = 0$

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

$\displaystyle \frac{d^3}{dx^3} (1+x^2)^{2/2} = 0$

$\displaystyle \frac{d^4}{dx^4} (1+x^2)^{3/2} = \frac{9}{(1+x^2)^{5/2}}$

$\displaystyle \frac{d^5}{dx^5} (1+x^2)^{4/2} = 0$

$\displaystyle \frac{d^6}{dx^6} (1+x^2)^{5/2} = \frac{225}{(1+x^2)^{7/2}}$

and so one can conjecture that one has

$\displaystyle \frac{d^{k+1}}{dx^{k+1}} (1+x^2)^{k/2} = 0$

when $k$ is even, and

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

when $k$ is odd. This is obvious in the even case since $(1+x^2)^{k/2}$ is a polynomial of degree $k$, but I struggled for a while with the odd case before finding a slick three-line proof. (I was first trying to prove the weaker statement that $\frac{d^{k+1}}{dx^{k+1}} (1+x^2)^{k/2}$ was non-negative, but for some strange reason I was only able to establish this by working out the derivative exactly, rather than by using more analytic methods, such as convexity arguments.) I thought other readers might like the challenge (and also I’d like to see some other proofs), so rather than post my own proof immediately, I’ll see if anyone would like to supply their own proofs or thoughts in the comments. Also I am curious to know if this identity is connected to any other existing piece of mathematics.