# Derive Differentiation Identity for Gegenbauer Polynomial

For $\lambda > 0$ and $n = 0,1,2,3,\ldots$, the following is an n-th order polynomial: $$P_{\lambda,n}(x)=(1-x^{2})^{-\lambda+1/2}\frac{d^{n}}{dx^{n}}(1-x^{2})^{n+\lambda-1/2}$$ Problem: Show that there are constants $C_{\lambda,n}$ such that $$P_{\lambda,n}'(x)=C_{\lambda,n}P_{\lambda+1,n-1}(x),\;\; \lambda > 0,\;\; n \ge 1.$$ I'd like to see a way to verify this using only basic calculus.

Background: The Gegenbauer polynomials, $G_{\lambda,n}$ are the unique polynomial coefficients in the power series expansion $$\frac{1}{(1-2rx+r^{2})^{\lambda}} = \sum_{n=0}^{\infty}G_{\lambda,n}(x)r^{n},\;\;\; |r| < 1, \; |x| \le 1,\; \lambda > 0.$$ (In the special case that $\lambda=1/2$, the Gegenbauer polynomials are the Legendre polynomials.) Differentiating the above with respect to $x$, and equating like powers of $r$, leads to the identity $$\frac{d}{dx}G_{\lambda,n}(x)=2\lambda G_{\lambda+1,n-1}(x),\;\;\; \lambda > 0,\; n\ge 1.$$ This identity is equivalent the one I have asked to be shown because the Gegenbauer polynomials have a Rodrigues Formula $G_{\lambda,n}(x)=c_{\lambda,n}P_{\lambda,n}$ for some constants $c_{\lambda,n}$. I would have thought that the first-order derivative relation $G_{\lambda,n}'=2\lambda G_{\lambda,n}$ would be obvious from the Rodrigues formula for $G_{\lambda,n}$, but I'm having trouble seeing it.

References:
1. A version of the Rodrigues formula $G_{\lambda,n}=c_{\lambda,n}P_{\lambda,n}$ and the constants $c_{\lambda,n}$ are found in
http://en.wikipedia.org/wiki/Gegenbauer_polynomials
2. A version of the stated recurrence $G_{\lambda,n}'=2\lambda G_{\lambda+1,n-1}$ is found in
http://en.wikipedia.org/wiki/Classical_orthogonal_polynomials#Gegenbauer_polynomials

-