I want to calculate the first derivative of the k-th Legendre-polynomial evaluated at s = 1 using the generalized product rule$$\frac{d^{n}}{ds^{n}}[F(s)G(s)]=\sum_{j=0}^n {n \choose j}F^{(n-j)}(s)G^{(j)}(s)$$ I started by using Rodrigues' formula$$P_k(s)=\frac{1}{2^{k}k!}(\frac{d}{ds})^{k}(s^2-1)^k$$ $$P^{'}_k(s)=\frac{1}{2^{k}k!}(\frac{d}{ds})^{k+1}(s^2-1)^k$$ By binomial theorem, I wrote $$P^{'}_k(s)=\frac{1}{2^{k}k!}(\frac{d}{ds})^{k+1}\sum_{j=0}^k {k \choose j}(s^2)^{(k-j)}(-1)^{(j)}$$ $$P^{'}_k(s)=\frac{1}{2^{k}k!}(\frac{d}{ds})^{k+1}(\frac{d}{ds})^{k}[(s^2)(-1)]$$ $$P^{'}_k(s)=\frac{1}{2^{k}k!}(\frac{d}{ds})^{2k+1}[-s^2]$$ I don't think this can be right, though. Can someone point out where I went wrong and how to further progress to the answer?
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The step where the binomial coefficient somehow is replaced by $(d/ds)^k$ is not by use of the generalized product rule because it uses derivatives not powers. – Somos Apr 24 '20 at 22:47
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Better would be to use the series expansion of Legendre polynomial around $s=1$. – Claude Leibovici Apr 25 '20 at 08:58