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Apr 24 at 14:44 comment added Iosif Pinelis @qifeng618 : Thank you for your comments. I obtained formula (10) by a simplification of this WolframAlpha output. The main (even if quite simple) idea in this answer, though, is what precedes (10): the separation of the $e^{1/u}$ factor and then using the Leibniz formula.
Apr 24 at 14:39 comment added Iosif Pinelis @NancyBoy : Thank you for your comment.
Apr 24 at 13:16 comment added qifeng618 More general formulas than (10) can be used to explicitly compute the $n$th order derivative of the functions like $f(\ln(ax+b))$ and $f\bigl((ax+b)^\alpha\bigr)$ as long as the $n$th order derivative of $f$ is explicitly computable. For details, please read the review articles at doi.org/10.1016/j.jmaa.2020.124382 and doi.org/10.32604/cmes.2022.019941.
Apr 24 at 12:13 comment added qifeng618 The formula (10) can be found in any one of the articles collected at qifeng618.wordpress.com/2018/05/10/…. This answer is almost same as the one at mathoverflow.net/a/469805.
Apr 24 at 7:13 comment added NancyBoy @IosifPinelis: very nice derivation! Thank you
Apr 24 at 1:26 comment added Iosif Pinelis @FredHucht : Thank you for your comment.
Apr 24 at 1:26 comment added Iosif Pinelis @PietroMajer : Thank you for your comment.
Apr 24 at 1:14 history edited Iosif Pinelis CC BY-SA 4.0
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Apr 23 at 6:52 comment added Fred Hucht @Iosif Pinelis Nice derivation! Note that for $n>0$, $\sum_{k=0}^n \frac{1}{k!}\binom{n-1}{k-1}(bx)^{-k-n}$ $=$ $\frac{1}{n\,(b x)^{n+1}}L_{n-1}^{(1)}(-\frac{1}{bx})$, with the generalized Laguerre polynomials $L$ from my answer.
Apr 23 at 3:26 comment added Pietro Majer I was starting the same computation! This seems the simplest way to connect the $P'_{\lambda,\mu}$'s with something known
Apr 22 at 21:18 history answered Iosif Pinelis CC BY-SA 4.0