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It is well-known that the power-exponential function $x^x$ and its first few derivatives are often taught in calculus.

Does the general formula for the $n$th derivative of the power-exponential function $x^x$ exist somewhere?

What is the general formula for the $n$th derivative of the power-exponential function $x^x$?

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  • $\begingroup$ To be clear, you are asking if there's some closed form for this rather than just describing it recursively? $\endgroup$
    – JoshuaZ
    Commented Dec 23, 2022 at 1:40
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    $\begingroup$ See Louis Comtet, Advanced Combinatorics. Reidel, Dordrecht, 1974, pp. 139–140 and D. H. Lehmer, Numbers associated with Stirling numbers and $x^x$, Rocky Mountain J. Math. 15 (1985), no. 2, 461–479, Number Theory (Winnipeg, Man., 1983). $\endgroup$
    – Ira Gessel
    Commented Dec 23, 2022 at 1:55
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    $\begingroup$ Just use the Faà di Bruno formula (en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula) for $x^x=f(g(x))$ with $f(u)=e^u$ and $g(x)=x\ln x$. $\endgroup$
    – Iosif Pinelis
    Commented Dec 23, 2022 at 4:26
  • $\begingroup$ @IraGessel Merry Christmas! Happy New Year 2023! This night I have derived a general and closed-form formula for the $n$th derivative of the function $x^x$ in terms of the Stirling numbers of the first kind. Later I will compare my result with those you told me. Thank you everybody! $\endgroup$
    – qifeng618
    Commented Dec 23, 2022 at 5:30
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    $\begingroup$ Related is the MSE question The $n$'th derivative of $x^x$. $\endgroup$
    – Dave L Renfro
    Commented Dec 23, 2022 at 14:03

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The Stirling numbers of the first kind $s(n,k)$ for $n\ge k\ge0$ can be analytically generated by \begin{equation*}%\label{1st-stirl-gen-funct} \frac{[\ln(1+t)]^k}{k!}=\sum_{n=k}^\infty s(n,k)\frac{t^n}{n!},\quad |t|<1, \end{equation*} which can be rearranged as Maclaurin's power series expansion of the power function \begin{equation}\label{1st-stirl-gen-funct-rewr} \biggl[\frac{\ln(1+t)}{t}\biggr]^k=\sum_{n=0}^\infty \frac{s(n+k,k)}{\binom{n+k}{k}}\frac{t^{n}}{n!}, \quad |t|<1, \quad k\ge0. \end{equation} Comtet's numbers $b(n,k)$ are generated by \begin{equation}\label{b(n-k)-gen-eq} \frac{[(1+t)\ln(1+t)]^k}{k!}=\sum_{n=k}^{\infty}b(n,k)\frac{t^n}{n!},\quad k\ge0. \end{equation} For $n\in\mathbb{N}_0=\{0,1,2,\dotsc\}$, we have \begin{equation}\label{power-exp-deriv-eq} (x^x)^{(n)}=n!x^{x-n}\sum_{k=0}^{n} x^{k} \sum_{j=0}^{k}\Biggl[\sum_{q=0}^{n-k} \frac{s(q+j,j)}{(q+j)!} \binom{j}{n-k-q}\Biggr]\frac{(\ln x)^{k-j}}{(k-j)!} \end{equation} and \begin{equation}\label{power-exp-deriv-b(n-k)} (x^x)^{(n)} =x^{x-n}\sum_{k=0}^{n} x^{k} \sum_{j=0}^{k}(-1)^{j}\binom{j-n-1}{j} b(n-j,k-j)(\ln x)^{j}. \end{equation} Consequently, Taylor's power series expansion around $x=1$ is \begin{equation}\label{power-exp-taylor-ser} x^x=\sum_{n=0}^\infty\Biggl[\sum_{k=0}^{n} \sum_{q=k}^{n} \frac{s(q,k)}{q!} \binom{k}{n-q}\Biggr](x-1)^n, \quad |x-1|<1 \end{equation} and \begin{equation}\label{b(n-k)-ser=1-eq} x^x=\sum_{n=0}^\infty\Biggl[\sum_{k=0}^{n} b(n,k)\Biggr]\frac{(x-1)^n}{n!}, \quad |x-1|<1. \end{equation} These texts are extracted from Theorem 1 of the following paper:

  1. Jian Cao, Feng Qi, and Wei-Shih Du, Closed-form formulas for the $n$th derivative of the power-exponential function $x^x$, Symmetry 15 (2023), no. 2, Article 323, 13 pages; available online at https://doi.org/10.3390/sym15020323.
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  • $\begingroup$ Just now I found the formula $$ [x^{ax}]^{(n)}=x^{ax}\sum_{k=1}^n \frac{n!}{k!}a^k x^{k-n} \sum_{j=0}^{k}{\left(-1\right)^{k-j}\,{{k}\choose{j}}\left(\sum_{i=0}^{n}{{{{{j}\choose{i}}\sum_{m=0}^{j}{\frac{m!}{(n-i)!}{{j }\choose{m}}{\left[n-i \atop m\right]}}}}}\left(\ln x\right)^{k-m}\right)} $$ on page 8 of the paper "Vladimir V. Kruchinin, Derivation of Bell polynomials of the second kind, arXiv (2011), available online at arxiv.org/abs/1104.5065". $\endgroup$
    – qifeng618
    Commented Jan 29, 2023 at 2:12
  • $\begingroup$ In the paper "Closed-form formulas for the $n$th derivative of the power-exponential function $x^x$", it was obtained that \begin{equation} \bigl[(1+x)^{t(1+x)}\bigr]^{(n)} =n!(1+x)^{t(1+x)-n}\sum_{k=0}^{n}t^k(1+x)^{k} \sum_{j=0}^{k} \Biggl[\sum_{q=0}^{n-k} \frac{s(q+j,j)}{(q+j)!} \binom{j}{n-k-q}\Biggr] \frac{[\ln(1+x)]^{k-j}}{(k-j)!}. \end{equation} Replacing $1+x$ by $x$ yields \begin{equation} \bigl(x^{tx}\bigr)^{(n)} =n!x^{tx-n}\sum_{k=0}^{n}t^kx^{k} \sum_{j=0}^{k} \Biggl[\sum_{q=0}^{n-k} \frac{s(q+j,j)}{(q+j)!} \binom{j}{n-k-q}\Biggr] \frac{(\ln x)^{k-j}}{(k-j)!}. \end{equation} $\endgroup$
    – qifeng618
    Commented Jan 29, 2023 at 2:28
  • $\begingroup$ See also Remark 4 in the paper "Wen-Hui Li, Dongkyu Lim, and Feng Qi. Expanding the function $\ln(1+\operatorname{e}^x)$ into power series in terms of the Dirichlet eta function and the Stirling numbers of the second kind. Carpathian Mathematical Publications Vol. 15, (2023), in press; available online at researchgate.net/publication/369476475" $\endgroup$
    – qifeng618
    Commented Mar 24, 2023 at 3:06
  • $\begingroup$ There is a related paper with this question: Jian Cao, Bai-Ni Guo, Wei-Shih Du, and Feng Qi, A sufficient and necessary condition for the power-exponential function $(1+\frac1x)^{\alpha x}$ to be a Bernstein function and related $n$th derivatives, Fractal and Fractional 7 (2023), no. 5, Article 397, 15 pages; available online at doi.org/10.3390/fractalfract7050397. $\endgroup$
    – qifeng618
    Commented Oct 25, 2023 at 22:03

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