<s>[INST] The $n$'th derivative of $x^x$ I want to know the $n$'th derivative of $f(x)=x^x$.
Then, I'll calculate $f(0)$ with Taylor expansion of $f(x)$ on $a=1$.
Here is my answer, but it is unfinished.
The derivative of $f(x)=x^x$
$$\begin{align}
f'(x)&=x^x(\log x+1)\\
f''(x)&=x^x(\log x+1)^2+x^{x-1}\\
f'''(x)&=x^x(\log x+1)^3+3x^{x-1}(\log x+1)-x^{x-2}\\[5pt]
f(x)^{(4)}&=x^x(\log x+1)^4+4x^{x-1}(\log x+1)^2-4x^{x-2}(\log x+1)+3x^{x-2}+2x^{x-3}\\
f(x)^{(5)}&=x^x(\log x+1)^5+10x^{x-1}(\log x+1)^3-10x^{x-2}(\log x+1)^2+15x^{x-2}(\log x+1)\\&\quad+10x^{x-3}(\log x+1)-10x^{x-3}-6x^{x-4}\\
f(x)^{(6)}&=x^x(\log x+1)^6+15x^x(\log x+1)^4-20x^{x-2}(\log x+1)^3+45x^{x-2}(\log x+1)^2\\&\quad+30x^{x-3}(\log x+1)^2-50x^{x-3}(\log x+1)+15x^{x-3}-46x^{x-4}(\log x+1)\\&\quad+40x^{x-4}+24x^{x-5}
\end{align}$$
Taylor expansion of $f(x)=x^x$ in $a=1$
$$\begin{align}
f(x)&=\sum_{i=0}^{n-1}\frac{f^{(i)}(1)}{i!}\\[5pt]
&\qquad=\frac1{0!}+\frac1{1!}(x-1)+\frac2{2!}(x-1)^2+\frac3{3!}(x-1)^3+\frac8{4!}(x-1)^4+\frac{12}{5!}(x-1)^5\\&\qquad+\frac{54}{6!}(x-1)^6+\cdots
\end{align}$$
 [/INST] $$\frac{d^n x^x}{dx^n}=\frac{d^ne^{x\ln(x)}}{dx^n}$$
Expanding $e^y$ as a series and then $\ln(x)$ as a limit:
$$\frac{d^n}{dx^n}\sum_{k=0}^\infty \frac{x^k \ln^k(x)}{k!}=\lim_{c\to0}\sum_{k=0}^\infty \frac1{k!}\frac{d^n}{dx^n}\frac{x^k(x^c-1)^k}{c^k}$$
Now use the general Leibniz rule:
$$\frac{d^n}{dx^n}x^k(x^c-1)^k=\sum_{m=0}^n\binom nm \frac{d^{n-m}}{dx^{n-m}}x^k\frac{d^m}{dx^m}(x^c-1)^k$$
and the binomial theorem:
$$\frac{d^m}{dx^m}(x^c-1)^k=\sum_{j=0}^k\binom kj (-1)^{k-j}\frac{d^m}{dx^m}x^{cj}$$
We now take the derivatives with factorial power $u^{(v)}$
$$\frac{d^v}{dx^v}x^r=x^{r-v}r^{(v)}$$
Therefore we only have a triple series:
$$\frac{d^n x^x}{dx^n}= \lim_{c\to0}\sum_{k=0}^\infty \sum_{m=0}^n\sum_{j=0}^k\frac{k!n!(-1)^{k-j}(c j)!x^{cj+k-n}}{j!m!(k-j)!(n-m)!(cj-m)!(k+m-n)!c^k}$$
Shown here when clicking “Approximate form” in the substitution section. The inner series likely has a hypergeometric answer. Additionally, $0\le j\le k\le \infty$ meaning that the $j,m$ series are interchangeable if both are infinite series. However, how would one take the limit?
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