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 $a=1$
$$\begin{align}
f(x)&=\sum_{i=0}^{n-1}\frac{f^{(i)}(1)}{i!}\\[5pt]
&\qquad=\frac1{0!}+\frac1{1!}(x-1)+\frac2{2!}(x-1)^2+\frac3{3!}(x-1)^3+\frac8{4!}(x-1)^4+\frac{12}{5!}(x-1)^5\\&\qquad+\frac{54}{6!}(x-1)^6+\cdots
\end{align}$$
 A: Extract from my note Derivatives of generalized power functions that appeared in the Reader Reflections column of Mathematics Teacher [Volume 103, Number 9; May 2010; pp. 630-631]:

Regarding the editor's note on higher derivatives of $x^{x},$ let $n$ be a positive integer, ${n \choose k}$ be the usual binomial coefficient, $f(x) = x^{x},$ and $g(x) = 1 + \ln{x}.$ Kulkarni (1984) gave the recursion formula
$$ f^{(n+1)}(x) \;\; = \;\; f^{(n)}(x)g(x) \;\; + \;\; \sum\limits_{k=1}^{n} \left[ {n \choose k} (-1)^{k-1}(k-1)! \right] f^{(n-k)}(x)x^{-k} $$
by observing $f^{(n+1)}(x)$ is the $n$th derivative of $f'(x) = x^x(1 + \ln{x})$ and then writing down the Leibniz formula for the $n$th derivative of a product, using the fact that the $k$th derivative of $1 + \ln x$ is $(-1)^{k-1}(k-1)!x^{-k}.$
S. B. Kulkarni, Solution to Problem 3977, School Science and Mathematics 84 #7 (November 1984), 629-630.

A: Hint
I think that you could simplify the calculations using sucessively logarithmic derivatives. This means that you will cascade the expressions and the $n^{th}$ derivative of $f(x)$ will be expressed as function of lower order derivatives. 
In fact I suppose that some of your derivatives are not correct since, after simplications, we get
$$f(x)=1+(x-1)+(x-1)^2+\frac{1}{2} (x-1)^3+\frac{1}{3} (x-1)^4+\frac{1}{12}
   (x-1)^5+\frac{3}{40} (x-1)^6-\frac{1}{120} (x-1)^7+\frac{59
   (x-1)^8}{2520}-\frac{71 (x-1)^9}{5040}+\frac{131
   (x-1)^{10}}{10080}+O\left((x-1)^{11}\right)$$
In order to check you expansion, I suggest you put is on the same graph as the original function plotting for $0 \leq x \leq 2$; the two curves are supposed to be very similar.
A: For $n\in\{0,1,2,\dotsc\}$, 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}
where $s(n,k)$ for $n\ge k\ge0$ denotes the Stirling numbers of the first kind. Consequently, we have
\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}
For more information and their proofs, please see the site https://mathoverflow.net/a/439188 and the paper

*

*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.

A: If you consider $f(u,v)=u^v$, and $u(x)=v(x)=x$, then $x^x=f(x)=f(u(x),v(x))$ an using the multivariate chain rule:
$$\frac{df}{dx}=\frac{\partial f}{\partial u}\frac{du}{dx}+\frac{\partial f}{\partial v}\frac{dv}{dx}=f_u+f_v$$
Here we use the fact that $\frac{du}{dx}=\frac{dv}{dx}=1$. And then again:
$$\frac{d^2f}{dx^2}=\frac{df}{dx}\left(f_u+f_v\right)=f_{uu}+f_{uv}+f_{vu}+f_{vv}=f_{uu}+2f_{uv}+f_{vv}$$
Inductively,
$$\frac{d^nf}{dx^n}=\sum_{k=0}^n\binom{n}{k}f_{\overbrace{u\cdots u}^k\overbrace{v\cdots v}^{n-k}}$$
Now
$$\begin{align}
f_{\overbrace{u\cdots u}^k\overbrace{v\cdots v}^{n-k}}
&=\frac{\partial^{k}}{\partial u^k}u^v(\ln u)^{n-k}\\
&=\sum_{j=0}^k\binom{k}{j}\left[\frac{\partial^{j}}{\partial u^j}u^v\right]\left[\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\right]\\
&=\sum_{j=0}^k\binom{k}{j}\left[j!\binom{v}{j}u^{v-j}\right]\left[\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\right]\\
\end{align}$$
It's gets messy after this (but I claim interesting!) so let's recap a bit before continuing. At this point we have:
$$f^{(n)}(x)=\sum_{k=0}^n\binom{n}{k}\sum_{j=0}^k\binom{k}{j}\left[j!\binom{v}{j}u^{v-j}\right]\left[\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\right]$$
And we may let $u=v=x$ wherever we do not yet need to take a derivative:
$$f^{(n)}(x)=\sum_{k=0}^n\sum_{j=0}^k\binom{n}{k}\binom{k}{j}\left[j!\binom{x}{j}x^{x-j}\right]\left[\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\right]$$
In fact we may let $x=1$ where we do not yet need to take a derivative, since ultimately you are after $f^{(n)}(1)$.
$$f^{(n)}(x)=\sum_{k=0}^n\sum_{j=0}^k\binom{n}{k}\binom{k}{j}\left[j!\binom{1}{j}\right]\left[\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\right]$$
Now we work on $\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}$. We have a composition of functions $g\circ h$, where $h(u)=\ln(u)$ and $g(u)=u^{n-k}$. It's $(k-j)$th derivative can be found with Faà di Bruno's formula.
$$\begin{align}
&\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\\
&=\sum_{m_1,\ldots,m_{k-j}}\frac{(k-j)!}{\prod_i m_i!i!^{m_i}}g^{\left(\sum m_i\right)}\left(\ln(u)\right)\prod_t\left(\frac{d^t}{du^t}\ln(t)\right)^{m_t}
\end{align}$$
where the first sum is over tuples of nonnegative integers such that $m_1+2m_2+\cdots+(k-j)m_{k-j}=k-j$. The function $g$ is a power function with nonnegative power $n-k$, so high derivatives of it are zero. Any tuple of the $m_i$ with $\sum m_i>n-k$ will contribute zero to the sum So we can use an additional constraint that $\sum m_i\leq n-k$.
We can simplify more, knowing that $g$ is a power function and knowing derivatives of $\ln(u)$.
$$\begin{align}
&\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\\
&=\sum_{m_1,\ldots,m_{k-j}}\frac{(k-j)!}{\prod_i m_i!i!^{m_i}}\left(\sum m_i\right)!\binom{n-k}{\sum m_i}(\ln u)^{n-k-\sum m_i}\prod_t\left(-(-1)^t(t-1)!u^{-t}\right)^{m_t}
\end{align}$$
The $i!$ in the denominator have some cancellation with $(t-1)!$ at the end. And then the ultimate power of $u$ is $-1m_1-2m_2-\cdots=-(k-j)$. And the product of all those $(-1)$s can be simplified:
$$\begin{align}
&\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\\
&=\sum_{m_1,\ldots,m_{k-j}}\frac{(k-j)!}{\prod_i m_i!}\left(\sum m_i\right)!\binom{n-k}{\sum m_i}(\ln u)^{n-k-\sum m_i}\left[\prod_t\left(-(-1)^t\right)^{m_t}\right]\left[\prod_t\left(u^{-t}\right)^{m_t}\right]\left[\prod_t\frac{\left((t-1)!\right)^{m_t}}{t!^{m_t}}\right]\\
&=\sum_{m_1,\ldots,m_{k-j}}\frac{(k-j)!}{\prod_i m_i!}\left(\sum m_i\right)!\binom{n-k}{\sum m_i}(\ln u)^{n-k-\sum m_i}(-1)^{k-j+\sum m_i}u^{j-k}\left[\prod_t\frac{1}{t^{m_t}}\right]\\
&=\sum_{m_1,\ldots,m_{k-j}}\frac{(k-j)!}{\prod_i i^{m_i}m_i!}\left(\sum m_i\right)!\binom{n-k}{\sum m_i}(-1)^{k-j+\sum m_i}u^{j-k}(\ln u)^{n-k-\sum m_i}
\end{align}$$
Regrouping from earlier, and letting $u=x=1$:
$$f^{(n)}(1)=\sum_{k=0}^n\sum_{j=0}^k\binom{n}{k}\binom{k}{j}\left[j!\binom{1}{j}\right]\sum_{m_i}\frac{(k-j)!}{\prod_i i^{m_i}m_i!}\left(\sum m_i\right)!\binom{n-k}{\sum m_i}(-1)^{k-j+\sum m_i}(0)^{n-k-\sum m_i}$$
Note that last power of $0$ is not $0$ when the exponent is $0$. So we proceed under the added constraint that $\sum m_i=n-k$.
$$
\begin{align}
f^{(n)}(1)&=\sum_{k=0}^n\sum_{j=0}^k\binom{n}{k}\binom{k}{j}\left[j!\binom{1}{j}\right]\sum_{m_i}\frac{(k-j)!}{\prod_i i^{m_i}m_i!}\left(n-k\right)!(-1)^{n-j}\\
&=n!\sum_{k=0}^n\sum_{j=0}^k\binom{1}{j}\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}(-1)^{n-j}
\end{align}$$
For $j>1$, the summand is $0$. So we consider the $j=0$ and $j=1$ terms only. But there is no $j=1$ term for $k=0$, so we also separate the $(k,j)=(0,0)$ term:
$$
\begin{align}
f^{(n)}(1)
&=(-1)^nn!\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}\\
&+(-1)^nn!\sum_{k=1}^n\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}\\
&+(-1)^{n-1}n!\sum_{k=1}^n\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}
\end{align}$$
Note we are only considering $n\geq 1$. We have accumulated some constraints on the tuples of $m_i$:

*

*It's a $(k-j)$-tuple: $m_1,m_2,\ldots,m_{k-j}$

*$\sum im_i=k-j$

*$\sum m_i=n-k$
In the first of the three terms above, $k=j=0$, so $k-j=0$. So we have an empty sum, $0$.
In the second term above, $j=0$. So we are summing over a $k$-tuple with $\sum im_i=k$.
In the third term, $j=1$. So we are summing over a $(k-1)$-tuple with $\sum im_i=k-1$.
$$
\begin{align}
f^{(n)}(1)=(-1)^nn!\sum_{k=1}^n\left[\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}-\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}\right]
\end{align}$$
In addition to the constraints above on the tuples, we also have $\sum m_i=n-k$.

Let's do check with $n=3$. When $k=1$, the first sum is over a $1$-tuple summing to $2$. So it's just $[2]$ but that does not satisfy $\sum im_i=k$. The second sum is over a $0$-tuple, so empty, contributing $0$.
When $k=2$, the first sum is over a $2$-tuple summing to $1$. The possibilities are $[1;0]$ and $[0;1]$. Only the latter satisfies $\sum im_i=k=2.$ And the second sum is over a $1$-tuple summing to $1$. So $[1]$. This does satisfy $\sum im_1=k-1=1$.
When $k=3$, the first sum is over a $3$-tuple summing to $0$. The only possibility is $[0;0;0]$. That does not satisfy $\sum im_i=k=3.$ And the second sum is over a $2$-tuple summing to $0$. So $[0;0]$. This does not satisfy $\sum im_1=k-1=2$.
So we have:
$$
\begin{align}
f^{(3)}(1)
&=-6\left[\frac{1}{1^{0}0!2^11!}-\frac{1}{1^{1}1!}\right]=3
\end{align}$$
This agrees with what you have for $f^{(3)}(1)$.

So we have a relatively concise expression for $f^{(n)}(1)$. However it sums over $k$, and then depends on partitions $n-k$ that satisfy an additional constraint. If you can simplify it further, that would be neat.
Here is an expression for the Taylor series centered at $1$.
$$T_1(x)=1+\sum_{n=1}^{\infty}(-1)^n\sum_{k=1}^n\left[\sum_{\begin{array}{c}m_1+\cdots+m_k=n-k\\1m_1+\cdots+km_k=k\end{array}}\frac{1}{\prod_{i=1}^k i^{m_i}m_i!}-\sum_{\begin{array}{c}m_1+\cdots+m_{k-1}=n-k\\1m_1+\cdots+(k-1)m_{k-1}=k-1\end{array}}\frac{1}{\prod_{i=1}^{k-1} i^{m_i}m_i!}\right](x-1)^n$$
And if this converges when $x=0$, you have
$$T_1(0)=1+\sum_{n=1}^{\infty}\sum_{k=1}^n\left[\sum_{\begin{array}{c}m_1+\cdots+m_k=n-k\\1m_1+\cdots+km_k=k\end{array}}\frac{1}{\prod_{i=1}^k i^{m_i}m_i!}-\sum_{\begin{array}{c}m_1+\cdots+m_{k-1}=n-k\\1m_1+\cdots+(k-1)m_{k-1}=k-1\end{array}}\frac{1}{\prod_{i=1}^{k-1} i^{m_i}m_i!}\right]$$
But I wouldn't know how to show that the sum converges, or independently find that it converges to $1$.
A: I tried the simplest approach to test myself. Let $u=x^{x-1}=e^{(x-1)\ln x}$, then $u=1+\sum_{n=1}^{\infty}\frac{1}{n!}(x-1)^n(\sum_{s=1}^{\infty}\frac{(-1)^{s-1}}{s}(x-1)^s)^n$ and expanding the binomial we have
$$u=1+\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n!}C_{n,k}(x-1)^{n+k}$$
where
$$C_{n,k}=\sum_{r_1+r_2+...+r_n=k}^{r_i\geq 1,\forall i}\frac{(-1)^{k-n}}{r_1r_2...r_n}.$$
Letting $n+k=m$, we have $u=1+\sum_{m=2}^\infty a_m(x-1)^m$ where
$$a_m=\sum_{n=1}^{\lfloor m/2\rfloor}\frac{C_{n,m-n}}{n!}.$$
Then we have
$$y=x^x=1+(x-1)+(x-1)^2+\sum_{m=3}^{\infty}(a_m+a_{m-1})(x-1)^m.$$
$a_2=C_{1,1}=1$, $a_3=C_{1,2}=-\frac12$, $a_4=C_{1,3}+\frac{C_{2,2}}{2}=\frac 13+\frac12=\frac56$, $a_5=C_{1,4}+\frac{C_{2,3}}{2}=-\frac 14+-\frac12=-\frac34$, $a_6=C_{1,5}+\frac{C_{2,4}}{2}+\frac{C_{3,3}}{6}=\frac15+\frac{11}{24}+\frac16=\frac{33}{40},.....$ and hence
$$y=1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3+\frac{1}{3}(x-1)^4+\frac{1}{12}(x-1)^5+\frac{3}{40}(x-1)^6+...$$
A: $$\frac{d^n x^x}{dx^n}=\frac{d^ne^{x\ln(x)}}{dx^n}$$
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?
