How would one find: $$\frac{\mathrm d}{\mathrm dx}{}^xx?$$ where ${}^ba$ is defined by $${}^ba\stackrel{\mathrm{def}}{=}\underbrace{ a^{a^{\cdot^{\cdot^{\cdot^a}}}}}_{\text{$b$ times}}$$

**Work so far**

The interval that I am working in is $(0, \infty)$. It doesn't make much sense to consider negative numbers. Although there exists no extension to the reals for tetration I am going to assume that it exists. My theory is that it shouldn't change the algebra involved; (correct me if I am wrong).

Some visual analysis on the curve and you can see that it diverges to $+\infty$ extremely rapidly. This means that the derivative is going to have similar properties as well.

Let $f(x, y):={}^yx$ so we can rewrite our tetration as $f(x, x)$. Now using the definition of the total derivative: $D\;g(x, y)=\partial_xg(x, y)+\partial_yg(x, y)$. This should allow us to differentiate $f$.

$$D\;f(x,y)=\frac{\partial}{\partial x}{}^yx+\frac{\partial}{\partial y}{}^yx$$

Let's focus on the first partial derivative $\partial_x{}^yx$. This is just the case of differentiating a finite power tower as $y$ is treated constant.

Firstly looking at some examples do derive a general formula for $D\;\;{}^nx$: $$ \begin{array}{c|c} n & D\;\;{}^nx\\ \hline 0 & 0\\ 1 & 1\\ 2 & {}^2x(\log x + 1)\\ 3 & {}^3x\times {}^2x\times x^{-1}(x\log x(\log x + 1)+1) \end{array} $$

It is easy to see that there is some pattern emerging however because of it's recursive nature I could not form a formula to describe it.

Edit$$\dfrac{d}{dx}\left(e^{{}^nx \log(x)}\right)={}^{n+1}x\dfrac{d}{dx}\left({}^nx \log(x)\right)={}^{n+1}x\left(({}^nx)' \log(x)+\frac{{}^nx}{x}\right)$$ The recursive formula for the partial was pointed out in comments however an explicit formula would be more useful for this purpose.

The second partial derivative is interesting and relies on properties of tetration.

I was hoping for it to be similar to exponention such that $$D_y \;x^y=D_y\;e^{y\log x}=e^{y \log x}\log x\;D_y\;y=x^y\log x$$ However I am not sure of an '$e$ for tetration' but I hope it would be something like this: $$D_y \;{}^yx=D_y\;{}^{y\;\text{slog} x}t={}^{y\;\text{slog} x}t\;\text{slog} x\;D_y\;y={}^yx\;\text{slog}\; x$$ Where $\text{slog}$ denotes the super logarithm (slogorithm), an inverse of tetration.

EditThis may as well be a possible identity which can easily be applied to the above: $$\text{slog}\;\left({}^yx\right)=\text{slog}^y\;(x)$$

I am unsure about using slogorithms and tetration in this way and I feel I might just be abusing notation.

**Work on Tetration**

I will update this section with more rigorous definitions and properties of tetration. I cannot prove all of them now.

For $x\in \Bbb R$ and $n \in \Bbb N$, $${}^nx:=\underbrace{ x^{x^{\cdot^{\cdot^{\cdot^x}}}}}_{\text{$n$ times}}\tag{1}$$

~~For $x\in\Bbb R$ and $a,b\in\Bbb N$?,~~
$${{}^b({}^ax)={}^{a^b}x\tag{$\not2$}}$$
Through simply algebra you can find that the above is not the case.

## Update:

This is just differentiating the pentation function.