What is the derivative of: $f(x)=x^{2x^{3x^{4x^{5x^{6x^{7x^{.{^{.^{.}}}}}}}}}}$?

It is not completed answer but I thought it can be an approach for such kind of questions. Thus I decided to post it. Let's define

$$f_n(x)=nx^{(n+1)x^{(n+2)x^{(n+3)x^{(n+4)x^{(n+5)x^{(n+6)x^{.{^{.^{.}}}}}}}}}}$$

Your function can be found by $$f_1(x)=f(x)$$ and you are looking for $$\frac {\partial f_n(x)}{\partial x} \bigg|_{n=1}=f'(x)$$

We can easily see a relation for $f_n(x)$ $$f_n(x)=nx^{f_{n+1}(x)}$$ $$\ln(f_n(x))=\ln(n) +f_{n+1}(x)\ln(x)$$

$$\frac {\partial \ln(f_n(x))}{\partial x}=\frac {\partial (f_{n+1}(x)\ln(x))}{\partial x}$$

$$\frac {\partial \ln(f_n(x))}{\partial x}=\frac {\partial (f_{n+1}(x)\ln(x))}{\partial x}$$

$$\frac{\partial f_n(x)}{\partial x} = f_n(x) \frac{\partial f_{n+1}(x)}{\partial x}\ln(x)+\frac {f_{n+1}(x)f_n(x)}{x}$$

----Let's put $n=1,2,3,....$

$$\frac{\partial f_1(x)}{\partial x}= \frac{\partial f_{2}(x)}{\partial x}f_1(x)\ln(x)+\frac {f_{2}(x)f_1(x)}{x}$$

$$\frac{\partial f_2(x)}{\partial x} = \frac{\partial f_{3}(x)}{\partial x}f_2(x)\ln(x)+\frac {f_{3}(x)f_2(x)}{x}$$

$$\frac{\partial f_3(x)}{\partial x} = \frac{\partial f_{4}(x)}{\partial x}f_3(x)\ln(x)+\frac {f_{4}(x)f_3(x)}{x}$$

$$.$$ $$.$$ $$.$$

$$\frac{\partial f_1(x)}{\partial x} = U(x)+\frac {f_{1}(x)f_2(x)}{x}+\frac {f_{1}(x)f_2(x)f_3(x) \ln(x) }{x}+\frac {f_{1}(x)f_2(x)f_3(x) f_4(x) \ln^2(x) }{x}+.... \tag{1}$$

Where $$U(x)=\lim\limits_{ n\to \infty } \frac{\partial f_n(x)}{\partial x} \ln^n(x)\prod_{k=1}^{n-1} f_n(x) $$

Finally we can express the derivative as

$$f'(x)=\frac{\partial f_1(x)}{\partial x} = U(x)+ \sum_{k=1}^{\infty} \frac {\ln^{k-1}(x)f_1(x)\prod_{n=2}^{k+1} f_n(x)}{x}$$

Note: Yet I have not found what $U(x)$ is. I estimate that $U(x)$ will vanish but I have not proved it yet. Maybe someone can help with some numerical values that if $U(x)=0$ or not. Thanks a lot for contributions and advice

An observation : I noticed a similar pattern with general formula $g'(x)$ in the question and $f'(x)$ that I wrote in (1). $$g(x)=h(x)^{h(x)^{h(x)^{h(x)^{h(x)^{h(x)^{h(x)^{.{^{.^{.}}}}}}}}}}$$ $$g'(x)=\frac{g^2(x)h'(x)}{h(x)\left[1-g(x)\ln(h(x))\right]}=$$

It can be rewritten as

$$g'(x)=\frac{g^2(x)h'(x)}{h(x)}(1+g(x)\ln(h(x))+g^2(x)\ln^2(h(x))+g^3(x)\ln^3(h(x))+....)$$

If $h(x)=x$ then we can obtain

$$g'(x)=\frac{g^2(x)}{x\left[1-g(x)\ln(x)\right]}=$$ $$g'(x)=\frac{g^2(x)}{x}(1+g(x)\ln(x)+g^2(x)\ln^2(x)+g^3(x)\ln^3(x)+....)$$ $$g'(x)=\frac{g^2(x)}{x}+\frac{g^3(x)}{x}\ln(x)+\frac{g^4(x)}{x}\ln^2(x)+....$$

It has similarities with my formula for $f'(x)$ that I wrote above. This result supports my idea that $U(x)$ may vanish.

Partial Answer / Observation
So it is clear that this function can be defined for $f(0)$ and $f(1)$, and is definitely defined only within some portion of this domain... accordingly, I chose to evaluate what happens with $0.5$ as you increase the tower height. Numerically, it appears that two limits are approached, one for even heights and one for odd heights (this is the nature of power towers evaluated between $0$ and $1$ for any I have ever calculated... there is probably a proof of this somewhere, at least for any sequence of heights that reach some limit). Regardless, it appears that the function is simply periodic between these two limits, and I would say that this should hold as you continue to increase the tower height. Now, this isn't a proof in any sense, (even for the value $0.5$), as numerical analysis alone won't cut this, but I think it provides some interesting insight, and is the only thing that got me any sort of result after hours of looking into this. (Note that I am not referencing iterated functions with the superscript, but the height of the tower.) As is pointed out in the comments, this function is probably only defined at a finite amount of points... One could probably find a way to show that a point such as $0.5$ diverges by analyzing the property that causes this dual limit (I am fairly certain it is due to the domain $[0,1]$), but I am not sure that such observations would be sufficient to prove this across the entire domain. $$f^1(0.5) \approx x = 0.5$$ $$f^2(0.5)\approx x^{2x} = 0.5$$ $$f^3(0.5)\approx x^{2x^{3x}} = 0.612...$$ $$f^4(0.5)\approx x^{2x^{3x^{4x}}} = 0.439...$$ $$f^5(0.5)\approx 0.679...$$ $$f^6(0.5)\approx 0.374...$$ $$f^9(0.5)\approx 0.804...$$ $$f^{10}(0.5)\approx 0.305...$$ $$f^{14}(0.5)\approx 0.3040559...$$ $$f^{18}(0.5)\approx 0.3040557...$$ $$f^{15}(0.5)\approx 0.81045144...$$ $$f^{19}(0.5)\approx 0.81045145968867...$$ $$f^{23}(0.5)\approx 0.81045145968869...$$