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

I happened to ponder about the differentiation of the following function:
$$f(x)=x^{2x^{3x^{4x^{5x^{6x^{7x^{.{^{.^{.}}}}}}}}}}$$
Now, while I do know how to manipulate power towers to a certain extent, and know the general formula to differentiate $g(x)$ wrt $x$, where $$g(x)=f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{.{^{.^{.}}}}}}}}}}$$
I’m still unable to figure out as to how I can adequately manipulate the function to differentiate it within its domain of convergence.

General formula: $$g'(x)=\frac{g^2(x)f'(x)}{f(x)\left[1-g(x)\ln(f(x))\right]}$$

#### Solutions Collecting From Web of "What is the derivative of: $f(x)=x^{2x^{3x^{4x^{5x^{6x^{7x^{.{^{.^{.}}}}}}}}}}$?"

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.
$$f^1(0.5) \approx x = 0.5$$
$$f^2(0.5)\approx x^{2x} = 0.5$$
$$f^3(0.5)\approx x^{2x^{3x}} = 0.612…$$
$$f^4(0.5)\approx x^{2x^{3x^{4x}}} = 0.439…$$
$$f^5(0.5)\approx 0.679…$$
$$f^6(0.5)\approx 0.374…$$
$$f^9(0.5)\approx 0.804…$$
$$f^{10}(0.5)\approx 0.305…$$
$$f^{14}(0.5)\approx 0.3040559…$$
$$f^{18}(0.5)\approx 0.3040557…$$
$$f^{15}(0.5)\approx 0.81045144…$$
$$f^{19}(0.5)\approx 0.81045145968867…$$
$$f^{23}(0.5)\approx 0.81045145968869…$$

Note: The below applies to a different function than OP intended. I was studying $$f(x)=f(x)=x^{(2x)^{(3x)^{(4x)^{(5x)^{(6x)^{(7x)^{.{^{.^{.}}}}}}}}}}$$ but don’t want to delete this.

Before we worry about derivatives, we need to find where the function is defined and continuous. $f(x)$ has to be seen as the limit of the sequence $x, x^{2x},x^{(2x)^{(3x)}}\dots$ If $x \gt 1$ this clearly diverges to infinity, if $x=0$ it is undefined. It might turn out to be defined at some negative integers, but will not be defined at other negative $x$, so we can concentrate on $0 \lt x \le 1$. If $x=1$, the sequence is just $1$ to higher and higher powers, so $f(1)=1$. If $\frac 12 \lt x \lt 1$ the sequence is a number less than $1$ to higher and higher powers, so the limit is $0$. If $x=\frac 12,$ the sequence is $\frac 12$ to ($1$ to higher and higher powers), so $f(\frac 12)=\frac 12$. If $\frac 13 \lt x \lt \frac 12,$ we have $2x \lt 1$, so $x^{(2x)^{\text { lots }}}$ goes to $0$. In general, if $x = \frac 1k$ we just evaluate up to $(k-1)x$. If $\frac 1{k+1} \lt x \lt \frac 1k$ we evaluate up to $(k-2)x$ because the powers of $kx$ will go to zero, so the powers of $(k-1)x$ go to $1$. Summing up $$f(x)=\begin {cases} 1&x=1\\0&\frac 12 \lt x \lt 1\\\frac 12&x=\frac 12\\1&\frac 13 \lt x \lt \frac 12\\ \frac 13^{\frac 23} &x=\frac 13 \\x&\frac 14 \lt x \lt \frac 13 \\\text {tower up to }(k-1)x&x=\frac 1k\\\text{tower up to }(k-2)x&\frac 1{(k+1)x} \lt x \lt \frac 1k\end {cases}$$

Now it is clear that $f(x)$ is differentiable at all points in $(0,1)$ that are not of the form $\frac 1k$ but the derivative will get very messy as $x$ gets small.

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
$$\frac {\partial f_n(x)}{\partial x} |_{n=1}=f'(x)$$

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.

Comment about $g(x)=f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{.{^{.^{.}}}}}}}}}}$

The comment is too long to be posted in the comments section. That is why it is posted in the answer section. But this is not an answer.

If the power tower is convergent $g(x)=f(x)^{g(x)}=e^{g\ln|f|}$

The derivative of $g(x)$ can be explicitly expressed thanks to the Lambert W function :

$ge^{-g\ln|f|}=1 \quad \to \quad -g\ln|f|e^{-g\ln|f|}= -\ln|f|$

$-g\ln|f|=W\left( -\ln|f|\right) \quad$ where $W$ is the Lambert W function.

$$g(x)=-\frac{1}{\ln|f(x)|}W\left( -\ln|f(x)|\right)$$

With condition $\quad 0<|f(x)|<e^{1/e}\quad$

We know that $\frac{dW(X)}{dX}=\frac{W(X)}{X\left(W(X)+1 \right)}$

With $X=-\ln|f(x)|\quad \to \quad g'(x)=\frac{d}{dX}\left(\frac{W(X)}{X} \right) \frac{dX}{dx}$

$g'(x)= \left( -\frac{W(X)}{X^2}+\frac{W(X)}{X^2\left(W(X)+1 \right)} \right) \left(-\frac{f’}{f} \right) = \left( -\frac{W(X)^2}{X^2\left(W(X)+1 \right)} \right) \left(-\frac{f’}{f} \right)$

$$g'(x)=\left(\frac{ W\left(-\ln|f(x)|\right) }{-\ln|f(x)| } \right)^2 \frac{1}{W\left(-\ln|f(x)|\right)+1}\:\frac{f'(x)}{f(x)}\quad \text{in} \quad 0<|f(x)|<e^{1/e}$$

Looking at this graph, it doesn’t appear to converge for values near $x=0$ and it may converge for values close and larger than $x=1$. Around $x=0.2$, it oscillates a lot and doesn’t appear to converge to a single value around $x<0.8$.

It also appears to have ‘good’ convergence for $0.8<x<1$.

It can be proved by induction that the iteration of $\lambda\cdot e^{x}$ is equivalent to the iteration of $e^{\lambda\cdot x}$ modulo the base. Therefore $g(x)$ is equivalent to:

$$g(x)=(x^2)^{(x^3)^{(x^4)^{\cdots}}}$$

For the above to make sense, using Barrow’s Theorem (here), we must have $\forall n>1\in\mathbb{N}$:

$$e^{-e}\le x^n\le e^{1/e}\Leftrightarrow$$

$$e^{-\frac{e}{n}}\le x \le e^{\frac{1}{ne}}$$

Precisely because the above must hold for all $n>1$, the applicable constraint reduces to:

$$\lim_{n\to\infty}e^{-\frac{e}{n}}\le x \le \lim_{n\to\infty}e^{\frac{1}{ne}}\Leftrightarrow$$

$$x=1$$

If $x=1$ however, then your expression is identically equal to 1, hence the derivative of your expression vanishes everywhere on its domain.