derivative of x^x^x... to infinity?

Solution 1:

By following this link you will see that you function is well-defined over the interval $I=\left[e^{-e},e^{\frac{1}{e}}\right]$ and the values in the endpoints are just $\frac{1}{e}$ and $e$. Over such interval, the functional equation: $$ f(x) = x^{f(x)} \tag{1}$$ leads to: $$ f'(x) = x^{f(x)}\left(f'(x)\log x+\frac{f(x)}{x}\right)\tag{2}$$ hence: $$ f'(x) = \frac{f(x)^2}{x\left(1-f(x)\log x\right)}\tag{3}$$ as you stated. We may also notice that: $$ f(x)=\frac{W(-\log x)}{-\log x}\tag{4}$$ where $W$ is the Lambert W-function.

Solution 2:

Not really an anser, rather an outline of what I would try to study this.

The first problem, at least to me, is the definition of the function (what you called "dealing with the infinite". The way I would attempt to define it is saying that the function $f$ is implicitly defined by the equation $$\begin{equation} f(x) = e^{f(x)\ln x}, \quad x > 0. \end{equation}$$ Now, the second problem is arguing why this function exists and is continuous, let alone differentiable. Note that if it exists, you automatically have a nice property: $f> 0$ on its domain of definition.

Assuming this is done, you would need to argue that $f$ is differentiable, before actually deriving it. As a side remark, looking at $x=e^{1/e}$ in the equation defining $f$, you get that $f(e^{1/e})$ must be equal to $e$, so that $f$ cannot be differentiable at $e^{1/e}$ (otherwise, the expression of the derivative you get will be infinite -- division by $0$).