How to differentiate $\lim\limits_{n\to\infty}\underbrace{x^{x^{x^{...}}}}_{n\text{ times}}$? Let $$f(x)=\lim\limits_{n\to\infty}\underbrace{x^{x^{x^{...}}}}_{n\text{ times}}$$
Is it possible to find $f'(x)$. If yes, please show all steps.
 A: This function is known as infinite tetration. The sequence is undefined for $x\leq 0$ except at some rational points, so there is no point talking about the derivative there. Ditto for large $x$, where the sequence diverges to $\infty$. 
Rather surprisingly for $x>0$ the limit only exists for $x\in[e^{-e},e^{1/e}]$ as shown by Euler, see here and here. On that interval $f(x)$ is the inverse function to $x^{1/x}$. This implies differentiability there, and that's where formal games with implicit differentiation actually work. At first glance it seems strange that the sequence converges even for some values $x>1$ since then $x^x>x$ and it is monotone increasing. However, as long as the value of $x$ is close enough to $1$ this is similar to $(1+\frac1n)^n$ that monotone increases but converges to $e$. The threshold value  $e^{1/e}$ is the maximum of $x^{1/x}$, there is nothing to be inverse to for larger values.
A: WLOG, let $y = f(x)$. You know that $$y =x^y$$
So
$$\ln(y) =y\ln(x)$$
$$\frac{dy}{dx}*\frac{1}{y} = \frac{dy}{dx}*\ln(x) + \frac{y}{x}$$
$$\frac{dy}{dx} = \frac{y^2}{x-xy\ln(x)}$$
A: $\textbf{hint:}$
$$f(x) = x^{f(x)}$$
Then proceed 
