TITLE: Differentiating $x^{x^{x^{...}}}$
QUESTION [5 upvotes]: How do I differentiate $$ x^{x^{x^{...}}}$$ with respect to $x$? (Note that $x$ is raised infinitely many times.)

My attempt: Let $y = x^{x^{x^{...}}}$. Taking logarithm of both sides we get $\ln y = y \ln x$ and let $f = y \ln x - \ln y$. Now $$\frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = \frac{y^2}{x(1 - y \ln x)}$$
Is this approach correct? If not how do I proceed ?
Help would be appreciated. Thanks.

REPLY [5 votes]: The way I would do this:
$$y=x^{x^{x^{\cdots}}}=x^y=\exp(y\ln x)$$
where I've used the notation $\exp(...)$ instead of $e^{\cdots}$ to make the working neater. Now, differentiating implicitly with respect to $x$, and using the fact that $\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$gives
$$\begin{align}\frac{dy}{dx}&=\left(\frac{dy}{dx}\ln x+\frac{y}{x}\right)\exp(y\ln x)\\
&=\left(\frac{dy}{dx}\ln x+\frac{y}{x}\right)x^y\end{align}$$
Rearrange to isolate $\frac{dy}{dx}$:
$$\frac{dy}{dx}\left(1-x^y\ln x\right)=\frac{yx^y}{x}\implies\frac{dy}{dx}=\frac{yx^y}{x(1-x^y\ln x)}=\frac{y^2}{x(1-y\ln x)}$$
which is the same result that you got; your way seems fine :)

I hope that was useful. If you have any questions please don't hesitate to ask :)