# Differentiating $x^{x^{x^{...}}}$

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.

• You should have a look to see if this map is well defined. Or a least to look at its domain. Mar 25, 2021 at 7:26
• Pauly covered the infinite tower problem in the remark of his answer starting with $\varphi_n(x)=\varphi_{n-1}(x)$. Mar 25, 2021 at 8:23
• Yes your answer is correct. Thought about sharing this, hope it helps: youtu.be/i_l1lz26C2M Mar 25, 2021 at 14:19

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 $$\frac{dy}{dx}$$: $$\frac{dy}{dx}\left(1-x^y\ln x\right)=\frac{yx^y}{x}\implies\frac{dy}{dx}=\frac{yx^y}{x(1-x^y\ln x)}=\frac{y^2}{x(1-y\ln x)}$$
The infinite tetration $$y=x^{x^{x^{...}}}$$ corresponds to $$y=-\frac{W(-\log (x))}{\log (x)}$$ where $$W(.)$$ is Lambert function. Use the chain rule knowing that $$\frac d{dt} W(t)=\frac{W(t)}{t (W(t)+1)}$$