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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.

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

2 Answers 2

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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 :)

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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 $$\frac d{dt} W(t)=\frac{W(t)}{t (W(t)+1)}$$

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