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During a quant interview with Banc One in 1996 post-physics doctorate, I choked on this interview question:

What is the derivative $\frac{dy}{dx}$ of $y=x^{x^{x^{.^{.^{.}}}}}$

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  • $\begingroup$ what is the derivative of $y = (f(x))^x$ is where I would start - doesn't get anywhere though $\endgroup$
    – Cato
    Commented Nov 1, 2016 at 16:08
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    $\begingroup$ Note that $y=x^y$ so $\ln y=y\ln x$. $\endgroup$
    – lulu
    Commented Nov 1, 2016 at 16:11
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    $\begingroup$ Try noting that $y = x^y$ and using implicit differentiation. (There is also the issue of whether $y$ is differentiable.) $\endgroup$
    – anomaly
    Commented Nov 1, 2016 at 16:11
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    $\begingroup$ That was $20$ years ago! So did you get the job? And what made you suddenly think of this question? $\endgroup$ Commented Nov 1, 2016 at 16:57
  • $\begingroup$ No, did not get the job, and likely not for just choking on this question. What made me think of it 20 yrs later ... not sure. $\endgroup$ Commented Nov 2, 2016 at 14:03

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Let me first state that I don't know if $y$ defined that way is meaningful or how it can be made mathematically meaningful. But since this came up in a quant interview the answer they were expecting had probably something to do with differentiation tricks. So here is one trick.

$$\log{y} = y\log{x}$$

Differentiate both sides with respect to $x$ to obtain

$$\frac{dy}{dx} = \frac{y^2}{x-xy\log{x}}$$

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  • $\begingroup$ You can make the notation meaningful in a rigorous sense by defining it as the limit as the partial, finite sequences of repeated exponentials. There is a fair amount of work to do in developing the theory to show that implicit differentiation should work in this circumstance (which I admittedly don't remember well). $\endgroup$
    – Mathily
    Commented Nov 1, 2016 at 16:21
  • $\begingroup$ @Mathily That would be my first guess as well. But it is far from clear to me under what conditions that limit exists, let alone if the the resulting object is differentiable with respect to its argument. $\endgroup$
    – Calculon
    Commented Nov 1, 2016 at 16:23
  • $\begingroup$ In a most unhelpful answer, a numeric check shows convergence is incredibly rapid for $x\in(0,1)$. There is obvious convergence at $x=1$ and nowhere else outside $(0,1)$; well maybe in the complex plane for $x\in(-1,0)$. $\endgroup$
    – Mathily
    Commented Nov 1, 2016 at 16:30
  • $\begingroup$ @Mathily I hope it converges to a $y$ such that $\frac{\log{y}}{y} = \log{x}$ $\endgroup$
    – Calculon
    Commented Nov 1, 2016 at 16:32
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    $\begingroup$ @Mathily Maybe I am missing a point but isn't $y$ equal to $1$ for any $x \in (0,1]$? $\endgroup$
    – Calculon
    Commented Nov 1, 2016 at 16:42
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I assume you'd like to know what the derivative is?

$y=x^{x^\cdots }$ gives $y=x^y=e^{\ln(x^y)}=e^{y\ln(x)}$. Now differentiating implicitly $$\frac{dy}{dx}=e^{y\ln(x)}\left(\frac{dy}{dx}\ln(x)+\frac{y}{x}\right)=x^y\ln(x)\frac{dy}{dx}+\frac{yx^y}{x}$$ So $$\left(1-x^y\ln(x)\right)\frac{dy}{dx}=yx^{y-1}$$ So $\frac{dy}{dx}=\frac{yx^{y-1}}{1-x^y\ln(x)}$.

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If $y = x^{x^{x^\cdots}}$ then a reasonable implicit definition of this function is to say that $y = x^y$; i.e., $y(x)$ is the fixed point of the iterative sequence $$\{y_n\}_{n \ge 0}, \quad y_{n+1} = x^{y_n}, \quad y_0 = x$$ whenever it is well-defined. Thus logarithmic implicit differentiation immediately yields $$\frac{y'}{y} = y' \log x + \frac{y}{x}$$ or $$y' = \frac{y/x}{(1/y) - \log x} = \frac{y^2}{x(1 - y \log x)} = \frac{(x^{x^\cdots})^2}{x(1- x^{x^\cdots} \log x)}.$$

A plot of this function (recursively defined as such) is shown below: enter image description here

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  • $\begingroup$ Just for my understanding, your implicit definition of $y$ and the limit definition of $y$ do not coincide, right? By the limit definition I mean taking the power of $x$ repeatedly. $\endgroup$
    – Calculon
    Commented Nov 1, 2016 at 17:17

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