quant interview question from 1996 at Banc One in Columbus Ohio 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 $y=x^{x^{x^{.^{.^{.}}}}}$

 A: 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
$$\frac{dy}{dx} = \frac{y^2}{x-xy\log{x}}$$
A: 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 $\frac{dy}{dx}=\frac{yx^{y-1}}{1-x^y\ln(x)}$.
A: 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:

