Rate of growth of an infinite, forever-growing power tower

First, let us start by doing some substitutions: $\displaystyle y=x^{x^{x^{x^{x^{...}}}}}$ $\displaystyle \implies y=x^y$ Now, it is clear we need to find the derivative of this function. To ease the differentiation of an exponential function, we know we have to use some sort of $e^x$ expression, so we rewrite as follows: $\displaystyle y=e^{y\ln x}$ By implicit differentiation: $\frac{\text{d}y}{\text{d}x}= \frac{\text{d}}{\text{d}x} \displaystyle{[e^{y\ln x}]}$ Solving it completely gives us: $\dfrac{\text{d}y}{\text{d}x}=\dfrac{y^2}{x(1-\ln x\cdot y)}$ Substituting for $y$ is the final answer: $\boxed{\dfrac{\text{d}y}{\text{d}x}=\dfrac{\left(x^{x^{x^{x^{x^{...}}}}}\right)^2}{x(1-\ln x\cdot x^{x^{x^{x^{x^{...}}}}})}}$

$\begin{aligned} y & = x^{x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}} \\ \implies y & = x^y = e^{y\ln x} \\ \frac {dy}{dx} & = \blue{e^{y\ln x}} \cdot \frac d{dx} (y\ln x) \\ & = \blue y \left(\ln x \frac {dy}{dx} + \frac yx\right) \\ \frac {dy}{dx} - y \ln x \frac {dy}{dx} & = \frac {y^2}x \\ \implies \frac {dy}{dx} & = \frac {y^2}{x(1-y\ln x)} = \boxed{\frac {\big(x^{x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}}\big)^2}{x \big(1-\ln x \cdot x^{x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}}\big)}} \end{aligned}$