Differentiation of $x^{{x}^{{x}^{...}}}$

Question

All of us experienced with this function : $$x^{x}$$ It's easy to differentiate it, so my question is: How to differentiate this function: $$x^{{x}^{...}}$$, maybe there is some recurrence equation? For example there is $n$, $x$ in "stairs". P.S. sorry for my English skills

Answer 1

Note that if $f(x) = x^{x^{x^{\cdots}}}$, we then have $$f(x) = x^{f(x)}$$ Taking the logarithm on both sides, we obtain $$\log(f(x)) = f(x) \log(x)$$ Differentiating we obtain $$\dfrac{f'(x)}{f(x)} = f'(x) \log(x) + \dfrac{f(x)}x$$ Now obtain $f'(x)$.

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If it is a finite stair, define $f_0(x) = x$, $f_1(x) = x^x$ and in general $f_{n+1}(x) = x^{f_n(x)}$, where $f_n(x)$ has $n$ $x$'s in the exponent. We can now compute the derivative inductively, taking the logarithm on both sides, we have $$\log\left(f_{n+1}(x) \right) = f_n(x) \log(x) \implies \dfrac{f_{n+1}'(x)}{f_{n+1}(x)} = f_n'(x) \log(x) + \dfrac{f_n(x)}x$$ Hence, we have $$f_{n+1}'(x) = f_{n+1}f_n'(x)\log(x) + \dfrac{f_{n+1}(x)f_n(x)}x$$ where we have $f_0'(x) = 1$, $f_1'(x) = x^x(1+\log(x))$.