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This question already has an answer here:

How to find the domain and range of $Y=x^{x^{x^{x^{x^{...}}}}}=x^Y$

I know how to differentiate the function. I don't know how to proceed further.

We should prove that domain:$[1/(e^e),e^{(1/e)}]$ and range:$[1/e,e]$

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marked as duplicate by Peter Foreman, max_zorn, callculus, Leucippus, José Carlos Santos Apr 3 at 7:50

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As you said: $$y=x^y$$ and so: $$\ln(y)=y\ln(x)$$ $$\ln(x)=\frac{\ln(y)}{y}\tag{1}$$ now if we take: $$L_1=\lim_{y\to\infty}\frac{\ln(y)}{y}=\lim_{y\to\infty}\frac{1}{y}=0$$ If you mean for a given domain that is the range, we can work backwards from $(1)$ to get the domain from the range. Differentiating we get: $$\frac1x=\left(\frac{1}{y^2}-\frac{\ln(y)}{y^2}\right)\frac{dy}{dx}$$ $$\frac{dy}{dx}=\frac{y^2}{x(1-\ln y)}$$ so our minimum/maximum will occur when $y=0$, but to work out $x$ you must evaluate: $$x=\exp\left(\lim_{y\to0}\frac{\ln(y)}{y}\right)$$

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