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Question Number 136656 by 0731619177 last updated on 24/Mar/21

	Answered by Ñï= last updated on 24/Mar/21

	$y = x^{x^{x^{. . .}}} = x^{y}$ $y^{'} = {(e^{y l n x})}^{'} = x^{y} (y^{'} l n x + \frac{y}{x}) = x^{y} y^{'} l n x + {y x}^{y - 1}$ $\Rightarrow y^{'} (1 - x^{y} l n x) = {y x}^{y - 1}$ $\Rightarrow y^{'} = \frac{{y x}^{y - 1}}{1 - x^{y} l n x} = \frac{x^{x^{x^{. . .}}} x^{x^{x^{x^{. . . .}}} - 1}}{1 - x^{x^{x^{x^{. . . .}}}} l n x}$
	Commented by 0731619177 last updated on 24/Mar/21

	$t h a n k s$
	Commented by snipers237 last updated on 24/Mar/21

	$S i r w h y n o t h a v i n g y = x^{x^{y}} ? o r y = x^{x^{x^{y}}} ?$ $t h a t u n f i n i s h e d e x p r e s s i o n {c a n}^{'} t d e f i n e d a r e a l o r f i n i t e n u m b e r .$
	Commented by MJS_new last updated on 25/Mar/21

	$y = x^{x^{x . . .}}$ $\ln y = \ln x^{x^{x . . .}}$ $\ln y = x^{x^{x . . .}} \ln x$ $\ln y = y \ln x$ $which can be solved for 0 < x ⩽ 1$
	Commented by snipers237 last updated on 24/Mar/21

	$i s y a r e a l n u m b e r ? D o e s i t e x i s t a s a f i n i t e n u m b e r ?$ $t h e l o g a r i t h m p r o p e r t i e s h o l d o n l y f o r p o s i t i v e r e a l n u m b e r s$ $i k n o w y c a n b e d e f i n e d a s \lim_{n \to \infty} u_{n} w i t h u_{0} = x > 0 a n d u_{n + 1} = x^{u_{n}}$ $h e n t h a t c o n v e r g e s . {A r e n}^{'} t t h e r e v a l u e s o f x > 0 w h o s e l e a d (u_{n}) t o d i v e r g e ?$
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