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Question Number 195020 by tri26112004 last updated on 22/Jul/23

$$y=x^{x^{x...}}$$$$=>\frac{dy}{dx}=¿$$

Answered by Frix last updated on 22/Jul/23

$$y=x^{x^{x...}}$$$$\ln y=x^{x^{x...}}\ln x$$$$\ln y=y\ln x$$$$\frac{1}{y}dy=\ln xdy+\frac{y}{x}dx$$$$\frac{1-y\ln x}{y}dy=\frac{y}{x}dx$$$$\frac{dy}{dx}=\frac{y^2}{x(1-y\ln x)}$$

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