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Question Number 66163 by aliesam last updated on 09/Aug/19

Answered by mr W last updated on 09/Aug/19

$$y=x^{x^{x^{....}}}$$$$y=x^y$$$$\ln y=y\ln x$$$$\frac{1}{y}\times \frac{dy}{dx}=\frac{y}{x}+\ln x\frac{dy}{dx}$$$$(1-y\ln x)\frac{dy}{dx}=\frac{y^2}{x}$$$$\Rightarrow \frac{dy}{dx}=\frac{y^2}{x(1-y\ln x)}$$$$\Rightarrow \frac{dy}{dx}=\frac{{\left(x^{x^{x^{...}}}\right)}^2}{x[1-\left(x^{x^{x^{...}}}\right)\ln x]}$$

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