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y-x-x-x-gt-dy-dx-




Question Number 195020 by tri26112004 last updated on 22/Jul/23
y=x^x^(x...)    => (dy/dx)=¿
$${y}={x}^{{x}^{{x}…} } \\ $$$$=>\:\frac{{dy}}{{dx}}=¿ \\ $$
Answered by Frix last updated on 22/Jul/23
y=x^x^(x...)    ln y =x^x^(x...)  ln x  ln y =yln x  (1/y)dy=ln x dy+(y/x)dx  ((1−yln x)/y)dy=(y/x)dx  (dy/dx)=(y^2 /(x(1−yln x)))
$${y}={x}^{{x}^{{x}…} } \\ $$$$\mathrm{ln}\:{y}\:={x}^{{x}^{{x}…} } \mathrm{ln}\:{x} \\ $$$$\mathrm{ln}\:{y}\:={y}\mathrm{ln}\:{x} \\ $$$$\frac{\mathrm{1}}{{y}}{dy}=\mathrm{ln}\:{x}\:{dy}+\frac{{y}}{{x}}{dx} \\ $$$$\frac{\mathrm{1}−{y}\mathrm{ln}\:{x}}{{y}}{dy}=\frac{{y}}{{x}}{dx} \\ $$$$\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} }{{x}\left(\mathrm{1}−{y}\mathrm{ln}\:{x}\right)} \\ $$

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