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Question-147135




Question Number 147135 by puissant last updated on 18/Jul/21
Answered by gsk2684 last updated on 18/Jul/21
let y=x^x^x^.    then y=x^y   ln y = y ln x  (1/y) (dy/dx)=y (1/x) + (dy/dx) ln x  ((1/y)−ln x) (dy/dx)=(y/x)  (dy/dx)=(y^2 /(x(1−y ln x)))
$${let}\:{y}={x}^{{x}^{{x}^{.} } } \:{then}\:{y}={x}^{{y}} \\ $$$$\mathrm{ln}\:{y}\:=\:{y}\:\mathrm{ln}\:{x} \\ $$$$\frac{\mathrm{1}}{{y}}\:\frac{{dy}}{{dx}}={y}\:\frac{\mathrm{1}}{{x}}\:+\:\frac{{dy}}{{dx}}\:\mathrm{ln}\:{x} \\ $$$$\left(\frac{\mathrm{1}}{{y}}−\mathrm{ln}\:{x}\right)\:\frac{{dy}}{{dx}}=\frac{{y}}{{x}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} }{{x}\left(\mathrm{1}−{y}\:\mathrm{ln}\:{x}\right)} \\ $$
Commented by puissant last updated on 18/Jul/21
thanks
$$\mathrm{thanks} \\ $$
Answered by puissant last updated on 18/Jul/21
posons  y=x^x^x^(x..)   ⇒ y=x^y   ⇒ln(y)=yln(x)  ⇒(dy/dx)×(1/y)=(dy/dx)ln(x)+(y/x)  ⇒(dy/dx)=(dy/dx)ln(x)×y+(y^2 /x)  ⇒(dy/dx)(1−yln(x))=(y^2 /x)  ⇒(d/dx)(x^x^x^.^.    )=(((x^x^x^x^.    )^2 )/(x(1−x^x^x^x^x    ln(x))))..
$$\mathrm{posons}\:\:\mathrm{y}=\mathrm{x}^{\mathrm{x}^{\mathrm{x}^{\mathrm{x}..} } } \Rightarrow\:\mathrm{y}=\mathrm{x}^{\mathrm{y}} \\ $$$$\Rightarrow\mathrm{ln}\left(\mathrm{y}\right)=\mathrm{yln}\left(\mathrm{x}\right) \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}×\frac{\mathrm{1}}{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{ln}\left(\mathrm{x}\right)+\frac{\mathrm{y}}{\mathrm{x}} \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{ln}\left(\mathrm{x}\right)×\mathrm{y}+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}} \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{1}−\mathrm{yln}\left(\mathrm{x}\right)\right)=\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}} \\ $$$$\Rightarrow\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{x}^{\mathrm{x}^{.^{.} } } } \right)=\frac{\left(\mathrm{x}^{\mathrm{x}^{\mathrm{x}^{\mathrm{x}^{.} } } } \right)^{\mathrm{2}} }{\mathrm{x}\left(\mathrm{1}−\mathrm{x}^{\mathrm{x}^{\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } } } \mathrm{ln}\left(\mathrm{x}\right)\right)}.. \\ $$
Answered by mr W last updated on 18/Jul/21
y=x^y   y^(1/y) =x  ((1/y))^(1/y) =(1/x)  (1/y)ln (1/y)=ln (1/x)  e^(ln (1/y)) ln (1/y)=ln (1/x)  ln (1/y)=W(ln (1/x))  (1/y)=e^(W(ln (1/x))) =((ln (1/x))/(W(ln (1/x))))  ⇒y=−((W(−ln x))/(ln x))    y=x^y   ln y=yln x  ((y′)/y)=y′ln x−(y/x)  ⇒y′=(y/(x(ln x−(1/y))))=−((W(−ln x))/(xln^2  x(1+(1/(W(−ln x))))))
$${y}={x}^{{y}} \\ $$$${y}^{\frac{\mathrm{1}}{{y}}} ={x} \\ $$$$\left(\frac{\mathrm{1}}{{y}}\right)^{\frac{\mathrm{1}}{{y}}} =\frac{\mathrm{1}}{{x}} \\ $$$$\frac{\mathrm{1}}{{y}}\mathrm{ln}\:\frac{\mathrm{1}}{{y}}=\mathrm{ln}\:\frac{\mathrm{1}}{{x}} \\ $$$${e}^{\mathrm{ln}\:\frac{\mathrm{1}}{{y}}} \mathrm{ln}\:\frac{\mathrm{1}}{{y}}=\mathrm{ln}\:\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{ln}\:\frac{\mathrm{1}}{{y}}={W}\left(\mathrm{ln}\:\frac{\mathrm{1}}{{x}}\right) \\ $$$$\frac{\mathrm{1}}{{y}}={e}^{{W}\left(\mathrm{ln}\:\frac{\mathrm{1}}{{x}}\right)} =\frac{\mathrm{ln}\:\frac{\mathrm{1}}{{x}}}{{W}\left(\mathrm{ln}\:\frac{\mathrm{1}}{{x}}\right)} \\ $$$$\Rightarrow{y}=−\frac{{W}\left(−\mathrm{ln}\:{x}\right)}{\mathrm{ln}\:{x}} \\ $$$$ \\ $$$${y}={x}^{{y}} \\ $$$$\mathrm{ln}\:{y}={y}\mathrm{ln}\:{x} \\ $$$$\frac{{y}'}{{y}}={y}'\mathrm{ln}\:{x}−\frac{{y}}{{x}} \\ $$$$\Rightarrow{y}'=\frac{{y}}{{x}\left(\mathrm{ln}\:{x}−\frac{\mathrm{1}}{{y}}\right)}=−\frac{{W}\left(−\mathrm{ln}\:{x}\right)}{{x}\mathrm{ln}^{\mathrm{2}} \:{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{W}\left(−\mathrm{ln}\:{x}\right)}\right)} \\ $$

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