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Question Number 147135 by puissant last updated on 18/Jul/21

Answered by gsk2684 last updated on 18/Jul/21

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

Commented by puissant last updated on 18/Jul/21

$$\mathrm{thanks}$$

Answered by puissant last updated on 18/Jul/21

$$\mathrm{posons}\mathrm{y}={\mathrm{x}}^{{\mathrm{x}}^{{\mathrm{x}}^{\mathrm{x}..}}}\Rightarrow \mathrm{y}={\mathrm{x}}^{\mathrm{y}}$$$$\Rightarrow \ln \left(\mathrm{y}\right)=\mathrm{yln}\left(\mathrm{x}\right)$$$$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\times \frac{1}{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dx}}\ln \left(\mathrm{x}\right)+\frac{\mathrm{y}}{\mathrm{x}}$$$$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dx}}\ln \left(\mathrm{x}\right)\times \mathrm{y}+\frac{{\mathrm{y}}^2}{\mathrm{x}}$$$$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}(1-\mathrm{yln}\left(\mathrm{x}\right))=\frac{{\mathrm{y}}^2}{\mathrm{x}}$$$$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{{\mathrm{x}}^{{\mathrm{x}}^{{.}^{.}}}}\right)=\frac{{\left({\mathrm{x}}^{{\mathrm{x}}^{{\mathrm{x}}^{{\mathrm{x}}^{.}}}}\right)}^2}{\mathrm{x}(1-{\mathrm{x}}^{{\mathrm{x}}^{{\mathrm{x}}^{{\mathrm{x}}^{\mathrm{x}}}}}\ln \left(\mathrm{x}\right))}..$$

Answered by mr W last updated on 18/Jul/21

$$y=x^y$$$$y^{\frac{1}{y}}=x$$$${\left(\frac{1}{y}\right)}^{\frac{1}{y}}=\frac{1}{x}$$$$\frac{1}{y}\ln \frac{1}{y}=\ln \frac{1}{x}$$$$e^{\ln \frac{1}{y}}\ln \frac{1}{y}=\ln \frac{1}{x}$$$$\ln \frac{1}{y}=W\left(\ln \frac{1}{x}\right)$$$$\frac{1}{y}=e^{W\left(\ln \frac{1}{x}\right)}=\frac{\ln \frac{1}{x}}{W\left(\ln \frac{1}{x}\right)}$$$$\Rightarrow y=-\frac{W(-\ln x)}{\ln x}$$$$$$$$y=x^y$$$$\ln y=y\ln x$$$$\frac{y^{\prime }}{y}=y^{\prime }\ln x-\frac{y}{x}$$$$\Rightarrow y^{\prime }=\frac{y}{x(\ln x-\frac{1}{y})}=-\frac{W(-\ln x)}{x{\ln }^{2}x(1+\frac{1}{W(-\ln x)})}$$

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