Question Number 75034 by chess1 last updated on 06/Dec/19
Answered by Kunal12588 last updated on 06/Dec/19
![y=x^x^x^x log y =x^x^x log x log (log y) = x^x log x + log(log x) (1/(y log y)) (dy/dx)=x^x (1+log x)log x+x^(x−1) +(1/(x log x)) (dy/dx)=x^(x^x^x +x^x ) log x [x^x (1+log x)log x + x^(x−1) +(1/(x log x)) ]](https://www.tinkutara.com/question/Q75036.png)
$${y}={x}^{{x}^{{x}^{{x}} } } \\ $$$${log}\:{y}\:={x}^{{x}^{{x}} } {log}\:{x} \\ $$$${log}\:\left({log}\:{y}\right)\:=\:{x}^{{x}} {log}\:{x}\:+\:{log}\left({log}\:{x}\right) \\ $$$$\frac{\mathrm{1}}{{y}\:{log}\:{y}}\:\frac{{dy}}{{dx}}={x}^{{x}} \left(\mathrm{1}+{log}\:{x}\right){log}\:{x}+{x}^{{x}−\mathrm{1}} +\frac{\mathrm{1}}{{x}\:{log}\:{x}}\: \\ $$$$\frac{{dy}}{{dx}}={x}^{{x}^{{x}^{{x}} } +{x}^{{x}} } {log}\:{x}\:\left[{x}^{{x}} \left(\mathrm{1}+{log}\:{x}\right){log}\:{x}\:+\:{x}^{{x}−\mathrm{1}} \:+\frac{\mathrm{1}}{{x}\:{log}\:{x}}\:\right] \\ $$
Answered by Kunal12588 last updated on 06/Dec/19
![y=x^x^x^x^x log y = x^x^x^x log x ⇒(1/y)(dy/dx)=x^(x^x^x −1) +log x (d/dx)(x^x^x^x ) ⇒(dy/dx)=x^x^x^x^3 (x^(x^x^x −1) +log x(x^(x^x^x +x^x ) log x [x^x (1+log x)log x + x^(x−1) +(1/(x log x)) ])) ⇒(dy/dx)=x^x^x^x^3 (x^(x^x^x −1) +x^(x^x^x +x^x ) (log x)^2 [x^x (1+log x)log x + x^(x−1) +(1/(x log x)) ])](https://www.tinkutara.com/question/Q75037.png)
$${y}={x}^{{x}^{{x}^{{x}^{{x}} } } } \\ $$$${log}\:{y}\:=\:{x}^{{x}^{{x}^{{x}} } } {log}\:{x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}={x}^{{x}^{{x}^{{x}} } −\mathrm{1}} +{log}\:{x}\:\frac{{d}}{{dx}}\left({x}^{{x}^{{x}^{{x}} } } \right) \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={x}^{{x}^{{x}^{{x}^{\mathrm{3}} } } } \left({x}^{{x}^{{x}^{{x}} } −\mathrm{1}} +{log}\:{x}\left({x}^{{x}^{{x}^{{x}} } +{x}^{{x}} } {log}\:{x}\:\left[{x}^{{x}} \left(\mathrm{1}+{log}\:{x}\right){log}\:{x}\:+\:{x}^{{x}−\mathrm{1}} \:+\frac{\mathrm{1}}{{x}\:{log}\:{x}}\:\right]\right)\right) \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={x}^{{x}^{{x}^{{x}^{\mathrm{3}} } } } \left({x}^{{x}^{{x}^{{x}} } −\mathrm{1}} +{x}^{{x}^{{x}^{{x}} } +{x}^{{x}} } \left({log}\:{x}\right)^{\mathrm{2}} \:\left[{x}^{{x}} \left(\mathrm{1}+{log}\:{x}\right){log}\:{x}\:+\:{x}^{{x}−\mathrm{1}} \:+\frac{\mathrm{1}}{{x}\:{log}\:{x}}\:\right]\right) \\ $$
Answered by Kunal12588 last updated on 06/Dec/19
$${y}={x}^{{x}^{…} } \\ $$$$\Rightarrow{y}={x}^{{y}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={x}^{{y}} \left(\frac{{y}}{{x}}+{log}\:{x}\frac{{dy}}{{dx}}\right) \\ $$$$\Rightarrow\left(\mathrm{1}−{x}^{{y}} {log}\:{x}\right)\frac{{dy}}{{dx}}={x}^{{y}−\mathrm{1}} {y} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\frac{{x}^{{y}} }{{x}}{y}}{\mathrm{1}−{y}\:{log}\:{x}}=\frac{{y}^{\mathrm{2}} }{{x}\left(\mathrm{1}−{ylog}\:{x}\right)} \\ $$$$=\frac{{x}^{\mathrm{2}{x}^{{x}^{….} } −\mathrm{1}} }{\left(\mathrm{1}−{x}^{{x}^{{x}^{…} } } {log}\:{x}\right)} \\ $$
Commented by chess1 last updated on 06/Dec/19
$$\mathrm{thanks} \\ $$