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 (1/y)×(dy/dx)=(y/x)+ln x (dy/dx) (1−y ln x)(dy/dx)=(y^2 /x) ⇒(dy/dx)=(y^2 /(x(1−y ln x))) ⇒(dy/dx)=(((x^x^x^(...) )^2 )/(x[1−(x^x^x^(...) ) ln x]))](Q66164.png)
$${y}={x}^{{x}^{{x}^{....} } } \\ $$$${y}={x}^{{y}} \\ $$$$\mathrm{ln}\:{y}={y}\:\mathrm{ln}\:{x} \\ $$$$\frac{\mathrm{1}}{{y}}×\frac{{dy}}{{dx}}=\frac{{y}}{{x}}+\mathrm{ln}\:{x}\:\frac{{dy}}{{dx}} \\ $$$$\left(\mathrm{1}−{y}\:\mathrm{ln}\:{x}\right)\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} }{{x}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} }{{x}\left(\mathrm{1}−{y}\:\mathrm{ln}\:{x}\right)} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\left({x}^{{x}^{{x}^{...} } } \right)^{\mathrm{2}} }{{x}\left[\mathrm{1}−\left({x}^{{x}^{{x}^{...} } } \right)\:\mathrm{ln}\:{x}\right]} \\ $$