Question

If $y={\sqrt{x}}^{{\sqrt{x}}^{{\sqrt{x}}^{\sqrt{x}}...\infty }}$ then show that, $x\frac{dy}{dx}=\frac{y^2}{2-y\log x}$

Answer 1

$y={\sqrt{x}}^{{\sqrt{x}}^{{\sqrt{x}}^{\sqrt{x}}...\infty }}$

$\Rightarrow y=(\sqrt{x}{)}^y$
$\therefore \log y=y\log \sqrt{x}=\frac{1}{2}y\log x.$
Now differentiating both side w.r.t $x$
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}.(\frac{dy}{dx}\log x+y.\frac{1}{x})$
$\Rightarrow (\frac{1}{y}-\frac{1}{2}\log x)\frac{dy}{dx}=\frac{1}{2}.\frac{y}{x}$

$\therefore x\frac{dx}{dy}=\frac{y^2}{2-y\log x}$