Question

Solve :
$y=x^{x^{x^{....\infty }}}$ ,then $\frac{dy}{dx}$

Answer 1

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

$\to y=x^y$

$\to \log y=y\log x$

$\to \frac{logy}{y}=\log x$

on differentiating, we get,

$\to \frac{dy}{dx}\frac{1-logy}{y^2}=\frac{1}{x}$

$\to \frac{dy}{dx}=\frac{y^2}{(1-logy)\left(x\right)}$