Question

If $y=x^{x^{x^{{\cdots }^{\infty }}}}$, find $\frac{dy}{dx}$

Answer 1

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

$y=x^y$

Taking $\log$ both sides, we have

$\log y=\log \left(x^y\right)$

$\log y=y\log x$

Differentiating above equation, w.r.t. $x$, we get

$\frac{1}{y}\frac{dy}{dx}=\frac{dy}{dx}.\log x+y.\frac{1}{x}$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}-\log x.\frac{dy}{dx}=\frac{y}{x}$

$\Rightarrow \frac{dy}{dx}(\frac{1}{y}-\log x)=\frac{y}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{y^2}{x(1-y\log x)}$

Hence $\frac{dy}{dx}=\frac{y^2}{x(1-y\log x)}$.