Question

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

• A. $\frac{y^2}{x(1-y\log x)}$
• B. $\frac{y}{x(1-\log x)}$
• C. $\frac{y^2}{x(y-\log x)}$
• D. None of these

Answer 1

The correct option is A $\frac{y^2}{x(1-y\log x)}$

The given function may be written as,
$y=x^y$
Taking $\log$ on both sides,
$\log y=\log x^y$
$\log y=y\log x$
Differentiate w.r. to $x$
$\frac{1}{y}\frac{dy}{dx}=y\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}y$
$\frac{1}{y}\frac{dy}{dx}=\frac{y}{x}+\log x.\frac{dy}{dx}$
$\frac{1}{y}\frac{dy}{dx}-\log x.\frac{dy}{dx}=\frac{y}{x}$
$\frac{dy}{dx}(\frac{1}{y}-\log x)=\frac{y}{x}$
$\frac{dy}{dx}\left(\frac{1-y\log x}{y}\right)=\frac{y}{x}$
$\therefore \frac{dy}{dx}=\frac{y^2}{x(1-y\log x)}$