If $f(x)=x^{x^{x ... \infty}}$ then $f'(x)=$

$\frac{(f(x))^2}{x(1-f(x) \log x)}$

The correct answer is Option (2) → $\frac{(f(x))^2}{x(1-f(x) \log x)}$

$y = x^{x^{x^{\cdot^{\cdot}}}}$

$y = x^y$

$\ln y = y \ln x$

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

$\frac{1}{y}\frac{dy}{dx} - \frac{dy}{dx}\ln x = \frac{y}{x}$

$\frac{dy}{dx}\left(\frac{1}{y} - \ln x\right) = \frac{y}{x}$

$\frac{dy}{dx} = \frac{y}{x\left(\frac{1}{y} - \ln x\right)}$

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

$f'(x) = \frac{y^2}{x(1 - y\ln x)} \text{ where } y = x^{x^{x^{\cdot^{\cdot}}}}$