Question

If $y = x^{x^{x....\infty} } , $ then $ x \frac{dy}{dx} $ is equal to

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

Answer 1

Answer: (A)

$y = x^{y} \Rightarrow \log y =y \log x $
$\Rightarrow \frac{1}{y} \frac{dy}{dx} = \frac{y}{x} + \log x \frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx} \left[ \frac{1-y\log x}{y}\right] = \frac{y}{x} $
$\Rightarrow x \frac{dy}{dx} = \frac{y^{2}}{1-y \log x} $