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)

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