If $y=x^{x^{x^{\dots }}}$ to infinity, then $\frac{dy}{dx}$ is equal to:

Solution

1. The expression can be written as $y=x^y$.
2. Taking the natural logarithm on both sides, $\ln y=y\ln x$.
3. Differentiating both sides with respect to $x$, $\frac{1}{y}\frac{dy}{dx}=\frac{dy}{dx}\ln x+\frac{y}{x}$.
4. Rearranging terms, $\frac{dy}{dx}(\frac{1}{y}-\ln x)=\frac{y}{x}$.
5. Solving for $\frac{dy}{dx}$, we get $\frac{dy}{dx}=\frac{y^2}{x(1-y\ln x)}$. Hence the answer is (A).