Question

Let $y=x^{x^{\cdots-\infty}}$, then $\frac{d y}{d x}$ is equal to

(A) $y x^{y-1}$

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

(C) $\frac{y}{x(1+y \log x)}$

(D) None of these

Answer 1

$\begin{array}{lrl}\text {  } \therefore  y=x^y \Rightarrow \log y=y \log x \\\Rightarrow \frac{1}{y} \frac{d y}{d x}=y \cdot \frac{1}{x}+\frac{d y}{d x} \log x \\\Rightarrow \frac{d y}{d x}\left[\frac{1}{y}-\log x\right] =\frac{y}{x} \\\Rightarrow  \frac{d y}{d x} =\frac{y^2}{x(1-y \log x)}\end{array}$

So, The correct option is (B).