Question

Let $y=x^{x^{x \ldots \in fty}}$ , then $\frac{d y}{d x \, } \, $ is equal to (given $x > 0$ )

• A. $yx^{y - 1}$
• B. $\frac{y^{2}}{x \left(\right. 1 - y ln x \left.\right)}$
• C. $\frac{y}{x \left(\right. 1 + y ln x \left.\right)}$
• D. None of these

Answer 1

Answer: (B)

Given, $y = x^{x^{x ... \in fty}}$ then $y = x^{y}$
Taking $ln$ on both the sides, we get
$ln y = y ln x$
$\Rightarrow \, \frac{1}{y}.\frac{d y}{d x}=y.\frac{1}{x}+\frac{d y}{d x}.ln x$
$\Rightarrow \frac{d y}{d x}\left[\frac{1}{y} - ln x\right]=\frac{y}{x}$
$\Rightarrow \, \frac{d y}{d x}=\frac{y^{2}}{x \left(\right. 1 - y ln x \left.\right)}$