Question

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

• A. $y{x}^{y-1}$
• B. $\frac{y^2}{x(1-ylnx)}$
• C. $\frac{y}{x(1+ylnx)}$
• 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
$lny=ylnx$
$\Rightarrow \frac{1}{y}.\frac{dy}{dx}=y.\frac{1}{x}+\frac{dy}{dx}.lnx$
$\Rightarrow \frac{dy}{dx}\left[\frac{1}{y}-lnx\right]=\frac{y}{x}$
$\Rightarrow \frac{dy}{dx}=\frac{y^2}{x(1-ylnx)}$