Q. Let $y = x^{x^{x\ldots}}$, then $\frac{dy}{dx\ }\$is equal to
A) $yx^{y - 1}$
B) $\frac{y^{2}}{x(1 - y\log{x)}}$
C) $\frac{y}{x(1 + y\log{x)}}$
D) None of these
Explanation
(b)
Given, $y = x^{y}$
$\Rightarrow {\ \ \ log}{y = y\log x}$
$\Rightarrow \ \frac{1}{y}.\frac{dy}{dx} = y.\frac{1}{x} + \frac{dy}{dx}.\log x\$
$\Rightarrow \frac{dy}{dx}\left\lbrack \frac{1}{y} - \log x \right\rbrack = \frac{y}{x}$
$\Rightarrow \ \frac{dy}{dx} = \frac{y^{2}}{x(1 - y\log{x)}}\ \$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.