Q. If $y = \sqrt{x\log_{e}{x,\ }}$then $\frac{dy}{dx}\ at\ x = e$ is
A) $\frac{1}{e}$
B) $\frac{1}{\sqrt{e}}$
C) $\sqrt{e}$
D) $e^{2}$
Explanation
(b)
Given, $y = \sqrt{x\log_{e}x}$
$\Rightarrow \ \ \frac{dy}{dx} = \frac{1}{\ 2\sqrt{x\log_{e}x}}\left( x \times \frac{1}{x} + 1 \times \log_{e}x \right)$
$\therefore\ \ \ \ \left( \frac{dy}{dx} \right)_{x = e} = \frac{1}{2\sqrt{e \times 1}}(1 + 1)\ \ \$
$= \ \frac{1}{\sqrt{e}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.