Question

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. None of these

Answer 1

The correct option is B $\frac{1}{\sqrt{e}}$

$y=\sqrt{x\log x}$
$\Rightarrow \frac{dy}{dx}=\frac{\frac{d}{dx}\left(x\log x\right)}{2\sqrt{x\log x}}$
$=\frac{\log x+x\frac{1}{x}}{2\sqrt{x\log x}}=\frac{\log x+1}{2\sqrt{x\log x}}$
At $x=e$
$\frac{dy}{dx}=\frac{1}{\sqrt{e}}$