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. $e^2$

Answer 1

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

Given that, $y=\sqrt{x{\log }_{e}x}$,

$\frac{dy}{dx}=\frac{1}{\sqrt{xlnx}}(x\frac{1}{x}+lnx.1)$

$\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{xlnx}}(1+lnx)$

At x=e ; $\frac{dy}{dx}=\frac{1}{2\sqrt{e.1}}(1+1)=\frac{1}{\sqrt{e}}$