Question

If $\mathrm{y}=\sqrt{(\mathrm{x}{\log }_{\mathrm{e}}\mathrm{x})}$ then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=\mathrm{e}$ is

• A. $\frac{1}{e}$
• B. $\frac{1}{\sqrt{e}}$
• C. $\sqrt{e}$
• D. $e^2$

Answer 1

The correct option is B

$\frac{1}{\sqrt{e}}$

Finding the derivative:

Step 1: Find the $\frac{dy}{dx}$:

Given that,

$\mathrm{y}=\sqrt{(\mathrm{x}{\log }_{\mathrm{e}}\mathrm{x})}$

Differentiate the above equation with respect to $x$

$$\begin{gathered} \frac{dy}{dx}=\frac{1({\log }_{e}x+x\times {\displaystyle \frac{1}{x}})}{2\sqrt{(\mathrm{x}{\log }_{\mathrm{e}}\mathrm{x})}}\left[\because \frac{d\left(\log x\right)}{dx}=\frac{1}{x}\right] \\ =\frac{({\log }_{e}x+1)}{2\sqrt{(\mathrm{x}{\log }_{\mathrm{e}}\mathrm{x})}} \end{gathered}$$

Step2: Finding out the value of $\frac{dy}{dx}atx=e$:

$\begin{array}{rcl}{\left(\frac{dy}{dx}\right)}_{x=e}& =& \frac{({\log }_{e}e+1)}{2\sqrt{(e{\log }_{\mathrm{e}}e)}}\\ & =& \frac{2}{2\sqrt{e}}\\ & =& \frac{1}{\sqrt{e}}\end{array}$

Hence, B is the correct option.