Question

If $\mathrm{y}={\mathrm{x}}^{\mathrm{x}}$ and $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\left(\frac{1}{\mathrm{y}}\right){\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2$ then $=$

• A. ${\mathrm{x}}^{\mathrm{y}}$
• B. ${\mathrm{x}}^{\mathrm{x}}$
• C. ${\mathrm{y}}^{\mathrm{x}}$
• D. $\mathrm{x}$

Answer 1

The correct option is B

${\mathrm{x}}^{\mathrm{x}}$

Explanation for the correct answer:

Find the $\frac{dy}{dx}$

The given function is $\mathrm{y}={\mathrm{x}}^{\mathrm{x}}$

Taking $\log$on both the sides,

$\log \mathrm{y}=\mathrm{x}\cdot \log \mathrm{x}$

Differentiate with respect to x,

$\begin{array}{rcl}\frac{1}{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \log \mathrm{x}+\mathrm{x}\frac{1}{\mathrm{x}}\\ & =& 1+\log \mathrm{x}\end{array}$

$\begin{array}{rcl}\frac{{\displaystyle \mathrm{dy}}}{{\displaystyle \mathrm{dx}}}& =& {\mathrm{x}}^{\mathrm{x}}[1+\log \mathrm{x}]\end{array}$

Hence, the correct option is B.