If $x>0, x^y=e^{x-y}$, then $\frac{d y}{d x}$ is equal to

• A

  $\frac{1}{(1+\log x)^2}$
• Correct answer
  B

  $\frac{\log x}{(1+\log x)^2}$
• C

  $\left(\frac{\log x}{1+\log x}\right)^2$
• D

  $\frac{(\log x)^2}{1+\log x}$

Hence, the correct answer is option: 'B'.