If $\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=2$,then $\frac{d y}{d x}$ is equal to

  • A
    $\frac{x^2+y^2}{x+y}$
  • B
    $\frac{x^2-y^2}{x+y}$
  • C
    $1$
  • D
    $2$

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