If $x=\sqrt{2^{\operatorname{cosec}^{-1} t}}$ and $y=\sqrt{2^{\sec ^{-1} t}},|t| \geq 1$ then $\frac{d y}{d x}=$

• A

  $\frac{x}{y}$
• B

  $\frac{y}{x}$
• Correct answer
  C

  $-\frac{y}{x}$
• D

  $-\frac{x}{y}$

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