Question: If $x=\sqrt{2^{\operatorname{cosec}^{-1} t}}$ and $y=\sqrt{2^{\sec ^{-1} t}}(|t| \geq 1)$, then $\frac{d y}{d x}$ is equal to :
$-\frac{y}{x}$
$\frac{9}{1+9 x^{3}}$
$-\frac{x}{y}$
$\frac{y}{x}$
Correct Option: 1
Solution:
Solution not required
