If $x=\sqrt{2^{{\mathrm{cosec}}^{-1}t}}$ and  $y=\sqrt{2^{{\sec }^{-1}t}}, \left(\left|t\right|\ge 1\right),$ then $\frac{dy}{dx}$ is equal to

Given, $x=\sqrt{2^{{\mathrm{cosec}}^{-1}t}}=2^{\frac{{\mathrm{cosec}}^{-1}t}{2}}$ and  $y=\sqrt{2^{{\sec }^{-1}t}}=2^{\frac{{\sec }^{-1}t}{2}},$ then we have

$x·y=2^{\frac{{\sec }^{-1}t+{\mathrm{cosec}}^{-1}t}{2}}$

And, we know that ${\sec }^{-1}x+{\mathrm{cosec}}^{-1}x=\frac{\mathrm{\pi }}{2},$

$\Rightarrow xy=2^{\frac{\pi }{4}}$

On differentiating, both sides with respect to $x,$ and applying product rule, we get

$x\frac{dy}{dx}+y=0$

$\Rightarrow \frac{dy}{dx}=-\frac{y}{x}.$