If x = 2 ^ cosec ^ - 1 t and y = 2 ^ ^ - 1 t , ( |t| ≥ 1), then...

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

$x \cdot y = 2\dfrac{\sec^{- 1}t + {cosec}^{- 1}t}{2}$

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

$⇒ xy = 2\dfrac{\pi}{4}$

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

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

$⇒ \dfrac{dy}{dx} = - \dfrac{y}{x} .$