If x=sqrt{2^{operatorname{cosec}^{-1} t}} and | vedclass

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(C) Given $x = \sqrt{2^{\operatorname{cosec}^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$.Squaring both sides,we get $x^2 = 2^{\operatorname{cosec}^{-1}

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$-\frac{y}{x}$

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(C) Given $x = \sqrt{2^{\operatorname{cosec}^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$.Squaring both sides,we get $x^2 = 2^{\operatorname{cosec}^{-1} t}$ and $y^2 = 2^{\sec^{-1} t}$.Multiplying the two equations: $x^2 y^2 = 2^{\operatorname{cosec}^{-1} t + \sec^{-1} t}$.Since $\operatorname{cosec}^{-1} t + \sec^{-1} t = \frac{\pi}{2}$ for $|t| \geq 1$,we have $x^2 y^2 = 2^{\pi/2}$.Differentiating both sides with respect to $x$: $\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(2^{\pi/2})$.$x^2 (2y \frac{dy}{dx}) + y^2 (2x) = 0$.$2x^2 y \frac{dy}{dx} = -2xy^2$.$\frac{dy}{dx} = -\frac{2xy^2}{2x^2 y} = -\frac{y}{x}$.

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

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