If x^{2} y^{2} = sin^{-1} sqrt{x^{2} + y^{2}} | vedclass

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(A) We know that for any $	heta \in [-1, 1]$,$\sin^{-1} 	heta + \cos^{-1} 	heta = \frac{\pi}{2}$.Given the equation $x^{2} y^{2} = \sin^{-1} \sqrt{

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$\frac{-y}{x}$

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(A) We know that for any $	heta \in [-1, 1]$,$\sin^{-1} 	heta + \cos^{-1} 	heta = \frac{\pi}{2}$.Given the equation $x^{2} y^{2} = \sin^{-1} \sqrt{x^{2} + y^{2}} + \cos^{-1} \sqrt{x^{2} + y^{2}}$,we can simplify the right side to get $x^{2} y^{2} = \frac{\pi}{2}$.Differentiating both sides with respect to $x$ using the product rule:$\frac{d}{dx}(x^{2} y^{2}) = \frac{d}{dx}(\frac{\pi}{2})$$x^{2} \cdot (2y \frac{dy}{dx}) + y^{2} \cdot (2x) = 0$$2x^{2}y \frac{dy}{dx} = -2xy^{2}$Dividing both sides by $2xy$ (assuming $x, y 
eq 0$):$\frac{dy}{dx} = \frac{-2xy^{2}}{2x^{2}y} = -\frac{y}{x}$.

If $x^{2} y^{2} = \sin^{-1} \sqrt{x^{2} + y^{2}} + \cos^{-1} \sqrt{x^{2} + y^{2}}$,then $\frac{dy}{dx} = $

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