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(D) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.Given equation is $x^2 y^2 = \frac{\pi}{2}$.Differentiating both

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$-2$

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(D) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.Given equation is $x^2 y^2 = \frac{\pi}{2}$.Differentiating both sides with respect to $x$ using the product rule:$\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(\frac{\pi}{2})$$2x y^2 + x^2 (2y \frac{dy}{dx}) = 0$Divide by $2x$ (assuming $x 
eq 0$):$y^2 + xy \frac{dy}{dx} = 0$Substitute $x = 1$ and $y = 2$:$(2)^2 + (1)(2) \frac{dy}{dx} = 0$$4 + 2 \frac{dy}{dx} = 0$$2 \frac{dy}{dx} = -4$$\frac{dy}{dx} = -2$

List-$I$	List-$II$
$A. x^2 + y^2 + 3xy = 7$	$I. \frac{x^2 + ay}{ax + y^2}$
$B. x^{2/3} + y^{2/3} = a^{2/3}$	$II. \frac{-(2x + 3y)}{3x + 2y}$
$C. x^3 + y^3 = 3axy$	$III. -(\frac{y}{x})^{1/3}$
$D. xy(x - y) = 2$	$IV. \frac{x^2 - ay}{ax - y^2}$
	$V. \frac{-y(2x + y)}{x(x + 2y)}$

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

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