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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$

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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