If y = sqrt{(1 - x)(1 + x)},then | vedclass

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(B) Given $y = \sqrt{(1 - x)(1 + x)}$.Squaring both sides,we get $y^2 = (1 - x)(1 + x) = 1 - x^2$.Differentiating both sides with respect to $x$,we ge

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$(1 - x^2)\frac{dy}{dx} + xy = 0$

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(B) Given $y = \sqrt{(1 - x)(1 + x)}$.Squaring both sides,we get $y^2 = (1 - x)(1 + x) = 1 - x^2$.Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(y^2) = \frac{d}{dx}(1 - x^2)$.Using the chain rule,$2y \frac{dy}{dx} = -2x$.Dividing by $2$,we get $y \frac{dy}{dx} = -x$.Multiply both sides by $y$,we get $y^2 \frac{dy}{dx} = -xy$.Since $y^2 = 1 - x^2$,substituting this gives $(1 - x^2) \frac{dy}{dx} = -xy$.Rearranging the terms,we get $(1 - x^2) \frac{dy}{dx} + xy = 0$.

If $y = \sqrt{(1 - x)(1 + x)}$,then

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