xsqrt{1 + y} + ysqrt{1 + x} = 0,then frac{dy} | vedclass

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(D) Given equation: $x\sqrt{1 + y} + y\sqrt{1 + x} = 0$Rearranging the terms: $x\sqrt{1 + y} = -y\sqrt{1 + x}$Squaring both sides: $x^2(1 + y) = y^2(1

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$-(1 + x)^{-2}$

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(D) Given equation: $x\sqrt{1 + y} + y\sqrt{1 + x} = 0$Rearranging the terms: $x\sqrt{1 + y} = -y\sqrt{1 + x}$Squaring both sides: $x^2(1 + y) = y^2(1 + x)$$x^2 + x^2y = y^2 + y^2x$$x^2 - y^2 + x^2y - y^2x = 0$$(x - y)(x + y) + xy(x - y) = 0$$(x - y)(x + y + xy) = 0$Since $x 
eq y$,we have $x + y + xy = 0$$y(1 + x) = -x$$y = -\frac{x}{1 + x}$Now,differentiate with respect to $x$ using the quotient rule:$\frac{dy}{dx} = -\frac{(1 + x)(1) - x(1)}{(1 + x)^2}$$\frac{dy}{dx} = -\frac{1 + x - x}{(1 + x)^2}$$\frac{dy}{dx} = -\frac{1}{(1 + x)^2} = -(1 + x)^{-2}$

$x\sqrt{1 + y} + y\sqrt{1 + x} = 0$,then $\frac{dy}{dx} = $

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