If x sqrt{1+y}+y sqrt{1+x}=0, then frac{d y}{ | vedclass

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(B) Given that $x \sqrt{1+y} = -y \sqrt{1+x}$.Squaring both sides, we get $x^2(1+y) = y^2(1+x)$.Rearranging the terms: $x^2 - y^2 + x^2y - xy^2 = 0$.F

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

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(B) Given that $x \sqrt{1+y} = -y \sqrt{1+x}$.Squaring both sides, we get $x^2(1+y) = y^2(1+x)$.Rearranging the terms: $x^2 - y^2 + x^2y - xy^2 = 0$.Factoring gives $(x-y)(x+y) + xy(x-y) = 0$, which implies $(x-y)(x+y+xy) = 0$.Since $x-y 
eq 0$ (as it does not satisfy the original equation), we must have $x+y+xy = 0$.Solving for $y$: $y(1+x) = -x$, so $y = -\frac{x}{1+x}$.Differentiating with respect to $x$ using the quotient rule:$\frac{dy}{dx} = -\frac{(1+x)(1) - x(1)}{(1+x)^2} = -\frac{1+x-x}{(1+x)^2} = -\frac{1}{(1+x)^2}$.

If $x \sqrt{1+y}+y \sqrt{1+x}=0$, then $\frac{d y}{d x}$ is equal to

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