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

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(A) Given equation is $x \sqrt{1+y} + y \sqrt{1+x} = 0$. Rearranging the terms,we get $x \sqrt{1+y} = -y \sqrt{1+x}$. Squaring both sides,we get $x^2(

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

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(A) Given equation is $x \sqrt{1+y} + y \sqrt{1+x} = 0$. Rearranging the terms,we get $x \sqrt{1+y} = -y \sqrt{1+x}$. Squaring both sides,we get $x^2(1+y) = y^2(1+x)$. Expanding this,$x^2 + x^2y = y^2 + xy^2$. Rearranging terms,$x^2 - y^2 + x^2y - xy^2 = 0$. Factoring,$(x-y)(x+y) + xy(x-y) = 0$. Since $x 
eq y$ (as $x \sqrt{1+y} = -y \sqrt{1+x}$ implies $x$ and $y$ have opposite signs),we can divide by $(x-y)$ to get $x+y+xy = 0$. Solving for $y$,$y(1+x) = -x$,so $y = \frac{-x}{1+x}$. Differentiating with respect to $x$ using the quotient rule,$\frac{dy}{dx} = \frac{(1+x)(-1) - (-x)(1)}{(1+x)^2}$. Simplifying,$\frac{dy}{dx} = \frac{-1-x+x}{(1+x)^2} = \frac{-1}{(1+x)^2}$.

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

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