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

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(B) Given that $x \sqrt{1+y} + y \sqrt{1+x} = 0$ ... $(i)$ Rearranging the terms,we get $x \sqrt{1+y} = -y \sqrt{1+x}$ On squaring both sides,we get $

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

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(B) Given that $x \sqrt{1+y} + y \sqrt{1+x} = 0$ ... $(i)$ Rearranging the terms,we get $x \sqrt{1+y} = -y \sqrt{1+x}$ On squaring both sides,we get $x^2(1+y) = y^2(1+x)$ $x^2 + x^2y = y^2 + xy^2$ $x^2 - y^2 + x^2y - xy^2 = 0$ $(x-y)(x+y) + xy(x-y) = 0$ $(x-y)(x+y+xy) = 0$ Since $x-y 
eq 0$ (as it does not satisfy the original equation),we must have $x+y+xy = 0$ $y(1+x) = -x$ $y = -\frac{x}{1+x}$ Now,differentiating with respect to $x$ using the quotient rule: $\frac{dy}{dx} = -\frac{(1+x)\frac{d}{dx}(x) - x\frac{d}{dx}(1+x)}{(1+x)^2}$ $\frac{dy}{dx} = -\frac{(1+x)(1) - x(1)}{(1+x)^2}$ $\frac{dy}{dx} = -\frac{1+x-x}{(1+x)^2}$ $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$

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

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