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If x1+y+y1+x=0,x≠yx\sqrt{1+y}+y\sqrt{1+x}=0, x\neq y, then dydx\frac{dy}{dx} is equal to

Solution

✅ Correct Option: 3

Shift and square: x2(1+y)=y2(1+x)x^2(1+y)=y^2(1+x), which factors as (x−y)(x+y+xy)=0(x-y)(x+y+xy)=0.

As x≠yx\neq y, x+y+xy=0x+y+xy=0, so y=−x1+xy=\frac{-x}{1+x}.

Differentiating, dydx=−(1+x)−x(1+x)2=−1(1+x)2\frac{dy}{dx}=-\frac{(1+x)-x}{(1+x)^2}=-\frac{1}{(1+x)^2}.

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