Given x1+y+y1+x=0 ⇒x1+y=-y1+x ⇒x2(1+y)=y21+x ⇒x2+x2y=y2+xy2 ⇒x2-y2+x2y-xy2=0 ⇒(x-y)(x+y)+xy(x-y)=0 ⇒x+y+xy=0 ⇒y=-x1+x differentiate w.r.to x on both sides dydx=-((1+x)1-x(1)(1+x)2) ⇒dydx=-1(1+x)2

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

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$\frac{1}{1 - x^{2}}$

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

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

$\frac{1}{1 + x^{2}}$

answer is B.

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Detailed Solution

$Given x \sqrt{1 + y} + y \sqrt{1 + x} = 0 \Rightarrow x \sqrt{1 + y} = - y \sqrt{1 + x} \Rightarrow x^{2} (1 + y) = y^{2} (1 + x) \Rightarrow x^{2} + x^{2} y = y^{2} + {xy}^{2} \Rightarrow x^{2} - y^{2} + x^{2} y - {xy}^{2} = 0 \Rightarrow (x - y) (x + y) + xy (x - y) = 0 \Rightarrow x + y + xy = 0 \Rightarrow y = \frac{- x}{1 + x} d i f f e r e n t i a t e w . r . t o x o n b o t h s i d e s \frac{d y}{d x} = - (\frac{(1 + x) 1 - x (1)}{{(1 + x)}^{2}}) \Rightarrow \frac{d y}{d x} = \frac{- 1}{{(1 + x)}^{2}}$