If x1+y+y.1+x=0 and x≠y then dydx=____

Given : x1+y=−y.1+xx2(1+y)=y2(1+x)⇒x2−y2=xy2−x2y⇒(x−y)(x+y)=xy(y−x)⇒x+y=−xy⇒y=−x(1+a)=−1+1(1+x)∴dydx=−1(1+x)2

If x1+y+y.1+x=0 and x≠y then dydx=____

Given : x1+y=−y.1+xx2(1+y)=y2(1+x)⇒x2−y2=xy2−x2y⇒(x−y)(x+y)=xy(y−x)⇒x+y=−xy⇒y=−x(1+a)=−1+1(1+x)∴dydx=−1(1+x)2

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If $x \sqrt{1 + y} + y . \sqrt{1 + x} = 0$ and $x \neq y$ then $\frac{d y}{d x} =____$

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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 : $x \sqrt{1 + y} = - y . \sqrt{1 + x}$
$\begin{array}{l} x^{2} (1 + y) = y^{2} (1 + x) \\ \Rightarrow (x^{2} - y^{2}) = x y^{2} - x^{2} y \\ \Rightarrow (x - y) (x + y) = x y (y - x) \\ \Rightarrow x + y = - x y \\ \Rightarrow y = - \frac{x}{(1 + a)} = - 1 + \frac{1}{(1 + x)} \\ ∴ \frac{d y}{d x} = - \frac{1}{(1 + x)^{2}} \end{array}$