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Question:

If $x \sqrt{1+y}+y \sqrt{1+x}=0$, for, $-1

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

Solution:

It is given that,

$x \sqrt{1+y}+y \sqrt{1+x}=0$

$\Rightarrow x \sqrt{1+y}=-y \sqrt{1+x}$

Squaring both sides, we obtain

$x^{2}(1+y)=y^{2}(1+x)$

$\Rightarrow x^{2}+x^{2} y=y^{2}+x y^{2}$

$\Rightarrow x^{2}-y^{2}=x y^{2}-x^{2} y$

$\Rightarrow x^{2}-y^{2}=x y(y-x)$

$\Rightarrow(x+y)(x-y)=x y(y-x)$

$\therefore x+y=-x y$

$\Rightarrow(1+x) y=-x$

$\Rightarrow y=\frac{-x}{(1+x)}$

Differentiating both sides with respect to x, we obtain

$y=\frac{-x}{(1+x)}$

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

Hence, proved.

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