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If $x \sqrt{1+y}+y \sqrt{1+x}=0$, for,$-1 < x < 1$, prove that $\frac{d y}{d x}=\frac{-1}{(1+x)^2}$
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Verified Answer
The given equation may be written as
$\mathrm{x} \sqrt{1+\mathrm{y}}=-\mathrm{y} \sqrt{1+\mathrm{x}}$
Squaring both sides, $x^2(1+y)=y^2(1+x)$
$\begin{aligned}
&\Rightarrow x^2-y^2=y^2 x-x^2 y \\
&\Rightarrow(x+y)(x-y)=-x y(x-y) \\
&\Rightarrow y=\frac{-x}{1+x} \\
&\therefore \quad \frac{d y}{d x}=-\left\{\frac{(1+x) \cdot 1-x \cdot(0+1)}{(1+x)^2}\right\}=\frac{-1}{(1+x)^2}
\end{aligned}$
$\mathrm{x} \sqrt{1+\mathrm{y}}=-\mathrm{y} \sqrt{1+\mathrm{x}}$
Squaring both sides, $x^2(1+y)=y^2(1+x)$
$\begin{aligned}
&\Rightarrow x^2-y^2=y^2 x-x^2 y \\
&\Rightarrow(x+y)(x-y)=-x y(x-y) \\
&\Rightarrow y=\frac{-x}{1+x} \\
&\therefore \quad \frac{d y}{d x}=-\left\{\frac{(1+x) \cdot 1-x \cdot(0+1)}{(1+x)^2}\right\}=\frac{-1}{(1+x)^2}
\end{aligned}$
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