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If $x \sqrt{1+y}+y \sqrt{1+x}=0$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\frac{-1}{(1+x)^2}$
Given, $x \sqrt{1+y}+y \sqrt{1+x}=0$
$\Rightarrow \quad x^2(1+y)=y^2(1+x) \Rightarrow x^2-y^2=x y(y-x)$
$\Rightarrow \quad x+y+x y=0 \Rightarrow y=\frac{-x}{1+x}$
On differentiating w.r.t ' $x$ ', we get
$\frac{d y}{d x}=\frac{(1+x)(-1)-(-x)(1)}{(1+x)^2}=\frac{-1-x+x}{(1+x)^2}=-\frac{1}{(1+x)^2}$
$\Rightarrow \quad x^2(1+y)=y^2(1+x) \Rightarrow x^2-y^2=x y(y-x)$
$\Rightarrow \quad x+y+x y=0 \Rightarrow y=\frac{-x}{1+x}$
On differentiating w.r.t ' $x$ ', we get
$\frac{d y}{d x}=\frac{(1+x)(-1)-(-x)(1)}{(1+x)^2}=\frac{-1-x+x}{(1+x)^2}=-\frac{1}{(1+x)^2}$
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