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# If $x\sqrt{1+y}+y\sqrt{1+x}=0\:, find \: \large\frac{dy}{dx}$

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## 1 Answer

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Toolbox:
• $\big(\large\frac{u}{v}\big)'=\large\frac{u'v-uv'}{v^2}$
Step 1:
The given equation may be written as $x\sqrt{1+y}=-y\sqrt{1+x}$
Squaring both sides we get
$x^2(1+y)=y^2(1+x)$
$x^2+x^2y=y^2+y^2x$
$x^2-y^2=y^2x-x^2y$
$(x+y)(x-y)=-xy(x-y)$
$x+y=-xy$
$x=-xy-y$
$x=-y(x+1)$

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