\[x\sqrt{1+y}+y\sqrt{1+x}=0\], then \[\frac{dy}{dx}=\] [RPET 1989, 96]

1.	\[x\sqrt{1+y}+y\sqrt{1+x}=0\], then \[\frac{dy}{dx}=\] [RPET 1989, 96]
A.	\[1+x\]
B.	\[{{(1+x)}^{-2}}\]
C.	\[-{{(1+x)}^{-1}}\]
D.	\[-{{(1+x)}^{-2}}\]
Answer» E.

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\[x\sqrt{1+y}+y\sqrt{1+x}=0\], then \[\frac{dy}{dx}=\] [RPET 1989, 96]

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