A) \[1+x\]
B) \[{{(1+x)}^{-2}}\]
C) \[-{{(1+x)}^{-1}}\]
D) \[-{{(1+x)}^{-2}}\]
Correct Answer: D
Solution :
\[x\sqrt{1+y}+y\sqrt{1+x}=0\]Þ \[{{x}^{2}}(1+y)={{y}^{2}}(1+x)\] Þ\[{{\left( \frac{{{d}^{2}}s}{d{{t}^{2}}} \right)}_{t=1/2}}=-8\]Þ \[x+y+xy=0\,,\,\,\,\,\,\,\left\{ \because x\ne y \right\}\] Þ \[\frac{dy}{dx}=\frac{-1}{{{(1+x)}^{2}}}\].
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