If \[y={{\left( x+\sqrt{1+{{x}^{2}}} \right)}^{n}},\] then \[(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}\] is [AIEEE 2002]

If \[y={{\left( x+\sqrt{1+{{x}^{2}}} \right)}^{n}},\] then \[(1+{{x}^{

1.	If \[y={{\left( x+\sqrt{1+{{x}^{2}}} \right)}^{n}},\] then \[(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}\] is [AIEEE 2002]
A.	\[{{n}^{2}}y\]
B.	\[-{{n}^{2}}y\]
C.	\[-y\]
D.	\[2{{x}^{2}}y\]
Answer» B. \[-{{n}^{2}}y\]

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If \[y={{\left( x+\sqrt{1+{{x}^{2}}} \right)}^{n}},\] then \[(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}\] is [AIEEE 2002]

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