If \[{{x}^{2}}+{{y}^{2}}=1\] then \[\left( {y}'=\frac{dy}{dx},{y}''=\frac{{{d}^{2}}y}{d{{x}^{2}}} \right)\] [IIT Screening 2000]

If \[{{x}^{2}}+{{y}^{2}}=1\] then \[\left( {y}'=\frac{dy}{dx},{y}''=\f

1.	If \[{{x}^{2}}+{{y}^{2}}=1\] then \[\left( {y}'=\frac{dy}{dx},{y}''=\frac{{{d}^{2}}y}{d{{x}^{2}}} \right)\] [IIT Screening 2000]
A.	\[y{y}''-2{{({y}')}^{2}}+1=0\]
B.	\[y{y}''+{{({y}')}^{2}}+1=0\]
C.	\[y{y}''-{{({y}')}^{2}}-1=0\]
D.	\[y{y}''+2{{({y}')}^{2}}+1=0\]
Answer» C. \[y{y}''-{{({y}')}^{2}}-1=0\]

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If \[{{x}^{2}}+{{y}^{2}}=1\] then \[\left( {y}'=\frac{dy}{dx},{y}''=\frac{{{d}^{2}}y}{d{{x}^{2}}} \right)\] [IIT Screening 2000]

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