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Integrating Factor

If x2+y2=1, t...

Question

If $x^{2} + y^{2} = 1$ , then $\frac{d^{2} y}{d x^{2}}$ is

A

0

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B

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C

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D

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Solution

The correct option is D

$\frac{d y}{d x} = \frac{- x}{y}, \frac{d^{2} y}{d x^{2}} = \frac{- 1}{y^{3}}$

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Similar questions

x^{2} + y^{2} = 1

(w h e r e y^{'} = \frac{d y}{d x}, y^{''} = \frac{d^{2} y}{d x^{2}})

{sin}^{- 1} (\frac{x^{2} - y^{2}}{x^{2} + y^{2}}) = log a

\frac{d^{2} y}{d x^{2}}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

\frac{d^{2} y}{d x^{2}} =

x^{2} + y^{2} + sin y = 4

, then the value of

\frac{d^{2} y}{d x^{2}}

(- 2, 0)

y = \frac{1}{x^{2} - a^{2}}

\frac{d^{2} y}{d x^{2}}

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