x-3-y-3-1-find-the-implceat-second-derivative-

Question Number 205827 by mustafazaheen last updated on 31/Mar/24

x^{3} + y^{3} = 1

find the implceat second derivative

Answered by cortano12 last updated on 31/Mar/24

3 x^{2} + 3 y^{2} y^{'} = 0

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

\frac{d}{d x} [\frac{d y}{d x}] = \frac{d}{d x} [- x^{2} y^{- 2}]

\frac{d^{2} y}{{d x}^{2}} = - 2 {x y}^{- 2} + 2 x^{2} y^{- 3} y^{'}

\frac{d^{2} y}{{d x}^{2}} = - \frac{2 x}{y^{2}} + \frac{2 x^{2}}{y^{3}} . (\frac{- x^{2}}{y^{2}})

= \frac{- 2 x}{y^{2}} - \frac{2 x^{4}}{y^{5}} = \frac{- 2 {x y}^{3} - 2 x^{4}}{y^{5}}

= \frac{- 2 x (x^{3} + y^{3})}{y^{5}} = - \frac{2 x}{y^{5}}

Answered by TonyCWX08 last updated on 31/Mar/24

x^{3} + y^{3} - 1 = 0

f (x, y) = x^{3} + y^{3} - 1

f_{x} (x, y) = 3 x^{2}

f_{y} (x, y) = 3 y^{2}

\frac{d y}{d x}

= - \frac{3 x^{2}}{3 y^{2}}

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

f (x, y) = - \frac{x^{2}}{y^{2}}

f_{x} (x, y) = - \frac{2 x}{y^{2}}

f_{y} (x, y) = \frac{2 x^{2}}{y^{3}}

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

= - \frac{- 2 x}{y^{2}} \div \frac{2 x^{2}}{y^{3}}

= \frac{2 x}{y^{2}} \times \frac{y^{3}}{2 x^{2}}

= \frac{y}{x}

Commented by Frix last updated on 31/Mar/24

Obviously this is wrong .

x^{3} + y^{3} = 1 \Leftrightarrow y = {(1 - x^{3})}^{\frac{1}{3}} \Rightarrow

y^{″} = - \frac{2 x}{{(1 - x^{3})}^{\frac{5}{3}}} = - \frac{2 x}{y^{5}} \neq \frac{y}{x}

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