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how do you find y'' of (x^3)+(y^3)=1?i got as far as y'=(-x^2)/y^2...help!  

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astreetman | Student, Grade 11 | eNoter

Posted October 26, 2011 at 2:22 PM via web

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how do you find y'' of (x^3)+(y^3)=1?
i got as far as y'=(-x^2)/y^2...help!

 

Tagged with implicit derivation, math

2 Answers | Add Yours

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted October 26, 2011 at 10:06 PM (Answer #1)

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We need to find the second derivative for the equation:

`x^3 + y^3 = 1`

`==> 3x^2 + 3y^2 y' = 0`

`==> 3y^2y' = -3x^2`

``Now we will differentiate again:

==> `(3y^2)'(y')+(3y^2)(y')' = (-3x^2)'`

`==> 6yy' + 3y^2y'' = -6x`

`==> 3y^2y'' = -6x-6yy'`

`==> y'' = (-6(x+yy'))/(3y^2)`

`==> y'' = (-2(x+yy'))/y^2`

`=> y'' = (-2(x+y(-x^2/y^2)))/y^2`

`==> y'' = (-2(x -x^2/y))/y^2 = (-2(xy-x^2)/y))/y^2 = (-2(xy-x^2))/y^3`

`==> y'' = (-2x(y-x))/y^3`

``

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jeyaram | Student , Undergraduate | Valedictorian

Posted October 26, 2011 at 3:52 PM (Answer #2)

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(x^3)+(y^3)=1

(y^3)=1-(x^3)

so y=3√{1-(x^3)}

 

Aliter

(x^3)+(y^3)=1

(y^3)=1-(x^3)

get log both side

3lg(y)=lg{1-(x^3)}

lg(y)=lg{1-(x^3)}/3

now get antilog both side

y=3√{1-(x^3)}

 

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