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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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`

``

(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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