calculus

Find
y''
by implicit differentiation.
3x3 + 4y3 = 1

asked by Anonymous
  1. 3x^3 + 4y^3 = 1

    9x^2 + 12y^2 y' = 0
    y' = -3x^2/4y^2

    now, using the quotient rule,

    y'' = (-24xy^2 + 3x^2(8yy'))/ 16y^4
    = (-3xy^2 + 3x^2yy')/2y^4
    = -3x (y - xy')/2y^3

    = -3x(y - x*(-3x^2/4y^2))/2y^3
    = -3x(4y^3 + 3x^3)/8y^5

    or, you can use implicit differentiation twice:

    9x^2 + 12y^2 y' = 0
    18x + 24y (y')^2 + 12y^2 y'' = 0
    3x + 4y y'^2 + 2y^2 y'' = 0
    2y^2 y'' = -(3x + 4y (y')^2)
    y'' = -(3x + 4y (y')^2)/2y^2
    = -(3x + 4y * 9x^4/16y^4)/2y^2
    = -3x(4y^3 + 3x^3)/8y^5

    posted by Steve

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