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Math Help - implicit differentiation 2nd derivative

  1. #1
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    implicit differentiation 2nd derivative

    i have to find y' and y" using implicit differentiation
    2x^3 - y^2 = 8

    what i did was

    d2x^3/dx - dy^2/dx = d8/dx

    d2x^3/dx = 6x^2
    dy^2/dx = 2y*y'
    d8/dx = 0

    6x^2 - 2yy' = 0
    -6x^2=-2yy'
    y'=6x^2/2y
    = 3x^2 / y

    but i don't know how to find 2nd derivative from here

    how do i do it?
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  2. #2
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    Quote Originally Posted by haebinpark View Post
    i have to find y' and y" using implicit differentiation
    2x^3 - y^2 = 8

    what i did was

    d2x^3/dx - dy^2/dx = d8/dx

    d2x^3/dx = 6x^2
    dy^2/dx = 2y*y'
    d8/dx = 0

    6x^2 - 2yy' = 0
    -6x^2=-2yy'
    y'=6x^2/2y
    = 3x^2 / y

    but i don't know how to find 2nd derivative from here

    how do i do it?
    This is correct so far, all you have to do is use the quotient rule and implicit differentiation.

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

    y''=\frac{6yx-3y'x^2}{y^2}
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  3. #3
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    would you show me a little detail please?
    like from y' to y"
    how the implicit differentiation is applied
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  4. #4
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    Quote Originally Posted by haebinpark View Post
    would you show me a little detail please?
    like from y' to y"
    how the implicit differentiation is applied

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

    Apply the quotient rule:

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

    =\frac{6yx-3y'x^2}{y^2}
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