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Math Help - second derivative

  1. #1
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    second derivative

    bk2 p50 ex17c q20
    given
    x^3 - 3xy +y^3 =4
    find
     \frac{d^2 y }{dx^2}

    my working:
    i use y' for dy/dx
    y' = \frac {x^2-y}{x-y^2}
    y''= \frac {(x-y^2)(2x-y')-(x^2-y)(1-2y y')}{(x-y^2)^2}
    y''= \frac {(-x+2x^2-y^2)y'+x^2-2xy+y}{(x-y^2)^2}
    stuck here. thanks

    p.s. why does that 'dash' -- y' look so bad in latex?
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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    bk2 p50 ex17c q20
    given
    x^3 - 3xy +y^3 =4
    find
     \frac{d^2 y }{dx^2}

    my working:
    i use y' for dy/dx
    y' = \frac {x^2-y}{x-y^2}
    y''= \frac {(x-y^2)(2x-y')-(x^2-y)(1-2y y')}{(x-y^2)^2}
    y''= \frac {(-x+2x^2-y^2)y'+x^2-2xy+y}{(x-y^2)^2}
    stuck here. thanks

    p.s. why does that 'dash' -- y' look so bad in latex?
    I haven't checked your work, but the next step is to insert your expression for y' in terms of x and y. It will be messy.

    -Dan
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by afeasfaerw23231233 View Post
    bk2 p50 ex17c q20
    given
    x^3 - 3xy +y^3 =4
    find
     \frac{d^2 y }{dx^2}

    my working:
    i use y' for dy/dx
    y' = \frac {x^2-y}{x-y^2}
    y''= \frac {(x-y^2)(2x-y')-(x^2-y)(1-2y y')}{(x-y^2)^2}
    y''= \frac {(-x+2x^2-y^2)y'+x^2-2xy+y}{(x-y^2)^2}
    stuck here. thanks

    p.s. why does that 'dash' -- y' look so bad in latex?
    It's ok for the first derivative.

    For the second derivative, I'd keep the formula you got by implicit differentiation :

    x^2-y-xy'+y'y^2=0

    Derivating :

    2x-y'-y'-xy''+2y'^2y+y''y^2=0

    2x-2y'-xy''+2y'^2y+y''y^2=0

    And then replacing...
    It's ugly, but it can be more friendly than using the quotient rule
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