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Need help with 2nd derivatives

  1. Oct 2, 2005 #1
    the probelm told me to take the 1st and 2nd derivatives of x^2-xy+y^2=9
    for the 1st derivative i got dy/dx = y-2x/2y-x
    but im stuck on taking the second derivative for impicit differenciation =(
     
  2. jcsd
  3. Oct 2, 2005 #2

    quasar987

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    Have you ever heard of the Theorem of implicit functions? It states that if you have an equation of the form F(x,y) = 0 where y is implicitely defined as a function of x, then

    [tex]dy/dx = -\frac{F'_x}{F'_y}[/tex]

    Use it on what you have found for dy/dx. It eases computation.
     
    Last edited: Oct 2, 2005
  4. Oct 2, 2005 #3

    lurflurf

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    Do you mean
    dy/dx = (y-2x)/(2y-x)

    It is easiest in my view to find the first and second derivatives of
    x^2-xy+y^2=9
    implicitly then express y' and y'' in terms of x and y if desired afterward.
    so
    (x^2-xy+y^2)'=9'
    2x-y-xy'+2yy'=0
    -(y-2x)+(2y-x)y'=0
    differentiate again
    (2x-y-xy'+2yy')'=0'
    or
    -(y-2x)'+[(2y-x)y']'=0'

    quasar987:
    Why do you have subscripts and primes?
    High school students often don't like theorems or partial derivatives.
    The subscipt means differentiate with respect to one variable holding others fixed so if
    F(x,y)=x^2-xy+y^2
    Fx=-(y-2x)
    Fy=(2y-x)
    and
    dF/dx=Fx+(Fy)y'
     
  5. Oct 2, 2005 #4

    quasar987

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    K. I pointed to that theorem because personally, by the time I got to doing implicit derivation in school, i was comfortable with theorems and partial derivatives.
     
  6. Oct 2, 2005 #5
    hmm, how exactly do i differenciate -(y-2x)+(2y-x)y'=0 ? i get y' and y''s in the derived equation....im so lost..
     
  7. Oct 2, 2005 #6

    lurflurf

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    [-(y-2x)+(2y-x)y']'=0'
    -(y-2x)'+[(2y-x)y']'=0 (lineararity)
    -(y'-2)+(2y-x)'y'+(2y-x)y''=0 (product rule)
    -(y'-2)+(2y'-1)y'+(2y-x)y''=0
    2+2(y'-1)y'+(2y-x)y''=0 (simplify)
    y''=[2+2(y'-1)y']/(x-2y)

    now if you want you can substitute in
    y'= (y-2x)/(2y-x)
    to get a (messy) expression for y'' in terms of x and y
     
  8. Oct 2, 2005 #7

    lurflurf

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    Right there with you. Many things would be simpler if partial derivatives were introduced early. I just wanted to say many people get confused when new theorems and concepts are dropped on them. The "standard" (read as bad) order of topics has implicit differentiation before partial differentiation. Also why do you have primes and subscripts?
     
  9. Oct 2, 2005 #8

    HallsofIvy

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    There's nothing wrong with getting y' in the formula for y". You can either go back to the formula you got for y', solve for y' and put that into y" or just leave with both y and y' in the formula. In most applications, its the value at a specific x you want and it's simplest to calculate y and y' as for that x, then substitute.
     
  10. Oct 2, 2005 #9

    quasar987

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    You cracked the code! [itex]F'_x[/itex] simply means the derivative of F wrt x. It is the notation use by Stewart. I prefer it to simply [itex]F_x[/itex] because in the first calculus class, we work with the notation [itex]f'[/itex] where diff. wrt x is implicit. It seems a more natural transition to keep the prime and add the specification wrt to what variable we differentiate, rather than drop the prime and have the subscript mean both "differentiate F" and "wrt x".
     
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