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Taking derivatives of DE to find higher derivatives

  1. Jul 6, 2012 #1
    1. The problem statement, all variables and given/known data
    http://gyazo.com/d875970963124b7d4a64acc887f168fa

    3. The attempt at a solution
    I dont understand the part where they take the derivative of a diff eq to find the higher derivatives can somebody explain it?
    Mainly the answer where they get
    y''=(y'')'=(1-y^2)'=-2yy'(this part)
    y^(4)=(y''')'=(-2yy')'=-2y'y'-2yy"(huh?)=-2(y')^2-2yy''
     
  2. jcsd
  3. Jul 6, 2012 #2

    tiny-tim

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    hi shemer77! :smile:

    (try using the X2 button just above the Reply box :wink:)
    since you have an equation for y'', you use it to simplify all the higher derivatives

    (1-y2)' = (-y2)' = (use the chain rule) (-2y)(y)'
    similarly, use the product rule for (-2yy')' :wink:
     
  4. Jul 6, 2012 #3
    Hmm ok I think I understand the second one but for the first one what I've been taught is
    (1-y2)'=1'-y2'=-2y
    where does that extra y' come from?
     
  5. Jul 6, 2012 #4

    tiny-tim

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    remember that ' means d/dx, not d/dy

    so the chain rule gives you

    y2' = d/dx (y2)

    = d/dy (y2) * dy/dx

    = 2y * y' :wink:
     
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