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D/dx g(df/dx)

  1. Oct 10, 2009 #1
    for example, isn't the derivative of the first derivative squared:

    d/dt (y')^2 = 2y'y''? why does susskind claim it is 2y'', in his classical lecture 3?
     
    Last edited: Oct 10, 2009
  2. jcsd
  3. Oct 11, 2009 #2

    Hepth

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    Gold Member

    Ahh, he's doing lagrangian mechanics.
    He is NOT doing
    [tex] \frac{d}{dt} y'^2[/tex]

    He is using euler's equations and doing instead:
    [tex] \frac{d}{dt} \frac{\partial}{\partial y'} (y'^2) = \frac{d}{dt} 2 y' = 2 y'' [/tex]
    The two dissappears from the kinetic energy term because its (1/2) m v^2.
     
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