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Time derivative

  1. Dec 2, 2009 #1
    I'm wondering, how does 2 multiplied by the first and second time derivatives of x equal the time derivative of the time derivative of x squared. Thanks.
     
  2. jcsd
  3. Dec 2, 2009 #2

    Born2bwire

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    Ummm... You mean,
    [tex]2\frac{d^2}{dt^2} f(t) = \frac{d}{dt} \left(\frac{d}{dt} f(t) \right)^2[/tex]
    This would be done by the chain rule. That is,
    [tex]\frac{d}{dx} (f(x))^2 = 2f(x)\frac{d}{dx}f(x)[/tex]
    In this sense, we take the two from the power of d/dt f(t) and take that as a coefficient, reduce the power by one, and then take the time derivative of d/dt f(t). I originally interpreted your question as
    [tex]2\frac{d}{dt}\frac{d^2}{dt^2} f(t) = \frac{d}{dt} \left(\frac{d}{dt} f(t) \right)^2[/tex]
    but I do not feel this expression is true.

    EDIT: Ok, let's fix this.

    Ummm... You mean,
    [tex]2\frac{d}{dt}f(t)\frac{d^2}{dt^2} f(t) = \frac{d}{dt} \left(\frac{d}{dt} f(t) \right)^2[/tex]
    This would be done by the chain rule. That is,
    [tex]\frac{d}{dx} (f(x))^2 = 2f(x)\frac{d}{dx}f(x)[/tex]
    In this sense, we take the two from the power of d/dt f(t) and take that as a coefficient, reduce the power by one, and then take the time derivative of d/dt f(t).
     
    Last edited: Dec 2, 2009
  4. Dec 2, 2009 #3

    nicksauce

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    Let v = dx/dt and a = dv/dt. Then I believe he means, why does 2va = d/dt (v^2)?

    It immediately follows from the chain rule:

    d/dt(v^2) = 2v*d/dt(v) = 2va
     
  5. Dec 2, 2009 #4

    Born2bwire

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    Ahh gotcha, I missed an f in there when I wrote it out on paper.

    [tex]2\frac{d}{dt}f(t)\frac{d^2}{dt^2} f(t) = \frac{d}{dt} \left(\frac{d}{dt} f(t) \right)^2[/tex]

    Of course that would make my original statement wrong too....

    I need more coffee.
     
  6. Dec 2, 2009 #5
    Awesome, thanks a lot for the help. I appreciate it.
     
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