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Partial derivatives question

  1. May 22, 2006 #1
    I'm trying to follow a derivation in given in a text book. Part of this derivation goes like this:

    [tex]\frac{d}{ds}\left(\frac{1}{c}\frac{dx}{ds}\right)=c\left(\frac{\partial^2\tau}{\partial x^2}\frac{\partial \tau}{\partial x} + \frac{\partial^2\tau}{\partial x \partial y}\frac{\partial \tau}{\partial y}\right)[/tex]
    [tex]=\frac{c}{2}\frac{\partial}{\partial x}\left[\left(\frac{\partial \tau}{\partial x}\right )^2 + \left (\frac {\partial \tau}{\partial y}\right )^2 \right][/tex]

    I worked through that and came up with the same answer, but without the factor of 1/2. Can anyone tell me where it comes from?
     
    Last edited: May 22, 2006
  2. jcsd
  3. May 22, 2006 #2

    J77

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    Differentiate out those last terms...

    d/dx(d tau/dx)^2=2d tau/dx * (d^2 tau/dx^2)
     
  4. May 22, 2006 #3
    Ah, gotcha! Thanks :)
     
  5. May 22, 2006 #4
    Hang on, after looking at it a bit more I'm not so sure...wouldn't that give you a factor of 2 out the front rather than 1/2?
     
  6. May 23, 2006 #5

    George Jones

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    Doesn't performing the differentiation in the bottom line result in the top line, since the 2 cancels the 1/2?

    If it does, then isn't everything OK?

    Regards,
    George
     
  7. May 24, 2006 #6

    J77

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    Yep - the '2' in my previous post cancels with the '2' of Blob's last term, giving the middle term...
     
  8. May 24, 2006 #7
    Haha, I see now...as you may have noticed, I'm not exactly the sharpest tool in the shed :)

    Thanks again.
     
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