Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivative identity

  1. Feb 14, 2010 #1
    When you study physics, you never really delve into the reasons behind some of mathematical identities, i was curious about this one as it occurs in Bloch's Theorem (correct me if I go wrong):

    [tex]\frac{d}{dt}(\frac{dE}{dk})=\frac{d^{2}E}{dtdk}=\frac{d^{2}E}{dkdt}=(\frac{d^{2}E}{dk^{2}})\frac{dk}{dt}[/tex]

    I checked this and the first and last part are equivalent.

    Does that mean you can interchange the numerators and denominators freely? (given that the derivative is an operator)
     
  2. jcsd
  3. Feb 14, 2010 #2
    Yes, and the last equality comes from the chain rule.

    Actually it's more accurate when we are talking about partial derivatives, since if you k & t has hidden relations between them, the full derivatives want necessarily commute. (But that depends on the nature of the problem, and sometimes this difference between partial and full derivatives is confusing)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook