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I can't believe the hint that is provided; can someone explain how it is true?

  1. Feb 20, 2012 #1
    I'm working on a question for a problem set that has the following hint:

    For any function A:

    [itex]\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

    I don't understand how this is true; this is how I try to show that both sides are equal but it doesn't work out...any help would be appreciated.

    [itex]\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

    [itex]\frac{d}{dr}\frac{dA}{dr} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

    [itex](\frac{d}{dr} + \frac{2}{r}) \frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

    [itex]\frac{d}{dr} + \frac{2}{r} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2})[/itex]

    [itex]\frac{d}{dr} + \frac{2}{r} = \frac{d}{dr}[/itex]

    [itex] \frac{2}{r} = \frac{d}{dr} - \frac{d}{dr}[/itex]

    [itex] \frac{2}{r} = 0[/itex]

    This is only true if r [itex]\rightarrow[/itex] [itex]\infty[/itex]. I think I must be doing something wrong - can anyone please explain what I'm doing wrong and show me how the hint is a true statement? Thanks!
     
  2. jcsd
  3. Feb 20, 2012 #2
    You're treating everything as if it's multiplication, that's not legal.
     
    Last edited by a moderator: Feb 20, 2012
  4. Feb 20, 2012 #3

    Mark44

    Staff: Mentor

    The step above makes no sense. d/dr(r2) does NOT mean d/dr * r2, so you can't divide by r2 to end up with d/dr.

    There are steps above that that also don't make sense, for essentially the same reason.
    The assumption here is that A = A(r); i.e., A is a function of r. Also, it is assumed that A is twice differentiable.

    Start on the right side and calculate (1/r2) * d/dr(r2 *dA/dr).
     
  5. Feb 20, 2012 #4
    Oh man, I can't believe I missed that - thanks so much it makes perfect sense now! It slipped my mind that the [itex]\frac{d}{dr}[/itex] would act on the [itex]r^{2}[/itex] and that I would have to use the chain rule. Thanks a lot!

    I see my mistake, thank you. I knew I was doing something fishy but I wasn't certain what it was. Thank you again for the prompt replies!
     
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