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Why is this true?

  1. Mar 17, 2007 #1
    1. The problem statement, all variables and given/known data
    (d/dx)c=(d/da)c(da/dx)

    where c=f(a(x))


    3. The attempt at a solution
    It seems correct because the da cancels but how do you get this from first principles?
     
  2. jcsd
  3. Mar 17, 2007 #2

    radou

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    You may want to investigate the chain rule.
     
  4. Mar 17, 2007 #3
    Could you possibly be a little bit more specific?
     
  5. Mar 17, 2007 #4
  6. Mar 17, 2007 #5

    HallsofIvy

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    "da/dx" and "dc/da" are NOT fractions so it is not correct to say that the "da" cancels!

    Are you saying you have never heard of the chain rule? You are being asked to prove the chain rule!
    [tex]\frac{dc}{dx}= \lim_{h\rightarrow 0} \frac{c(x+h)- c(x)}{h}[/tex]
    "Before" the limit, this is a fraction- you can cancel before the limit but be careful, this is not a trivial proof.
     
    Last edited: Mar 19, 2007
  7. Mar 17, 2007 #6
    I have heard of the chain rule.

    In fact I realise that the chain rule is just what I stated (d/dx)c=(d/da)c(da/dx)
     
  8. Mar 18, 2007 #7
    But physicists and applied mathematicians like to treat them as fractions in the limit. Is it okay to treat them as fractions and specify "in the limit"?
     
  9. Mar 19, 2007 #8

    HallsofIvy

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    Yes, because the derivative is the limit of a fraction, you can always treat them "like" a fraction- that's one of the advantages of the dy/dt notation over f '. And, in fact, it is motivation for defining the "differentials", dx and dy= f '(x) dx.

    My point here was that, if you are being asked to prove the chain rule, you can't just "treat dy/dx like a fraction" since the chain rule is part of what allows us to do that.
     
  10. Mar 19, 2007 #9

    Gib Z

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    If i was a pedantic bastard, I would say perhaps pivoxa15 was speaking of using hyperreals? Thank god im not :D
     
  11. Mar 19, 2007 #10

    HallsofIvy

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    If you do, I'll beat you around the head and shoulders with a 2 by 4!
     
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