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Differentiation - Chain Rule

  1. Nov 1, 2014 #1
    In one physics problem if $$r^2= \lambda^2(1+\frac{m}{2\lambda})^2$$

    what is ##dr^2 ?##

    Should I find ##dr## starting from ##r= \lambda(1+\frac{m}{2\lambda})## first and then square or find ##dr^2## starting from r^2? I know this is a basic question in differentiation using chain rule but it seems I am stuck over this..
     
    Last edited by a moderator: Nov 1, 2014
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  3. Nov 1, 2014 #2

    SteamKing

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    A little more context would be helpful here.

    Are you trying to find (dr)2 or dr2. I think there's a difference.
     
  4. Nov 1, 2014 #3
    It is written as $$dr^2$$ I was just wondering if I should derive once and then after finding the answer, derive twice.. What do you say?
     
  5. Nov 1, 2014 #4

    SteamKing

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    This notation is ambiguous to me. I can't advise you further without more information about this problem where you found it.
     
  6. Nov 2, 2014 #5

    Fredrik

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    I would interpret ##dr^2## as ##(dr)^2##, not as ##d(r^2)##. The straightforward way is to solve for r, compute ##dr=\frac{dr}{d\lambda}d\lambda##, and then square the result.
     
  7. Nov 2, 2014 #6
    It is the same $$dr^2$$ found in the 3 sphere metric for example..
     
  8. Nov 2, 2014 #7

    PeroK

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    ##r^2 = λ^2(1 + \frac{m}{λ} + \frac{m^2}{4λ^2}) = λ^2 + mλ + \frac{m^2}{4}##

    ##r = λ(1 + \frac{m}{2λ}) = λ + \frac{m}{2}##

    ##\frac{d(r^2)}{dλ} = 2λ + m##

    ##d(r^2) = (2λ + m)dλ \ \ (A)##

    ##\frac{dr}{dλ} = 1##

    ##dr = dλ##

    ##(dr)^2 = (dλ)^2 \ \ (B)##

    A or B, take your pick.
     
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