Differentiation - Chain Rule

  • Thread starter PhyAmateur
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  • #1
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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..
 
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  • #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.
 
  • #3
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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?
 
  • #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.
 
  • #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.
 
  • #6
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It is the same $$dr^2$$ found in the 3 sphere metric for example..
 
  • #7
PeroK
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It is the same $$dr^2$$ found in the 3 sphere metric for example..
##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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