Wikipedia shows a proof of product rule using differentials by

  • Thread starter PhDorBust
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  • #1
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Wikipedia shows a proof of product rule using differentials by Leibniz. I am trying to correlate it to the definition of a differential and am having no success.

cee7a4c95c74d87aa466510391727d61.png


Differential Definition: http://eom.springer.de/D/d031810.htm
 
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  • #2
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Look at how the derivative of a product of functions is derived using the limit of the difference function. The numerator would be u(x + [itex]\Delta x[/itex]) [itex]\cdot[/itex] v(x + [itex]\Delta x[/itex]) - uv.
 
  • #3
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You are right up to this step:

[tex] d u\cdot v = u \cdot dv + v \cdot du + du \cdot dv[/tex]

However, if differential is small, differential multiplying to another differential is much smaller, which can be neglected (In limit sense). Therefore:

[tex] d u\cdot v = u \cdot dv + v \cdot du[/tex]
 
  • #4
HallsofIvy
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You are right up to this step:

[tex] d u\cdot v = u \cdot dv + v \cdot du + du \cdot dv[/tex]

However, if differential is small, differential multiplying to another differential is much smaller, which can be neglected (In limit sense). Therefore:

[tex] d u\cdot v = u \cdot dv + v \cdot du[/tex]
I would think that the whole purpose of using differentials (which is basically using "non-standard analysis") and developing all the machinary necessary to even define differentials is to avoid saying "in limit sense"! In terms of differentials, the product of two differentials is 0: [itex]du\cdot v= u\cdot dv+ v\cdot du[/itex] directly.
 
  • #5
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I agree that differential has the meaning of limit already. But without saying so, it is rather difficult to explain why du dv = 0 in PhDorDust's example. Furthermore, I wonder if product of differentials must be zero, since we sometimes have dx dy in double integral. I think the reason is that, we only need to take the dominating term, which is u dv and v du, but not dv du.
 

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