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dy/dx, dx, dy (1 Viewer)

Trebla

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From Preliminary calculus, it was suggested and formally derived that 'dy/dx' is a notation for the limiting value as h approaches zero of [f(x + h) - f(x)]/h. My teacher said that dy/dx is a notation of this limit and hence dy and dx cannot exist separately on their own using this definition.
Also in notation of the chain rule in calculus: dy/dx = (dy/du)(du/dx), it 'appears' that the du "cancels" out, but my teacher states that this is NOT what happens because du does not exist on its own. Also with relations like (dy/dx)(dx/dy) = 1, many have interpreted that as dx's and dy's 'cancelling out' but once again we are reminded that this is not the case. I have also encountered notations like d(x²) = 2x, where the "d(...)" appears to be just the same notation for the derivative.
However, when we learnt integration by substitution and integration by parts it appears that the notation appears to be "split" (even in the integration notation) and my teacher said that this "splitting" is fine. e.g. u = x², du/dx = 2x hence du = 2x dx. My teacher said that the concept behind this is university level material and tells us to just assume it, and that we are only allowed to do this "split" in integration by substitution or integration by parts.
Well, since I have now completed my HSC, and plan to study Mathematics at a tertiary level, (and also from curiosity) can someone please explain that concept? And are the interpretations of "cancelling" actually incorrect?
 

Affinity

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There's something called nonstandard analysis, where you can work with dy and dx's, otherwise the notation just happens to be good

Regarding to splitting in integration by substitution, the splitting amounts to proving that:

if u=g(x)

{Int[x=a..T] f(x) g'(x) dx} - {Int[u=g(a) .. g(T)] f(g^-1(u)) du} = 0
This you can do by showing that when T=a the LHS = 0 and the derivative of the LHS = 0. you can do this with 2 Unit knowledge,with appropriate regularity conditions

You can also think of du and dx as some kind of measure over the real line, and du/dx their relative density.

Remember that notation is there to convey ideas and suggest correct things.
 
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Trebla

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Can you please explain non-standard analysis?
 

Affinity

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Hmm.. I won't prove that it works, but you can use first order logic to build it. Roughly one extends the real numbers by an element c such that
c>0 and for all real numbers r>0, c< r.
call this universe the hyperreals, which consists of real numbers, infinitesmal c and the elements in the for r+c where r is real.
call r the standard part of r+c.
Then one can define f'(x) as the standard part of [f(x+c) - f(x)]/c

see how c works as dx here?

well up to now it's not found to be too useful
 
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