Are dy and dx in dy/dx independent? Please help

In summary, the conversation discusses the treatment of dy/dx as a ratio of dy and dx, which can be manipulated independently. This may be seen as an abuse of notation, but it works and can be justified by considering dy and dx as small finite changes before reaching the limit where they become infinitesimal. However, mathematicians may create new structures to make this notation work, such as defining differentials and using differential forms in abstract spaces.
  • #1
Mesmerized
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A quote from the book of T.L. Chow 'Mathematical methods for physicists'

The reader may notice that dy/dx has been treated as if it were a ratio of dy and
dx, that can be manipulated independently. Mathematicians may be unhappy
about this treatment. But, if necessary, we can justify it by considering dy and
dx to represent small finite changes (delta)y and (delta)x, before we have actually reached the
limit where each becomes infinitesimal.
(instead of delta in parenthesis in the last sentence the latin letter delta is intended, just didn't know how to put it)

This part was from the chapter about differential equations. Can someone elaborate on this a little. I generally understand that the derivative is different than just the ratio of function change to it's argument change, but in which cases can we take it as a ratio of dy to dx and treat them independently without getting a wrong solution for the differential equation.
thanks
 
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  • #2
Mathematicians are unhappy with this little abuse of notation but... it works! You can take it as a mnemonic of the corresponding procedures taken "tha rigorous way". For example, when you write

[tex]\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}[/tex]

you actually mean

[tex]\frac{dy^{-1}}{dx}(x)=\frac{1}{y'(y^{-1}(x))}[/tex]

and theese exact formulations must be kept in mind.

I don't undertand what you say about dx and dy being independent. They can be dependent. For example, on a circumference

[tex]x^2+y^2=1[/tex]

you have dy = -dx, so they are not independent.
 
  • #3
thanks Petr. yeah you are right about your last example, I forgot to tell tell that I mean that y is the function of x, y=y(x), my fault. And for that function a differential equation is to be solved, and that's where my question - can dy/dx be treated as a fraction, and, thus for example if dy/dx=f(x,y)/g(x,y) then can we write g(x,y)dy=f(x,y)dx. Is it right always, or by doing so we can come to a wrong solution in some cases?
 
  • #4
Provided you don't divide by things that can be zero, they give true results. The first times you use theese tricks, try also writing "the good way", and after a while you'll convince yourself that Leibniz was not a fool.
 
  • #5
Mathematicians are all that unhappy with it. Or, rather, they do what they always do when physicists start playing "fast and loose" with notation- they create a new structure in which the notation does work. Any calculus book with start with the derivative dy/dx (which is NOT a fraction) and then define the "differentials" dy and dx so that it can be treated like a fraction. And, of course the whole idea of "differential forms" in more abstract spaces is an important field of mathematics.
 
  • #6
thank you very much, it helped a lot
 

1. What does dy/dx stand for?

Dy/dx represents the derivative of a function, or the rate of change of a function with respect to its independent variable.

2. Are dy and dx in dy/dx separate entities?

Yes, dy and dx are separate entities in the notation dy/dx. Dy represents the change in the dependent variable, while dx represents the change in the independent variable.

3. Can dy/dx be simplified?

Yes, dy/dx can be simplified using differentiation rules and algebraic manipulation. However, the resulting expression will still involve both dy and dx.

4. Is dy/dx a fraction?

In the context of calculus, dy/dx is not a fraction. It is a notation used to represent the derivative of a function.

5. Are dy and dx in dy/dx independent?

No, dy and dx are not independent in the notation dy/dx. They are both part of the same expression and cannot be treated separately.

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