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Derivative question

  1. Feb 21, 2013 #1
    If ##f=f(x)## why then is ##df=\frac{df}{dx}dx##?
  2. jcsd
  3. Feb 21, 2013 #2
    While there are several approaches to differentials Wikipedia has a way of explaining this through linear mapping.
  4. Feb 21, 2013 #3
    I didn't asked that. I asked why we can multiply and devide with ##dx##.
  5. Feb 21, 2013 #4


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    Well first of all [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex] is a map and [itex]f(x)\in \mathbb{R}[/itex] is a value. You don't equate them. Secondly, the link Ferramentarius gave was precisely what you were asking. You don't "divide" the differentials; the wikipedia subsection linked by Ferramentarius has the pertinent details. That whole "division" of differentials thing is a very hand wavy technique that I myself have only seen in certain physics texts.
  6. Feb 21, 2013 #5
    You have to realize that [itex]\frac{df}{dx}[/itex] and [itex]dx[/itex] are usually very different kind of objects. While [itex]\frac{df}{dx}dx[/itex] looks like an ordinary multiplication, it really isn't. And [itex]\frac{df}{dx}[/itex] is not a division of [itex]df[/itex] and [itex]dx[/itex].

    In fact, in most calculus texts, things like [itex]df[/itex] and [itex]dx[/itex] don't even exist. There, the notation [itex]\frac{df}{dx}[/itex] is just the notation for a derivative. It is nothing more than a notation. A very handy notation, at that. But it's no division.

    As you progress in mathematics, you might encounter formal definitions of [itex]dx[/itex]. But even then, you still can't interpret things like [itex]\frac{df}{dx}[/itex] as a division!!

    The only way you can see [itex]\frac{df}{dx}[/itex] as some kind of division, is by nonstandard calculus. But this is very nonstandard. Only a very very minor number of books use it.
  7. Feb 21, 2013 #6
    Is then in eq
    ##y'(x)=y(x)##, ##y(x)## value of function or function? :D Question for WannabeNewton.

    micromass ok called whatever you want. Why isn't equal ##df=\frac{df}{dx}##?
  8. Feb 21, 2013 #7
    Value of function.

    Well, it's hard to explain because you never really saw a definition of df, I guess.

    It's clear that [itex]\frac{df}{dx}[/itex] is a function [itex]\mathbb{R}\rightarrow \mathbb{R}[/itex] if [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex].

    However, [itex]df[/itex] is rigorously defined as a function [itex]\mathbb{R}\rightarrow \mathbb{R}^*[/itex], where [itex]\mathbb{R}^*[/itex] is the dual space of [itex]\mathbb{R}[/itex]. So for any [itex]p\in \mathbb{R}[/itex], we can make sense of [itex]df_p[/itex] and this will send tangent vectors to a real number.

    Maybe you understood some of the above, maybe not. If you didn't, then I have to refer to books which explain it much better than me. A good and basic book is "Calculus on Manifolds" by Spivak. He says some things about this.
  9. Feb 22, 2013 #8
    Thanks for you're answer. I suppose like
    ##df(x,y)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy##
    special case
  10. Feb 22, 2013 #9


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    Yes, but are you willing to just accept [itex]df(x,y)= \partial f/\partial x dx+ \partial f/\partial y[/itex] when you were NOT willing to accept [itex]df= f'(x)dx[/itex]?

    To really explain either one, you need a precised definition of "df" which is what ferramentarius linked to.

    You then responded with "I didn't asked that. I asked why we can multiply and devide with dx."
    Well, no, you did NOT ask "why we can multiply an divide by dx". That is NOT what we are doing in writing "dy= f'(x)dx". Whether or not we can tread "dx" as if it were a number, depends on how we define "differentials", going back to ferramentarius' link.\

    "y" is the function, "y(x)" is a specific value of that function at the given value of x. Similarly, y' is the derivative function, y'(x) is a specific value of that derivative at the given value of x.
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