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I was always taught do not say "$dy$ divided by $dx$", instead "$dy$ by $dx$" because it's not really dividing.

I then studied differentiation from first principles, where one takes two points on a curve: eg. $(x, y)$ and $(x+\delta x, y+\delta y)$

$$\therefore \text{ gradient} = \frac{(y+\delta y)- y}{(x+\delta x) - x}$$

I'll skip the continuation but one gets the derivative of the equation through some algebra. If you simply expand the brackets instantly you simplify it to $\frac{\delta y}{\delta x}$ which is a division.

Firstly, why is the sign different - is it just because it is easier to right $d$ than $\delta$?

Secondly, why can one not say $dy$ divided by $dx$?

Thanks

Mahidevran
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Cobbles
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  • $\delta y$ and $\delta x$ are finite, so they can be divided. $\frac {dy}{dx}$ is not a fraction -- we just use that notation because it behaves like a fraction in some formulas --, so it's not technically "$dy$ divided by $dx$", though of course, there is a division going on in the background (in the limit definition). –  Dec 09 '14 at 21:35
  • For the difference between $dx$ and $\partial x$ please see: http://math.stackexchange.com/a/183330/90932 – wilsnunn Dec 09 '14 at 21:38
  • @DanielWilson-Nunn OP's not talking about the partial derivatives, he's talking about a finite change $\delta y$ along the tangent line. So $\dfrac {\delta y}{\delta x}$ is literally a division of two finite numbers -- as opposed to $\dfrac {\partial y}{\partial x}$ which is just formal notation. –  Dec 09 '14 at 21:39
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    This might be interesting: http://math.blogoverflow.com/2014/11/03/more-than-infinitesimal-what-is-dx/ – Thomas Dec 09 '14 at 21:42
  • $dy$ could mean the differential, and in that context both $dy$ and $dx$ are finite numbers and could be divided. $dy=f'(x) dx= \dfrac{dy}{dx} dx$ so $\dfrac{dy}{dx}=\dfrac{dy}{dx}$ ... could we interpret the two sides (of the latter) in a different way, e.g. as a fraction (LHS), or not a fraction (RHS), what would the equality mean in this context? – Mirko Dec 10 '14 at 02:41

3 Answers3

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One can certainly say it. The question is whether $dy$ and $dx$ actually mean anything by themselves, such that $dy/dx$ is their quotient. And the answer is yes: they refer to the $x$ and $y$ coordinates of a displacement along the tangent line to the curve.

Robert Israel
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  • (...) an infinitesimal displacement (...) – AlexR Dec 09 '14 at 21:39
  • No, not an infinitesimal displacement, an arbitrary displacement. $dx$ is a new variable, and $dy = (dy/dx) dx$. – Robert Israel Dec 09 '14 at 21:42
  • It is critical here that the displacement is along the tangent line rather than along the curve itself. – Ian Dec 09 '14 at 21:46
  • @RobertIsrael Well, it obviously depends on which formalism you're using. I think it would be better to use the notation $\delta y = \frac {dy}{dx} \delta x$ OR $dy = f'(x) dx$ to reduce ambiguity, but that's just my $2$ cents. –  Dec 09 '14 at 21:47
  • @Ian Okay this analogy will fail as soon as we go higher. – AlexR Dec 09 '14 at 21:47
  • @Bye_World Usually $\mathrm d x$ is a short for the differential equivalent to the lebesgue measure in integration for the variable $x$, but getting that right is quite an advanced topic. – AlexR Dec 09 '14 at 21:48
  • $dx$ means a whole bunch of related things in different fields. It has little to do with integration when it is used in differential geometry, for example. – Ian Dec 09 '14 at 21:49
  • @AlexR I haven't yet studied measure theory, but I have seen this formalism, which seems to be the one that Robert is using here. –  Dec 09 '14 at 21:50
  • For reference, this is the notation I refer to. – AlexR Dec 09 '14 at 21:54
  • @AlexR I'm aware of differential forms, I've been studying differential geometry for a while now. I just haven't done measure theory. But differential forms are not infinitesimals (as your first comment would suggest). And I'm still pretty sure Robert was using the definition from the Wikipedia article I linked to. :) –  Dec 10 '14 at 03:38
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I don't think there's any harm in saying "$dy$ divided by $dx$". But${}\ \ldots$

I was taught that $$ \frac{dy}{dx} = \lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}. $$

Certainly $\Delta y$ and $\Delta x$ are numbers.

The differentials $dy$ and $dx$ are thought of as infinitely small but nonzero increments of $y$ and $x$. This raises a question of whether one can really make sense of such a concept as infinitely small but nonzero quantities, and whether one can then do calculus with them, using them in just the way in which the notations $dy$ and $dx$ are used. Robinson's nonstandard analysis does make use of infinitely small quantities, but not in just exactly the way in which Leibniz (who introduced the notation) and Euler (who used it more extensively than anyone else) used them.

The whole theory of calculus can be justified without proving that infinitely small increments can be taken literally. That justification was done in the 19th century by "epsilons and deltas".

So the fact that one doesn't take them literally is why some people say you shouldn't speak of dividing when it's not clear that it's really dividing.

However, speaking of infinitely small increments is an immensely useful heuristic.

Michael Hardy
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It is helpful, particularly once you start studying multivariable calculus, to recognize the derivative $\frac{dy}{dx}$ as not a real number but a linear map from tangent vectors on the domain to tangent vectors on the image.

From this, some misguided people get the impression that one "shouldn't" call $\frac{dy}{dx}$ in ordinary one-variable calculus a ratio. At a given value of $x$, tangent vectors $dx$ at $x$ and their pushforwards $dy$ can be canonically identified with real numbers, and the derivative of $y$ at $x$ is 100% rigorously the quotient $\frac{dy(dx)}{dx}$ (when $dx\neq 0$), which one can show does not depend on $dx$. As Robert points out, this has a very helpful and still 100% rigorous geometric interpretation when you intepret $y$ as a graph, and $dx$ and $dy$ as infinitesimal horizontal and vertical displacements.

user7530
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