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$dx$ appears in differential equations, such us derivatives and integrals.

For example, a function $f(x)$ its first derivative is $\dfrac{d}{dx}f(x)$ and its integral $\displaystyle\int f(x)dx$. But I don't really understand what $dx$ is.

J. M. ain't a mathematician
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Garmen1778
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    good question, but you can find lot about it in web as well ! – Theorem May 09 '12 at 21:39
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    Although the title is not exactly the same as your question, I believe your question is answered quite thoroughly in this similar post: http://math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio – mboratko May 09 '12 at 21:39
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    Also: http://math.stackexchange.com/questions/23902/what-is-the-practical-difference-between-a-differential-and-a-derivative/23914#23914 – mboratko May 09 '12 at 21:48
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    And http://math.stackexchange.com/questions/46530/can-i-ever-go-wrong-if-i-keep-thinking-of-derivatives-as-ratios – Robert Israel May 09 '12 at 22:01
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    @Garmen: Please take a look at my comments under the answer you accepted. That answer is quite misleading; one of the limits is wrong, and the concept "infinitesimally small" is being used informally without a definition. While there is an interesting branch of mathematics called non-standard analysis that defines infinitesimal quantities, standard analysis (which is presumably what you're asking about) has no such concept. – joriki May 10 '12 at 21:42
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    dx is a differential form. Start your search there. –  May 16 '12 at 05:37
  • @Garmen1778 : there is a new answer by Carl Mummert that is the best so far and is infinitely better than the one you accepted. – Stefan Smith Dec 13 '13 at 16:36
  • @Stefan, did you happen to notice that Garmen asked this question a year-and-a-half ago, and hasn't been seen here in over half a year? I'm afraid the damage, if any, has already been done. – Gerry Myerson Dec 13 '13 at 21:59
  • @GerryMyerson : Thanks for noticing that. Apparently this is a lost cause. But Carl Mummert's excellent answer from today is up to 10 votes, vs. 11 votes for the accepted answer. Maybe Carl's answer will at least end up with the most votes. – Stefan Smith Dec 13 '13 at 22:09
  • http://blog.cambridgecoaching.com/but-what-is-dx-really-calculus-terms-explained – Aatmaj May 10 '21 at 11:35

6 Answers6

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The formal definition of an expression such as $$ \int_0^1 x^2\,dx $$ will depend on the setting. This is because there is not just one "theory of integration" - there are several different theories in different areas.

I like the presentation at the beginning of this note by Terence Tao. The key point is that there are really at least three different viewpoints on integration in elementary calculus:

  • Indefinite integration, which computes antiderivatives

  • An "unsigned definite integral" for finding areas under curves and masses of objects

  • A "signed definite integral" for computing work and other "net change" calculations.

The value of an expression such as $\int_0^1 x^2\,dx$ comes out the same under all these interpretations, of course.

In more general settings, the three interpretations generalize in different ways, so that the "dx" comes to mean different things. In the setting of measure theory, "dx" is interpreted as a measure; in the context of differential geometry, it is interpreted as a 1-form.

But, for the purposes of elementary calculus, the only role of the "dx" is to tell which variable is the variable of integration. In other words, it lets us distinguish $$ \int_0^1 uv\,du = v/2 $$ from $$ \int_0^1 uv\,dv = u/2 $$

Carl Mummert
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    there is also a fourth interpretation of $dx$ as an infinitesimal quantity under the nonstandard setting of the hyperreal field of Robinson. Also there are other interpretations of other nonstandard settings as in smooth infinitesimal analysis, just to name a phew for the curious reader – Masacroso Jun 11 '19 at 14:14
  • Though this answer is years old, I just want to comment that differential geometry doesn't only use the differential form generalization. It also uses a measure-theoretic analog: densities, which make integration possible on manifolds even without orientation. (These are important e.g. in relativity theory.) The rough intuition here is that a form + an orientation = a density. – symplectomorphic Dec 22 '20 at 06:46
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Formally, $dx$ does not mean anything. It's just a syntactical device to tell you the variable to differentiate with respect to or the integration variable.

lhf
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    IMHO this is just one, very unenlightening way to look at it. – Michael Bächtold May 10 '12 at 10:01
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    @Michael, sorry, can't tell my math-for-poets classes that it's a differential one-form; I can (and do) give them lhf's answer, and they get some mileage out of it. – Gerry Myerson May 10 '12 at 13:18
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    your answer is not clear and not answer the question, because if just we consider tow function $F(x,y)$ and $G(x,y)$ defined on $\mathbb{R}^2$, what would be the meaning of $F(x,y)dx-G(x,y)dy=0$ by the use of your definition !!!? – Abdelmajid Khadari May 11 '12 at 00:48
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    The question is about the meaning not the utility of $dx$. – Abdelmajid Khadari May 11 '12 at 00:54
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    your answer is not correct. If you see a equation with dx and dy then they mean something and it is not simply a part of a derivative or an integral description. And you can't always simply split them to solve this equation – Gargo May 16 '12 at 07:24
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    @GerryMyerson: I've never taught math-for-poets, but I wonder if teaching such people formal laws for manipulating symbols, without telling them about the meaning of it is more valuable than trying to explain in poetic terms what a differential one form is. (Maybe something like infinitesimal coordinates?) – Michael Bächtold May 23 '12 at 09:44
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    @Michael, I encourage you to try that approach some time and report back on the results. Actually, math-for-poets isn't really what I deal with; it's math-for-accountants and other business students. – Gerry Myerson May 23 '12 at 10:07
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    @lhf : your answer is incorrect. $dx$ is a differential form. You should delete your answer. – Stefan Smith Dec 13 '13 at 05:42
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    @GerryMyerson : you don't have to tell your math-for-poets classes anything about differential one-forms, but if you tell them that formally, $dx$ does not mean anything, it's a lie. – Stefan Smith Dec 13 '13 at 05:43
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    @Stefan, $dx$ is different things to different people. I'm guessing that for the person who asked the question, the answer posted by lhf is about right, and any attempt to tell that person about differential forms would be counterproductive. – Gerry Myerson Dec 13 '13 at 08:58
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    @GerryMyerson : regardless of what one thinks $dx$ means, this answer begins with a falsehood. The OP should be least be informed that $dx$ has a precise meaning. – Stefan Smith Dec 13 '13 at 14:34
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    @GerryMyerson : also, you can tell your math-for-poets orally that it's a differential one-form in about five seconds, and then move on, and never mention differential forms again. If you write down lhf's answer verbatim, you've lied to them. You can easily reword lhf's idea in a form that avoids oversimplification. – Stefan Smith Dec 13 '13 at 15:05
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    @Stefan Smith: I think it is quite debatable whether the "dx" in calculus 1 is a one-form. For example, a typical calculus textbook will have a formal definition of the Riemann integral, but not any mention of one-forms. This holds even for books like Rudin's Principles of Mathematical Analysis. So it could appear to be somewhat revisionist to say that dx is a one-form in that context. And, even if we want to be revisionist, what is to say that "dx" is not a measure, instead of a one-form? – Carl Mummert Dec 13 '13 at 15:08
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    The problem with "$dx$ as pure syntax" is that it leaves change-of-variables as yet another formula that must be applied by rote. A little geometric intuition ($dx$ as an infinitesimal length, which gets scaled when you change variables) goes a long way. – user7530 Dec 13 '13 at 15:29
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    @CarlMummert : the accepted answer begins with a blanket statement that $dx$ does not mean anything. That is incorrect. And of course a calculus textbook might not mention one-forms, because differential forms are a very difficult topic that most calculus students can't handle. – Stefan Smith Dec 13 '13 at 16:03
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    @Stefan Smith: what surprises me about the overall answers to this question is that I would expect to see other answers with blanket statements: "$dx$ is a differential form", "$dx$ is a measure", "$dx$ is an infinitesimal", etc. In other words there are several "correct" answers, and this is just one of them. In the context of symbolic integration, which is after all a purely syntactic enterprise, the only role of the $dx$ is to specify the variable of integration. Why not write another answer beginning with "$dx$ is a differential form"? – Carl Mummert Dec 13 '13 at 16:08
  • @lhf : I made a comment that $dx$ is a differential one-form. $dx$ can also be interpreted as a measure. I don't know if there are other ways. Either way, the first sentence of your answer is incorrect. – Stefan Smith Dec 13 '13 at 16:39
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    How's this for a compromise: "For the purposes of Math 130 at Macquarie University, $dx$ doesn't mean anything." – Gerry Myerson Dec 13 '13 at 21:56
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    @GerryMyerson : I find that acceptable. It is completely different from the wording in the accepted answer. But if you are going to tell students that, you should tell them why they should write $dx$ at all., – Stefan Smith Dec 13 '13 at 22:04
  • It's obviously not "meaningless," since with $dx$ replaced with $dy$ you get a different result. At the very least, the $dx$ indicates with variable is changing... – Thomas Andrews Dec 13 '13 at 22:23
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    Obviously it's not meaningless. It means the product of $d$ with $x$... which is my somewhat snarky way of saying that, yes, of course context determines meaning, and in this context, there is no meaning (so the answer is correct). – Ben Millwood Dec 14 '13 at 14:40
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    @carlmummert sorry to reply to a 7 year old comment, but the dx of freshman calculus is a 1 form, and not a measure, because the integral from a to b is the opposite of the integral from b to a. We would just integrate a measure over an interval, and the sign wouldn't change. – Steven Gubkin Apr 28 '20 at 22:44
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As Silvanus Thompson put it in his book Calculus made easy: $\mathrm dx$ means "a little bit of $x$".

If that is not satisfying, there are various more precise explanations. One of them is: $\mathrm dx$ is a differential one-form.

Michael Bächtold
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    I think "a little bit of $x$" holds many, many advantages for engineering maths. – JP McCarthy Jun 24 '13 at 21:10
  • What is meant by a little bit of x?I mean suppose we want to calculate the value of the slope at point (x,y).Then dx means a step to the right.But this dx is not in the line segment x.So how does it mean a little bit of x?It is outside of the line segment. – a_i_r Jan 30 '21 at 07:15
  • @AritraBarua X is usually a variable number, which is pictured along the x axis, not along the graph of the function, so it is along the line segment of the x axis. Does that answer your question? – Michael Bächtold Jan 30 '21 at 07:35
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    I mean imagine a line of length x.Then dx is an extension of the line to the right.But here why is dx a little bit of x when it is on the extension? – a_i_r Jan 30 '21 at 07:58
  • You could also shrink the line, instead of extending it (since the "sign" of dx could also be negative). But I agree that the expression "a little bit of" can be confusing, as you explain. Probably a better way of capturing the idea is to say that d means "a little change of", instead of "a little bit of". – Michael Bächtold Feb 19 '21 at 12:33
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$dx$ means a very very small quantity, $dx=x_2-x_1$ where $x_1$ and $x_2$ very very near to $x$ (in geometry a very small distance), when you derive $\frac{d}{dx}f(x)$ it means you calculate the propinquity of $df(x)$ and $dx$, when you integrate, the sign $\int$ means a continuous sum, so $\int f(x) dx$ means a continuous sum of all the quantities $f(x) dx$ (geometrically very very small rectangles), in graduate language $dx$ is a linear map (differential form).

MJD
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Abdelmajid Khadari
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The d$x$ comes from approximating the area under a curve by a discrete sum of narrow rectangular slices of heights $f(x_i)$ and equal widths $\Delta x = x_{i+1}-x_i$. Look up Riemann sum for more details. So the area is then approximately $\sum^n_{i=1} f(x_i) \Delta x$. This approximation becomes exact when $\Delta x$ becomes arbitrarily small, which is symbolized by replacing $\Delta x$ by d$x$ (and $\sum$ by $\int$). For derivatives, similar story; just replace "area" in the above by "slope" or "gradient", where the approximation is now a chord of length d$x$ along X-direction. NB: correct notation is d$x$, not $dx$.

Lucozade
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I have a relevant blog post for those comfortable with multivariate calculus, found here.

Though treating $\mathrm{d}x$ as simply "that thing" works, there is hidden meaning behind it. To be clear, in this case, we are using $x$ to refer to the identity function $x(t)=t$, where $t\in\mathbb{R}$.

Imagine an arrow tangent to the real line. This can be represented by a vector (magnitude and direction) and a real number (position). We call such a vector a tangent vector. If the vector is $v$ and the position is $p$, then we denote the corresponding tangent vector* by $v_p$.

Mathematicians define $\mathrm{d}x$ as a type of function, called a differential $1$-form, that takes in a position $p$ and outputs a so-called "dual" tangent vector, or cotangent vector, which we call $\mathrm{d}x_p$. This $\mathrm{d}x_p$ is, somewhat confusingly, a function on the tangent vectors positioned at $p$. In the case of real numbers, though, the expression for $\mathrm{d}x_p$ is easy: $$\mathrm{d}x_p(v_p)=v.$$

Though many people will prefer to work with the (arguably more rudimentary) notion of "infinitesimals", differential forms have many advantages over the "old way". For example, change of variables, commonly known as $u$-substitution, has a simple formula in terms of something called the "pullback". As another example, when we move on to calculus on spaces that aren't strictly Euclidean, these differential forms give powerful information about the space itself, such as (in some sense) how many holes it has.

(* For the professionals, what I am meaning here is that $v_p=v\left.\frac{d}{dx}\right|_p$. Feel free to heckle in the comments.)

Robin Goodfellow
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