# What does $dx$ mean?

$dx$ 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.

• good question, but you can find lot about it in web as well ! – Theorem May 9 '12 at 21:39
• Although the title is not exactly the same as your question, I believe your question is answered quite thoroughly in this similar post: math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio – process91 May 9 '12 at 21:39
• – process91 May 9 '12 at 21:48
• – Robert Israel May 9 '12 at 22:01
• @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

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.

• IMHO this is just one, very unenlightening way to look at it. – Michael Bächtold May 10 '12 at 10:01
• @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
• @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 9:44
• @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 8:58
• @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

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 $$\int_0^1 uv\,dv = u/2$$

• 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 6:46

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.

• I think "a little bit of $x$" holds many, many advantages for engineering maths. – JP McCarthy Jun 24 '13 at 21:10

$dx$ 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).

• If I didn't already know what $dx$ was, I don't think this answer would clear anything up for me. – Gerry Myerson May 10 '12 at 23:48

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 $dx$.

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.)

lhf's answer is fine if you just need to solve calculus problem but is incorrect. Assuming x is an independant variable, dx is the real that satisfies for all x > 0, dx < x where x is real.

• This is true in nonstandard analysis but in standard analysis - like one does in a calculus course - $dx$ doesn't satisfy such relations. – Cameron Williams Jun 24 '13 at 20:42
• There is no real number that has this property. – littleO Dec 15 '13 at 19:09
• Not all hyperreals are real. – Robin Goodfellow Dec 23 '14 at 14:32