38

In school, I was taught that we use $\dfrac{du}{dx}$ as a notation for the first derivative of a function $u(x)$. I was also told that we could use the $d$ just like any variable.

After some time we were given the notation for the second derivative and it was explained as follows:

$$ \frac{d\left(\frac{du}{dx}\right)}{dx} = \frac{d^2 u}{dx^2} $$

What I do not get here is, if we can use the $d$ as any variable, I would get the following result:

$$ \frac{d\left(\frac{du}{dx}\right)}{dx} =\frac{ddu}{dx\,dx} = \frac{d^2 u}{d^2 x^2} $$

Apparently, it is not the same as the notation we were given. A $d$ is missing.

I have done some research on this and found some vague comments about "There are reasons for that, but you do not need to know..." or "That is mainly a notation issue, but you do not need to know further."

So what I am asking for is: Is this really just a notation thing? If so, does this mean we can actually NOT use d like a variable? If not, where does the $d$ go?

I found this related question, but it does not really answer my specific question. So I would not see it as a duplicate, but correct me if my search has not been sufficient and there indeed is a similar question out there already.

TShiong
  • 1,257
Numenkok Balok
  • 491
  • 19
    $d$ is not a variable; in other words $dxdx$ is not $d$ times $x$ times $d$ times $x$. At best, you can think of $dx$ as one object (with a two letter name), an infinitesimal. – Michael Burr Jan 14 '16 at 19:06
  • 16
    $d$ cannot be used just like any variable. Otherwise you will have $du/dx=u/x$ for example. – velut luna Jan 14 '16 at 19:08
  • 3
    I have a hard time to believe somebody told you this. Maybe they said it about the entity "$dx$" – quid Jan 14 '16 at 20:24
  • 3
    It's because the dx is "in parentheses", so to speak. – user541686 Jan 14 '16 at 21:20
  • 27
    I kinda assumed $dx^2$ meant $(dx)^2$; i.e. $dx$ is basically one variable. – Akiva Weinberger Jan 15 '16 at 00:16
  • I asked about the origin of this notation on https://hsm.stackexchange.com/questions/3323/who-invented-the-leibnitz-notation-fracd2ydx2-for-the-second-derivati. – Federico Poloni Jan 15 '16 at 15:28
  • @IwillnotexistIdonotexist I don't think I've seen a multivariable calculus book that does that. I've only seen books write things like $\partial x\partial y$ in the denominator, and I believe for good reason. – Mark S. Jan 16 '16 at 15:03
  • 1
    @IwillnotexistIdonotexist First, it destroys the nice "multiplication" of factors of the form $\frac{\partial}{\partial x}$ by saying something like "$\partial\partial=\partial^2$ on top, but $(\partial x)(\partial y)=\partial xy$ on bottom". More importantly, IMO, is the fact that, in general: $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ don't commute, but $x$ and $y$ do. So to use the notation you proposed would require that $\partial xy$ in a denominator can't be replaced with $\partial yx$, even though in all other real-anaylsis contexts $xy\to yx$ would be fine. – Mark S. Jan 16 '16 at 15:30
  • @MarkS. Fair point; By Clairaut's Theorem reversing the order of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ is valid for many functions, but definitely not always. – Iwillnotexist Idonotexist Jan 16 '16 at 15:37
  • 1
    You really can't use $d$ as a variable; however, you can use $dx$ as a variable. (This works well when doing, for example, integration by substitution.) Possibly this is what you were told (or what they meant to tell you.) – Toby Bartels May 02 '19 at 18:45
  • Mistake is that $d(du)=d(\frac{du}{dx})$ is false. I wrote a some answer https://math.stackexchange.com/questions/477507/is-d2y-dx-a-valid-mathematical-notation/4830775#4830775 which can be useful also here. – zkutch Dec 20 '23 at 02:57

5 Answers5

46

where does the $d$ go?

Physicist checking in. All the other answers seem to focus on whether $d$ is a variable and are neglecting the heart of your question.

Simply put, $dx$ is the name of one thing, so in your example

$$\frac{d^2u}{dx^2}=\frac{d^2u}{\left(dx\right)^2}$$

In your words, the "second $d$" is inside the implied parentheses.

user1717828
  • 979
  • 9
    +1. I'm surprised that so many other answers missed this aspect of the question. – mweiss Jan 15 '16 at 16:22
  • 2
    I guess the same could occur with a delta, for example with the formula $U=\frac12 k \Delta x^2$ one might interpret it as $U=\frac12 k (\Delta x)^2$ (the elastic potential energy in a Hooke spring). – Jeppe Stig Nielsen Jan 15 '16 at 22:10
  • 1
    Thanks for this short but good answer. Using the $dx$ as one unit and not as two separate things $d$ and $x$ clears the things up a lot. – Numenkok Balok Jan 18 '16 at 07:50
  • 3
    Gotta love physicists. – Arrow Jan 18 '16 at 11:23
  • 1
    The problem is that if dx is meant to be a single symbol, then I'll start to doubt the meaning of lots and lots of other symbols. Which ones are one and which ones are two? – Juan Perez Jun 23 '21 at 13:53
43

Gottfried Wilhelm Leibniz, who introduced this notation in the 17th century, intended $dx$ to be an infinitely small change in $x$ and $du$ to be the corresponding infinitely small change in $u$, so that if, for example, $du/dx=3$ at a particular point that means $u$ is changing $3$ times as fast as $x$ is changing at that point.

The notation $\dfrac{d^2u}{dx^2}$ actually means $\dfrac{d\left(\dfrac{du}{dx}\right)}{dx}$, the infinitely small change in $du/dx$ divided by the corresponding infinitely small change in $x$. Thus the second derivative is the rate of change of the rate of change.

Notice that if $u$ is in meters and $x$ in seconds, then $du/dx$ is in $\dfrac{\text{m}}{\text{sec}}$, i.e. meters per second, and $d^2 u/dx^2$ is in $\dfrac{\text{m}}{\text{sec}^2}$, i.e. meters per second per second. Thus $dx^2$ means $(dx)^2$, so the units of measurement of $x$ get squared, and $d^2y$ is in the same units of measurement that $y$ is in, consistently with the fact that $y$ is not a part of what gets squared in the numerator.

Michael Hardy
  • 1
  • 1
    Now I think of it, using $dx^2$ instead of $(dx)^2$ does not seem strange given that we write $\mathrm{cm}^2$ rather than $(\mathrm{cm})^2$. – Joe May 27 '21 at 09:26
37

$d$ is not a variable, and neither is $dx$ for that matter.

It is confusing because in some case, like the chain rule, differentials act like variables which can cancel:

$$\frac{dy}{dx}\frac{dx}{dt}=\frac{dy}{dt}$$

However, it is most appropriate to think of $\frac{d}{dx}$ as an operator that does something.

Thus, $\frac{d}{dx}(\frac{d}{dx} y)=\frac{d^2}{dx^2}y$.

Somewhat similarly, you wouldn't say that $\sin^2 x=s^2i^2n^2x$

Edit: In case it isn't from the example, you cannot separate $dx$. That is, $dx$ is not $d$ times $x$. This is very much analogous to chemistry when we say things like $\Delta H$. This isn't $\Delta$ times $H$. It is $\Delta$ (change) of $H$.

pancini
  • 19,216
  • 5
    Are you sure this is a good example? I wouldn't say $\sin^2 x = (\sin x)^2$ either, if I didn't know that's the conventional meaning. What I would rather say is $\sin x^2 = (\sin x)^2$, which is not commonly understood so. – leftaroundabout Jan 14 '16 at 22:48
  • 4
    @leftaroundabout I agree. I would more likely mistake $\sin^2 x$ to mean $\sin(\sin(x))$, which would actually be similar to the reasoning for $\frac{d}{dx}(\frac{d}{dx}y) = \left(\frac{d}{dx}\right)^2(y)=\frac{d^2}{dx^2}y$. And of course the issue with $\sin x ^ 2 = (\sin x)^2 $ is that it could easily be mistaken for $\sin\left(x^2\right)$ – David Etler Jan 15 '16 at 01:42
  • @leftaroundabout I don't see how somebody who didn't know what $\sin^2 x$ means could reasonably come to the conclusion taht it would mean $\sin(x^2)$. The "squared" is applied to the sine, so it could only reasonably mean "take the sine of $x$ and then square it" or "take the sine of $x$ twice (i.e., $\sin(\sin x)$." – David Richerby Jan 15 '16 at 05:31
  • 1
    @DavidRicherby: I didn't say anything about $\sin(x^2)$. In fact that was the point of my comment: it would, analogously to $\mathrm{d}x^2 \equiv (\mathrm{d}x)^2$, make some sense to have the convention $\sin! x^2 \equiv (\sin x)^2$, i.e. to parse the function application tighter than exponentiation. However, what's actually done is a weirder convention, namely $\sin^2 x \equiv (\sin x)^2$. – leftaroundabout Jan 15 '16 at 08:27
  • 5
    -1 This answer doesn't seem to address the main issue of the question ("A $d$ is missing.") at all. The OP seems to think that $dx^2$ means $d(x^2)$, not $(dx)^2$, so I don't see how he could agree with the notation $\frac{d}{dx}\frac{d}{dx} = \frac{d^2}{dx^2}$ used here without explanation. – JiK Jan 15 '16 at 10:02
  • Ok; I added another line in case it was still ambiguous. Thanks for the feedback. – pancini Jan 15 '16 at 17:37
8

${\rm d}(A)$ means an infinitesimally small change in $A$. The ${\rm d}$ is an operator and you better look at it as a function and not a value.

If anything we drop the parenthesis from ${\rm d}x$ for brevity as it should be ${\rm d}(x)$ as in $$\frac{{\rm d}(y)}{{\rm d}(x)}$$ and $$\frac{{\rm d}(\frac{{\rm d}(y)}{{\rm d}(x)})}{{\rm d}(x)} = \frac{ \frac{1}{{\rm d}(x)} {\rm d}({\rm d}(y))}{{\rm d}(x)} = \frac{{\rm d}({\rm d}(y))}{({\rm d}(x))^2} = \frac{{\rm d}^2(y)}{({\rm d}x)^2} = \frac{{\rm d}^2 y}{{\rm d}x^2}$$

John Alexiou
  • 13,816
  • 3
    The derivative $\frac{dy}{dx}$ is not the ratio of a small change of $y$ to a small change in $x$. Even in nonstandard analysis it's not defined that simply. –  Jan 14 '16 at 19:19
  • 6
    Actually it is $$ \frac{{\rm d}A}{{\rm d}x} = \lim_{h\rightarrow 0} \frac{ \left(A(x+h) - A(x)\right)}{\left( (x+h)-x \right)} $$ This is the definition of a derivative (https://en.wikipedia.org/wiki/Derivative). – John Alexiou Jan 14 '16 at 19:24
  • 1
    I'm well aware but your answer makes it seem like $\frac{dy}{dx} = \frac{d(y)}{d(x)}$ where you define $d(y)$ as a "small change in $y$" and likewise for $d(x)$. To be clearer, your answer makes it seem like $$\frac{dA}{dx} = \frac{A(x+h)-A(x)}{((x+h)-x)}$$ without the limiting process! –  Jan 14 '16 at 19:26
  • Ok, I added the word "infinitesimally" to the answer. Happy now? The point of the answer is to view the ${\rm d}(\square)$ as an operator. – John Alexiou Jan 14 '16 at 19:33
  • 3
    There are several contexts in which $d$ should be considered an operator -- chief among them being as the exterior derivative of a differential $k$-form -- but IMO the derivative in scalar calculus is not one of them. An "infinitesimally small number" not equal to zero doesn't exist in $\Bbb R$. This is fine as a "heuristic" but the claim that $\frac{dy}{dx}$ actually is a ratio -- whether of infinitesimals or finite differences -- is just not true. –  Jan 14 '16 at 19:37
  • A value can be infinitesimally small without being zero, the same way the Dirac function can be infinite but evaluate to a finite value over an integral. So the result of ${\rm d}(A)$ exists in $\Bbb R$ although undefined. I think you imply that ${\rm d}x \equiv 0$ whereas I imply that ${\rm d}x \neq 0$ – John Alexiou Jan 14 '16 at 20:56
  • 3
    That's false. Such a quantity would violate the Archmedean property of the real numbers. You have to extend the reals to the hyperreals to make use of nonzero infinitesimals. This is the closest formalization of what you're talking about that exists in mathematics and it still doesn't define the derivative as a fraction of infinitesimals, but as the standard part of a fraction of infinitesimals. Note that is not a part of standard analysis. –  Jan 14 '16 at 21:00
  • 1
    Also the formalization of the Dirac "function" is as a distribution or a measure -- i.e. something that only makes sense under an integral sign. It's not a function which evaluates to infinity at $x=0$ -- that's just the picture that we give students so that they have some way to visualize it. –  Jan 14 '16 at 21:04
  • 3
    Arguments about axiomatisation aside, this answer is the best intuitive answer to "where does the d go?", in my opinion. If you wanted to make it precise, you could simply say that $df$ is defined as $f(x+h)-f(x)$, and then have an implicit convention that we always take the $h\to 0$ limit whenever we write down an expression involving $d$. – N. Virgo Jan 16 '16 at 08:06
  • @Bye_World: There are places for squeemish nonsense and this one isn't one of them. $dy/dx$ might not be 'formally' a ratio, but it is BY DEFINITION indiscernable from a finite, well-defined ratio - you give me a desired level of precision and I will give you a ratio that is, up to precision, identical to $dy/dx$ at the given point. "This is intuitionists physics, not real math!!!!" you cry. But obviously, this is exactly what we're doing when we are taking limits. – Benjamin Lindqvist Jan 19 '16 at 22:08
5

Think of the meaning of $d/dx$. The $d$ in the numerator is an operator: it says, "take the infinitesimal difference of whatever follows $d/dx$". In contrast, the $dx$ in the denominator is just a number (yes, I know; mathematicians, please don't cringe): it is the infinitesimal difference in $x$.

So $d/dx$ means "take the infinitesimal difference of whatever follows, and then divide by the number $dx$."

Similarly, $d^2/dx^2$ means "take the infinitesimal difference of the infinitesimal difference of whatever follows, and then divide by the square of the number $dx$."

In short, the $d$ in the numerator is an operator, whereas in the denominator, it is part of a symbol. A slightly less ambiguous notation, as suggested by user1717828, would be to put the $(dx)$ in the denominator in parenthesis, but it really isn't necessary in practice.

Viktor Toth
  • 626