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In school I was taught that we use $\frac{du}{dx}$ 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:

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

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

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

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.

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    $\begingroup$ $d$ 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. $\endgroup$ – Michael Burr Jan 14 '16 at 19:06
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    $\begingroup$ $d$ cannot be used just like any variable. Otherwise you will have $du/dx=u/x$ for example. $\endgroup$ – velut luna Jan 14 '16 at 19:08
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    $\begingroup$ I have a hard time to believe somebody told you this. Maybe they said it about the entity "$dx$" $\endgroup$ – quid Jan 14 '16 at 20:24
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    $\begingroup$ It's because the dx is "in parentheses", so to speak. $\endgroup$ – Mehrdad Jan 14 '16 at 21:20
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    $\begingroup$ I kinda assumed $dx^2$ meant $(dx)^2$; i.e. $dx$ is basically one variable. $\endgroup$ – Akiva Weinberger Jan 15 '16 at 0:16
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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.

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    $\begingroup$ +1. I'm surprised that so many other answers missed this aspect of the question. $\endgroup$ – mweiss Jan 15 '16 at 16:22
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    $\begingroup$ 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). $\endgroup$ – Jeppe Stig Nielsen Jan 15 '16 at 22:10
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    $\begingroup$ 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. $\endgroup$ – Numenkok Balok Jan 18 '16 at 7:50
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    $\begingroup$ Gotta love physicists. $\endgroup$ – Arrow Jan 18 '16 at 11:23
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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.

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$d$ 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 $H$.

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    $\begingroup$ 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. $\endgroup$ – leftaroundabout Jan 14 '16 at 22:48
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    $\begingroup$ @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 $\sin\left(x^2\right)$ $\endgroup$ – David Etler Jan 15 '16 at 1:42
  • $\begingroup$ @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., $\sin(\sin x)$." $\endgroup$ – David Richerby Jan 15 '16 at 5:31
  • $\begingroup$ @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 $\sin^2 x \equiv (\sin x)^2$. $\endgroup$ – leftaroundabout Jan 15 '16 at 8:27
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    $\begingroup$ -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. $\endgroup$ – JiK Jan 15 '16 at 10:02
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${\rm d}(A)$ 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 $$\frac{{\rm d}(\frac{{\rm d}(y)}{{\rm d}(x)})}{{\rm d}(x)} = \frac{ \frac{1}{{\rm d}(x)} {\rm d}({\rm d}(y))}{{\rm d}(x)} = \frac{{\rm d}({\rm d}(y))}{({\rm d}(x))^2} = \frac{{\rm d}^2(y)}{({\rm d}x)^2} = \frac{{\rm d}^2 y}{{\rm d}x^2}$$

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    $\begingroup$ 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. $\endgroup$ – user137731 Jan 14 '16 at 19:19
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    $\begingroup$ 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 (en.wikipedia.org/wiki/Derivative). $\endgroup$ – ja72 Jan 14 '16 at 19:24
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    $\begingroup$ 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. $\endgroup$ – user137731 Jan 14 '16 at 19:37
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    $\begingroup$ 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. $\endgroup$ – user137731 Jan 14 '16 at 21:00
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    $\begingroup$ 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 $d$. $\endgroup$ – Nathaniel Jan 16 '16 at 8:06
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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.

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