Official notation for double derivative is:

$$\frac{\mathrm d^2s}{\mathrm dt^2}$$

This term seems inconsistent. Two considerations:

  1. We have infinitesimal change in distance $\mathrm ds$ per infinitesimal change in time $\mathrm dt$: $\mathrm ds/\mathrm dt$. Both terms are a tiny value/interval. Because the $\mathrm d$ symbolizes difference, I would as a change of the change of the distance to time intuitively write: $$\frac{\mathrm d(\mathrm ds/\mathrm dt)}{\mathrm dt}=\frac{(\mathrm ds^2/\mathrm dt^2)}{\mathrm dt}=\frac{\mathrm ds^2}{\mathrm dt^3}$$ where the extra $\mathrm d$ says that both terms are now "double" infinitesimal differences.

  2. Maybe more properly following mathematical logic and not my intuition, the $\mathrm d$ could be considered a "free" variable in itself that can be multiplied onto this $\mathrm ds/\mathrm dt$ fraction numerator: $$\frac{\mathrm d(\mathrm ds/\mathrm dt)}{\mathrm dt}=\frac{(\mathrm d^2s/\mathrm dt)}{\mathrm dt}=\frac{\mathrm d^2 s}{\mathrm dt^2}$$ That agrees with the actual notation but doesn't really make physical sense now. $\mathrm d$ means (infinitesimal) difference, so that $\mathrm ds=s_{final}-s_{start}$, and therefore it makes no physical sense to consider the $\mathrm d$ and the $s$ separate. The $\mathrm ds$ is physically just a "name"/"symbol" for one term, which could just as well have been called $x$ or $a$ or anything else.

Now, while searching for an explanation, the answers always tend to consider $\frac{\mathrm d}{\mathrm dt}$ as one symbol in itself, so that a double derivative is $\frac{\mathrm d}{\mathrm dt}\frac{\mathrm d}{\mathrm dt}s=\frac{\mathrm d^2}{\mathrm dt^2}s=\frac{\mathrm d^2s}{\mathrm dt^2}$ - which makes even less physical sense, since the $\mathrm dt$ term has to be a separable term before we can treat $\frac{\mathrm ds}{\mathrm dt}$ as a normal fraction (as done in integration e.g.). $\frac{\mathrm d}{\mathrm dt}$ can't possibly be just "a symbol".

Why is $\frac{\mathrm d^2s}{\mathrm dt^2}$ the correct one in a physical context, where $\mathrm ds$ actually means the infinitesimal difference in $s$? Are my considerations in point 2 correct, and I just can't figure out that splitting $\mathrm d$ and $s$ is allowed?

  • $\begingroup$ I think the last one makes the most sense $\endgroup$ – Karl Feb 22 '17 at 20:48
  • $\begingroup$ @Karl I think "sense" then has a different meaning for me when thinking in physical context. How can $\mathrm d$ and $s$ be seperated and treated like were they two distinct variables? $\endgroup$ – Steeven Feb 22 '17 at 20:51
  • $\begingroup$ I'm not totally sure they are. Are you referring to solving variable separable equations? The notations conveniently hide the behind the scenes maths. $\endgroup$ – Karl Feb 22 '17 at 20:55
  • $\begingroup$ $\dfrac d {dt}$ is a symbol: it is the name of the operation of "differentiation". See Second derivative : Notation. $\endgroup$ – Mauro ALLEGRANZA Feb 22 '17 at 21:01

The mathematical answer, as has been noted by other in comments and answers, is that $d/dt$ is a single symbol in modern usage, and so it makes sense that the operator applied twice would be $(d/dt)^2 = \frac{d^2}{dt^2}$.

The physicist does often treat something like $ds/dt$ as a ratio between small quantities - at least intuitively. Strictly speaking mathematically this is somewhere between sloppy and wrong. In a lot of cases, however, it works functionally and the physicist goes along happily. This seems to be more along the lines of what you're asking.

In this case, (again noting that this is not mathematically rigorous at all), you could look at it this way. Maybe you have

$v(t) = \frac{ds}{dt} \\ a(t) = \frac{dv}{dt}$

So far so good since each one has the form that your intuition demanded - The first "looks" like a ratio between small changes in position with small changes in time. The second "looks" like a ratio between small changes in velocity with small changes in time. If you substitute, then you get

$a(t) = \frac{d}{dt} \frac{ds}{dt} = \frac{d^2 s}{dt^2}$

I'll again emphasize that thinking of these as separate symbols is mathematically sloppy and/or wrong, but it is consistent with what you were calling "physical intuition" in your question. The $d^2$ in the numerator came because you "simplified" the fraction - If you want to see it in terms of infinitesimals, you need run that last equation right-to-left to put it in a different form.


$\frac {d^2s}{dt^2}$ is the rate of change of the rate of change of distance. It is a differential operator acting on something that is already a derivative.

While Leibnitz originally thought of $\frac {ds}{dt}$ as a ratio of infinitesimals that was effectively a fraction in all senses, that is no longer what the notation means today.

$\frac {d}{dt}$ is indeed one symbol!

$\frac {ds}{dt}$ is the differential operator, $\frac {d}{dt}$ applied to the function $s(t)$ i.e. $\frac {ds}{dt} = \frac {d}{dt} s(t)$

$\frac {d^2s}{dt^2}$ is the differential operator applied to $\frac {ds}{dt}$ or $\frac {d}{dt}\frac{ds}{dt}$

so $\frac {d^2s}{dt^2} = (\frac {d}{dt})(\frac {d}{dt}) s(t)$

Futhermore, $\frac {ds^2}{dt^2}$ could be interpereted to be $(\frac {ds}{dt})^2$

  • $\begingroup$ Aha, thank you for this answer. A follow-up: So $\frac{ds}{dt}$ has become intepretted as $\frac{d}{dt}s$ - why not something like $\frac{d}{d}\frac{s}{t}$, which would then cause the second derivative to be $\frac{d}{d}\frac{d}{d}\frac{s}{t}=\frac{d^2s}{d^2t}$? Such notation would at least look consistent allover. $\endgroup$ – Steeven Feb 22 '17 at 21:05
  • $\begingroup$ It would not be any more a linear operator, which by definition operates on functions $\endgroup$ – Bernard Feb 22 '17 at 21:09

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