Official Leibniz notation for double derivative is:
$$\frac{\mathrm d^2s}{\mathrm dt^2}$$
This term seems inconsistent. Two considerations:
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.
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?
Update
The answers already given at this time both indicate the use of $\mathrm d/\mathrm dt$ as merely a symbol. So, neither of my two suggestions mentioned above are the case. Sure, I can accept that. But the question still remains of why as well of how come we still treat them as variables then, e.g. in integration?
Let me clarify those two points:
Firstly, if it indeed is the case that $\mathrm d/\mathrm dt$ is merely a symbol and should be thought of as just a symbol, then I do not understand the motivation for this symbol.
- Why did Leibniz choose $\mathrm d/\mathrm dt$ as a symbol, which causes the confusion and inconsistency described in the question above?
- Why not, say, $\mathrm d/\mathrm d$, in which case we would get a writing-style that at least looks a bit more "consistent": $$\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm ds}{\mathrm dt}\qquad \frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm d^2s}{\mathrm d^2t}\qquad \frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm d^3s}{\mathrm d^3t}\qquad \cdots$$
- Or even better yet, if the two $\mathrm d$ involved in this $\mathrm d/\mathrm dt$ symbol have no meaning as neither a variable nor an indicator of a change in the parameter, then why use this letter at all? Why not stick to, say, the prime-notation throughout and never jump into the Leibniz notation: $$s_t'\qquad s_t''\qquad s_t'''\qquad \cdots$$
And secondly, if the $\mathrm d/\mathrm dt$ really just is a symbol, and that's it, then how come we suddenly can treat it as a fraction again containing a set of variables $\mathrm ds$ and $\mathrm dt$ that we can split apart during for instance integration? Such as here: $$\frac{\mathrm ds}{\mathrm dt}=v\quad\Leftrightarrow\quad \mathrm ds=v\,\mathrm dt\quad\Leftrightarrow\quad \int 1 \,\mathrm ds=\int v\,\mathrm dt \quad\Leftrightarrow\quad s=\int v\,\mathrm dt$$
I hope to get this notion cleared out and appreciate all comments and answers that can help.