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I try to make sense of notations like $$\frac{df}{dx},\frac{d^2 f}{dx^2},\text{etc.}$$ My idea is the following: Let $d_\epsilon$ denote the operand $f\mapsto(x\mapsto f(x+\epsilon)-f(x))$. Then I interpret a term $t$ containing the symbol $d$ as $\lim_{\epsilon\rightarrow 0}t_\epsilon$, where $t_\epsilon$ is the result of replacing all occurences of $d$ in $t$ with $d_\epsilon$. Applying this to $\frac{d^2 f}{dx^2}$, while treating $d^2 f$ as $d(d(f))$ and $dx^2$ as $d(\text{id})^2$, gives me $$x\mapsto\lim_{\epsilon\rightarrow 0}\frac{f(x+2\epsilon)-2f(x+\epsilon)+f(x)}{\epsilon^2}.$$ This does not meet the definition of the second derivative of $f$, however, I suspect the two approaches may coincide under certain smoothness assumptions on $f$, maybe being $C^2$ is enough? How about the general case $\frac{d^n f}{dx^n}$?

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  • $\begingroup$ Under reasonable regularity assumptions ($C^k$ for the $k$th derivative will do), your interpretation is correct. The other way is to think of just $d/dx$ as one indivisible operator, and then interpret $d^n/dx^n$ as $(d/dx)^n$. $\endgroup$ – Ian Jun 10 '18 at 15:29
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I think it comes from: $$f'(x)=\frac{\mathrm{d}f(x)}{\mathrm{d}x}$$ $$f''(x)=(f')'(x)=\frac{\mathrm{d}f'(x)}{\mathrm{d}x}=\frac{\mathrm{d}\frac{\mathrm{d}f(x)}{\mathrm{d}x}}{\mathrm{d}x}$$ And if we forget that they're not fractions for a second we can get something like this: $$f''(x)=\frac{\mathrm{d}\mathrm{d} f(x)}{\mathrm{d}x \mathrm{d}x}=\frac{\mathrm{d}^2f(x)}{\mathrm{d}x^2}$$ And you can do the same for the higher order derivatives as well.

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