Is it mathematically correct to write $$f'(x)=\lim_{dx\to0}\frac{f(x+dx)-f(x)}{dx},$$ rather than $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}?$$ If not, what is the difference? If so, why isn't this notation used from the beginning? My feeling for the latter is that it would align the derivative more with the inverse of the indefinite integral.
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3$\begingroup$ For educators: follow the notation in the textbook. Even if both of these are correct, it does not help students at this level to see variant notations. $\endgroup$– Gerald EdgarOct 21, 2015 at 14:48
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4$\begingroup$ In smooth infinitesimal analysis one writes $f(x+dx)=f(x)+f'(x)dx$, where $dx$ is an infinitesimal, without any limits. $\endgroup$– Dag Oskar MadsenOct 21, 2015 at 22:04
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6$\begingroup$ @GeraldEdgar I would be more convinced by that if I ever saw any evidence that my students read the textbook, or if I found a textbook that I could wholeheartedly recommend that they read. $\endgroup$– Mike ShulmanOct 27, 2015 at 5:41
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4 Answers
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To use $$f'(x)=\lim_{dx\to0}\frac{f(x+dx)-f(x)}{dx}$$ is mathematically correct if $dx$ is the name for a real variable. (If it should be something else it needs to be made clear what it should be.)
It would also be correct to say $$f'(x)=\lim_{\text{small}\to 0}\frac{f(x+\text{small})-f(x)}{\text{small}}$$ with the understanding that $\text{small}$ is the name of a real variable.
An issue I see with what you propose is though that $dx$ is not a common notation for a real variable, but rather something else. What exactly $dx$ means, depends on the context, but typically it is not used to denote a real number.
A very related notation that is more common is $$f'(x)=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$
The idea is that the $\Delta x$ is a finite difference, as opposed to an infinitesimal difference that might be denoted by $dx$ (where the latter notion might or might not be made precise).
Personally I prefer to use just an $h$ or something similar, to emphasis that it is just some real parameter there nothing special or mysterious. To name it $dx$ or $\Delta x $ goes counter this so I would not do it.
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1$\begingroup$ Agreed that it is mathematically correct if $dx$ is the name for a real variable. I do use it that way, despite all the $h$s in the textbook, because I think it is important to emphasize that the role played by this variable is that of a *small change in* $x$. And I think insisting on writing $\Delta x$ when the small change is "finite" is misguided pedantry, especially in textbooks that a few sections later on when talking about "differentials" define $dx = \Delta x$! $\endgroup$ Oct 27, 2015 at 5:35
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In the other examples, one can make sense of $\mathrm{d}x$ in a consistent way: e.g. if we model it as differential forms, then $\frac{\mathrm{d}y}{\mathrm{d}x}$ is a ratio between two differential forms, $\int \sin x \, \mathrm{d} x$ is the anti-derivative of a differential form, and $\int_a^b \sin x \, \mathrm{d} x$ is the integral of a differential form along a (directed) path. One can even reasonably extend differentiable functions to the exterior algebra, and make literally true statements like $f(x + \mathrm{d}x) - f(x) = f'(x) \mathrm{d}x$.
However, if one were to write the calc-1 definition of derivative as $$ \lim_{\mathrm{d}x \to 0} \frac{f(x + \mathrm{d}x) - f(x)}{\mathrm{d} x} $$ the $\mathrm{d}x$ here doesn't represent anything resembling a differential form at all. The superficial similarity to my last remark in the previous paragraph makes things worse, not better, since it would conflate the distinct ideas, and thus make it harder to intuitively arrive at what $\mathrm{d} x$ means.
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1$\begingroup$ This definitely sheds some light on the comparison between $h$ and $dx$. I suppose that to use the latter in the sense of the definition of the derivative, and defining $dx$ as the same differential form as integration, the definition would not make sense since $dx$ is already infinitesimally small. That is, we'd get $0/0$ for every derivative when making $dx\to 0$. Is this a right conclusion? $\endgroup$ Oct 23, 2015 at 11:38
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$\begingroup$ I disagree that $dx$ here doesn't represent anything resembling a differential form. In fact, it resembles a differential form quite a lot! If we free ourselves from insisting that a "differential form" must always depend linearly on $dx$, then we can consider $f(x+dx)-f(x)$ to indeed be a differential form (a function depending on both a variable and a small change in that variable) that is linearly approximated by another differential form $dy$ (essentially by definition of the latter). $\endgroup$ Oct 27, 2015 at 5:38
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A sitenote: When you leave classical analysis and you take non-standard analysis, then you have infinitesimals in your underlying theory (they are part of the so called hyperreals). In this theory you can write $$\frac{f(x+dx)-f(x)}{dx}$$ with $dx$ being an infinitesimal. This gives a number which difference to $f'(x)$ is only an infinitesimal (see the comments to this question). No hocus pocus necessary ;-)
answered Oct 24, 2015 at 14:16
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1$\begingroup$ Yes, in non-standard analysis you need no limit to define the derivative. The reason is, that you have infinitesimals which don't exist in classical analysis and which you can use in the definition of the derivative. $\endgroup$ Oct 24, 2015 at 14:21
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1$\begingroup$ See math.stackexchange.com/questions/1271476/… for more details $\endgroup$ Oct 24, 2015 at 14:22
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3$\begingroup$ As a side note, for differentiability, the (standard part of the) quotient can't depend upon which non-zero infinitesimal is used, so it takes at least one explicit quantifier to define the derivative and to make sure it exists in the first place. $\endgroup$ Oct 24, 2015 at 16:17
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3$\begingroup$ You still need hocus pocus to define the derivative: NSA just picks a different magic (albeit, arguably conceptually simpler magic). $\endgroup$– user797Oct 25, 2015 at 9:48
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2$\begingroup$ +1, but: let E be the expression written in this answer. Note that in NSA, E is not the derivative f'(x). The derivative equals the standard part of E. $\endgroup$– user507Oct 25, 2015 at 21:32
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When Leibniz (co)invented the calculus, he also introduced the symbol $dx$ for an infinitesimal increment of the variable $x$. Today when the calculus is presented there is an immediate asymmetry between the independent variable $x$ and the dependent variable $y$ but to Leibniz $x$ and $y$ had equal rights; he was mostly working with curves in the $x,y$ plane. When you are dealing with more than one variable, it is important to indicate which variable is undergoing the increment. Therefore the symbols $dx,dy$ are indispensable.
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2$\begingroup$ This does not seem to answer the question. Can you elaborate on how this is related to possibly writing $dx$ for $h$ in the definition? $\endgroup$ Feb 11, 2016 at 17:56