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So that's a question about something I want to deeply understand. Please correct me if anything of the following is wrong.

First of all for differentiation, the symbol $\frac{dy}{dx}$ came from the ratio $\frac{\Delta y}{\Delta x}$, and indeed this turned out to be a very suggestive notation, as you can find a great discussion about that issue here is $\frac{dy}{dx}$ not a ratio?. For example, the chain rule is almost trivial using this notation, because $\frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta u} * \frac{\Delta u}{\Delta x}$, and so in the limiting process the equation holds. (ignoring for the moment the issue of dividing by zero when $\Delta u$ is zero for arbitrarily small values of $\Delta x$). And that explains why it works in the chain rule and the differentials appears to cancel out, this is because indeed the deltas before the limiting process cancel out.

For definite integral the notation $\int_a^b f(x)dx$ again mimics that fact that the definite integral is a summation process(Leibniz considered it a sum of infinite number of infinitesimals quantities), that is the limit of a Riemmann sum $\sum f(x) \Delta x$, and if you think of integration by substitution, then Riemann sums makes the famous formula acceptable, for if you let $x = g(u)$ then for small $\Delta x$ we have $\Delta x \approx g'(u) \Delta u$ so the original Riemann sum will roughly be equal to $\sum f(g(u)) g'(u) \Delta u$, and these two sums can be rigorously proved to be equal in the limiting process, and the latter sum approximates $\int_{g^-1(a)}^{g^-1(b)} f(g(u)) g'(u)du$, so this explains why the substitution rule in the definite integral really works when treating $dx$ as differential, without resorting to the fundamental theorem of calculus and chain rule to justify it and then saying we put $dx$ to remember the chain rule, it works because $dx$ mimics $\Delta x$

What I don't understand is why the differential notation in the indefinite integral case also works. It works again when integrating by subtitution to find an antiderivative, it just works like magic like in the definite integral case, just treat it as differential and all the substitution rules will work, whether normal u-substitution or inverse trigonometric substitutioin. Yes I know we can prove it works in the indefinite integral case in both directions(forward and inverse substitution) like here teaching integration by substitution, but we could also do the same with differentiation and definite integral and prove that this differential notation work, without understanding why it did really turn out to work, that is because the differential mimics $\Delta x$ in both differerntiation and definite integral and working with $\Delta x$ we can arrive at the chain rule for differentiation and substitution rules for integration.

I hope I made my question clear.

Loai Ghoraba
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    Notations don't "work". They express what we know when we know what we are doing. – Kurt G. Jun 12 '24 at 11:48
  • @KurtG. I agree, and in the case of the derivative and the definitie integral I mentioned above, we expressed our knowledge by using differential notation because we know from where it came from and that this notation will be useful, but I don't know what knoweldge do we have about indefinite integral that made us use differential notation, knowing in advance it will be of our benefit – Loai Ghoraba Jun 12 '24 at 14:56
  • Any notation that expresses the relationship between $F$ and $f$ in Joshua Tiley's answer is as good as another. Newton used a few different ones that became out of fashion. Popularity is it that makes us prefer one notation over the other, not knowledge. If you think that the compatibility of Leibniz' notation with the chain rule for $F$ is an estetic advantage I don't disagree with that. – Kurt G. Jun 12 '24 at 16:04
  • What I mean, the compatibility of Leibniz' notation with the chain rule for $F$ is not a mere coincidence that it is estetic, it has a reason. So is there a similar reason for the indefinite integral, or it is just an estetic coincidence? – Loai Ghoraba Jun 12 '24 at 17:06
  • We are evolving around the question if the chain rule $$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$$ is an arithmetic cancellation of $dx,.$ This is clearly not the case but Leibniz' notation does us the estetic favour of helping to remember the rule. That's all. I quit now because that question if $dy/dx$ is a fraction has been discussed a zillion times here on MSE. – Kurt G. Jun 12 '24 at 17:10

1 Answers1

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An indefinite integral $\int f(x)\,dx$ can be defined as:

  1. A definite integral $\int_a^x f(t)\,dt$ for arbitrary $a$
  2. An antiderivative, i.e. any $F$ with $F'\equiv f$.

According to FTC these are equivalent. Well, if the Leibniz notation works for you for definite integration, then use 1.

Joshua Tilley
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