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In differential calculus for beginners, Joseph Edwards alerts the reader against the fallacious notion of $\frac{dy}{dx}$ being a ratio.

But then in Integral calculus they play around with $\ dy$ and $\ dx$ as if $\frac{dy}{dx}$ was a ratio. They do all kinds of crazy stuff like writing $\ dy$ as a function of $\ dx$. Furthermore, in "Calculus made Easy" by Silvanus Thompsun, he completely uses it as a ratio. In fact, he proves chain rule and other theorems in Calculus by treating $\frac{dy}{dx}$ just like the ratio of two real numbers and he gets away with it without any problems.

For example, he proves the chain rule somewhat like this:

Let $\ y = f(g(x))$. Then let $\ g(x) = u$. Therefore $\ y = f(u)$, so $\frac{dy}{du} = f'(u)$. And $\frac{du}{dx} = g'(x)$, since $\frac{dy}{du} • \frac{du}{dx}= \frac{dy}{dx}, \frac{dy}{dx} = f'(u) • g'(x) = f'(g(x)•g'(x)$.

Isn't that a fallacious way? But if it is, why does it work out perfectly? Some readers have referred to a previous question, but it doesn't solve my problem because what i specifically want to know is why when the derivative is not exactly a ratio, treating it like one doesnt lead to contradictions? How can you get the same chain rule proved with formal, logical, airtight proofs by playing with the derivative like a drunkard?

Asaf Karagila
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Vasus Deus
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    This has been asked – FShrike Dec 10 '23 at 18:04
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    https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio – FShrike Dec 10 '23 at 18:05
  • True, but I couldn't get what I was looking for. What concerns me is when it is not a ratio, why doesn't it lead to contradictions when people use it like a ratio? – Vasus Deus Dec 10 '23 at 18:06
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    This has been asked, but to answer your question: not really. The derivative is a "ratio of infinitesimals" and is formally defined as a limit. In particular, $\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$. – codeing_monkey Dec 10 '23 at 18:07
  • Just because it doesn't lead to a wrong answer means it's correct – Andrew Dec 10 '23 at 18:07
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    @VasusDeus It's a very successful notation, which has properties of a ratio. But they are all justified by theorems and formal proofs. It's not like you can treat it as a ratio and get a proof of the chain rule or something. That wouldn't be correct. – Mark Dec 10 '23 at 18:09
  • @Andrew I guess a typo made the difference? – Vasus Deus Dec 10 '23 at 18:09
  • this has been asked and it's also the second question with more upvotes on MSE – Sine of the Time Dec 10 '23 at 18:10
  • @VasusDeus : I am proving to you now why it does no lead to contradictions: As we know we can solve certain ODEs by treating $dy/dx$ as a ratio, for example $dy/dx=y,.$ That may not be rigorous but we can check very rigorously that the solution thus obtained solves the ODE. No contradiction. I will happily do this until I die. – Kurt G. Dec 10 '23 at 18:13
  • @JohnDouma because in other cases it leads to a contradiction. Till date, I don't know of any case where treating dy/dx as a ratio caused contradictions. I'd really love to know if there are any – insipidintegrator Dec 10 '23 at 18:22
  • @VasusDeus One reason to not use it as a ratio, is that people would then also use partial derivatives as ratios, and this definitely leads to contradictions. E.g. $df = \frac {\partial f} {\partial x} dx + \frac {\partial f} {\partial y} dy$: you cannot simplify $\partial x$ with $dx$. Which is the reason why the symbol used is not the same, but many people (including me) often write $dx$ instead of $\partial x$. – Jean-Armand Moroni Dec 10 '23 at 18:25
  • I haven't read Thompson, so I'm not sure how careful the book's treatment is, but one should use derivatives' definitions as limits, rather than pretend they're fractions, to prove $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$. – J.G. Dec 10 '23 at 18:29

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Actually from one point of view it is a ratio, but from another is not.

It can be viewed as a ratio, because:

$$\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x)-f(x)}{\Delta x}$$

is a limit of this ratio.

But it can also be viewed as not a ratio, because the outcome is finally either a fixed contant or infinity or undefined rather than a ratio.

Hope this helps you.

ZYX
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