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I thought I knew that we use differentials just because they made calculus intuitive and there was math rigor behind all of these. But when I saw this 3B1B video about the topic this problem came out.

For example, deriving implicitly: $$ x^2+y^2 = 5 $$ we get $$ 2x\mathrm{d}x +2y\mathrm{d}y = 0 $$ $$ 2xdx+2y\frac{dy}{dx}dx=0 $$ and then to find the slope of the graph we solve for dy/dx $$ \frac{dy}{dx}=-\frac{x}{y} $$

Does't that imply that $dx \neq 0$ ?

And how would it be in the general case whe differentiating implicitly

Aatmaj
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zazke
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    $dx$ is not zero , it is just the limit as $\Delta x \to 0 $ meaning it is infinitesimally small but not zero. – The Integrator May 29 '18 at 05:10
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    @TheIntegrator: That makes absolutely no sense. If it's the limit of $\Delta x$ as $\Delta x\to 0$, then it's $0$. – Ted Shifrin May 29 '18 at 05:15
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    @TedShifrin how does that make no sense? if one says $x$ approaches $a$ then it means that one picks values closer and closer to $a$ but not exactly $a$. Explain to me how that makes no sense – The Integrator May 29 '18 at 05:34
  • @TheIntegrator We would have $\lim_{x\to a} x =a$. See the possible duplicates top answer for why viewing $dx$ as an "infinitesimal" is a bad view -- especially point made regarding Archimedian Property. – Andrew Tawfeek May 29 '18 at 05:49
  • 3b1b, in his calculus videos, explicitly says he thinks about the $dx$'s and $dy$'s as actual non-zero numbers (which are just very close to $0$, i think the phrase "some tiny amount" is said several times each video), which is a problematic view if you want to go into rigorous details, but it's wonderful for intuition about what the different terms signify (it's how Newton and Leibniz first developed the field, and how it was taught for over a century). – Arthur May 29 '18 at 06:14
  • @TheIntegrator thanks for your answer, but I think more correct would be factor out the dx and say dx can be cero, but it would be a trivial answer to my formula. – zazke Jun 04 '18 at 05:33
  • @Arthur Thank you, and yes it's kinda problematic if one wants mathematical rigor. Also, the (ε, δ)-definition of limit says just that but in other words. "close but not it" – zazke Jun 04 '18 at 05:41
  • I want to make a couple comments about why TheIntegrator's comment above that $dx$ is "just the limit as $\Delta x \to 0$" does not make sense. First of all, this phrase does not state what we are taking the limit of. "just the limit as $\Delta \to 0$" of what function? Secondly, if the function we're taking a limit of is $\Delta x$, then of course $\lim_{\Delta x \to 0} \Delta x = 0$. Finally, there is no such thing as a real number that is "infinitesimally small" but not zero. – littleO May 10 '21 at 06:53

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