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The operator $\frac{d}{dx}$ sometimes behaves like a fraction.For eg. when we use chain rule to write $\frac {d}{dy}=\frac {d}{dx}\frac {dx}{dy}$ it apparently seems the $dx$ cancel out to give the previous thing.Also when differentiating parametric functions we use $\frac {dy}{dx}=\frac {dy/dt}{dx/dt}$ it seems that the $dt$ cancel out as if they were fractions.

I referred a few books but could not find any explanations (rather how the thing works).I will be great if someone explains this.

Thanks for any help!!

Soham
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  • They are fractions, but this interpretation has gone out of style. A. Ya. Khinchin's book Eight Lectures on Mathematical Analysis explains this. – MJD Jun 15 '17 at 03:58
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    You might want to look at previous questions on this topic, such as https://math.stackexchange.com/questions/1784671 – David K Jun 15 '17 at 04:02
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    similar questions (here, here) – Dando18 Jun 15 '17 at 04:05
  • 3blue1brown has a video that may help with this. The link takes you to the relevant part of the video, but I recommend watching the whole thing if any part of it seems confusing. – WB-man Jun 15 '17 at 04:09
  • They are fractions. Note that the derivative is defined as the limit of a fraction! – Tucker Jun 15 '17 at 04:13
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    More precisely, they were fractions, before taking the limit. – quasi Jun 15 '17 at 04:29
  • @quasi But when I write dx I have already taken the limit and carrying out my operations on basis of what I obtained after taking the limit... – Soham Jun 15 '17 at 05:30
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    @tatan -- Yes, but when you roll back to the before-limit context, the laws of multiplying fractions were applicable. So (informally), roll back, multiply the fractions, take the new limit, and compare against taking the limits first, and multiplying later. – quasi Jun 15 '17 at 06:28
  • https://en.wikipedia.org/wiki/Chain_rule#First_proof – quasi Jun 15 '17 at 06:40
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    I reread the Khinchin thing last night, it's brilliant. (Section 32 if you are looking for it.) I can't do it justice by summarizing, because it's so pithy. But Khinchin says, let $y$ be a function of $x$, and let $\Delta x$ be a small increment in $x$; this produces a corresponding small increment in $y$ which we can call $\Delta y$. This $\Delta y$ is in general hard to calculate, but if we define $dy = y'(x) \Delta x$, then $dy$ is fairly easy to calculate. The notation is a little odd, because $dy$ is actually a function of both $x$ and $\Delta x$. But… – MJD Jun 16 '17 at 12:35
  • One can show that $dy$ is a good approximation to $\Delta y$ in the following sense: The quantity $\alpha = \Delta y - dy$ is infinitesimal as $\Delta x\to 0$, and moreover it is an infinitesimal of a higher order than $\Delta y$, so if you replace $\Delta y$ with $dy$, not only does the absolute error introduced go to zero as $\Delta x\to 0$ but also the relative error. – MJD Jun 16 '17 at 12:38
  • Moreover, the quantity $dy$ is completely characterized by that property and by the property of being a linear function of $\Delta x$; it is the unique linear approximation to $\Delta y$ that gives an infinitesimal relative error. Khinchin observes that one can use those two characterizing properties as a definition of $dy$ independent of the derivative, and then define the derivative in terms of differentials as $\lim_{\Delta x\to 0}\frac{dy}{dx}$ if one wants; one can also prove the chain rule based on quotients of differentials defined in this way. – MJD Jun 16 '17 at 12:39
  • @MJD Thanks for your help!!! – Soham Jun 16 '17 at 12:57

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