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Suppose $f:\mathbb{R}^n\to\mathbb{R}$ is differentiable, we write its total differential as $$df = \sum_{i=1}^n \frac{\partial f}{\partial x_i} dx_i.$$ I am a bit confused on what the notation $dx_i$ means in this case? I know when each $x_i$ depends on some $t$ (i.e. $x_i:\mathbb{R}\to\mathbb{R}$), we have $$df = \sum_{i=1}^n \frac{\partial f}{\partial x_i} \frac{dx_i}{dt}$$ but the first case confuses me. Wouldn't each $dx_i=1$? If somebody could elaborate on this a bit that would be great.

varpi
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    Who is "we"? What book did you find that expression in? What was the chapter about? I'd like to have some context, please. – Jackozee Hakkiuz Aug 02 '21 at 04:24
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    Some notes that my instructor used, but here's a wiki link (I also meant total differential, not derivative, will edit that but I could also use some clarification on the differences): https://en.wikipedia.org/wiki/Differential_of_a_function#Differentials_in_several_variables – varpi Aug 02 '21 at 04:26
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    The second paragraph starts with the note "The precise meaning of the variables dy and dx depends on the context of the application and the required level of mathematical rigor", so what is your context of application? – Jackozee Hakkiuz Aug 02 '21 at 04:27
  • For example, if we wish to find some $x^$ such that $df=0$ at $x^$, how would we interpret it? – varpi Aug 02 '21 at 04:32
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    That sounds like you should be looking for a place where all partial derivatives of $f$ vanish. You can think of $dx_1,\dots,dx_n$ as linearly independent quantities so that the equation $$ 0 = df = \partial_1 f dx_1 + \dots + \partial_n f dx_n $$ imples that every $\partial_i f=0$. – Jackozee Hakkiuz Aug 02 '21 at 04:34
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    One place to look for a perfectly rigorous explanation of the meaning of this equation is a book that explains calculus on manifolds or differential forms, such as Hubbard and Hubbard (which is a fairly readable example of such a book). – littleO Aug 02 '21 at 04:36
  • So each of the $dx_i$ does not contribute to the problem? I just wish there is a more rigorous way this is explained. Intuitively I understand it as "$df$ is the amount $f$ changes, which depends on how much $f$ changes with respect to each $x_i$ and how much each $x_i$ changes". But is there a more rigorous explanation? Also feel free to let me know if this intuition is incorrect – varpi Aug 02 '21 at 04:36
  • Here's an attempt to interpret this equation without using too much of the language of differential forms. For each $x \in \mathbb R^n$, $df(x)$ is a function that takes as input a vector $\Delta x = \begin{bmatrix} \Delta x_1 \ \vdots \ \Delta x_n \end{bmatrix} \in \mathbb R^n$ and returns as output the number $\sum_{i=1}^n \frac{\partial f(x)}{\partial x_i} \Delta x_i$. We care about $df$ because it tells us how to approximate the change in $f$ when the input to $f$ changes from $x$ to $x + \Delta x$. If $\Delta x$ is "small", then $df(x)(\Delta x) \approx f(x + \Delta x) - f(x)$. – littleO Aug 02 '21 at 04:38
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    For your specific problem you can just think of $df=0$ as a shorthand for "all partial derivatives vanish". If you really want to give the $df$ and the $dx_i$ a meaning you can follow littleO's suggestion and find a book about differential forms, which is a possible formalization of what the objects $df$ and $dx_i$ are. – Jackozee Hakkiuz Aug 02 '21 at 04:38
  • @littleO I will take a look at that, thanks – varpi Aug 02 '21 at 04:38
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    See this answer for the single-variable special case, or this one for the multivariable case for the explanation of 1-forms on $\Bbb{R}^n$ which simply amounts to thinking of derivatives as linear transformations. – peek-a-boo Aug 02 '21 at 04:42
  • I am voting to close this question as too broad; the tl;dr here is that $dx_i$ represents an arbitrary, infinitesimal first-order change in $x_i$, and $df$ is the resulting first-order change in $f$. A more rigorous treatment is beyond the scope of a single Math.SE answer; littleO and peek-a-boo have given good starting points. – user7530 Aug 02 '21 at 05:59
  • the $dx_i$ are differential forms. See here – Masacroso Aug 02 '21 at 15:51

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