Undergrad Need help with derivative notation

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The discussion revolves around finding a compact notation for the derivative of a scalar function when the variable is a vector rather than a single variable. The gradient operator, denoted as ∇f, is suggested as a suitable compact notation for this multivariable case. There is a debate about the correct notation for indexing vector components, with some arguing for using subscripts instead of superscripts to avoid confusion. Ultimately, the original poster realizes that their notational issue led to an error in their proof related to differential geometry. The conversation highlights the importance of notation in mathematical proofs and clarity in vector calculus.
orion
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If I have a scalar function of a variable ##x## I can write the derivative as: ##f'(x)=\frac{df}{dx}##.

Now suppose ##x## is no longer a single variable but a vector: ## x=(x^1, x^2, ..., x^n)##. Then of course we have for the derivative ##(\frac{\partial f}{\partial x^1}, ..., \frac{\partial f}{\partial x^n})##.

But for a proof I need a compact notation like ##\frac{df}{dx}## for this multivariable case. Does such a compact notation exist? I mean, a notation without making explicit reference to components.

Thanks in advance.
 
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orion said:
If I have a scalar function of a variable ##x## I can write the derivative as: ##f'(x)=\frac{df}{dx}##.

Now suppose ##x## is no longer a single variable but a vector: ## x=(x^1, x^2, ..., x^n)##. Then of course we have for the derivative ##(\frac{\partial f}{\partial x^1}, ..., \frac{\partial f}{\partial x^n})##.

But for a proof I need a compact notation like ##\frac{df}{dx}## for this multivariable case. Does such a compact notation exist? I mean, a notation without making explicit reference to components.

Thanks in advance.
How about the ##∇f## operator?
(https://en.wikipedia.org/wiki/Gradient)
 
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Thanks, fresh 42. I'm sorry I'm late in responding, but I forgot I wrote this question. It turns out that after I wrote this, I realized a mistake I was making in the proof and you are right, the gradient works. Thanks again.
 
orion said:
If I have a scalar function of a variable ##x## I can write the derivative as: ##f'(x)=\frac{df}{dx}##.

Now suppose ##x## is no longer a single variable but a vector: ## x=(x^1, x^2, ..., x^n)##. Then of course we have for the derivative ##(\frac{\partial f}{\partial x^1}, ..., \frac{\partial f}{\partial x^n})##.
This -- ## x=(x^1, x^2, ..., x^n)## -- should probably be written as ## x=(x_1, x_2, ..., x_n)## to avoid confusion. Although I have seen a few textbooks that use superscripts as indexes, most use superscripts to denote exponents rather than indexes.

Also, this -- ##(\frac{\partial f}{\partial x^1}, ..., \frac{\partial f}{\partial x^n})## -- should be written as ##(\frac{\partial f}{\partial x_1}, ..., \frac{\partial f}{\partial x_n})## for the same reason.
orion said:
But for a proof I need a compact notation like ##\frac{df}{dx}## for this multivariable case. Does such a compact notation exist? I mean, a notation without making explicit reference to components.

Thanks in advance.
 
Mark44 said:
This -- ## x=(x^1, x^2, ..., x^n)## -- should probably be written as ## x=(x_1, x_2, ..., x_n)## to avoid confusion. Although I have seen a few textbooks that use superscripts as indexes, most use superscripts to denote exponents rather than indexes.

Also, this -- ##(\frac{\partial f}{\partial x^1}, ..., \frac{\partial f}{\partial x^n})## -- should be written as ##(\frac{\partial f}{\partial x_1}, ..., \frac{\partial f}{\partial x_n})## for the same reason.

No, it has to be written the way I wrote it. Otherwise, the Einstein summation convention does not work and also there is a need to distinguish contravariant components from covariant components.

I realize that I posted in a calculus forum but that was because I wanted input on a derivative notation from vector calculus. It's actually a proof in differential geometry. But in the end I found out that my notational problem was pointing a way to an error in my proof.
 

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