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My question is very simple, I'm a beginner student in real analysis in several variables and I don't understand the meaning of the $\cdot$ in the expression: $(g\circ f)'(a)=g'(f(a))\cdot f'(a):\mathbb R^m\to\mathbb R^p$. Does it mean a composition? Why doesn't the author denote this using the classical $\circ$ symbol?

Chain Rule: Let $U\in \mathbb R^m, V\in \mathbb R^n$ be open sets and $f:U\to\mathbb R^n$ differentiable at the point $a$, with $f(U)\subset V$ and $g:V\to \mathbb R^p$ differentiable at the point $f(a)$. Then $g\circ f:U\to \mathbb R^p$ is differentiable at the point $a$ with $(g\circ f)'(a)=g'(f(a))\cdot f'(a):\mathbb R^m\to\mathbb R^p$

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    It means multiplication, like for example $x\cdot x=x^2$. – 2017-02-09
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    @MJD But they aren't numbers. Note that $g'(f(a)$ and $f'(a)$ are functions. – 2017-02-09
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    Just to be sure: what is your book's definition of $f'(a)$? Can we say that $f'(a)$ is a matrix? – 2017-02-09
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    The product of two functions, say $f$ and $g$, is another function, say $h=f\cdot g$, which has $h(x) = f(x)\cdot g(x)$ for each $x$. – 2017-02-09

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The $\cdot$ actually stands for matrix multiplication. Recall that $f'(a)$ is an $m \times n$ matrix for any $a$.

Recall that $f'(a)$ is actually a linear map from $\Bbb R^n$ to $\Bbb R^m$. That is, if $\mathcal L(\Bbb R^n,\Bbb R^m)$ denotes the space of such maps, then $$ f': \Bbb R^n \to \mathcal L(\Bbb R^n,\Bbb R^m) $$ Consequently, $f'(a)$ actually has a suppressed argument: if $h \in \Bbb R^n$, then $[f'(a)](h) \in \Bbb R^m$.

With that, we may write the chain rule properly as follows: $$ [(g \circ f)'(a)](h) = [g'(f(a))]([f'(a)](h)) $$ In other words, there are two levels of composition here. One is denoted by $\circ$, the other is denoted by $\cdot$. For linear maps, the tendency is to think of composition as "matrix multiplication", hence the answers given.

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    Or, to be extra correct and confusing, it's composition of linear maps. Matrices only appear once you choose a basis. – 2017-02-09
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    @Arthur I would then write "the pointwise composition of linear maps" to emphasize that we should "plug in $a$" before composing. That being said, who are we kidding? $f$ is a function on $\Bbb R^n$, of course there's a canonical basis. – 2017-02-09
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    We have chosen / been given / fixed $a$ from the start, so there is no need for the "pointwise" part. Anyways, I agree that this is pedantic. Let's just say it's a matrix multiplication. – 2017-02-09
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    @Arthur You're right; I just remember getting hung up on that point when I was looking at Frechet derivatives. In any case, agreed. – 2017-02-09
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    @Arthur so if it's a composition of linear transformations, why doesn't the author use the $\circ$ symbol? – 2017-02-09
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    @Arthur... and there it is – 2017-02-09
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    @user42912 I could answer that. However, I would need you to add (to your question) the definition of a derivative that you're primarily working with. – 2017-02-09
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    @Omnomnomnom Sorry about that. Good luck. – 2017-02-09
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    Do you mean derivatives of functions $f:\mathbb R\to \mathbb R$ or the definition of derivatives at higher dimensions? – 2017-02-09
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    The higher dimensions definition. Or, if you prefer not to type it, let's use links. Give me a sec. – 2017-02-09
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    @user42912 Does your definition look like that given [here](https://en.wikipedia.org/wiki/Derivative#Total_derivative.2C_total_differential_and_Jacobian_matrix), or is it just defined to be the [Jacobian matrix](https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant)? – 2017-02-09
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    It's a linear transformation as shown in the link you've just sent. The Jacobian matrix is just this transformation in the usual bases, but the definition is the linear transformation. – 2017-02-09
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    ...hence your confusion. Okay, I'll add something to my question. – 2017-02-09
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    See my latest edit. – 2017-02-09
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    @Arthur I guess it was good that you brought out the fundamental confusion here. So thanks – 2017-02-09