I'm reading the total derivative as a linear map definition on Wikipedia. The function in the definition is $f:U \mapsto \mathbb{R}^m$, where $U$ is an open set in $\mathbb{R}^n$. However, the derivative is defined as a linear map that maps from $\mathbb{R}^n$ to $\mathbb{R}^m$.

Does that mean the linear map has a different domain (i.e. $\mathbb{R}^n$) than the function itself (the function has domain $U$)? How is that possible? In my understanding, the total derivative, when given a point from the domain of the function itself (i.e. $U$), tells us the speed and direction of the function at that point. If so, it doesn't make sense to put a point that's in $\mathbb{R}^n$ but not in $U$ into the total derivative, does it?


No, the derivative is defined just in the domain of the function. In other words: how you could define the derivative of $f$ where there is no $f$?

You are mixing the concept of derivative of a function with the concept of derivative of a function at a point. Observe that if $f:U\to\Bbb R^m$ for some open $U\subset\Bbb R^n$ is differentiable then

$$\partial f: U\to\mathcal L(\Bbb R^n,\Bbb R^m)$$

and $\partial f(x)\in\mathcal L(\Bbb R^n,\Bbb R^m)$ for some $x\in U$. You see the difference?

However you can expand the domain of the function that defines a derivative, if you wish, whenever the expansion defines a function properly. But this new function is not anymore the derivative of the original function.

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  • $\begingroup$ great answer! A follow up question: using the same notation before, let $D_jf_i(x)$ be the partial derivative of the ith component function with respect to the jth variable at $x$. Then are domain and codomain of $D_jf_i$ (not $D_jf_i(x)$) $U$ and $\mathcal{L}(\mathbb{R},\mathbb{R})$ respectively? $\endgroup$ – user1691278 Mar 11 '18 at 6:58
  • $\begingroup$ yes, this is correct. $\endgroup$ – Masacroso Mar 11 '18 at 7:57
  • $\begingroup$ Why is the codomain not $\mathbb{R}$? For each input (i.e. $x \in U$), you get a real number. $\endgroup$ – user1691278 Mar 11 '18 at 8:16
  • $\begingroup$ @user1691278 the codomain of $D_jf_k(x)$ is $\mathcal L(\Bbb R,\Bbb R)$ what means that $D_j f_k(x)h\in\Bbb R$, that is, it defines the multiplication of two real numbers, then you can see $\mathcal L(\Bbb R,\Bbb R)$ as just $\Bbb R$ because there is the isomorphism between both defined by $D_j f_k(x)\cdot 1$ $\endgroup$ – Masacroso Mar 11 '18 at 9:31
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    $\begingroup$ that is, the map defined by $$H:\mathcal L(\Bbb R,\Bbb R)\to\Bbb R,\quad A\mapsto A\cdot 1$$ is an isomorphism (and it is also a topological isometry). Then we can identify $\mathcal L(\Bbb R,\Bbb R)$ with just $\Bbb R$. This is the isomorphism that we wse when we talk about derivatives on the real line, that is, if $f:\Bbb R\to\Bbb R$ then you can see also $f'(x)\in\mathcal L(\Bbb R,\Bbb R)$ as the map $y\mapsto f(x)\cdot y$ for any $y\in\Bbb R$. $\endgroup$ – Masacroso Mar 11 '18 at 9:38

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