In linear algebra one is able to prove that for any positive integer $n$, the determinant function $\det: F^{n\times n}\to F$(where $F$ is a field) is the unique alternating $n$-linear function(of the rows) that satisfies $\det (\textbf{I}_n)=1$. Similarly, can we use some of the fundamental properties of the derivative of a function(in the special case) to build, or characterize the definition of derivative for any function $f:\mathbb R^m\to \mathbb R^n$?

Here is how I started: First, for any function $f:\mathbb R^m\to\mathbb R^n$, the derivative $D_f$ of $f$ should assign an element $D_f(\textbf{v})$ for every point $\textbf{v}$ in $\mathbb R^m$. Then, since linearity is a very fundamental property, one should expect that the codomain of $D_f$ is a vector space over $\mathbb R$ and $D_f$ is a linear transformation between them. We then define total derivative of $f$ to be a tuple $(D_f, V_f)$, where $V_f$ is a vector space.

Clearly that is not enough for proving $(D_f, V_f)$ to be unique(for $V_f$ we only require the uniqueness up to isomorphism). Is it possible to add more properties(such as the product rule) so that we can obtain a characterization?

Since I do not have a strong analysis background, please do not hesitate to criticize if the question can be stated in a more elegant way.

  • $\begingroup$ The determinant is a notion of pure algebraic structure. The derivative is a notion of analysis on an algebraic structure. There is, however, the definition of a derivative in algebra, more specifically, in ring theory where you define the derivative of a polynomial to be, well, what it ought to be. $\endgroup$ – Will M. Mar 5 at 5:26

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