# Characterize the Notion of (Total) Derivative.

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.

• 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. – Will M. Mar 5 at 5:26