# Are derivations as general as derivatives in $\Bbb R^n$?

Assuming we start with the derivations, $w_p(f)$ that are linear and satisfy the product rule.

Is it possible to show that what $w(f) = D_v(f)$ without first defining what a directional derivative, Taylor's theorem, etc.

In other words, can one show that $w(f) = \lim_{t \rightarrow 0} \frac{f(p + vt) - f(p)}{t}$ for some $v$ without first defining derivatives.

• The space of derivations is a $C^\infty(\mathbb{R}^n)$ (or similar) module, so you're going to have to impose some extra conditions to get uniqueness. On the other hand, $w(x^n) = nx^{n-1} w(x)$, so you can at least recover the derivative for analytic functions (which may be begging the question) with some reasonable continuity assumptions on $w$. – anomaly Aug 22 '18 at 13:07
• I am interested in smooth functions. I am not sure I get your point about uniqueness. There is a bijection from a derivation at point p to the directional derivatives at that point. I am interested in derivations that operate on $C^\infty(R^n)$ – Jeff Aug 23 '18 at 3:45
• Thinking about this more, it doesn't seem possible. One needs to be able to write $f(x) = x g_1(x) + x^2 g_2(x)$, which is what we get from Taylor's theorem. – Jeff Aug 23 '18 at 15:56