The standard definition is that a function $f:\mathbb{R}^n\to \mathbb{R}$ is *differentiable* at a point $x$ if there exists a linear map $\mathrm{d}f_x: \mathbb{R}^n \to \mathbb{R}$ such that

$$f(x+h) = f(x) + \mathrm{d}f_x(h) + \epsilon \|h\|$$

where $\epsilon\to 0$ as $h\to 0$. This is stronger than the existence of all partial (or directional) derivatives, but weaker than their continuity. However, when talking about higher differentiability, one usually switches to talking about partial derivatives, asking them to be continuous in order to prove basic properties.

Suppose that instead we define $f$ to be "twice differentiable" at $x$ if in addition to $\mathrm{d}f_x$ as above there exists a quadratic form $\mathrm{d}^2f_x$ such that

$$f(x+h) = f(x) + \mathrm{d}f_x(h) + \frac{1}{2}\mathrm{d}^2f_x(h) + \epsilon \|h\|^2$$

where $\epsilon\to 0$ as $h\to 0$. This is true if $f$ has continous second-order partials (it's the multidimensional Taylor expansion, with $\mathrm{d}^2f_x$ the Hessian matrix).

- Does this imply that all second-order partial derivatives of $f$ exist?
- If so, does it imply that the mixed second-order partials are equal?

exactlylike the 1-dimensional case (with the mess of factorials hidden away within a clean formalism); of course, one can bust out coordinates and recover the usual messier explicit version from that. – user76758 May 6 '14 at 21:47