Perhaps the point is that existence of partial derivatives is not enough for a function of several variables to be differentiable in the multivariable sense, or even necessarily continuous: see this MO question for a standard example of a function of two variables all of whose partial (and even all directional) derivatives exist but is not even continuous at the origin.
[Edit: the following paragraph was added after reading Qiaochu's comment.]
It is also possible for all the partial derivatives in a given coordinate system to exist -- e.g. $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$ -- but for directional derivatives in other directions not to exist. (For instance, let $f: \mathbb{R}^3 \rightarrow \mathbb{R}$ be the function which is $0$ at $(x,y,z)$ if at least two of the variables are zero or if all of $x$,$y$,$z$ have rational coordinates and which is $1$ at all other points.) Here the divergence will be defined with respect to the standard coordinate system but not with respect to a different coordinate system.
Anyway, in general a function which has a continuous derivative (or partial derivatives) is much better behaved than a merely differentiable function. Inserting the hypothesis that a function is $C^1$ -- unless you really need to be considering weaker hypotheses than that for a specific application -- is generally a prudent practice.