Is this really a typo? Let $U \subseteq \mathbb R^n$ and $F: U \to \mathbb R^m$ a function with coordinate functions $f_i$. My notes say that:
If $F$ is differentiable on $U$ the Jacobian of $F$ is defined at each point in $U$, its $nm$ entries are functions on $U$. These functions need not be continuous on $U$; they are continuous on $U$ if and only if $F$ is of class $C^1$. 
My problem is with the only if part of last part of the statement: it should be $C^k$ with  $k \ge 1$, right? (I'm merely asking because there is this nagging doubt I have that it might not be a typo after all.)
 A: No, it's not a typo. To say $F$ is $C^1$ on $U$ is to say that it is continuously differentiable on $U$. This means the mapping $p \mapsto dF_p$ is continuous as a mapping from $U$ to the space of operators on smooth functions at $p$. It is a very nice result that this continuity of operators reduces to the much easier to understand criteria of continuity of the component functions of the Jacobian. 

Again, I think you are misinterpreting the term "differentiable" to mean that derivatives of all orders exist. Of course, if it was the case that $f \in C^k$ for $k>1$ then $f \in C^2$ implies $f' \in C^1$. But, I don't believe that is the intended use of the term "differentiable". Consider the example below: for $x \neq 0$ define
$$ f(x) = x^2\sin(1/x) $$
and for $x=0$ define $f(0)=0$. Differentiate for $x \neq 0$ and obtain,
[ f'(x) = 2x\sin(1/x)-\cos(1/x) ]
thus clearly $f'$ is not continuous at $x=0$. However, $f'(0)$ does exist and it can be shown to be zero by careful arguments from the definition of the derivative. Therefore, $f$ is differentiable at $x=0$ however, $f \notin C^1(0)$. Here's a picture I show my calculus I class to show how this happens geometrically but it's for a slightly different example $f(x) = \frac{x}{2}+x^2\sin(1/x)$ to give slope $1/2$ instead of zero at $(0,0)$

Incidentally, your suffering this day is precisely the reason I think we should use the term smooth instead of "differentiable" to describe a function which supports arbitrarily many continuous derivatives. But, minds much greater than mine disagree, so the suffering must continue.

