Mild Error on Page 32 Spivak's Calculus of Manifolds that I couldn't find online. I'm just asking if this is actually an error, as I could not find it in any errata online, math stackexchange questions etc.
In Spivak's Calculus on Manifolds, page 32, I believe there is a mild error in the statement of Theorem 2-9.
The theorem states:
"Let $g_{1} ,..., g_{m}$:$\Bbb{R}^{n} \rightarrow \Bbb{R}$ be continuously differentiable at $a$ and let $f:\Bbb{R}^{m} \rightarrow \Bbb{R}$ be differentiable at $(g_{1}(a), ... , g_{m}(a)) $. Define the function $F:\Bbb{R}^{n} \rightarrow \Bbb{R}$ by $F(x) = f(g_{1}(x), ... , g_{m}(x)). $ Then
$D_{i}F(a) = \sum_{j=1}^m D_{j}f(g_{1}(a), ... ,g_{m}(a))\cdot D_{i}g_{j}(a).$"
I believe it is an error that the $g_{i}$ must be assumed to be continuously differentiable (as opposed to just differentiable), as he proves in Theorem 2-3 on page 20 that the function $g:\Bbb{R}^{n} \rightarrow \Bbb{R}^{m}, x\rightarrow(g_{1}(x), ... , g_{m}(x))$ is differentiable iff the $g_{i}$ are just differentiable, with no continuity requirement.
Normally I would just ignore this and assume it is an error, but he explicitly states after the proof that this theorem is weaker than the chain rule because the $g_{i}$ must be continuously differentiable.
Am I correct in assuming that by Theorem 2-3 they need not be?
 A: You are right, it is not needed that the $g_i$ are continuously differentiable. If you look at the proof of Theorem 2-9, you see that Spivak's argument is

Since $g_i$ is continuously differentiable at $a$, it follows from Theorem 2-8 that $g$ is differentiable at $a$.

In the next step he uses the chain rule (Theorem 2-2) to complete the proof. This seems in a sense absurd: Spivak invokes the general form of the chain rule to prove a special case of the chain rule. This is not an error (not even a mild one), but why does he do that?
The only explanation that I have is that the general chain rule only states that $D (g \circ f) (a) = Dg(f(a)) \circ Df(a)$ which is less concrete than the formula in Theorem 2-9. Theorem 2-7 gives a description of $Df(a)$ via partial derivatives, but its proof is based on the general chain rule. On the top of p.32 Spivak says

Although the chain rule was used in the proof of Theorem 2-7, it could easily have been eliminated. With Theorem 2-8 to provide differentiable functions, and Theorem 2-7 to provide their derivatives, the chain rule may therefore seem almost
superfluous. However, it [the chain rule] has an extremely important corollary concerning partial derivatives.

In my opinion he only wants to say that the chain rule is not superfluous if you want to calculate partial derivatives of composed functions. However, Theorem 2-8 is definitely not needed to prove Theorem 2-9 in the form "$g_i$ differentiable". Therefore I would say Spivak is unnecessarily confusing his readers. However, I guess that he wanted to, but erroneously did not state Theorem 2-9 with the stronger assumption that also $f$ is continuously differentiable at $(g_1(a), \ldots ,g_m(a))$. What is the benefit of this variant? The best approach to show that a function is differentiable at a point $p$ is to verify that all partial derivatives exist in a neigborhood of $p$ and are continuous at $p$. This is a sufficient criterion, but it is not necessary. In case it is not satisfied, it may be a very unpleasant task to check that the function is differentiable at $p$. See the examples on Spivak's book. Thus for practical applications Theorem 2-9 in the above form is the most suitable variant of the chain rule although it is weaker than Theorem 2-2.
