[proofplan] The proof is an unpacking of the definition of a $C^1$ map between open subsets of finite-dimensional Euclidean spaces. In one direction, membership in $C^1(U;\mathbb{R}^m)$ gives differentiability at every point and continuity of the derivative map. In the other direction, those two properties are exactly the defining requirements for $f$ to be continuously differentiable. [/proofplan]

[step:Unpack the definition of a $C^1$ map]By definition, the assertion $f \in C^1(U;\mathbb{R}^m)$ means that $f$ is differentiable at every point $a \in U$ and that the map \begin{align*} Df: U &\to \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m) \end{align*} defined by $a \mapsto Df_a$ is continuous. Here $Df_a: \mathbb{R}^n \to \mathbb{R}^m$ denotes the Fréchet derivative of $f$ at $a$; for an increment vector $h \in \mathbb{R}^n$ with $a+h \in U$, the first-order approximation error is negligible compared with $|h|$ as $h \to 0$. The space $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ is equipped with its operator norm topology.[/step]

[guided]The notation $f \in C^1(U;\mathbb{R}^m)$ has two pieces of content. First, for each point $a \in U$, the map $f: U \to \mathbb{R}^m$ must be differentiable at $a$ in the Fréchet sense. This means that for each increment vector $h \in \mathbb{R}^n$ with $a+h \in U$, there is a [linear map](/page/Linear%20Map) \begin{align*} Df_a: \mathbb{R}^n &\to \mathbb{R}^m \end{align*} such that the first-order approximation error is negligible compared with $|h|$ as $h \to 0$. Second, the derivative must vary continuously with the base point. Thus the assignment \begin{align*} Df: U &\to \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m) \end{align*} given by $a \mapsto Df_a$ is required to be continuous, where the target is the [normed vector space](/page/Normed%20Vector%20Space) of linear maps from $\mathbb{R}^n$ to $\mathbb{R}^m$ with the operator norm. Therefore the definition of $C^1(U;\mathbb{R}^m)$ is precisely the conjunction of differentiability at every point and continuity of the derivative map.[/guided]

[step:Derive the two stated properties from $f \in C^1(U;\mathbb{R}^m)$] Assume $f \in C^1(U;\mathbb{R}^m)$. Applying the definition from the previous step, $f$ is differentiable at every point $a \in U$, and the derivative map $Df: U \to \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ is continuous. This proves the forward implication. [/step]

[step:Recognize the stated properties as the definition of $C^1$] Conversely, assume that $f$ is differentiable at every point $a \in U$ and that the derivative map \begin{align*} Df: U &\to \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m) \end{align*} given by $a \mapsto Df_a$ is continuous. These are exactly the two conditions in the definition of $f \in C^1(U;\mathbb{R}^m)$. Hence $f \in C^1(U;\mathbb{R}^m)$. Combining the forward and reverse implications proves the equivalence. [/step]