[proofplan] We write the derivative $Df_a$ in the standard bases of $\mathbb{R}^m$ and $\mathbb{R}^n$. Its matrix entries are the component partial derivatives $\partial_j f_i(a)$, so the Frobenius norm is the sum of the squares of all these entries. For each scalar component $f_i$, the Euclidean norm of $\nabla f_i(a)$ is the sum of the squares of the entries in the $i$-th row of the Jacobian. Summing those row identities gives the desired formula. [/proofplan] [step:Write the derivative matrix in component form] Fix $a\in U$. Since $f\in C^1(U;\mathbb{R}^n)$, each component map \begin{align*} f_i:U\to\mathbb{R} \end{align*} is continuously differentiable for $1\leq i\leq n$. Let $A\in\mathbb{R}^{n\times m}$ denote the standard matrix of the [linear map](/page/Linear%20Map) $Df_a:\mathbb{R}^m\to\mathbb{R}^n$ with respect to the standard bases of $\mathbb{R}^m$ and $\mathbb{R}^n$. By the component formula for the derivative of a vector-valued $C^1$ map, the entries of $A$ are \begin{align*} A_{ij}=\partial_j f_i(a) \end{align*} for every $1\leq i\leq n$ and every $1\leq j\leq m$. [/step] [step:Expand the Frobenius norm of the derivative] By the definition of the Frobenius norm on standard matrices, \begin{align*} \|Df_a\|_F^2=\sum_{i=1}^n\sum_{j=1}^m |A_{ij}|^2. \end{align*} Substituting the component formula $A_{ij}=\partial_j f_i(a)$ gives \begin{align*} \|Df_a\|_F^2=\sum_{i=1}^n\sum_{j=1}^m |\partial_j f_i(a)|^2. \end{align*} [guided] The Frobenius norm is defined by squaring every entry of the standard matrix of the linear map and summing those squares. Here the relevant linear map is \begin{align*} Df_a:\mathbb{R}^m\to\mathbb{R}^n. \end{align*} Its standard matrix is $A\in\mathbb{R}^{n\times m}$, and the previous step identified each entry as \begin{align*} A_{ij}=\partial_j f_i(a). \end{align*} Therefore the definition of the Frobenius norm gives \begin{align*} \|Df_a\|_F^2=\sum_{i=1}^n\sum_{j=1}^m |A_{ij}|^2. \end{align*} Replacing each $A_{ij}$ by the corresponding [partial derivative](/page/Partial%20Derivative) produces \begin{align*} \|Df_a\|_F^2=\sum_{i=1}^n\sum_{j=1}^m |\partial_j f_i(a)|^2. \end{align*} This expression groups the squared entries of the [Jacobian matrix](/page/Jacobian%20Matrix) by rows: the $i$-th row contains exactly the first partial derivatives of the scalar component $f_i$ at $a$. [/guided] [/step] [step:Identify each row sum with a component gradient norm] For each $1\leq i\leq n$, the Euclidean gradient of the scalar function $f_i$ at $a$ is \begin{align*} \nabla f_i(a)=(\partial_1 f_i(a),\ldots,\partial_m f_i(a))\in\mathbb{R}^m. \end{align*} By the definition of the Euclidean norm on $\mathbb{R}^m$, \begin{align*} |\nabla f_i(a)|^2=\sum_{j=1}^m |\partial_j f_i(a)|^2. \end{align*} [/step] [step:Sum the component gradient identities] Summing the preceding identity over $i=1,\ldots,n$ gives \begin{align*} \sum_{i=1}^n |\nabla f_i(a)|^2=\sum_{i=1}^n\sum_{j=1}^m |\partial_j f_i(a)|^2. \end{align*} The right-hand side is exactly the expression already obtained for $\|Df_a\|_F^2$. Hence \begin{align*} \|Df_a\|_F^2=\sum_{i=1}^n |\nabla f_i(a)|^2. \end{align*} Since $a\in U$ was arbitrary, the formula holds for every $a\in U$. [/step]