[proofplan] The differential $Du_a$ is a [linear map](/page/Linear%20Map) from $\mathbb{R}^m$ to $\mathbb{R}$, so its standard matrix is a $1\times m$ row matrix. Its entries are precisely the first partial derivatives of $u$ at $a$. The Frobenius norm of this row matrix and the Euclidean norm of the gradient vector are both the square root of the sum of the squares of those same partial derivatives. [/proofplan] [step:Write the standard matrix of the scalar differential] Let $A\in\mathbb{R}^{1\times m}$ denote the standard matrix of the linear map $Du_a:\mathbb{R}^m\to\mathbb{R}$. Since $u\in C^1(U;\mathbb{R})$, each [partial derivative](/page/Partial%20Derivative) $\partial_{x_j}u(a)$ exists for $j\in\{1,\ldots,m\}$, and the [Jacobian matrix](/page/Jacobian%20Matrix) is \begin{align*} A=\begin{pmatrix}\partial_{x_1}u(a)&\partial_{x_2}u(a)&\cdots&\partial_{x_m}u(a)\end{pmatrix}. \end{align*} Equivalently, for every $j\in\{1,\ldots,m\}$, \begin{align*} A_{1j}=\partial_{x_j}u(a). \end{align*} [guided] The object $Du_a$ is the total derivative of $u$ at $a$. Because $u$ is scalar-valued, this derivative is a linear map \begin{align*} Du_a:\mathbb{R}^m\to\mathbb{R}. \end{align*} With respect to the standard basis of $\mathbb{R}^m$ and the standard basis of $\mathbb{R}$, such a linear map is represented by a $1\times m$ matrix. Let $A\in\mathbb{R}^{1\times m}$ be that standard matrix. The entries of the standard matrix of the derivative are the first partial derivatives of the component functions. Here there is only one component function, namely $u$ itself. Therefore, for each column index $j\in\{1,\ldots,m\}$, \begin{align*} A_{1j}=\partial_{x_j}u(a). \end{align*} Thus \begin{align*} A=\begin{pmatrix}\partial_{x_1}u(a)&\partial_{x_2}u(a)&\cdots&\partial_{x_m}u(a)\end{pmatrix}. \end{align*} The important point is that the differential is represented as a row matrix, while the gradient is usually regarded as a vector in $\mathbb{R}^m$; the upcoming norm computation only uses the same list of scalar entries. [/guided] [/step] [step:Compute the Frobenius norm from the matrix entries] By the definition of the Frobenius norm of a linear map through its standard matrix, \begin{align*} \|Du_a\|_F^2=\sum_{j=1}^m A_{1j}^2. \end{align*} Using $A_{1j}=\partial_{x_j}u(a)$ for each $j\in\{1,\ldots,m\}$ gives \begin{align*} \|Du_a\|_F^2=\sum_{j=1}^m \bigl(\partial_{x_j}u(a)\bigr)^2. \end{align*} Since $\|Du_a\|_F\geq 0$, it follows that \begin{align*} \|Du_a\|_F=\left(\sum_{j=1}^m \bigl(\partial_{x_j}u(a)\bigr)^2\right)^{1/2}. \end{align*} [/step] [step:Compute the Euclidean norm of the gradient] By the definition of the gradient of a scalar $C^1$ function, \begin{align*} \nabla u(a)=\bigl(\partial_{x_1}u(a),\partial_{x_2}u(a),\ldots,\partial_{x_m}u(a)\bigr)\in\mathbb{R}^m. \end{align*} Therefore the Euclidean norm of $\nabla u(a)$ satisfies \begin{align*} |\nabla u(a)|=\left(\sum_{j=1}^m \bigl(\partial_{x_j}u(a)\bigr)^2\right)^{1/2}. \end{align*} Comparing this expression with the formula obtained for $\|Du_a\|_F$ gives \begin{align*} \|Du_a\|_F=|\nabla u(a)|. \end{align*} This proves the claim. [/step]