Why is $\|D_{v}g(x)\|=\Bigl\|\sum\limits_{j=1}^n\frac{\partial g}{\partial x_j}v_j\Bigr\|$?

Question

Can someone explain why for a continuously differentiable function $g:E \to \mathbb{R}^{n}$, $E \subset \mathbb{R}^{n}$, and vector $v \in \mathbb{R}^{n}$, $\Vert D_{v}g(x) \Vert = \bigg\Vert \displaystyle\sum_{j=1}^{n} \dfrac{\partial g}{\partial x_{j}} v_{j}\bigg\Vert$, where $j \in \{1,2,\cdots, n\}$, and $D_{v}$ is the directional derivative with respect to $v$?

Answer 1

The following is a standard result in every textbook on Real Analysis:

If $f: E\rightarrow \mathbb{R}^n$ (with $E\subset \mathbb{R}^n$ open) is differentiable in $x_0\in E$. Then the partial derivatives $\frac{\partial f}{\partial x_i}$ exist in $x_0$ and $$Df(x_0) v = \sum_{i=1}^n \frac{\partial f}{\partial x_i}v_i$$

This follows immediately from, e.g., Theorem 9.17 in Rudin's Real and Complex analysis and the subsequent discussion. Since this is a standard result I don't think I should create a copy of the proof.

This should answer your question.