TITLE: Directional derivative in the direction of a sum of two vectors.
QUESTION [13 upvotes]: Q. Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$. For each vector $\mathbf{v} \in \mathbb{R}^{n}$, we define
$$
D_{\mathbf{v}} f(\mathbf{a})=\lim _{t \rightarrow 0} \frac{f(\mathbf{a}+t \mathbf{v})-f(\mathbf{a})}{t}
$$
if the limit exists. $D_{\mathbf{v}} f(\mathbf{a})$ is the directional derivative of $f$ with respect to $v$ at $a$.
Show that for vectors $\mathbf{v}, \mathbf{w} \in \mathbb{R}^{n},$ one has
$$
D_{\mathbf{v}+\mathbf{w}} f(\mathbf{a})=D_{\mathbf{v}} f(\mathbf{a})+D_{\mathbf{w}} f(\mathbf{a})
$$
My attempt:
$$
\begin{array}{l}\lim _{t \rightarrow 0} \frac{f(a+t(\mathbf v+\mathbf w))-f(a)}{t} \\ =\operatorname{lim}_{t \rightarrow 0}\frac{ f(a+t\mathbf v+t\mathbf w)-f(a)}{t}\\
=\operatorname{lim}_{t \rightarrow 0}\frac{ f(a+t\mathbf v+t\mathbf w)-f(a+t\mathbf v)}{t}+\frac{ f(a+t\mathbf v)-f(a)}{t}
\end{array}
$$
Now, I have to prove that
$$\operatorname{lim}_{t \rightarrow 0}\frac{ f(a+t\mathbf v+t\mathbf w)-f(a+t\mathbf v)}{t}=D_{\mathbf{w}} f(\mathbf{a})
$$
But how to?

REPLY [2 votes]: Let $f:\mathbb{R}^2 \to \mathbb{R}$ be one on the axes and zero everywhere else.
Then $D_{e_k} f(0) = 0$ but the directional derivative in any other direction does not exist.