Directional derivative in the direction of a sum of two vectors. Q. Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be a map. 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?
 A: Assuming that $f$ is differentiable at $x=a$ we have
$$f(\mathbf a+t\mathbf v+t\mathbf w)=f(\mathbf a)+\nabla f(\mathbf a)\cdot(t\mathbf v+t\mathbf w)+o(|t\mathbf v+t\mathbf w|)=\\=f(\mathbf a)+\nabla f(\mathbf a)\cdot t\mathbf v+\nabla f(a)\cdot t\mathbf w+o(|t\mathbf v+t\mathbf w|)$$
therefore
$$\frac{ f(\mathbf a+t\mathbf v+t\mathbf w)-f(\mathbf a)}t=\nabla f(\mathbf a)\cdot \mathbf v+\nabla f(\mathbf a)\cdot \mathbf w+\frac{o(|t\mathbf v+t\mathbf w|)}{t}$$
and
$$\lim _{t \rightarrow 0} \frac{f(\mathbf a+t\mathbf v+t\mathbf w)-f(\mathbf{a})}{t}=\nabla f(\mathbf a)\cdot \mathbf v+\nabla f(\mathbf a)\cdot \mathbf w=D_{\mathbf{v}} f(\mathbf{a})+D_{\mathbf{w}} f(\mathbf{a})$$
A: If $\;f\;$ is differentiable at $\;a\;$  it is pretty easy, since then by linearity of the scalar product (or inner product, to name it with other words) we get
$$D_{v+w}f(a)=\nabla f(a)\cdot(v+w)=\nabla f(a)\cdot v+\nabla f(a)\cdot w=D_vf(a)+D_wf(a)$$
If $\;f\;$ isn't differentiable then it may not be true...but I can't produce a counterexample right now.
A: As a counter example when $f$ is not differentiable, suppose $f(0) = 0$. Then take $f(x) = 0$ along the lines $\{ te_1 \in \mathbb{R}^n \ | \ t\in\mathbb{R}\}$ and $\{t e_2\in \mathbb{R}^n \ | \ t\in\mathbb{R} \}$ then
$$D_{e_1}f(0) = 0, \ D_{e_2}f(0) = 0$$
Set $f(x) = \text{sgn}(x_1)|x|$ for  $x\in \{ t(e_1+e_2)\in \mathbb{R}^n \ | \ t\in\mathbb{R} \}$ then
\begin{align}
D_{e_1+e_2}f(0) &= \lim_{t\rightarrow 0}\frac{f(t(e_1+e_2)) - f(0)}{t} 
\\
&= \lim_{t\rightarrow 0}\frac{\sqrt{2}|t|\text{sgn}(t) - 0}{t}
\\ &= \sqrt{2} 
\\
&\neq D_{e_1}f(0) + D_{e_2}f(0) 
\end{align}
A: 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.
A: The counter-example is made of pieces of a cone:
$$z=f(x,y)=\text{sgn}(x)\sqrt{|xy|}.$$
Clearly $f(x,0)=f(0,y)=0$ for all $x$ or $y$, so the directional derivatives along $\mathbf v=(1,0)$ and $\mathbf w=(0,1)$ are both $0$.
But the directional derivative along $\mathbf v+\mathbf w=(1,1)$ is
$$\lim_{t\to0}\frac{f(t,t)-f(0,0)}{t}=\lim_{t\to0}\frac{\text{sgn}(t)\sqrt{t^2}}{t}$$
$$=\lim_{t\to0}\frac{\text{sgn}(t)\,|t|}{t}=\lim_{t\to0}\frac{t}{t}=1.$$
More generally, the directional derivative along $(a,b)$ is
$$\lim_{t\to0}\frac{f(at,bt)-f(0,0)}{t}=\lim_{t\to0}\frac{\text{sgn}(a)\text{sgn}(t)\sqrt{|ab|}\,|t|}{t}$$
$$=\text{sgn}(a)\sqrt{|ab|}.$$
