# Is $f'((0,0);d)=0$ (directional derivative) for every direction $d \in \Bbb R^{n}$? [closed]

If $$\frac{\partial f}{\partial x}(0,0) = \frac{\partial f}{\partial y}(0,0) = 0$$, then $$f'((0,0);d)=0$$ (directional derivative) for every direction $$d \in \mathbb{R}^n$$.

Is this true? I'm trying to find a counterexample to prove it false, but nothing comes to mind.

• confusing double use of the letter $d$ found here: $d \in \mathbb R^d$ Jun 1 at 15:23

Consider the following example. Let $$f:\mathbb{R}^2\to \mathbb{R}$$ be such that $$f(x,0)=0$$, $$f(0,y)=0$$ for all $$x,y\in\mathbb{R}$$ but $$f(x,y)=\sqrt{x^2+y^2}$$ whenever $$xy\neq 0$$. It is obvious that both $$\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$$.
Now let $$\vec{v}=(a,b)$$ be a unit vector in $$\mathbb{R}^2$$ with $$ab\neq 0$$. So you can try to compute $$\frac{\partial f}{\partial \vec{v}}(0,0)=\lim_{t\to 0}\frac{f(ta,tb)-f(0,0)}{t}=\lim_{t\to 0}\frac{t-0}{t}=1\neq 0.$$
What you might have in mind is the formula $$\frac{\partial f}{\partial \vec{v}}=\vec{v}\cdot \nabla f(x)=a\frac{\partial f}{\partial x}+b\frac{\partial f}{\partial y}$$. But you should note that this only holds if you suppose some sort of smoothness condition for $$f$$.
Let $$d=2$$ and $$f=\begin{cases}\frac{xy}{x^2+y^2},&x^2+y^2\neq0\\0, &x^2+y^2=0\end{cases}$$ It's easy to see that $$\frac{\partial f}{\partial x}f(0,0) = \frac{\partial f}{\partial y}f(0,0) = 0$$ but $$\triangle f(0,0)\neq o\left(\sqrt{x^2+y^2}\right)$$