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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.

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

2 Answers 2

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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 $f$.

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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 $$\triangle f(0,0)\neq o\left(\sqrt{x^2+y^2}\right)$$

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