# Can I say $f$ is differentiable at $c$ if $D_u(c) = \nabla f(c) \cdot u$ for all unit vectors $u$?

Can I say $f$ is differentiable at $c$ if $D_u(c) = \nabla f(c) \cdot u$ for all unit vectors $u$?

I think I can because it guarantees a tangent plane.

But I don't know how to prove this precisely.

Anyone can give an advice or a proof?

Thanks.

Now you cannot say this. Consider $f \colon \mathbb R^2 \to \mathbb R^2$ given by $$f(x) = \begin{cases} 1 & x_1^2 = x_2, x_2 > 0\\ 0 & \text{otherwise} \end{cases}$$ at $c=0$. Then for any ray $[0,\infty) \cdot u$, $f$ is constant on some intervall $[0,\epsilon) \cdot u$, that is $D_u f(0) = 0$. But $f$ is not continuous at 0, hence not differentiable.