TITLE: Why is any arbitrary directional derivative always recoverable from the gradient?
QUESTION [11 upvotes]: I understand what partial derivatives, directional derivatives, and the gradient are. I can even follow symbolically from their definitions why:
$$D_\mathbf{u} f = \nabla f \cdot \mathbf{u}$$
But nonetheless I find it surprising that knowing the derivative in just 3 directions at a point (the gradient) is sufficient to figure out what it must be in any direction. The equation states it must be the case that the derivative in an arbitrary direction must be a weighted combination (dot product) the gradient with the weights determined by how x-axis'y and how y-axis'y the direction u is. What prevents me from constructing a function where this isn't true? Why can't a simultaneous increase in x and y give a dramatically different result than either alone? (e.g. a function that rises in the x+ direction and the y+ direction, but falls dramatically along the diagonal?)

REPLY [3 votes]: Geometric explanation
It is because for a differentiable function $f:\Bbb R\to\Bbb R^n$ an amount of $n$ vectors is enough to know the tangent plane. All tangent vectors lie in the tangent plane and an $n$-dimensional plane is uniquely determined by $n$ (linearly independent) vectors.



If you have a function with a "sharp bend" (e.g. think about a roof), where all directional derivatives exist, but no unique tangent plane, then this will not work because not all tangent vectors lie in a plane.
Think about a differentiable function as a function which has unique tangent planes everywhere. Your counterexamples are not differentiable then.