Divergence theorem applied to Neumann boundary condition

Question

I frequently see the following stated without proof in notes of PDE :

$$\frac{\delta u}{\delta \nu} = <\nabla u, \nu >$$ both evaluated at a point $p$, and $\nu$ denotes outward normal of boundary

But I can't seem to figure out how to prove this.

Answer 1

Unless I am misunderstanding your question, this is actually just the definition of the normal derivative of a function. If a function $f$ is differentiable at $p$, then the directional derivative exists along any vector $v$ and is equal to $\nabla f \cdot v$.