# Intuition behind inner product of gradient

Let $f: \mathbb{R}^{n} \to \mathbb{B}$ be continuously differentiable and take the points $x, p \in \mathbb{R}^{n}$. I am familiar with the idea of a directional derivative being given as an inner product of the gradient of a point with another vector in $\mathbb{R}^{n}$. For example, the directional derivative of $f$ in the direction $p$ with respect to the point $x$ can be denoted as $\langle \triangledown f(x), p \rangle$. However, I cannot seem to wrap my mind around the idea of the inner product of the gradient of a point with respect to itself itself. More succinctly, what can we learn about the function $f$ at $p$ given the identity: $$\langle \triangledown f(p), p \rangle$$ This identity can be rewritten as $\langle \triangledown f(p), p \rangle = \sum\limits_{i = 1}^{n}\frac{\partial{f}}{\partial{p_{i}}}p_{i}$, and I'm not quite sure what we can gather from this. I know this is somewhat vague, but if anyone could give me some insight onto the mechanics of what is going on here, I would be very appreciative.

## 1 Answer

A "direction" is generally given by a unit vector, not just any vector.
If $p$ is not necessarily a unit vector, $\langle \nabla f(x), p \rangle$ would be the rate of change of $f$ as you pass through $x$ with velocity vector $p$.

Now you're looking at the case $x=p$. So your velocity vector is the same as the position vector: the direction is directly away from the origin.

• Your explanation is enlightening and much appreciated, sir. – Decave Nov 10 '13 at 20:12