Proof that $D_{\vec{u}}f(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{u}$

Question

I'm sure this is possibly a duplicate, but I haven't been able to find the original and have been looking for a solid while (maybe I'm just missing the name?). If you know the name of this shoot it in my direction.

I'm studying for my finals and realized that I don'thave a proof for this identity in my notes despite knowing that I need to be able to apply it and prove it.

$D_{\vec{u}}f(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{u}$

Thanks in advance :)

Answer 1

Let $f$ be differentiable at $ \vec{x}$ and put $g(h):=f(\vec{x}+h \vec{u})$. Then $D_{\vec{u}}f(\vec{x})=g'(0)$. Now compute $g'(0)$ with the chain rule.