Directional Derivative and Gradients

Theorem

$Df(x)v=\nabla f\cdot v$

Proof

1. With $f:{\mathbb{R}}^k\to \mathbb{R}$
2. With $v\in {\mathbb{R}}^k$
3. The function $h(t)=f(x+tv)=h(t)=f(c(t))$ where $c(t)=x+tv$
4. Then, $Df(x)v=\frac{d}{dx}f(x+tv){|}_{t=0}$
5. $=D(f\circ c)$
6. $=\nabla f(c(t))c^{\prime }(t)$
7. $=\nabla f(c(t))\cdot v$
8. $=\nabla f\cdot v$