Why is the directional derivative $D_v(F)$ equal to $\nabla F \cdot v$? Why is the directional derivative $D_{\bf v}(F)$ equal to $\nabla F \cdot {\bf v}$?  It doesn't seem obvious from the definition of the directional derivative, $$\lim_{h \to 0} \frac{f({\bf x} + h{\bf v}) - f({\bf x})}{h}$$
 A: If you accept the multi-variate chain rule (and you can look up the proof for it to see that it makes sense), then you have that along a given trajectory in direction $v$ from point $x$, parametrized by parmaeter $t$,
$$(D_v(f))(x) = \sum_i (\partial f / \partial x_i) (d x_i / dt) = ((\nabla f)(x)) \cdot v$$  
A: The gradient is by definition the vector $w$ satisfying $\langle w, v\rangle = \nabla_v F.$ That such a vector must exist, and is equal to $(\partial F/\partial x_1, \partial F/\partial x_2, \ldots)$, follows from linearity of the directional derivative and the fact that
$$\nabla_{e_i} F = \lim_{h\to 0} \frac{f(x+he_i) - f(x)}{h} = \frac{\partial f}{\partial x_i}.$$
Notice that different, non-trivial metrics will give you different formulas for $\nabla F$.
A: How about this (in 2-D for notational simplicity):
\begin{eqnarray*}
D_vF(x,y)
&=&
\lim_{h\to0}\frac{f(x+hv_1,y+hv_2)-f(x,y)}{h}\\
&=&
\lim_{h\to0}\frac{f(x+hv_1,y+hv_2)-f(x,y+hv_2)}{h}
+
\lim_{h\to0}\frac{f(x,y+hv_2)-f(x,y)}{h}\\
&=&
v_1 \partial_xf(x,y)
+
v_2 \partial_yf(x,y) ,
\end{eqnarray*}
as needed.
A: Still another proof:-
If $f\in \mathcal{C^1}$, we can write, by Taylor series, $$f(\mathbf{x}+h\mathbf{d})=f(\mathbf{x})+h\mathbf{d}^T \nabla F(\mathbf{x})+o(h^2)$$ Hence from the definition of directional derivative it follows that $$D_{\mathbf{d}}(F)(\mathbf{x})=\mathbf{d}^T\nabla F(\mathbf{x})$$
