Prove that $D_{\mathbf{v}}{f}(\mathbf{x}) = \nabla f(\mathbf{x}) \cdot{} \mathbf{v}$ Given the directional derivative
$$D_{\mathbf{v}}{f}(\mathbf{x}) = \lim_{h \rightarrow 0}{\dfrac{f(\mathbf{x} + h\mathbf{v}) - f(\mathbf{x})}{h}}$$
and that $f$ is differentiable at $\mathbf{x}$, how do I prove that
$$D_{\mathbf{v}}{f}(\mathbf{x}) = \nabla f(\mathbf{x}) \cdot{} \mathbf{v}$$
where $\nabla f(\mathbf{x}) = $ grad $(f(\mathbf{x}))$?
My attempt:
$$(D_{\mathbf{v}}{f}(\mathbf{x}))_{x_1} = \lim_{h \rightarrow 0}{\dfrac{f(x_1 + hv_1, 0, 0, ...) - f(x_1, 0, 0, ...)}{h}} = \lim_{h \rightarrow 0}{\dfrac{f(x_1 + hv_1, 0, 0, ...) - f(x_1, 0, 0, ...)}{hv_1}} v_1$$
Then $(D_{\mathbf{v}}{f}(\mathbf{x}))_{x_1} = \lim_{h \rightarrow 0}{\dfrac{f(x_1 + hv_1) - f(x_1)}{h}} = \lim_{hv_1 \rightarrow 0}{\dfrac{f(x_1 + hv_1) - f(x_1)}{hv_1}} v_1 = \dfrac{\partial f}{\partial x_1} v_1$

So all I have to prove is:
$$D_{\mathbf{v}}{f}(\mathbf{x}) = \sum_i (D_{\mathbf{v}}{f}(\mathbf{x}))_{x_i}$$

 A: By assumption the (scalar) function $f$ is differentiable at ${\bf x}$. This means that there is a linear function $df({\bf x})=:\phi$ such that
$$\lim_{{\bf x}\to{\bf 0}}{f({\bf x}+{\bf X})-f({\bf x})-\phi({\bf X})\over|{\bf X}|}=0\ .$$ 
This can be written denominator free as
$$f({\bf x}+{\bf X})-f({\bf x})=\phi({\bf X})\ +o\bigl(|{\bf X}|\bigr)\qquad({\bf X}\to{\bf 0})\  .\tag{1}$$
Let a "direction vector" ${\bf v}$ be given. Putting ${\bf X}:= h{\bf v}$ in $(1)$ we obtain
$$f({\bf x}+h{\bf v})-f({\bf x})=h\phi({\bf v})\ +o\bigl(|h{\bf v}|\bigr)\qquad(h\to 0)\  ,$$
or
$${f({\bf x}+h{\bf v})-f({\bf x})\over h}=\phi({\bf v})+o(1)=\nabla f({\bf x})\cdot{\bf v}+o(1)\qquad(h\to0)\ ,$$
which proves the claim.
A: Let $g(h):=f(x+hv)$ where $x:=(x_1,...,x_n)$ and $v:=(v_1,...,v_n)$ and $h\in\mathbb{R}$. Then since $f$ is differentiable we have $g(h)$ is differentiable. Therefore 
$$g'(h)=\frac{d}{dh}f(x+hv)=\frac{d}{dh}f(x_1+hv_1,...,x_n+hv_n)=\frac{d}{dh}f(y_1,...,y_n)$$
where $y_k:=x_k+hv_k$. Applying the rule of total differentiation we have
$$g'(h)=\frac{\partial f}{\partial y_1}\frac{dy_1}{dh}+...+\frac{\partial f}{\partial y_n}\frac{dy_n}{dh}=\frac{\partial f}{\partial y_1}v_1+...+\frac{\partial f}{\partial y_n}v_n=\langle \nabla_y f,v\rangle$$
As $h\to 0$ we get $y\to x$ and therefore $$g'(0)=\langle \nabla_xf,v\rangle$$
