Why does Continuous Partial Differentiability Imply Total Differentiability? Let $f: \mathbb{R}^d \to \mathbb{R}$ be such that the partial derivatives $\frac{\partial f}{\partial x_i}:\mathbb{R}^d \to \mathbb{R}$ exist everywhere and are continuous. Then show that $f$ is totally differentiable everywhere, which in particular implies that the gradient is given by 
$\nabla f(x_0)= \left( \frac{\partial f}{\partial x_1}(x_0), \ldots, \frac{\partial f}{\partial x_d}(x_0) \right)$ and the directional derivatives are given by
$D_vf(x_0) = v \cdot \nabla f(x_0)$.
I am trying to understand why this result can be used to prove Rademacher's Differentiation Theorem since I have an upcoming presentation.
As always, any advice would be greatly appreciated.  
 A: The function $f:\mathbb{R^2} \mapsto \mathbb{R}$ has a total derivative at a point $x$ if there exists a linear operator $Df(x)(\cdot)$ such that for every $\epsilon >0$ there is a $\delta > 0$ such that if $0 < ||h|| < \delta$, then
$$|f(x+h) -f(h) - Df(x)(h)| < \epsilon||h||.$$
Define the operator as
$$Df(x)(h) = \partial_1f(x_1,x_2)h_1+\partial_2f(x_1,x_2)h_2$$
Now consider the following path from $x = (x_1,x_2)$ to $x+h =(x_1+h_1,x_2+h_2)$:
$$ (x_1,x_2) \rightarrow(x_1+h_1,x_2) \rightarrow(x_1+h_1,x_2+h_2).$$
Using the mean value theorem,
$$|f(x_1+h_1,x_2+h_2) - f(x_1,x_2)- \partial_1f(x_1,x_2)h_1 - \partial_2f(x_1,x_2)h_2| \\
=|f(x_1+h_1,x_2+h_2) - f(x_1+h_1,x_2) +f(x_1+h_1,x_2)- f(x_1,x_2)-\partial_1f(x_1,x_2)h_1 - \partial_2f(x_1,x_2)h_2| 
\\=|\partial_2f(x_1+h_1,\xi)h_2 + \partial_1f(\eta,x_2)h_1 -\partial_1f(x_1,x_2)h_1 - \partial_2f(x_1,x_2)h_2|
\\ \leq|\partial_1f(\eta,x_2)-\partial_1f(x_1,x_2)||h_1|+|\partial_2f(x_1+h_1,\xi)-\partial_2f(x_1,x_2)||h_2|$$
where $x_1 < \eta < x_1 + h_1$ and $x_2 < \xi < x_2 + h_2.$
Since partial derivatives are continuous at $x = (x_1,x_2)$, there exists $\delta >0 $ such that if $||h|| < \delta$, then
$$|\partial_1f(\eta,x_2)-\partial_1f(x_1,x_2)|< \frac{\epsilon}{\sqrt{2}},\\|\partial_2f(x_1+h_1,\xi)-\partial_2f(x_1,x_2)|< \frac{\epsilon}{\sqrt{2}}.$$
Applying Cauchy-Schwarz we get
$$|f(x_1+h_1,x_2+h_2) - f(x_1,x_2)- \partial_1f(x_1,x_2)h_1 - \partial_2f(x_1,x_2)h_2|\\< \sqrt{(\epsilon/\sqrt{2})^2+(\epsilon/\sqrt{2})^2}||h||= \epsilon ||h||.$$
It is straightforward to generalize the proof for $d > 2$.
