TITLE: If $f$ is differentiable at $a$, then all first order partial derivatives of $f$ exist at $a$
QUESTION [7 upvotes]: Let $f$ be a vector function. If $f$ is differentiable at $a$, then all first order partial derivatives of $f$ exist at $a$ Moreover, the total derivative of $f$ at $a$ is unique and can be compute by the following: 
$$
Df(a) = \left[\frac{\partial f_i}{\partial x_j}(a) \right]_{m \times n} := \begin{bmatrix}
  \frac{\partial f_1}{\partial x_1}(a)  &\ldots &\frac{\partial f_1}{\partial x_n}(a) \\
\vdots &\ddots &\vdots \\
\frac{\partial f_m}{\partial x_1}(a) &\ldots &\frac{\partial f_m}{\partial x_n}(a)
\end{bmatrix}
$$
Please help me to prove? 
This is in fact a theorem. But I need to learn its proof. Please can somebody prove tags step by step. Thank you so much. I Will be happy if one can teachs me its proof.

REPLY [5 votes]: Suppose $Df(\textbf{a})$ exists. Let $\textbf{h}(s) = s \textbf{e}_1$; by the definition of $Df$ we have that
\begin{align*}
\lim_{s\to 0} \frac{f[\textbf{a} + \textbf{h}(s)] - f(\textbf{a}) - Df(\textbf{a})\textbf{h}(s)}{\|\textbf{h}(s)\|} &= 0\\
\lim_{s\to 0} \frac{f[\textbf{a} + s\textbf{e}_1] - f(\textbf{a}) - sDf(\textbf{a})\textbf{e}_1}{s} &= 0\\
\lim_{s\to 0} \frac{f[\textbf{a} + s\textbf{e}_1] - f(\textbf{a})}{s} &= Df(\textbf{a})\textbf{e}_1\\
\frac{\partial f}{\partial x_1}(\textbf{a}) &= Df(\textbf{a})\textbf{e}_1,
\end{align*}
where the last step uses the usual definition of the partial derivative of a multivariable function. By repeating the argument for different basis vector $\textbf{e}_i$ you get the other columns of $Df$.