# If $f$ is differentiable at $a$, then all first order partial derivatives of $f$ exist at $a$

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:

$$Df(a) = \left[\frac{\partial f_i}{\partial x_j}(a) \right]_{m \times n} := \begin{bmatrix} \frac{\partial f_1}{\partial x_1}(a) &\ldots &\frac{\partial f_1}{\partial x_n}(a) \\ \vdots &\ddots &\vdots \\ \frac{\partial f_m}{\partial x_1}(a) &\ldots &\frac{\partial f_m}{\partial x_n}(a) \end{bmatrix}$$

• $\frac{f(a+h)-f(a))-Df(a)h}{\vert\vert h\vert\vert} \to 0 as h \to 0$ @user7530 – Mathlover4 Jul 15 '13 at 19:08
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 $Df$.