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.

  • 1
    $\begingroup$ What is your definition of differentiability? $\endgroup$
    – user7530
    Jul 15, 2013 at 18:48
  • $\begingroup$ $\frac{f(a+h)-f(a))-Df(a)h}{\vert\vert h\vert\vert} \to 0 as h \to 0$ @user7530 $\endgroup$
    – Mathlover4
    Jul 15, 2013 at 19:08

1 Answer 1


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 $Df$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .