Trying to understand "derivative or Jacobian of smooth map" From some lecture notes I am trying to puzzle through ....
"... the derivative or Jacobian of a smooth map $f: \mathbb{R}^m \rightarrow \mathbb{R}^n$ at a point $x$ is a linear map $Df: \mathbb{R}^m \rightarrow \mathbb{R}^n$.  In terms of partial derivatives,  $Df_x(X) = (\sum_j\partial_{x_j}f_1 \cdot X_j,
\sum_j \partial_{x_j}f_2\cdot X_j, ...)$ ... "
I'm so confused I'm not even sure where to begin.  Well, first, shouldn't the derivative be a map $Df:\mathbb{R}^m\rightarrow \mathbb{R}^m\times\mathbb{R}^n$?  Third, I am familiar with 3D integral calculus, and the only Jacobian I heard discussed there doesn't look like this at all, except, of course, that they both involve partial derivatibes.  Also, I don't even know what $f_1 \cdot X_j$ means.
Thanks.
 A: The best way to think about the derivative is:
\begin{equation*}
\tag{$\spadesuit$}f(x + \Delta x) \approx f(x) + f'(x) \Delta x.
\end{equation*}
The approximation is good when $\Delta x$ is small.  This equation expresses the fact that $f$ is "locally linear" at $x$.
How can we make sense of ($\spadesuit$) when $f:\mathbb R^n \to \mathbb R^m$?
\begin{equation*}
f(\underbrace{x}_{n \times 1} + \underbrace{\Delta x}_{n \times 1}) \approx \underbrace{f(x)}_{m \times 1} + \underbrace{f'(x)}_{?} \underbrace{\Delta x}_{n \times 1}.
\end{equation*}
It appears that $f'(x)$ should be something that, when multiplied by an $n \times 1$ column vector, returns an $m \times 1$ column vector.  In other words, $f'(x)$ should be an $m \times n$ matrix.
If we prefer to think in terms of linear transformations rather than matrices, we can write
\begin{equation*}
f(x + \Delta x) \approx f(x) + Df(x) \Delta x.
\end{equation*}
Here $Df(x)$ is a linear transformation that takes $\Delta x$ as input, and returns $f'(x) \Delta x$ as output.  This equation is what it means to be "locally linear" in the multivariable case.
Taking this as our starting point, it's not too hard to show that
\begin{equation*}
f'(x) = 
\begin{bmatrix} 
\frac{\partial f_1(x)}{\partial x_1} & \cdots & \frac{\partial f_1(x)}{\partial x_n} \\
\vdots & \ddots & \vdots \\
\frac{\partial f_m(x)}{\partial x_1} & \cdots & \frac{\partial f_m(x)}{\partial x_n}
\end{bmatrix}.
\end{equation*}
(The functions $f_i$ are the component functions of $f$.)
If 
\begin{equation*}
X = \begin{bmatrix} X_1 \\ \vdots \\ X_n \end{bmatrix},
\end{equation*}
then
\begin{equation*}
f'(x) X = 
\begin{bmatrix}
\sum_{j=1}^n \frac{\partial f_1(x)}{\partial x_j} X_j \\
\vdots \\
\sum_{j=1}^n \frac{\partial f_m(x)}{\partial x_j} X_j
\end{bmatrix},
\end{equation*}
as you can see just by doing the matrix-vector multiplication.
This is the equation given in your question.
A: If $f: \Bbb R^m \to \Bbb R^n$ is a differentiable mapping then at every $x \in \Bbb R^m$ its derivative $Df_x$ is a linear mapping from $\Bbb R^m$ to $\Bbb R^n$, therefore $Df_x \in \operatorname{Lin}(\Bbb R^m, \Bbb R^n)$. The elements of the matrix representing $Df_x$ are the partial derivatives of the partial functions of $f$. When you have $m=n$ you have the case of the Jacobian matrix.
Consider the implications of your question: if $Df$ were a mapping from $\Bbb R^m$ to $\Bbb R \times \Bbb R^n$ then for $m=n=1$ we'd have that the derivative of real functions of one variable is a point in $\Bbb R^2$.
Applying a vector $X \in \Bbb R^m$ to $Df_x$ you obtain a vector in $\Bbb R^n$ whose entries are the product of rows in $Df_x$ by the vector $X$. In symbols,
$$\begin{bmatrix} \partial_1 f_1 & \cdots & \partial_m f_1 \\ \vdots & \ddots & \vdots \\ \partial_1 f_n & \cdots & \partial_m f_n \end{bmatrix} \begin{bmatrix} X_1 \\ \vdots \\ X_m \end{bmatrix} = \begin{bmatrix} \sum_{j=1}^m \partial_j f_1 X_j \\ \vdots \\ \sum_{j=1}^m \partial_j f_n X_j \end{bmatrix}.$$
