Problem about differential of a linear map Please can you tell how to solve this problem clearly? Please solve this explanatorily. Thank you  

 A: Say that $F: N^n \to M^m$ is a smooth function, mapping
$$
p = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}
\mapsto
\begin{bmatrix} y_1 \\ \vdots \\ y_m \end{bmatrix}
= 
\begin{bmatrix} F_1(p) \\ \vdots \\ F_m(p) \end{bmatrix}
= F(p).
$$
Then, the differential is linear map $F_{*,p}: T_p(N) \to T_{F(p)}(M)$ between the tangent spaces.  Choosing the basis $\{ \frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_n} \}$ for $T_p(N)$ and the basis $\{ \frac{\partial}{\partial y_1}, \dots, \frac{\partial}{\partial y_m} \}$ for $T_{F(p)}(M)$, we can represent the differential by the $(m \times n)$-matrix
$$
\begin{bmatrix}
\frac{\partial y_1}{\partial x_1} & \cdots & \frac{\partial y_1}{\partial x_n} \\
\vdots & \ddots & \vdots \\
\frac{\partial y_m}{\partial x_1} & \cdots & \frac{\partial y_m}{\partial x_n} 
\end{bmatrix}
=
\begin{bmatrix}
\frac{\partial F_1(x)}{\partial x_1} & \cdots & \frac{\partial F_1(x)}{\partial x_n} \\
\vdots & \ddots & \vdots \\
\frac{\partial F_m(x)}{\partial x_1} & \cdots & \frac{\partial F_m(x)}{\partial x_n} 
\end{bmatrix}
$$
In the special case that $L: \mathbb{R}^n \to \mathbb{R}^m$ is a linear map, we can write $L(x) = Ax$, where $A$ is an $(m \times n)$-matrix.
What do the partial derivatives look like?  Put $A = [a_{ij}]$, so for any $1 \le i \le m$,
$$
y_i = F_i(x) = a_{i1}x_1 + \cdots + a_{in}x_n,
$$
and for any $1 \le i \le m$, $1 \le j \le n$,
$$
\frac{\partial F_i(x)}{\partial x_j} = \frac{\partial}{\partial x_j}\big( a_{i1}x_1 + \cdots + a_{in}x_n \big) = a_{ij}.
$$
Thus, the matrix of partial derivatives, representing the differential of $L$ in coordinates, is nothing other than the matrix of $L$ itself, once we make the identification $T_p(\mathbb{R}^n) \overset{\sim}{\to} \mathbb{R}^n$.
