The derivative of a linear transformation Suppose $m > 1$. Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a smooth map. Consider $f + Ax$ for $A \in \mathrm{Mat}_{m\times n}, x \in \mathbb{R}^n$. Define $F: \mathbb{R}^n \times \mathrm{Mat}_{m\times n} \rightarrow \mathrm{Mat}_{m\times n}$ by $F(x,A) = df_x + A$.
So what is $dF_x$?
(A) Is it $dF(x,A) = d(df_x + A) = d f_x + A$? And therefore,
$$dF(x,A) =\left( \begin{array}{ccc}
\frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_m}{\partial x_1} \\
\vdots &  & \vdots \\
\frac{\partial f_1}{\partial x_n} & \cdots & \frac{\partial f_m}{\partial x_n}\end{array} \right) + 
\left( \begin{array}{ccc}
a_{11} & \cdots & a_{1m} \\
\vdots &  & \vdots \\
a_{n1} & \cdots & a_{nm}\end{array} \right)$$
(B) Or should it be $dF(x,A) = d(df_x + A) = d^2 f_x + I$? And therefore,
$$dF(x,A) =\left( \begin{array}{ccc}
\frac{\partial^2 f_1}{\partial x_1^2} & \cdots & \frac{\partial^2 f_m}{\partial x_1^2} \\
\vdots &  & \vdots \\
\frac{\partial^2 f_1}{\partial x_n^2} & \cdots & \frac{\partial^2 f_m}{\partial x_n^2}\end{array} \right) + 
\left( \begin{array}{ccc}
1 & \cdots & 0 \\
\vdots &  & \vdots \\
0 & \cdots & 1\end{array} \right)$$
Does this look right? Thank you very much.

Since, by your definition, $F$ is a matrix-valued function, $DF$ would be a rank-3 tensor with elements
$$
(DF)_{i,j,k} = \frac{\partial^2 f_i(x)}{\partial x_j \partial x_k}
$$
Some authors also define matrix-by-vector and matrix-by-matrix derivatives differently be considering $m \times n$ matricies as vectors in $\mathbb{R}^{mn}$ and "stacking" the resulting partial derivatives.
See this paper for more details.
