Let us write $[a_{ij}:i=1,...,m;\,j=1,...,n]$ to mean the $m\times n$ matrix which has $a_{ij}$ in the $i$th row of the $j$th column. When $A_{ij}\,$ ($i=1,...,m;\,j=1,...,n$) is a $p\times q$ matrix, then we accordingly write $[A_{ij}:i=1,...,m;\,j=1,...,n]$ for the $m\times n$ block matrix whose $(i,j)$th (block) entry is $A_{ij}$; this can also be interpreted as an $mp\times nq$ matrix with scalar entries. We treat a (column) $n$-vector $\pmb x=(x_1,...,x_n)$ as an $n\times 1$ matrix, and the transpose of such a vector is a $1\times n$ (row) matrix $\pmb x\!^\top=[x_1,...,x_n]$.
The notation works best if the derivative with respect to a vector variable $\pmb x=(x_1,...,x_n)$ is taken to be a row matrix:$$\frac{\partial y}{\partial\pmb x}=\left[\frac{\partial y}{\partial x_1},...,\frac{\partial y}{\partial x_n}\right].$$The derivative with respect to the corresponding row matrix $\pmb x\!^\top$ is defined dually as the (column) vector $$\frac{\partial y}{\partial\pmb x^{\!\top}}=\left(\frac{\partial y}{\partial x_1},...,\frac{\partial y}{\partial x_n}\right).$$Accordingly, the derivative with respect to an $m\times n$ matrix $X=[\pmb x_1,...,\pmb x_n]$, where $\pmb x_i$ is the $m$-vector $(x_{i1},...,x_{im})$, is$$\frac{\partial y}{\partial X}=\left(\frac{\partial y}{\partial\pmb x_1},...,\frac{\partial y}{\partial\pmb x_n}\right),$$namely the $n\times m$ matrix$$\frac{\partial y}{\partial X}=\left[\frac{\partial y}{\partial x_{ji}}:i=1,...,n;\,j=1,...,m\right].$$Finally, the corresponding matrix derivative of a $p\times q$ matrix $Y$ is$$\frac{\partial Y}{\partial X}=\left[\frac{\partial Y}{\partial x_{ji}}:i=1,...,n;\,j=1,...,m\right],$$which is an $n\times m$ ($p\times q$)-block matrix (i.e. an $np\times mq$ matrix).
In the case of your question, $A= [a_{ij}:i=1,...,m;\,j=1,...,n]$, $\pmb x=(x_1,...,x_n)$, $Y=A\pmb x$, and we are treating $A=X$ as the derivative variable, with $\pmb x$ constant. If you plug this in to the above expression for the matrix derivative, you end up with the $mn\times m$ matrix$$\frac{\partial(A\pmb x)}{\partial A}=\pmb x\otimes\mathrm I_m,$$where $\mathrm I_m$ is the $m\times m$ identity matrix. See the Wikipedia articles for Kronecker (tensor) product and matrix calculus.
Best Answer
You can use Jacobi's identity
\begin{eqnarray} \frac{\partial}{\partial A_{ij}}f(A) &=& \frac{\partial }{\partial A_{ij}} \left( 2 + \log |\det(A)|\right) \\ &=& \frac{1}{|\det(A)|}\frac{\partial}{\partial A_{ij}}\det (A) = C_{ij}(A) \end{eqnarray}
where $C(A)$ is the cofactor matrix of A