Let $\phi:M_2(\mathbb{R})\rightarrow \mathbb{R}$ be the determinant function. Identify $M_2(\mathbb{R})$ with $\mathbb{R}^4$ by $\begin{bmatrix} x & y\\z & w\end{bmatrix}\mapsto (x,y,z,w)$.
Then (following convention of Munkres' Calculus on manifolds), for $A=\begin{bmatrix} a & b\\c & d\end{bmatrix}\leftrightarrow (a,b,c,d)$, the matrix of the total detivative of $\phi$ at $A$ is the $1\times 4$ matrix $$D_A=\begin{bmatrix} d & -c & -b & a\end{bmatrix}$$ Hence the derivative map is given by $$\begin{bmatrix} p\\q\\r\\s\end{bmatrix} \mapsto D_A\begin{bmatrix} p\\q\\r\\s\end{bmatrix}=dp-cq-br+sa$$
On the other hand, it was asked in a problem to show that the derivative of determinant map is by $D_A(B)=\mbox{Tr}((\mbox{adj}\,A)^tB)$. But $$ \mbox{Tr}((\mbox{adj}\,A)^tB)= \mbox{Tr}\Big{(} \begin{bmatrix}d & -c \\-b & a\end{bmatrix} \begin{bmatrix}p & q\\r & s\end{bmatrix}\Big{)}=dp-cr-bq+as.$$
So the answers are looking different!Can one point out my mistake and also suggest standard book reference for the result of total derivative of determinant map?
$$\mbox{adj}\,A = \begin{bmatrix}d & -c \\ -b & a\end{bmatrix}$$ while $$\mbox{adj}\,A^t = \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$$ It is the latter that is used in the formula.