Problem. Let $\mathcal{M}_{2 \times 2}(\mathbb{R})$ the space of square matrix of order $2$ and consider the fuction $f: \mathcal{M}_{2 \times 2}(\mathbb{R}) \to \mathcal{M}_{2 \times 2}(\mathbb{R})$ defined by: $$A=\begin{pmatrix} a&b\\c&d\\ \end{pmatrix} \longmapsto f(A) =\begin{pmatrix} d&-b\\-c&a\\ \end{pmatrix}.$$ Show that $$\det(A + H) = \det(A) + \operatorname{tr}(f(A)H) + \det(H),$$ and conclude that $\det$ is differentiable, with $\det'(A) = f(A)^{t}$. Note that if $\det(A) \neq 0$, then $f(A) = \det(A)A^{-1}$.
Notation. $f(A)^{t}$ is the transposed.
I proved that $\det(A + H) = \det(A) + \operatorname{tr}(f(A)H) + \det(H)$, but I don't know to use it for show that $\det'(A) = f(A)^{t}$. I tried to write $\det(A+H) - \det(A) = f(A)^{t} \cdot H + o(H)$ using the equality, and to show that $\displaystyle \lim_{H \to 0}\frac{o(H)}{||H||} = 0$, but I couldn't. I appreciate any hints!
Not exactly an answer, but may illuminate slightly.
I guess I would note that the functions $h \mapsto \det(A+h E_{ij})$ are all polynomials in $h$ and hence smooth. Hence the partial derivatives exist and are continuous and so $\det$ is differentiable.
To find a formula, we can start with special cases.
Note that $\det(I+ t E_{ij}) = 1 + t \delta_{ij}$ and so $D \det(I) (E_{ij}) = \delta_{ij}$.
Hence $D \det (I)(H) = \sum_{ij} H_{ij} \det(I)(E_{ij}) = \operatorname{tr} H$.
Now suppose $A$ is invertible. Then $\det(A+H) = \det A \det (I+ A^{-1} H)$, and it follows from this that $D \det(A)(H) = (\det A )D \det(I)(A^{-1}H) = \det A \operatorname{tr} (A^{-1}H)$. Since $A \operatorname{adj}(A) = (\det A )I$, we have $D \det(A)(H) = \operatorname{tr}(\operatorname{adj}(A) H)$.
Since both sides of the latter formula are continuous and the invertible matrices are dense in the space of matrices, this formula is true for all $A$.
To obtain the other representation of the derivative, we use the inner product $\langle A , B \rangle = \operatorname{tr}(A^TB)$. Then $D \det(A)(H) = \langle \operatorname{adj}(A)^T , H \rangle$.