I have shown that the Frechet derivative at $\mathbf{I}$ of the determinant map is $\text{tr}\,\mathbf{H}$. In notation:
$$D \det \mathbf{A}\big|_{\mathbf{I}} (\mathbf{H})=\text{tr}\,\mathbf{H}$$
Is it possible, using this fact, to deduce the derivative of $\det$ at arbitrary $\mathbf{A}?$ i.e. to easily show: $$D\det\mathbf{A}\,(\mathbf{H})=\det\mathbf{A}\,\text{tr}\,(\mathbf{A}^{-1}\mathbf{H})$$
This answer suggests that we can, but I don't understand the hint that is given: I think they mean take $f:\;(\mathbf{X}, \mathbf{Y})\mapsto \mathbf{XY}$ and use the chain rule on $\det f$ but I don't understand how to do this, because I get $Df\cdot D\det (f)$ which I cannot really make sense of.
Otherwise I am tempted to write: $$D \det \mathbf{A}\big|_{\mathbf{I}} (\mathbf{A}^{-1}\mathbf{H})=\text{tr}\,(\mathbf{A}^{-1}\mathbf{H})$$ but then I somehow need to squeeze out $(\det\mathbf{A})^{-1}$ from the left hand side, and have $D\det \mathbf{A}\,(\mathbf{H})$ left... if that is possible and/or makes sense.
Pleeease can someone help me?
The derivative of the determinant map at $A$ is the coefficient $a(H)$ in the expansion $$ \det(A+tH)=\det A + t \, \color{blue}{a(H)} + O(t^2) . $$ But $$\begin{split} \det(A+tH) &= \det\bigl(A(I+tA^{-1}H)\bigr) \\&= \det A \, \det(I+tA^{-1}H) \\&= [\text{according to the fact that you already know}] \\&= \det A \cdot \bigl( 1 + t \, \operatorname{tr}(A^{-1}H) + O(t^2) \bigr) \\&= \det A + t \, \color{blue}{\det A \, \operatorname{tr}(A^{-1}H)} + O(t^2) \bigr) , \end{split}$$ so $$ a(H) = \det A \, \operatorname{tr}(A^{-1}H) . $$