Let $f(A)=detA$ for all A in $L(\mathbb{R}^n\rightarrow \mathbb{R}^n)$
Prove:
a) $f(A)=Det(A)$ is a continuously differentiable function.
b)$(Df)_{Id}(H)=tr(H)$ for all $H\in L(\mathbb{R}^n\rightarrow \mathbb{R}^n) $
c)$(Dlogf)_{A}(H)=tr(A^{-1}H)$ for all $H\in L(\mathbb{R}^n\rightarrow \mathbb{R}^n)$and invertible A in $L(\mathbb{R}^n\rightarrow \mathbb{R}^n)$
So for a) I used the fact that Det(A) is a polynomial of the entries of A so it is differentiable $\infty$ times
for b) I found this formula which comes from Jacobi's formula: $\det(A + \epsilon X) - \det(A) = \operatorname{tr}(\operatorname{adj}(A) X) \epsilon + O(\epsilon^2) = \det(A) \operatorname{tr}(A^{-1} X) \epsilon + O(\epsilon^2)$, and by using A=I that basically proves it but we haven't proved Jacobi's formula yet so if there is any proof to that which doesn't come from Jacobi's formula I'd love to see it.
c) I don't know yet but my intuition is it comes from$\det(A + \epsilon X) - \det(A) = \operatorname{tr}(\operatorname{adj}(A) X) \epsilon + O(\epsilon^2) = \det(A) \operatorname{tr}(A^{-1} X) \epsilon + O(\epsilon^2)$ and the chain rule