$D_{Id_n}\det(X)= \det(x_1 \ e_2 \ e_3 \ ... \ e_n) \ + \ ... \ + \det(e_1 \ ... \ e_{n-1} \ x_n)$, $n \geq 1$
where $X=(x_1 \ x_2 \ ... \ x_n) \in M_{n,n}, \\ det:M_{n,n} \to \mathbb{R} $
I want to find the total derivative of $\det$ at $id_n$, evaluated at any matrix $X$ , and this is what I finally got. However, I am not sure if I can simplify this result.
Any help would be appreciated.
I'm guessing you arrived at this formula by using the fact that $\det$ is a multilinear function of its $n$ columns. There is indeed a way to simplify this answer. If you use the cofactor expansion formula (or whatever it's called), it's easy to see that for all $j \in \{1, \dots, n\}$ \begin{align} \det(e_1, e_2, \dots,e_{j-1}, x_j, e_{j+1}, \dots e_n) = (x_j)_j \end{align} (write this out as a matrix and expand along the $j^{th}$ row). Hence, \begin{align} (D_{\text{id}_n}\det)(X) &= \sum_{j=1}^n (x_j)_j \\ &:= \text{trace}(X) \end{align}