TITLE: Derivative of a determinant of a matrix field
QUESTION [8 upvotes]: Let $A(x_1,...,x_n)$ be an  $n\times n$ matrix field over $R^n$. 
I am interested in the partial derivative determinant of $A$ in respect to $x_i$. In can be shown that:
$\frac{\partial{\det(A)}}{\partial{x_i}} = \det(A)\cdot\sum_{a=1}^{n}{\sum_{b=1}^{n}{ A^{-1}_{a,b} \cdot \frac{ \partial{A_{b,a}} }{ \partial{x_i} }}}$
I was able to prove this using induction and careful, boring calculations, but I was wondering if there was any intuition behind this formula?

REPLY [6 votes]: Another proof, using the characteristic polynomial
$$
\det(A+tI) = t^n+t^{n-1}\text{Tr}(A) + t^{n-2}c_2(A) + \dots+ t c_{n-1}(A) + \det(A)
$$
where $c_i(A) = \text{Tr}(\Lambda^i A)$ is the $i$-th characteristic coefficient.
Namely, assume that $A$ is invertible. Then
$$
\det(A+tX) = t^n\det(t^{-1}A+X) = t^n\det(A(A^{-1}X + t^{-1}I)) = t^n\det(A)\det(A^{-1}X+t^{-1}I)
$$
$$
 = t^n \det(A) \Big( t^{-n}+t^{1-n}\text{Tr}(A^{-1}X) + t^{2-n}c_2(A^{-1}X) + \dots+ t^{-1} c_{n-1}(A^{-1}X) +    \det(A^{-1}X)\Big)
$$
$$
 = \det(A)\Big(1+t\text{Tr}(A^{-1}X) + O(t^2)\Big)
$$
Thus
$$
d\det(A)X = \partial_t\big|_0 \det(A+tX) = \partial_t\big|_0  \det(A)\Big(1+t\text{Tr}(A^{-1}X) + O(t^2)\Big)
$$
$$ 
 = \det(A)\text{Tr}(A^{-1}X) = \text{Tr}(\text{Adj}(A)X)
$$
Where $\text{Adj}(A)$ is the adjugate of $A$ which satifies Cramer's rule $\text{Adj}(A).A=A.\text{Adj}(A)=\det(A).I$.
Since invertible matrices are dense, the formula follows. 
This poof can be simplified a little: Prove it for $A=I$ first, and then use that $\det:GL(n)\to (\mathbb R\setminus 0,\cdot)$ is a group homomorphism.
Finally note, that Jacobi's formula
$$ 
\det(e^A) = e^{\text{Tr}(A)}
$$ 
is a consequence, by inserting $tA$ instead of $A$ and differentiating.
Both sides are 1-parameter subgroups and satisfy the same ODE with the same initial condition. 
EDIT: Misprints corrected; Thanks to Peter Kravchuk for pointing them out.