Derivative of a determinant of a matrix field 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? 
 A: 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.
A: Suppose that you have an $n$-dimensional vector space $V$, equipped with a standard volume n-form $(x_1,\dots,x_n)$. Let $\{b_1,\dots,b_n\}$ be a standard basis such that $(b_1,\dots,b_n)=1$.
For a linear operator $A:V\to V$ we have:
$$
(x_1,\dots,x_n)\mathrm{tr}A=\sum_{i=1}^n(x_1,\dots,Ax_i,\dots,x_n).
$$
You can check that it agrees with the standard $\mathrm{tr}A=\sum_iA_{ii}$ by setting $x=b$ and expanding $Ab_i$ in the basis.
Also, we have
$$
(x_1,\dots,x_n)\mathrm{det}A=(Ax_1,\dots,Ax_n),
$$
can be easily compared with your favorite definition of $\det$.
Now, consider an $(n-1)$-form $\omega$. There exists a unique vector $x_\omega$ s.t.
$$
\omega(x_1,\dots,x_{n-1})=(x_\omega,x_1,\dots,x_{n-1}).
$$
Fix some $x_0$ and let $\omega$ be 
$$
\omega(x_1,\dots,x_{n-1})=(x_0,Ax_1,\dots,Ax_{n-1}),
$$
which defines some $x_\omega$
$$
(x_\omega,x_1,\dots,x_{n-1})=(x_0,Ax_1,\dots,Ax_{n-1}).
$$
This procedure defines a linear map $A^{*}:V\to V$ by $x_0\mapsto x_\omega$. Let me write this as
$$
(A^*x_1,x_1,\dots,x_n)=(x_1,Ax_2,\dots,Ax_n),
$$
now if you let $x_1=Ax_0$ you immediately obtain $AA^* =I\det A$, so by continiuty $A^*$ coincides with your favourite definition of the adjugate $\mathrm{adj}A$.
All that said, here is a nice proof of the Jacobi's formula:
For any set of $x_1,\dots,x_n$,
\begin{align}
\frac{\partial}{\partial\alpha}(Ax_1,\dots,Ax_n)=&\sum_{i=1}^n(Ax_1,\dots,\frac{\partial A}{\partial\alpha}x_i,\dots,Ax_n)=\\\\
\sum_{i=1}^n(x_1,\dots,A^*\frac{\partial A}{\partial\alpha}x_i,\dots,x_n)=&\mathrm{tr}\left(A^*\frac{\partial A}{\partial\alpha}\right)(x_1,\dots,x_n).
\end{align}
Now we only have to recall the above definition of $\det$, and immediately obtain
$$
\frac{\partial}{\partial\alpha}\det A=\mathrm{tr}\left(A^* \frac{\partial A}{\partial\alpha}\right).
$$
Of course, for an invertible $A$ we have $A^* = A^{-1}\det A$.
A: your identity follows simply by using $\log({\rm det}\; A)= {\rm tr}\; (\log A)$, so 
$$\frac{\partial}{\partial x_i}{\rm det}\;A=
\frac{\partial}{\partial x_i} \exp({\rm tr}\;\log A)=
({\rm det}\;A) \frac{\partial}{\partial x_i}{\rm tr}\;\log A=
({\rm det}\;A)\;{\rm tr}\;\left(A^{-1}\frac{\partial}{\partial x_i}A\right)$$
this identity is known as Jacobi's formula.
