Derivative of a determinant

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How can I get that? Is it some application of the Chain Rule considering the determinant as a function of the vector function $f$?

Notation $d_{i j}$ is the cofactor of $\frac{\partial f_i}{\partial x_j}$ in the Jacobian matrix and $\partial_i$ stands for the partial derivative with respect to $x_i$

The highlighted part is just because I believe that it’s a typo!!!

An important consequence of this, which i can obtain with a little bit of work is the following:

Let $f$ be an infinitely differentiable function of $x_0,x_1,..,x_n$, $f: \mathbb{R}^{n+1} \rightarrow \mathbb{R}^n$

Let $A_i$ be the matrix with columns the partial derivatives $f_{x_0},..,f_{x_{i-1}},f_{x_{i+1}},..,f_{x_{n}}$ and let, for all $i\not = j$, $C_{ij}$ be the matrix with columns $\displaystyle f_{x_i x_j},f_{x_0},..,f_{x_n}$ excluding $f_{x_i}$ and $f_{x_j}$


$\displaystyle \frac{\partial{\det(A_i)}}{{\partial x_i}} = \sum_{j<i}(-1)^j \det(C_{ij})+\sum_{j>i}(-1)^{j-1} \det(C_{ij})$

Solutions Collecting From Web of "Derivative of a determinant"

Finding $ \det'(A) $ for $ \det : \mathbb{R}^{n^2} \rightarrow \mathbb{R}$ is easy if $ A $ is invertible. Observe that $ f(x) = \det(xI + B) $ is the characteristic polynomial for $-B$. Hence if $f(x) = x^n + ax^{n-1} + …. $ then
$-a$ is sum of roots i.e. $ -a = tr(-B) $, hence $ a = tr(B) $. So it is easy to see that for any matrix norm $\|.\| $
$$ f(1) = \det(I + B) = 1 + tr(B) + O(\|B\|^2) $$
So if $ A $ is invertible, then $\det(A+H) = \det(A)\det(I + A^{-1}H) $, using above relation you get
$$ \det(A+H) = \det(A) + \det(A)tr(A^{-1}H) + O(\|H\|^2) $$
Hence $ \det'(A)(H) = \det(A)tr(A^{-1}H) $.