The Derivative of the Determinant of a function matrix This problem has been bugging me for a while and I can’t find any help online on this problem so I’ll make a new thread.

Given $n^2$ functions $f_{i,j}$, each differentiable on the interval (a,b). Define $F(x) = det[f_{i,j}(x)]$ for each x in (a,b). Prove that the derivative F’(x) is a sum of n determinants
$$F’(x) = \sum_{i=1}^n det(A_i(x))$$
Where $A_i(x)$ is the matrix obtained by differentiating the functions in the ith row of $[f_{i,j}(x)]$

I have tried proving this statement by induction upon n (where n is the dimension of the original matrix of functions) but got stuck on something that just felt like I wasn’t doing it correctly
$$ F’(x) = \sum_{i=1}^n (-1)^{1+j} \biggl( \bigl( f’_{1,j}(x) det(min_{1,j}[f_{i,j}(x)])\bigr) + f_{1,j}(x)(det(min_{1,j}[f_{i,j}(x)]))’\biggr)$$
Where $min_{i,j}A is the ith row and jth column minor of A
 A: The determinant ${\rm det}(A)$ of a square matrix $A=[a_{ij}]_{1\leq i\leq n,\>1\leq j\leq n}$ is a homogeneous $n^{\rm th}$ degree polynomial $p({\bf a})$ in the $n^2$ variables $a_{ij}$. For each given $i\in[n]$ one has 
$$p({\bf a})=\sum_{j=1}^n a_{ij}A_{ij}$$
(development of ${\rm det}(A)$ with respect to the $i^{\rm th}$ row), whereby the cofactors $A_{ij}$ depend only on the $a_{i'k}$ with $i'\ne i$. It follows that
$${\partial p({\bf a})\over\partial a_{ij}}=A_{ij}\qquad\bigl((i,j)\in[n]^2\bigr)\ .$$
In the case at hand the $a_{ij}$ are functions $t\mapsto f_{ij}(t)$ of an exterior variable $t$, and we want to compute the derivative of the function
$$\psi(t):=p\bigl({\bf f}(t)\bigr)\ .$$
By the chain rule we have to differentiate "through" all $n^2$ intermediate variables $a_{ij}$ and obtain
$$\psi'(t)=\sum_{1\leq i\leq n,\>1\leq j\leq n}{\partial p({\bf f}(t))\over\partial a_{ij}}\>f_{ij}'(t)=\sum_{i=1}^n\left(\sum_{j=1}^n A_{ij}\bigl({\bf f}(t)\bigr)\> f_{ij}'(t)\right)=\sum_{i=1}^n {\rm det}\bigl(A_i(t)\bigr)\ ,$$
whereby the $A_i(t)$ are the matrices described in the text of the question.
