Let $n \in \mathbb{N}$. For every $1 \leq i, j, \leq n$, let $f_{ij}(x)$ be differentiable. Define the $n \times n$ matrix $A(x)$ whose $(i, j)^{\text{th}}$ entry equals $f_{ij}(x)$.
Let $F(x) = \text{det}(A(x))$. We wish to find a formula for $F'(x)$.
For every $1 \leq i \leq n$, we denote $A_{i}(x)$ to be the matrix obtained from $A(x)$ by replacing the $i^{\text{th}}$ row of $A(x)$ with the derivative of the $i^{\text{th}}$ row. For example, when $n = 2,$ we have
$$A(x) = \begin{bmatrix}f_{11}(x) & f_{12}(x) \\ f_{21}(x) & f_{22}(x) \end{bmatrix},$$
and $$A_{1}(x) = \begin{bmatrix} f'_{11}(x) & f'_{12}(x) \\ f_{21}(x) & f_{22}(x) \end{bmatrix}.$$
When $n = 2$, show $F'(x) = \text{det}(A_{1}(x)) + \text{det}(A_{2}(x))$.
Prove by induction on $n$ that
$$F'(x) = \text{det}(A_{1}(x)) + \text{det}(A_{2}(x)) + \cdots + \text{det}(A_{n}(x)). $$
I'm really not too sure how to approach this problem. For the first part, at least, I think the best way is to work from the right-hand side. So, we have
$$\text{det}(A_{1}(x)) + \text{det}(A_{2}(x)) = \left(f'_{11}(x)f_{22}(x) - f'_{12}(x)f_{21}(x)\right) + \left(f_{11}(x)f'_{22}(x) - f_{12}(x)f'_{21}(x) \right)$$
I don't know how to show this equals $F'(x)$, though.
Case $n=2$ is simple:
$$F = \det(A) = f_{11}f_{22} - f_{12}f_{21}$$
$$F' = f_{11}'f_{22} + f_{11}f_{22}' - f_{12}'f_{21} - f_{12}f_{21}' = $$ $$f_{11}'f_{22} - f_{12}'f_{21} + f_{11}f_{22}' - f_{12}f_{21}'$$
For the induction step, you'd need to expand $\det(A)$ by the first row, and differentiate (using the inductive assumption to compute the derivative of minors).