Prove that if D(x) = det [f_ij(x) ]_(n × n), then D'(x) =

Prove that if $D(x) = \det\bigl[f_{ij}(x)\bigr]_{n\times n}$, then $D'(x) = \sum_{k=1}^{n} D_{k}(x)$ where $D_{k}(x)$ is obtained from $D(x)$ by replacing the $k$-th row with its derivative $\bigl(f'_{k1},\,f'_{k2},\ldots,f'_{kn}\bigr)$, all other rows unchanged.

Step 1: Write the determinant as a Leibniz sum

By definition, the determinant is a signed sum over all permutations $\sigma$ of $\{1,\ldots,n\}$, each term picking exactly one entry from every row: $D(x) = \sum_{\sigma\in S_{n}}\operatorname{sgn}(\sigma)\prod_{i=1}^{n} f_{i,\sigma(i)}(x)$.

This is a finite sum of finite products, so it can be differentiated term by term.

Step 2: Apply the product rule to each term

For $n$ factors the product rule reads $\bigl(g_{1}g_{2}\cdots g_{n}\bigr)' = \sum_{k} g'_{k}\prod_{i\neq k}g_{i}$: differentiate one factor at a time and leave the rest alone. Hence

\[ \frac{d}{dx}\prod_{i=1}^{n}f_{i,\sigma(i)} = \sum_{k=1}^{n} f'_{k,\sigma(k)}\prod_{i\neq k}f_{i,\sigma(i)}, \]

and therefore

\[ D'(x) = \sum_{\sigma\in S_{n}}\operatorname{sgn}(\sigma)\sum_{k=1}^{n} f'_{k,\sigma(k)}\prod_{i\neq k}f_{i,\sigma(i)}. \]

Step 3: Interchange the two sums

Both sums are finite, so the order may be swapped: \[D'(x) = \sum_{k=1}^{n}\underbrace{\sum_{\sigma\in S_{n}}\operatorname{sgn}(\sigma)\,f'_{k,\sigma(k)}\prod_{i\neq k}f_{i,\sigma(i)}}_{\text{a determinant}}.\]

Step 4: Identify the inner sum

For each fixed $k$, the inner sum is exactly the Leibniz expansion of the matrix whose $k$-th row has been replaced by the derivatives $f'_{k1},\ldots,f'_{kn}$ while rows $i\neq k$ are untouched. That matrix's determinant is $D_{k}(x)$, so

\[ D'(x) = \sum_{k=1}^{n}D_{k}(x). \]

Step 5: Verification in the 2 × 2 case

Here the rule should reproduce the ordinary product rule twice over: \[D = \begin{vmatrix} f_{11} & f_{12} \\ f_{21} & f_{22}\end{vmatrix} = f_{11}f_{22} - f_{12}f_{21}\]

\[D' = f'_{11}f_{22} + f_{11}f'_{22} - f'_{12}f_{21} - f_{12}f'_{21}\]

\[D_{1}+D_{2} = \begin{vmatrix} f'_{11} & f'_{12} \\ f_{21} & f_{22}\end{vmatrix} + \begin{vmatrix} f_{11} & f_{12} \\ f'_{21} & f'_{22}\end{vmatrix} = \bigl(f'_{11}f_{22}-f'_{12}f_{21}\bigr) + \bigl(f_{11}f'_{22}-f_{12}f'_{21}\bigr) \;\;\checkmark\]