The derivative of the determinant (Jacobi's formula)

Idea. The determinant is a polynomial in $t$; its linear coefficient is read off from the identity permutation. Jacobi's formula follows by factoring out $A$, and extends to singular $A$ because both sides are continuous in $A$.

Step 1: $A=I$. By Leibniz,
$$\det(I+tB)=\sum_\sigma\operatorname{sgn}(\sigma)\prod_{i=1}^n(\delta_{i\sigma(i)}+tb_{i\sigma(i)}).$$
For $\sigma\ne\mathrm{id}$, at least two indices have $\sigma(i)\ne i$, so the product contains at least two factors $tb_{i\sigma(i)}$: the term is $O(t^2)$. For $\sigma=\mathrm{id}$, $\prod_i(1+tb_{ii})=1+t\sum_ib_{ii}+O(t^2)$. Hence
$$\det(I+tB)=1+t\operatorname{tr}B+O(t^2),$$
and the derivative at $0$ is $\operatorname{tr}B$.

Step 2: $A$ invertible. $A+tB=A(I+tA^{-1}B)$, so
$$\det(A+tB)=\det A\cdot\det(I+tA^{-1}B)=\det A\,\big(1+t\operatorname{tr}(A^{-1}B)+O(t^2)\big).$$
The derivative at $0$ is $\det A\operatorname{tr}(A^{-1}B)=\operatorname{tr}\big(\det A\cdot A^{-1}B\big)=\operatorname{tr}(\operatorname{adj}(A)B)$, because $\operatorname{adj}A=\det A\cdot A^{-1}$.

Step 3: all $A$. Fix $B$. Both sides $D(A):=\frac d{dt}\det(A+tB)|_{0}$ and $\operatorname{tr}(\operatorname{adj}(A)B)$ are polynomial in the entries of $A$ (the first because $\det(A+tB)$ is a polynomial in $t$ and the entries, and its $t$-coefficient is then polynomial in $A$). They agree on the dense set $GL_n$ (Part 2) and are continuous, so they agree everywhere.

Numeric check. For random $3\times3$ matrices, the difference quotient at $t=10^{-6}$ gave $3.24429$ against $\operatorname{tr}(\operatorname{adj}(A)B)=3.24429$.

Remark.
- Equivalent form: $\frac{d}{dt}\det A(t)=\det A\operatorname{tr}(A^{-1}A')$, the basis of *Liouville's formula* for linear ODE.
- Taking $B=I$: $\frac{d}{dt}\det(A+tI)|_0=\operatorname{tr}\operatorname{adj}A$ is the coefficient $c_1$ of the characteristic polynomial up to sign.