The strategy is to factor the perturbed determinant $\det(A(t+h))$ as $\det(A(t))\det(I + hA(t)^{-1}\dot{A}(t) + O(h^2))$, and then expand the second factor to first order using the multilinearity of the determinant. The trace emerges as the only first-order contribution. **Step 1: Factorisation.** Since $A(t)$ is invertible, write \begin{align*} \det(A(t+h)) &= \det\!\bigl(A(t) + h\dot{A}(t) + O(h^2)\bigr) = \det(A(t))\,\det\!\bigl(I + hA(t)^{-1}\dot{A}(t) + O(h^2)\bigr), \end{align*} where the second equality uses $\det(BC) = \det(B)\det(C)$ with $B = A(t)$ and $C = I + hA(t)^{-1}\dot{A}(t) + O(h^2)$. **Step 2: First-order expansion of $\det(I + hB)$.** Set $B := A(t)^{-1}\dot{A}(t)$ and consider $\det(I + hB)$ for small $h$. The determinant is a multilinear [function](/page/Function) of its columns. Writing $I + hB = (e_1 + hBe_1 \mid \cdots \mid e_n + hBe_n)$ where $e_j$ denotes the $j$-th standard basis vector, multilinearity gives \begin{align*} \det(I + hB) &= \det(e_1 \mid \cdots \mid e_n) + h\sum_{j=1}^{n}\det(e_1 \mid \cdots \mid Be_j \mid \cdots \mid e_n) + O(h^2). \end{align*} The first term equals $1$. In the $j$-th summand, the matrix differs from the identity only in column $j$, where it has $Be_j$. Expanding along column $j$, the only surviving entry is the diagonal one $(Be_j)_j = B_{jj}$, so $\det(e_1 \mid \cdots \mid Be_j \mid \cdots \mid e_n) = B_{jj}$. Hence \begin{align*} \det(I + hB) &= 1 + h\sum_{j=1}^{n}B_{jj} + O(h^2) = 1 + h\,\operatorname{tr}(B) + O(h^2). \end{align*} **Step 3: Conclusion.** Substituting back into Step 1 and taking the [limit](/page/Limit), \begin{align*} \frac{d}{dt}\det(A(t)) &= \lim_{h \to 0}\frac{\det(A(t+h)) - \det(A(t))}{h} = \det(A(t))\lim_{h \to 0}\frac{\det(I + hB) - 1}{h} = \det(A(t))\,\operatorname{tr}(B), \end{align*} which gives $\frac{d}{dt}\det(A(t)) = \det(A(t))\,\operatorname{tr}(A(t)^{-1}\dot{A}(t))$.