The fundamental matrix satisfies $\dot{F}(t) = A(t)F(t)$. The key identity needed is: **Jacobi's identity:** $\dfrac{d}{dt}\det F(t) = \operatorname{tr}(A(t))\,\det F(t)$. To verify this in the $2 \times 2$ case: write $\dot{F}_{ij} = A_{i1}F_{1j} + A_{i2}F_{2j}$. Differentiating $\det F = F_{11}F_{22} - F_{12}F_{21}$ and substituting: \begin{align*} \frac{d}{dt}\det F &= (A_{11}F_{11} + A_{12}F_{21})F_{22} + F_{11}(A_{21}F_{12} + A_{22}F_{22}) \\ &\quad - (A_{11}F_{12} + A_{12}F_{22})F_{21} - F_{12}(A_{21}F_{11} + A_{22}F_{21}) \\ &= (A_{11} + A_{22})\det F = \operatorname{tr}(A)\,\det F. \end{align*} Since $A(t) = Jf_{\hat{x}(t)}$, we have $\operatorname{tr}(A(t)) = \nabla \cdot f(\hat{x}(t))$. The identity $\frac{d}{dt}\det F = (\nabla \cdot f)\,\det F$, together with the initial condition $\det F(0) = \det I = 1$, integrates to \begin{align*} \det F(t) = \exp\!\left(\int_0^t \nabla \cdot f(\hat{x}(\tau))\, d\tau\right). \end{align*} Setting $t = P$ gives $\mu_2 = \det F(P) = \exp\!\left(\int_0^P \nabla \cdot f(\hat{x})\, dt\right)$.