The strategy is to differentiate the determinant of the deformation gradient $A(t) := \hat{J}(t,x)$ using Jacobi's identity for the derivative of a determinant, and then to identify the resulting trace as the divergence of the velocity field. The key ingredients are the chain rule applied to the flow equation and the [Jacobi Formula For Differentiating Determinants](/theorems/719). **Step 1: Derivative of the determinant.** Fix $x \in M$. Define the $n \times n$ matrix \begin{align*} A(t) &:= \hat{J}(t,x), \quad \text{so that } A_{ij}(t) = \frac{\partial \Phi_i}{\partial x_j}(t,x). \end{align*} Since $\Phi(0,x) = x$, we have $A(0) = I$. We claim that $A(t)$ is invertible for all $t \in [0,T]$. To see this, note that Step 2 below shows $A$ satisfies the linear matrix ODE $\dot{A}(t) = B(t) A(t)$ where $B(t) := \nabla u(t, \Phi(t,x))$. Define $C: [0,T] \to \mathbb{R}^{n \times n}$ as the unique solution of the companion ODE $\dot{C}(t) = -C(t)\, B(t)$, $C(0) = I$. Differentiating the product gives $\frac{d}{dt}(C A) = \dot{C} A + C \dot{A} = -C B A + C B A = 0$, so $C(t) A(t) = C(0) A(0) = I$ for all $t$. Hence $A(t)^{-1} = C(t)$ exists for all $t \in [0,T]$. The [Jacobi Formula For Differentiating Determinants](/theorems/719) gives \begin{align*} \frac{d}{dt}\det(A(t)) &= \det(A(t))\operatorname{tr}\bigl(A(t)^{-1}\dot{A}(t)\bigr). \end{align*} **Step 2: Computing $\dot{A}$.** Differentiating the flow equation $\frac{d}{dt}\Phi(t,x) = u(t, \Phi(t,x))$ with respect to $x_j$ and exchanging the order of [differentiation](/page/Derivative) (justified by the regularity $u \in C([0,T]; W^{1,\infty})$) yields, by the chain rule, \begin{align*} \dot{A}_{ij}(t) &= \frac{\partial}{\partial x_j}\bigl[u_i(t, \Phi(t,x))\bigr] = \sum_{k=1}^{n} \frac{\partial u_i}{\partial y_k}(t, \Phi(t,x)) \cdot A_{kj}(t). \end{align*} Denoting by $\nabla u(t, \Phi(t,x)) \in \mathbb{R}^{n \times n}$ the spatial gradient matrix of $u$, with entries $(\nabla u)_{ik} = \frac{\partial u_i}{\partial y_k}$, this reads in matrix form $\dot{A}(t) = \nabla u(t, \Phi(t,x)) \cdot A(t)$. **Step 3: Identifying the trace.** Substituting into the [Jacobi Formula For Differentiating Determinants](/theorems/719) gives \begin{align*} \operatorname{tr}\bigl(A^{-1}\dot{A}\bigr) &= \operatorname{tr}\bigl(A^{-1} \, \nabla u(t, \Phi(t,x)) \, A\bigr) = \operatorname{tr}\bigl(\nabla u(t, \Phi(t,x))\bigr) = (\nabla \cdot u)(t, \Phi(t,x)), \end{align*} where the second equality uses the [Invariance Of Trace And Determinant](/theorems/401) (cyclic property $\operatorname{tr}(A^{-1}BA) = \operatorname{tr}(B)$), and the third uses $\operatorname{tr}(\nabla u) = \sum_i \partial_i u_i = \nabla \cdot u$. Combining with Step 1 yields \begin{align*} \partial_t \mathcal{J}(t,x) &= \mathcal{J}(t,x)\,(\nabla \cdot u)(t, \Phi(t,x)), \end{align*} as claimed.