We say that a family of pairs measures/vector fields $(\rho_t,v_t)$ with $v_t\in L^1(\rho_t;\mathbb{R}^d)$ and $\int_0^T\|v_t\|_{L^1(\rho_t)}dt=\int_0^T\int_{\Omega}|v_t|d\rho_tdt<+\infty$ solves the continuity equation on (0,T) in the distributional sense if for any bounded and Lipschitz test function $\phi\in C_c^1((0,T\times\overline{\Omega})$, we have $$\int_0^T\int_{\Omega}(\partial_t\phi)d\rho_tdt+\int_0^T\int_{\Omega}\nabla\phi\cdot v_t d\rho_tdt=0$$
We can also define a weak solution of the continuity equation through the following condition: we say that $(\rho_t,v_t)$ solves the continuity equation in the weak sense if for any test function $\psi\in C_c^1(\overline{\Omega})$, the function $t\mapsto\int\psi d\rho_t$is absolutely continuous in t and, for a.e. $t$, we have $$\frac{d}{dt}\int_{\Omega}\psi d\rho_t=\int_{\Omega}\nabla\psi\cdot v_t d\rho_t$$
Now apparently the two notions of solutions are equivalent, i.e. every weak solution is a distributional solution and every distributional solution is a weak solution.
If I consider a distributional solution $(\rho_t,v_t)$ and I take the particular "separated" test functions $\phi(t,x)=a(t)\psi(x)$ then I get $$\int_0^T a'(t)\int_{\Omega}\psi(x)d\rho_t(x)dt=-\int_0^1 a(t)\int_{\Omega}\nabla\psi(x)\cdot v_t d\rho_t(x)dt$$ which means that the distributional derivative of $t\mapsto\int_{\Omega}\psi(x)d\rho_t(x)$ is $t\mapsto\int_{\Omega}\nabla\psi(x)\cdot v_t d\rho_t(x)$. But then I don't understand why the actual derivative (not just the distributional derivative) of $t\mapsto\int_{\Omega}\psi(x)d\rho_t(x)$ is $t\mapsto\int_{\Omega}\nabla\psi(x)\cdot v_t d\rho_t(x)$.