Let $\Phi$ be the unique solution of $$\begin{cases} \frac{d}{dt}\Phi(x,t) = f(\Phi(x,t),t) \quad t >0 \\ \Phi(x,0) = x \quad x \in \mathbb{R}^N \end{cases}$$ where we have assumed $f$ smooth.

How do you prove that $\Phi(\cdot, t)$ is smooth and in particular that the following holds?

$$\begin{cases} \frac{d}{dt} \nabla \Phi(x,t) = \nabla_1 f(\Phi(x,t),t)\nabla \Phi(x,t) \quad t>0 \\ \nabla\Phi(x,0) = 1 \quad x \in \mathbb{R}^N \end{cases}$$ and $$\begin{cases} \frac{d}{dt} J \Phi(x,t) = \mathrm{div} f(\Phi(x,t),t)J \Phi(x,t) \quad t>0 \\ \nabla\Phi(x,0) = 1 \quad x \in \mathbb{R}^N \end{cases}$$ where $Jf = det \nabla f$.