Derivative and Jacobian determinant of solution of ODE 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 $Jf = det \nabla f$.
 A: The smoothness of $\Phi$ usually is addressed in the textbooks on ODEs (some of them are discussed on MO and on MSE as well).
An application of the chain rule yields the equation for $\nabla \Phi$, as Deane Yang commented. The equation for $\det \nabla \Phi$ is also well-known, but I cannot immediately recall a book where its proof is presented. But the standard proof was already sketched by Deane Yang.
For any differentiable time-dependent matrix $A(t)$ by Jacobi's formula $(\det A(t))' = \mathop{\mathrm{tr}} (\mathop{\mathrm{adj}(A(t)) \cdot A'(t)})$.
In particular, if $A'(t) = B(t) A(t)$ for some matrix $B$ then $$\det(A(t))' = \mathop{\mathrm{tr}} (\mathop{\mathrm{adj}(A(t)) \cdot B(t) \cdot A(t)}) = \mathop{\mathrm{tr}} (B(t) \cdot A(t) \cdot \mathop{\mathrm{adj}(A(t))}) 
= \det(B(t)) \det(A(t)),$$ where a standard property of adjugate matrix was used in the second equality. It then remains to substitute $A_{ij}(t) = \nabla_j \Phi_i(x,t)$ and $B_{ik}(t) = \partial_k f_i(\Phi(x,t),t)$.
