Differentiating the determinant of the Jacobian of a diffeomorphism (don't understand a proof)

You need the formula $$ \frac{d}{dt} \det(G(t)) = \det(G(t)) \text{trace}[G(t)^{-1} G'(t)] .$$

Let's prove it for $t = t_0$. Let $H(t) = G(t_0)^{-1} G(t)$. Then $H'(t) = G(t_0)^{-1} G'(t)$. Since $H(t_0) = I$, it is relatively straightforward to show $$ \frac d{dt} \det(H(t))\big|_{t = t_0} = \text{trace}(H'(t_0)) .\tag1$$ Now substitute for $H(t)$, and the formula drops out for $t = t_0$.

Since $J_{A_t}$ is Lipschitz, it is differentiable almost everywhere (by Rademacher's Theorem), and I think this should be enough to make the whole proof work.

To see $(1)$, start with $H(t) = I + (t-t_0) H'(t_0) + o(t-t_0)$. Then if you write out the formula for the determinant, you get $\det(H(t)) = 1 + (t-t_0)\text{trace}(H'(t_0)) + o(t-t_0)$.