Let $d\in\mathbb N$, $\tau>0$, $U\subseteq\mathbb R^d$ be open and $T_t$ be a $C^1$-diffeomorphism from $U$ onto an open subset of $\mathbb R^d$ for $t\in[0,\tau)$ with $T_0=\operatorname{id}_U$. Note that $$V:=\bigcup_{t\in[0,\:\tau)}T_t(U)$$ is open.
Assume $[0,\tau)\ni t\mapsto T_t(x)$ is differentiable for all $x\in U$. Can we show (under suitable additional assumptions, if necessary) that there is a $v:[0,\tau)\times V\to\mathbb R^d$ with $$v\left(t,T_t(x)\right)=\frac{\partial T}{\partial t}(t,x)\tag1$$ for all $t\in[0,\tau)\times V$?
If $U=\mathbb R^d$ (and hence $V=\mathbb R^d$), we may simply set $$v(t,x):=\frac{\partial T}{\partial t}\left(t,T_t^{-1}(x)\right)\tag2.$$
EDIT 1: I want to choose $v$ such that it is (jointly) continuous. By assumption, $$[0,\tau)\times U\ni(t,x)\mapsto T_t(x)\tag3$$ is partially differentiable in both the first and the second variable. So, it should be differentiable and hence (jointly) continuous.
EDIT 2: I wonder whether any differentiability properties of $v$ with respect to the second variable carry over to $v$. I've found the following excerpt in a book, which seems to indicate this, but I actually don't understand how they conclude (2.76):