Show that the solution of $\frac{\rm d}{{\rm d}t}X^x(t)=v(t,X^x(t))$, $X^x(0)=x$, is differentiable in $x$ Let $X^x$ be the solution of$^1$ \begin{align}\frac{\rm d}{{\rm d}t}X^x(t)&=v(t,X^x(t))\\ X^x(0)&=x\end{align} and $$T_t(x):=X^x(t).$$

Assuming that $v$ is differentiable in the second argument, can we show that $T_t$ is differentiable?

I'm only able to prove this when $v$ is twice differentiable in the second argument, since then Taylor's theorem is applicable.

$^1$ Assume $v:[0,T]\times\mathbb R^d\to\mathbb R^d$ is Lipschitz continuous in the second argument uniformly with respect to $t$ and continuous in the first variable.

Define an augmentation of the ODE system via
$$
\frac{d}{dt}U^x(t)=\frac{\partial v}{\partial x}(t,X^x(t))\,U^x(t), ~~~ U^x(0)=I.
$$
Then use Grönwall or similar to find a bound for $E^{x,Δx}(t)=X^{x+\Delta x}(t)-X^x(t)-U^x(t)\Delta x$.

Assume that the following considerations are restricted to a compact domain so that $X^x(s)$, $X^{x+Δx}(s)$, and the connecting segment are inside the domain for all $s\in[0,t]$.
Obviously, $E^x(0)=0$ by construction. Then
\begin{align}
E^{x,Δx}(t)&=\int_0^t\frac{d}{dt}E^{x,Δx}(s)\,ds
\\
&=\int_0^t\left(v(s,X^{x+Δx}(s))-v(s,X^{x}(s))-∂_xv(s,X^{x}(s))U^x(s)Δx\right)\,ds
\\
&=\int_0^t\left[v(s,X^{x+Δx}(s))-v(s,X^{x}(s))-∂_xv(s,X^{x}(s))\left(X^{x+Δx}(s)-X^{x}(s)\right)\right]\,ds
\\&\qquad
 +\int_0^t∂_xv(s,X^{x}(s))\left[X^{x+Δx}(s)-X^{x}(s)-U^x(s)Δx\right]\,ds
\\
&=\int_0^t\rho_v\left(X^{x}(s),X^{x+Δx}(s)-X^{x}(s)\right)\,ds 
+ \int_0^t∂_xv(s,X^{x}(s))E^{x,Δx}(s)\,ds
\end{align}
Now one could argue that the first integrand is uniformly $o(Δx)$ by continuity of $X^x$ in $x$ and the definition of differentiability of $v$, while $∂_xv$ in the second integrand is bounded by the Lipschitz constant $L$ of $v$. This allows to find $δ>0$ for some given $ε>0$ so that for all $\|Δx\|<δ$ one gets the integral inequality with its solution per Grönwall
$$
\|E^{x,Δx}(t)\|\le ε\|Δx\|t+L\int_0^t\|E^{x,Δx}(s)\|\,ds
\implies 
\|E^{x,Δx}(t)\|\le ε\|Δx\|\frac{e^{Lt}-1}{L}
$$
from where $E^{x,Δx}(t)=o(Δx)\implies\dfrac{∂X^x(t)}{∂x}=U^x(t)$ follows.
