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.

Let me add the assumption that there is some uniformity in the differentiabilty of $v(t,x)$ in that
if we write
$$ f(t,x,\Delta x) = v(t,x+\Delta x) - v(t,x) - \frac{\partial v}{\partial x}(t,x), $$
then given $\epsilon>0$, there exists $\delta>0$ such that
$$ | f(t,x,\Delta x)| \le \epsilon |\Delta x| \quad \text{ for $|\Delta x| < \delta$} $$
(that is to say, the choice of $\delta$ is uniform).
Since $v$ is Lipschitz in the second variable, the partial derivatives of $v$ are uniformly bounded:
$$ \left|\frac{\partial v}{\partial x}\right| \le M $$
Also, by an application of Grönwall's inequality, we can see that $T_t(x)$ is Lipschitz with constant at worst $e^{tM}$.
It seems to me that we only need to show $T_t(x)$ is differentiable at $x = 0$.  And then we can assume without loss of generality that $v(t,0) = 0$.
Consider the ODE
$$ \frac{d}{dt} Y^x = \frac{\partial v}{\partial x} v(t,0) Y^x , \quad Y^x(0) = x ,$$
and write $U_t(x)$ for $Y^x(t)$.  This ODE is linear, so $U_t(x)$ is clearly differentiable with respect to $x$, with derivative $U_t$.
Now look at $X^x(t) - Y^x(t)$:
\begin{align} \frac d{dt} (X^x(t) - Y^x(t)) &= v(t,X^x) - \frac{\partial v}{\partial x} v(t,0) X^x + \frac{\partial v}{\partial x} v(t,0) (Y^x(t) - X^x(t)) 
\\&= f(t,0,X^x) + \frac{\partial v}{\partial x} v(t,0) (Y^x(t) - X^x(t)) .
\end{align}
So as long as $X^x(t)$ and $Y^x(t)$ are less than $\delta$, which is the case if $|x| \le e^{-tM} \delta$, then
\begin{align} \left|\frac d{dt} (X^x(t) - Y^x(t)) \right| 
&\le \epsilon |X^x(t)| + M|X^x(t) - Y^x(t)|
\\ &\le \epsilon e^{tM}|x| + M|X^x(t) - Y^x(t)|
\end{align}
So by using a version of Grönwall's inequality that looks like the variations of parameters formula, we see that if $|x|$ is sufficiently small, then
$$ T_t(x) - U_t(x) = o(|x|) .$$
Thus $$ \nabla T_t(0) = U_t .$$
