Arnold's ODE computation of phase velocity

Question

In Arnold's ODE he defines "phase velocity vector of the flow" as

$v(x) = \dfrac{d}{dt}g^t(x)\lvert_{t=0}$

As an example, he asked the readers to find the flows on the lines for $g^t(x) = x+t$, the answer provided is $1$.

Question, why can't we differentiate the $x$ term as well?

So in the above example we would have:

$\dfrac{dg^t(x)}{dt} = \dfrac{dx+t}{dt} = \dot x + 1$

Doesn't $x$ depend on t? That's the definition of the solution isn't it?

Answer 1

In the flow $g^t(x)$, $x$ denotes the initial value which is constant along the trajectory. The derivative of a constant is, of course, zero.