Let $W$ be a closed symplectic manifold, $H:W\to \Bbb R$ a Hamiltonian, $x\in W$ a critical point of $H$, $X_H$ the associated Hamiltonian vector field, and $\psi^t:W\to W$ the (global) flow of $X_H$. Since $x$ is a critical point we have $\psi^t(x)=x$ for all $t$. Suppose that $x$ is nondegenerate as a periodic solution of the Hamiltonian system. This means that the differential $(d\psi^1)_x:T_xW\to T_xW$ does not 1 as an eigenvalue. In particular $(d\psi^1)_x(Z)\neq Z$ for every nonzero $Z\in T_xW$.
I am reading Proposition 5.4.5 of Audin, Damian's book Morse Theory and Floer Homology, and it is written as follows: Since $\psi^0=\text{id}$, there must be a $t$ such that $$\frac{d}{dt}(d\psi^t)_x(Z)\neq0,$$ and this derivative equals $(d\psi^t)_x([X_H,Z]).$
Why does there exist such $t$? I can see that the map $t\mapsto (d\psi^t)_x(Z)$ is nonconstant, so its derivative cannot be identically zero. But why do we need "Since $\psi^0=\text{id}$"?
Why does the derivative equals $(d\psi^t)_x([X_H,Z])$?
Any hints will be appreciated