I am reading a paper, and it says the following:
Let $V = V(x(t))$, i.e., a function of state $x(t)$ but not depend explicitly on $t$.
Now consider a mapping $$\tilde{t} = t+s, $$for any $s\in \mathbb{R}$. So a time shift. The paper says that "under this change of variables, $V_{\tilde{t}}=V_t$." So my question is
How to show $$\frac{\partial V}{\partial t} = \frac{\partial V}{\partial \tilde{t}}$$
What I am thinking is $$\frac{\partial V}{\partial \tilde{t}} = \lim_{\Delta t \rightarrow0} \frac{ V\big(x(t+\Delta t)\big)- V\big(x(t)\big)}{\Delta t+s}$$ However, I cannot see how it can become $\frac{\partial V}{\partial t}$.
By the chain rule, \begin{align*}V_{\tilde{t}} &= V_{x}x_{\tilde{t}} \\ &= V_{x}x_{t}t_{\tilde{t}} \\ &= V_{x}x_{t}(1) \qquad \text{(since $t = \tilde{t}-s$ and $s$ is constant)} \\ &= V_{x}x_t \\ &= V_t\end{align*} as claimed.