Time derivative of inverse of flow diffeomorphisms Suppose I have a time dependent vector field $V(t)$ on a Riemannian manifold $\mathcal{M}$ which is generated by a diffeomorphism group $\{\varphi_t\}_{t \in I} \in \operatorname{Diff}(\mathcal{M})$. How can I prove that:
$$\frac{\mathrm{d} }{\mathrm{d} t} (\varphi_{t}^{-1})=-\left(\varphi_{t}^{-1}\right)_{*}\left(\frac{\mathrm{d}}{\mathrm{d} t} \varphi_{t}\right)$$
I don't know precisely how to make sense of this time derivative. I think this might be due to some form of the chain rule but I'm lost on how to prove the equality above and I've been trying it for a while now. Would really appreciate some help. Further context can be found here (I found the equality on page $16$). 

Regarding my comments in Ivo's answer, I'll make some considerations that I think might justify the two equalities. In the first one, we want to compute (for a fixed $s_0$) the derivative of the curve $t \mapsto \tilde{f}(t) \doteq \varphi_{t}^{-1}(\varphi_{s_0}(p))$. We have:
$$ \frac{\partial f}{\partial t}(t_0,s_0) = \left. \left( \frac{\rm d}{\mathrm{d}t} \tilde{f} \right) \right|_{t = t_0} = \mathrm{d}\tilde{f}_{t_0}(1) = \left(\frac{{\rm d}}{{\rm d}t}\varphi_t^{-1}\right)\Bigg|_{\varphi_{t_0}^{-1}(\varphi_{s_0}(p))} \in T_{\tilde{f}(t_0)} M = T_{\varphi_{t_0}^{-1}(\varphi_{s_0}(p))} M$$
And for the second one, we want to compute the derivative of the curve $s \mapsto \bar{f}(s) = \varphi_{t_0}^{-1}(\varphi_{s}(p)) = (\varphi_{t_0}^{-1} \circ \alpha)(s)$ (where $\alpha$ is defined obviously). By the chain rule we get:
$$\frac{\partial f}{\partial s}(t_0,s_0) = \mathrm{d} \bar{f}_{s_0}(1) = d(\varphi_{t_0}^{-1})_{\alpha(s_0)}(\mathrm{d} \alpha_{s_0}(1)) = (\varphi_{t_0}^{-1})_\ast\left(\left(\frac{{\rm d}}{{\rm d}s}\varphi_s\right)(p)\right)$$
 A: You don't have the group property $\varphi_{t+s} = \varphi_t\circ \varphi_s$ if $V$ is time-dependent. This holds only if $V$ is autonomous. The only thing you have is that $\varphi_0 = {\rm Id}$. It's also good to note that $({\rm d}/{\rm d}t)(\varphi_t) = V \circ \varphi_t$ by definition. But in any case, this is indeed an instance of the chain rule, as follows: fixed $p \in M$, we have the equality $$\varphi_t(\varphi_t^{-1}(p)) = p.$$Of course we want to take the derivative of both sides with respect to $t$. For the right side life is great and one gets zero. For the left side, the classical trick applies: let $f(t,s) = \varphi_t^{-1}(\varphi_s(p))$ and note that what we want is $$\frac{{\rm d}}{{\rm d}t} f(t,t) = \frac{\partial f}{\partial t}(t,t) + \frac{\partial f}{\partial s}(t,t).$$Now, we have that $$\frac{\partial f}{\partial t}(t,s) = \left(\frac{{\rm d}}{{\rm d}t}\varphi_t^{-1}\right)\Bigg|_{\varphi_t^{-1}(\varphi_s(p))} \implies \frac{\partial f}{\partial t}(t,t) = \left(\frac{{\rm d}}{{\rm d}t}\varphi_t^{-1}\right)\Bigg|_p,$$and also that $$\frac{\partial f}{\partial s}(t,s) = (\varphi_t^{-1})_\ast\left(\left(\frac{{\rm d}}{{\rm d}s}\varphi_s\right)\Bigg|_p\right) \implies \frac{\partial f}{\partial s}(t,t) = (\varphi_t^{-1})_\ast\left(\left(\frac{{\rm d}}{{\rm d}t}\varphi_t\right)\Bigg|_p\right)$$Omitting $p$, we have that $$\frac{{\rm d}}{{\rm d}t}\varphi_t^{-1} = -(\varphi_t^{-1})_\ast \left(\frac{{\rm d}}{{\rm d}t}\varphi_t\right)$$as wanted.
