References to understand $\frac{d}{dt} d(x,y) = \frac 12 \inf \int_\gamma g'(S,S) ds$, for flows on manifolds Let $M$ a 2 dimensional manifold with a Riemann metric $g_t$, depending smoothly on a parameter (time) $t$. For each $t$ there is a distance $d_t$. I want to prove the following:
$$\dfrac{d}{dt}\Bigg|_{t = 0} d_t(x,y) =  \dfrac 12 \inf_{\gamma} \int_\gamma \dot g(S, S) ds $$
the infimum is over all geodesic (for the metric $g_0$) $\gamma$ with endpoints $x$ and $y$ and
$$ S = \frac{\gamma'(u)}{\sqrt{g(\gamma'(u), \gamma'(u))}}, \ \ \ dx =\sqrt{g(\gamma'(u), \gamma'(u))} du. $$
This is equation (18) in here. I am assuming $\dot = \dfrac d{dt}\Bigg|_{t = 0} g_t$ and $g_0 = g$.
I am aware of some variational formulas, but I don't know how can I use these since I don't know if there is a smooth  variation $f(t,s)$ such that $\gamma_t(s) = f(t,s)$ is a geodesic for $g_t$ with endpoints $x,y$.
I only have a introductory course in Riemann geometry, so I would like some references for this, and more generally, for flows of the metric in a manifold. Any tips/ideas for this specific problem are welcome!
Thanks and sorry about my english

Edit.:
Following a hint in a comment here:
Let $\gamma_t:[0,1] \to M$ a minimizing geodesic for $g_t$, $\gamma_t(0) = x$, $\gamma_t(1) = y$, and $L(\alpha)$  the length with respect $g_t$ of a curve $\alpha$. So $d_t(x,y) = L_t(\gamma_t)$. First we notice
$$L_t(\gamma_0) \geq L_t(\gamma_t) = d_t(x,y)$$
On other hand, writing $g_t = g + t \dot g + o(t)$ we get
$$ \begin{align}
d_t(x,y) = L_t(\gamma_t) &= \int_0^1 \sqrt{g_t(\gamma_t'(u), \gamma_t'(u))} du \\
&= \int_0^1 g(\gamma_t'(u), \gamma_t'(u)) + t\dfrac 12 \dfrac{\dot g(\gamma_t'(u), \gamma_t'(u))}{\sqrt{g(\gamma_t'(u), \gamma_t'(u))}} + o(t) du 
\\ &= L_0(\gamma_t) + \dfrac t2 \int_0^1 \dfrac{\dot g(\gamma_t'(u), \gamma_t'(u))}{\sqrt{g(\gamma_t'(u), \gamma_t'(u))}} du + o(t)
\end{align} $$
Since $L_0(\gamma_t) \geq L_0(\gamma_0) = d_0(x,y)$:
$$d_t(x,y) \geq d_0(x,y) + \dfrac{t}{2}\int_{\gamma_t} \dot g(S,S)ds + o(t)$$
Then
$$\dfrac{L_t(\gamma_0) - L_0(\gamma_0)}{t}\geq \dfrac{d_t(x,y)-d_0(x,y)}{t} \geq \dfrac 12 \int_{\gamma_t} \dot g(S,S)ds + \dfrac{o(t)}{t}$$
Following Artic Char we get the upper bound  but I don't know why $\int_{\gamma_t} \dot g(S,S)ds $ goes to $ \int_{\gamma} \dot g(S,S) ds$ to get the lower bound.
(by $o(t)$ I mean a function $f$ such that $ \displaystyle \lim_{t \to 0} \frac{f(t)}{t} = 0$)
 A: In general, $d_t(x, y)$ is not differentiable in $t$. Let's assume that $x, y$ are so chosen so that it is differentiable at $t=0$.
Let $\gamma$ be any minimizing geodesic in $(M, g_0)$ so that $\gamma(0)=x$, $\gamma(1) = y$. Then
\begin{align}
\frac{d}{dt} \bigg|_{t=0} L_{g_t} (\gamma)&= \frac{d}{dt}\bigg|_{t=0} \int_0^1 g_t( \gamma'(u) , \gamma'(u)) du \\
&= \frac 12 \int_0^1 \frac{\dot{g}_0 (\gamma'(u), \gamma'(u))}{\sqrt{g_0(\gamma'(u), \gamma'(u))}} du \\ 
&= \frac 12 \int_{\gamma} \dot g_0 (S, S) ds. 
\end{align}
Since $d_0(x, y) = L_{g_0} (\gamma)$ and $d_t(x, y) \le L_{g_t}(\gamma)$, for $t>0$ we have
$$ \frac{d_t(x, y) - d_0(x, y)}{t} \le  \frac{L_{g_t} (\gamma) - L_{g_0}(\gamma)}{t}$$
Taking $t\to 0^+$ on both sides,
$$ \frac{d}{dt}\bigg|_{t=0} d_t(x,y) \le \frac 12 \int_{\gamma} \dot g_0 (S, S) ds.$$
On the other hand, for $t <0$,
$$ \frac{d_t(x, y) - d_0(x, y)}{t} \ge  \frac{L_{g_t} (\gamma) - L_{g_0}(\gamma)}{t}$$
Taking $t\to 0^-$,
$$ \frac{d}{dt}\bigg|_{t=0} d_t(x,y) \ge \frac 12 \int_{\gamma} \dot g_0 (S, S) ds.$$
Thus we have
$$ \frac{d}{dt}\bigg|_{t=0} d_t(x,y) = \frac 12 \int_{\gamma} \dot g_0 (S, S) ds$$
if $d_t(x, y)$ is differentiable at $t=0$.
Since this is true for all such minimizing geodesics, we can take the infimum among all such geodesics (Obviously we have shown something stronger, that the equality holds for each minimizing geodesic $\gamma$ with the need to take infimum).
Remark Just for fun, I am including a simple example where $d_t(x, y)$ is not differentiable at $t=0$. Let $M = \mathbb R/\mathbb Z$, $x = 1/4$, $y = 3/4$ and let $b$ be a nonnegative, nonzero bump function supported in $[1/3, 2/3]$ and let $g_t = 1 + t b$. At $t=0$, $x, y$ is joined by two geodesics in $(M, g_0)$, $\gamma_1(t) =  1/4+ t/2$, and $\gamma_2(t) =1/4 - t/2$. When $t<0$, $\gamma_1$ is shorter than $\gamma$ and when $t>0$, $\gamma_2$ is shorter. Thus
$$ d_t(x, y) = \begin{cases} \frac 12 - t\int_0^1 b & \text{ when } t<0, \\ \frac 12 & \text{ when } t\ge 0
\end{cases}$$
and $d_t(x, y)$ is not differentiable at $t=0$.
