In General Relativity if $(M,g,\nabla)$ is spacetime being $\nabla$ the Levi-Civita connection, then we can define on a geodesically convex set $U\subset M$ the map $\sigma : U\times U\to \mathbb{R}$
$$\sigma(x,y)=\dfrac{1}{2}g_x(\exp_x^{-1}(y),\exp_x^{-1}(y))$$
in other words, $\sigma(x,y)$ is a half of the geodesic distance squared between two points $x,y\in U$.
Given a chart $(x,V)$ (here we are thinking that $U\subset V$), we can then consider the covariant derivative $\nabla_\mu$ with respect to $\frac{\partial}{\partial x^\mu}$. I want to compute objects like
$$\nabla_\mu \sigma(x,y)$$
$$\nabla_\mu \nabla_\nu \sigma(x,y)$$
with the understanding that the derivative keeps $x$ fixed. So $x$ is a parameter here.
The first derivative is obviously the partial, since $\nabla_X$ is $X$ itself on $C^\infty(M)$.
The issue is that we need to differentiate $\exp_x^{-1}$ and this doesn't seem trivial. All I know about $\exp_x$ is that it is a map $\exp_x : T_x M\to U$ which given $X\in T_xM$, if $\gamma_{x,X} : I\subset\mathbb{R}\to U$ is the unique geodesic with $\gamma_{x,X}(0)=x$ and $\gamma_{x,X}'(0)=X$ then $\exp_x(X)=\gamma_{x,X}(1)$.
I have no idea on how to differentiate this map, let alone its inverse.
So how do we compute $\nabla_\mu \sigma(x,y)$ with $x$ fixed? How the exponential map is differentiated?
The differential of the exponential map is known for the Riemannian case (or at least we may have some information about it).
Proposition (Gallot, Hulin, Lafontaine - Riemannian Geometry, 3.46) Let $m$ be a point of a Riemannian manifold $(M,g)$, and $u,v$ be two tangent vectors at $m$. Let $c$ be the geodesic $r \mapsto \exp_mrv$, and $Y$ be the Jacobi field along $c$ such that $Y(0) = 0$ and $Y'(0) = u$. Then $$ Y(r)=d_{tv}\exp_m\cdot ru. $$
Looking at the proof of this proposition and the others needed to arrive at this point, I would say that there is no use of the positivity of the metric, but I could be wrong.