Does the Riemannian distance function obey the Leibniz rule?

Let $M$ be a Riemannian manifold, and let $
alpha,beta:[0,delta) to M$ be two smooth paths satisfying $alpha(t) neq beta(t)$ for every $t$.

Suppose that for every $t$ there is a unique length-minimizing geodesic from $alpha(t)$ to $beta(t)$. This implies that the Riemannian distance function $d:M times M to mathbb R$ is differentiable at the point $left(
alpha(t),beta(t)right)$.

Now, for a given (fixed) point $p in M$, denote by $d_p:M to mathbb R$ the distance function from $p$, that is $d_p(q)=d(p,q)$. Its differential at a point $q in M$ is denoted by $d(d_p)_q:T_qM to mathbb R$.

Question:
Does
$$frac{d}{{dt}}dleft(alpha(t),beta(t)right)=d(d_{alpha(t)})_{beta(t)}(dot beta(t))+d(d_{beta(t)})_{alpha(t)}(dot alpha(t))$$
hold?

This is a sort of "Leibniz-rule"…I tried to play with the formula described here, using exponential maps and initial velocities of geodesics, but so far without success.

This Leibniz-rule clearly holds for the case of $mathbb R^n$, endowed with the standard Euclidean metric.