Does the Riemannian distance function obey the Leibniz rule?

Question

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 $$\frac{d}{{dt}}d\left(\alpha(t),\beta(t)\right)=d(d_{\alpha(t)})_{\beta(t)}(\dot \beta(t))+d(d_{\beta(t)})_{\alpha(t)}(\dot \alpha(t))$$ 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.

Answer 1

Actually, this isn't related the Riemannian geometry at all. The result follows from the following general assertion:

Let $f:M \times M \to \mathbb R$ be smooth. Then $$ \frac{d}{{dt}}f\left(\alpha(t),\beta(t)\right)=d(f_{\alpha(t)}^L)_{\beta(t)}(\dot \beta(t))+d(f_{\beta(t)}^R)_{\alpha(t)}(\dot \alpha(t)), $$ where $f_{p}^L$ and $f_{q}^R$ are the functions $M \to \mathbb R$ obtained by fixing the first and second variables to be $p,q$ respectively.

This is easily proved via observing the splitting of $T(M \times M)$ as a direct sum.