*I posted this same question on [MSE](https://math.stackexchange.com/questions/3912520/points-of-differentiability-of-squared-distance-from-a-point-in-metric-spaces) with no answer.*

Let $I:=(0, + \infty)$ and let $(X,d)$ be a complete and separable metric space. In this setting we say that $u : I \to X$ is *absolutely continuous* if there exists $g \in L^1(I)$, $g \ge0$ s.t.
$$ \tag{1}\label{1} d(u_s,u_t) \le \int_s^t g(r)dr \quad \forall \, s,t \in I, \quad s \le t.$$
It is well known (see [Duda - Absolutely Continuous Functions with Values in a Metric Space][1] ([DOI](http://dx.doi.org/10.14321/realanalexch.32.2.0569))) that such a $u$ is metrically differentiable.

Now for an absolutely continuous $u : I \to X$ and $v \in X$, consider the real-valued function $f_{u,v} : I \to [0, + \infty)$ defined by
$$f_{u,v}(t):=d^2(u_t,v) \quad t \in I.$$
Then $f$ is also absolutely continuous, so it is differentiable for almost all $t \in I$. Let $A_{u,v}$ be the full-measure domain of differentiability for $f_{u,v}$.

**Can we show that $A_{u,v}$ depends only on $u$?**

For example, this is true if $X$ is a real separable Hilbert space, and even in Wasserstein spaces. Is this true in general?


  [1]: https://www.researchgate.net/publication/2124495_Absolutely_Continuous_Functions_with_Values_in_a_Metric_Space