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**Update and answer: The distance function $f_p(x)=d(p,x), p, x \in M$ cannot be smooth at any point $q \in C(p)$, where $C(p)$ is the cut locus of $p$. It is not differentiable at $q \in C(p)$ when there are at least 2 minimal geodesics connection $p$ and $q$. But it can be differentiable if $q \ne p$ is a first conjugate point with a unique minimal geodesic connecting $p$ and $q$; in this case, it cannot be differentiable up to the 2nd order though.

Update and answer: The distance function $f_p(x)=d(p,x)$, $p, x \in M$ cannot be smooth at any point $q \in C(p)$, where $C(p)$ is the cut locus of $p$. It is not differentiable at $q \in C(p)$ when there are at least 2 minimal geodesics connection $p$ and $q$. But it can be differentiable if $q \ne p$ is a first conjugate point with a unique minimal geodesic connecting $p$ and $q$; in this case, it cannot be differentiable up to the 2nd order though.

Update and answer: On a Riemannian manifold $(M,d)$, the distance function from a fixed point $p$, i.e., $f_p(x)=d(p,x), p, x \in M$can still be differentiable at a first conjugate point $q \ne p$ of $p$ when there is a unique minimizing geodesic connecting $p$ and $q$

**Update and answer: The distance function $f_p(x)=d(p,x), p, x \in M$ cannot be smooth at any point $q \in C(p)$, where $C(p)$ is the cut locus of $p$. It is not differentiable at $q \in C(p)$ when there are at least 2 minimal geodesics connection $p$ and $q$. But it can be differentiable if $q \ne p$ is a first conjugate point with a unique minimal geodesic connecting $p$ and $q$; in this case, it cannot be differentiable up to the 2nd order though.

Update and answer: I asked a math professor at Princeton University. Sigh .. This answer is "No", and based on his explanation to me, hereunder is the quickest proof: suppose $d(x,y)$, $x \ne y$ and $x$ fixed, is differentiable at $y$, then $\vert \nabla d(x,y)\vert = 1$ has to hold. But when $y$ is a conjugate point, we have $\nabla d(x,y) = 0$, a contradiction. In summary, $d(x,y)$ is differentiable at $y, y \ne x$ if and only if there is a unique minimal geodesic connecting $x$ and $y$ and $y$ is not a conjugate point to $x$, i.e., $d(x,y)$ is differentiable at $y, y \ne x$ if and only if $y$ is not in the cut locus of $x$. Thank you Leo Moos for your suggestion to put this as an answer.

Update and answer: On a Riemannian manifold $(M,d)$, the distance function from a fixed point $p$, i.e., $f_p(x)=d(p,x), p, x \in M$can still be differentiable at a first conjugate point $q \ne p$ of $p$ when there is a unique minimizing geodesic connecting $p$ and $q$