Let $(M,g)$ be a Riemannian manifold, geodesically complete, and assume logarithms are well defined and smooth. 


Let $c: I\to M $ be a smooth path in $M$, and $x\in M$. **Can we say something about** $$\Vert\nabla\_{\dot{c}(t)}\log\_{c(t)}(x)\Vert\_{c(t)} ?$$


I can easily prove that $$\Vert\nabla\_{\dot{c}(t)}\log\_{x}(c(t))\Vert\_{x}=\Vert \dot{c}(t)\Vert\_{c(t)},$$ by differentiating the equality $$\exp\_{x}(\log\_x(c(t)))=c(t),$$ and using the Gauss lemma to justify that $d(\exp\_x)\_{\log(c(t))}:T\_x M\to T\_{c(t)}M$ is a radial isometry. 


**Is it also true that $$\Vert\nabla\_{\dot{c}(t)}\log\_{c(t)}(x)\Vert\_{c(t)}=\Vert \dot{c}(t)\Vert\_{c(t)}?$$**



Differentiating Riemannian logarithmic map
I believe the answer to your question is "no". I am not familiar with the term "Riemannian logarithmic map," but I imagine you mean the inverse of the Riemannian exponential map. Your condition is thus that the exponential map from any point of $M$ is a diffeomorphism (in particular this implies that $M$ is diffeomorphic to $\mathbb{R}^n$).


Since the exponential map from $x$ is a diffeomorphism, there exist geodesic normal co-ordinates about $x$ for the whole manifold, and so the corresponding *distance function* $r=d(x,\cdot)$ is smooth on the whole manifold (except at $x$).


At each point $y$ in $M\setminus\{x\}$, one has that $\operatorname{exp}\_y(-r\nabla r) =x$. So, in your terminology, $\operatorname{log}\_y(x)=-r\nabla r$.


You ask essentially whether, at each point $c$ in $M$, for each vector $v$ in $T\_cM$,
$$\lVert\nabla\_{v}(-r\nabla r)\rVert\_c=\lVert v\rVert\_c;$$
equivalently whether the map from $T\_cM$ to $T\_cM$,
$$v\mapsto \nabla\_v(-r\nabla r),$$
is always an isometry (of vector spaces). In general it is not. For example, in the hyperbolic plane, if I calculate correctly, 
$$\nabla\_{\frac{\partial}{\partial\theta}}(r\nabla r)=r\coth(r)\frac{\partial}{\partial\theta}.$$


This is closely related to the famous question of the "Hessian of the square of the distance function" (since $\nabla (-rdr)=-\tfrac{1}{2}\operatorname{Hess}(r^2)$), which is equal to the identity only in flat manifolds.


