Difference between parallel transport and derivative of the exponential map Given a Riemannian manifold $M$, let $c(t) = \exp_p(tX)$ be the geodesic emanating from $p \in M$ with initial value $X$. Let $t_0$ be small enough, then we have to ways to map $T_pM$ to $T_{c(t_0)} M$ isomorphically. One is the parallel transport along $c$, let's call it $P_{c, 0, t_0}$ and the other is given by
$$ d \exp_p|_{tX}: T_{tX}T_pM \cong T_p M \longrightarrow T_{\exp_p(tX)}M = T_{c(t_0)}.$$
My question is: What is the relation between those two? 
The parallel transport is a linear isometry (in all dimensions), and the derivative of the exponential map is a radial isometry by the Gauss lemma (also in all dimensions), meaning
$$ \langle d \exp_p|_{tX} \cdot Y, \dot{c}(t) \rangle = \langle Y, X \rangle $$
for all $Y \in T_pM$. 
In two dimensions, this means that the two mappings coincide up to scaling, as there is only on orthogonal direction to the radial one. However, in higher dimensions, this is not true, i guess, though I find it hard to actually compute examples...
Are there formulas which relate the two concepts with curvature terms?
\Edit: Computed that on $S^3$, conincides with the parallel transport except that vectors orthogonal to the direction of parallel transport are multiplied by $\frac{\sin r}{r}$.
 A: This question has been asked and answered on MathOverflow. I have replicated the accepted answer by J. GE below.

To understand the realationship between $Q$ and $P$, it suffices to study how they act on an orthornormal(o.n.) basis. Here is the detailed argument:
Let $J_i(t)$ be the Jacobi field along $c$, with $J_i(0)=0$ and $J_i'(0)=e_i$, where $e_1, \cdots, e_{n-1}, X$ is an o.n basis of $T_p(M)$. Then one can show that $J_i(t)=Q_t(e_i)$,
  since
  $$
Q_t(e_i)=dexp_{tX} (te_i).
$$
Now extend $e_i$ to $E_i(t)$ along $c$ via parallel transport $P$. One can write the $J_i$ (i.e. $Q_t$) in terms of the o.n. basis $E_i(t)$:
  $$
Q_t(e_i)=J_i(t)=\sum_j a_{ij}E_j(t)=\sum_j a_{ij}P_t(e_j).
$$
Let $A=(a_{ij})$. Using Jacobi equation, one can show that 
  $$
A''(t)+R(t)A(t)=0,
$$
  where $R(t)=(R_{ij}(t))$ is the curvature term defined by $R_{ij}(t)=\langle R(\dot{c}, E_i) \dot{c}, E_i \rangle$.
This is essentially Jacobi equation. So roughly speaking, $Q$ and $P$ satisfy the Jacobi equation.
remark: It seems my definition of $Q$ differs by a rescaling fact $t$. So if we use $N$ to denote your map, then $Q_t(e_i)= N(t e_i)$.

A: The relation between the differential of the exponential map $\text{d } \text{exp}$ and parallel transport $P$ (along geodesics) is expounded in the two papers (in the context of optimization on manifolds):
1 ''An accelerated first-order method for non-convex optimization on manifolds'' and
2 ''Curvature-Dependant Global Convergence Rates for Optimization on Manifolds of Bounded Geometry''
Both also provide bounds on $||\text{d } \text{exp}||$, see Proposition A.3 in 1 or Theorem 3.12 in 2.  The following is from 1:

Let $$f_{K_{\text{low}}}(t) =  \left\{ \begin{array}{lII} r^2\left(1 -
 \frac{\sin(t/r)}{t/r}\right) & \quad K_{\text{low}} = 1/r^2 > 0  \\
 \frac{t^2}{6} & \quad K_{\text{low}} = 0  \\
 r^2\left(\frac{\sinh(t/r)}{t/r} - 1\right) & \quad K_{\text{low}} =
 -1/r^2 < 0  \end{array} \right\}.$$
Let $(M, g)$ be a Riemannian manifold with sectional curvatures
bounded below by $-K$ and above by $K$.  Let $x \in M$, $s \in T_x M$,
and $\gamma(t) = \text{exp}_x (t s)$.  If $\gamma$ is defined and has
no interior conjugate point on the interval $[0,1]$, then $$||(\text{d
 } \text{exp})_x (s) - P_s)[\dot{s}]|| \leq K ||\dot{s}_{\perp}||
 f_{-K}(||s||) \quad \forall \dot{s} \in T_x M$$ where $\dot{s}_{\perp}
 = \dot{s} - \frac{\langle s, \dot{s} \rangle}{||s||^2} s$ is the component of $\dot{s}$ perpendicular to $s$, and $P_{t s}$ denotes
parallel transport along $\gamma$ from $\gamma(0)$ to $\gamma(t)$.

