How to differentiate exponential map with parameter dependent basepoint Let $(M,g)$ be a  Riemannian manifold, $\gamma:I\rightarrow M$ a geodesic. Let $c:(-\varepsilon,\varepsilon)$ is defined to be another geodesic with 
$$c(0)= \gamma(0),\ \ \ c^\prime(0)=X.$$
Let $V(s)$,$W(s)$ be parallel vector fields along $c$.
The derivative that I want to compute is 
$$\frac{\partial}{\partial s}\exp_{c(s)} \big(t\cdot(V(s)+sW(s))\big)$$
at $s=0$.
I looked at examples, but there is no such thing as the $c(t)$ there so I can not figure it out.
Thank you very much, I hope it is not a too simple calculation question for this forum.
 A: I encountered this problem several weeks ago when I read the paper 
"A Riemannian interpolation inequality a la Borell, Brascamp and Lieb"
There is more or less a way to get rid of the dependence to the base point. The drawback is that we have to calculate the derivative of the distance function and consider $t$ not so big here (So it might not be very useful).
Let $y = y(t) =\exp_x tV$, where $x = c(0)$. If $x$ is not in the cut locus of $y$ (So $t$ is not big), then the function 
$$ d^2_{y} (z) = d^2(y, z)$$
is differntiable at a neigborhood of $x$. Moreover, we can use Gauss lemma to show 
$$ (*) \ \ \ \exp_{c(s)} \bigg( - \frac 12 \nabla d^2_{y} (c(s))\bigg) = y$$
For simplicity call $Z(s) = t(V(s) + sW(s))$. Then write 
$$\exp_{c(s)}(Z(s)) = \exp_{c(s)} \bigg(- \frac 12 \nabla d^2_{y} (c(s)) + \frac 12 \nabla d^2_{y} (c(s)) + Z(s) \bigg)$$
Then using (*) and chain rule we have
$$\frac\partial{\partial s}\exp_{c(s)}(Z(s))\big|_{s=0} = (d\exp_x)_{tV} \bigg(\frac 12 \nabla_X  \nabla d^2_{y} (x) + (\nabla_X Z)(0)\bigg)$$ 
In your case $V, W$ are parallel along $c$, so 
$$\frac\partial{\partial s}\exp_{c(s)}(Z(s))\big|_{s=0} = (d\exp_x)_{tV} \bigg(\frac 12 \nabla_X  \nabla d^2_{y} (x) + tW \bigg)$$ 
