Derivative of the exponential map I'm new to Riemannian geometry and always confused by the derivatives there. Consider on a Riemannian manifold $M$, define the following map
$$u(r,\theta)=\text{exp}_{x} (rf(\theta))$$
where $x \in M$, $r \in \mathbb{R}$ and $f(\theta)\in T_xM$.
What is $\partial_{r}u$? I know it is equal to $f(\theta)$ when$ r=0$, what about other $r$? Also, how to compute $\partial_{\theta}u$ ?
 A: If $x \in M$ is fixed, and you want to compute the partial derivative $(\partial_ru)(r,\theta)$, you compute the derivative of the map $$r \mapsto rf(\theta) \mapsto \exp_x(rf(\theta)),$$which is the composition $\exp_x \circ (r \mapsto f(\theta))$, where $\theta$ is fixed. Thus $$(\partial_ru)(r,\theta) = {\rm d}(\exp_x)_{rf(\theta)}(f(\theta)),$$where one thinks of $f(\theta)$ as an element of $T_{rf(\theta)}(T_xM) \cong T_xM$. At $r = 0$, of course this boils down to $(\partial_ru)(0,\theta) = f(\theta)$.
Similarly, for $(\partial_\theta u)(r,\theta)$, you compute the derivative of the map $$\theta \mapsto rf(\theta) \mapsto \exp_x(rf(\theta)),$$which is the composition $\exp_x \circ (\theta \mapsto rf(\theta))$, where $r$ is fixed, and this yields $(\partial_\theta u)(r,\theta) = {\rm d}(\exp_x)_{rf(\theta)}(rf'(\theta))$. This time you see that $(\partial_\theta u)(0,\theta) = 0$, which makes sense geometrically. So essentially, the question boils down to what you know about the derivative ${\rm d}(\exp_x)_v$ for $v \neq 0$. Even if you don't know much, you may be able to combine the above computations with results such as the Gauss lemma to achieve whatever is it that you're trying to do here.
