Why is $D(\exp_{y})_{\dot{\sigma}(0)}(\dot{\sigma}(0))=\dot{\sigma}(1)$ Pardon my ignorance, the following question came up in a proof of a paper I was reading. My background in Riemannian geometry is a bit weak, so I apologize in advance if this is a trivial question
Let $M$ be a compact Riemannian manifold with points $x$ and $y$ and $\sigma$ be a distance minimizing smooth curve connecting $x$ and $y$ with $\sigma(0)=y$ and $\sigma(1)=x$. With this, why is the following true?

$D(\exp_{y})_{\dot{\sigma}(0)}(\dot{\sigma}(0))=\dot{\sigma}(1)$

 A: First, by definition of differential, $D(exp_y)_{\dot{\sigma}(0)}\dot{\sigma}(0)$ is computed as follows.  Pick a curve $\alpha(t)$ in $T_yM$ for which $\alpha(0) = \dot{\sigma}(0)$ and for which $\alpha'(0) = \dot{\sigma}(0)$.  Then $$D(\exp_y)_{\dot{\sigma}(0)}\dot{\sigma}(0) := \left.\frac{d}{dt}\right|_{t=0} \exp_y(\alpha(t)).$$
Let's use $\alpha(t) = \dot{\sigma}(0) + t\dot{\sigma}(0) = (1+t)\dot{\sigma}(0)$.
Then we claim that $\exp_y(\alpha(t)) = \sigma(1+t)$.
Indeed, both are geodesics so it's enough to show that they agree at $t=-1$ and that they have the same derivative when $t=-1$.
Well, $\exp_y(\alpha(-1)) = \exp_y(0) = y = \sigma(1-1)$, so they agree when $t=-1$.  Further, $\left.\frac{d}{dt}\right|_{t=-1} \exp_p(\alpha(t)) = (D\exp_y)_0 (\dot{\sigma}(0)) = \dot{\sigma}(0)$ since $(d\exp_y)_0 = Id$.  This is the same as $\left.\frac{d}{dt}\right|_{t=-1} \sigma(1+t)$ by the chain rule (since the derivative of $1+t$ is $1$).
Putting all this together, $D(\exp_y)_{\dot{\sigma}(0)}\dot{\sigma}(0) = \left.\frac{d}{dt}\right|_{t=0}\sigma(1+t) = \sigma'(1),$ as claimed.
