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How to show $$ d(\exp_q)_0(v)=\frac{d}{dt}\Big|_{t=0} (\exp_q(tv)) $$ where $q\in M$, $M$ is a smooth Riemannian manifold, and $v\in T_qM$. $\exp$ is exponential map, defined as $$ \exp_q(v)=\exp(q,v)=\gamma(1,q,v) $$ where $\gamma(t)$ is a geodesic satisfying $\gamma(0)=q, \gamma'(0)=v$.

This problem is from the Proposition 2.9 of Do Carmo's Riemannian Geometry.

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Enhao Lan
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    Let $f\colon M\to N$ be a smooth map between two smooth manifolds and $m\in M,\ v\in T_m M$. Let $\gamma\colon (-1,1)\to M$ be a smooth curve such that $\gamma(0)=m,\ \gamma'(0)=v$. Then, $df_m(v)=(f\circ \gamma)'(0)$. – Sumanta May 26 '21 at 12:35

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