Let $\gamma:[0,1]\rightarrow M$ be a geodesic and let $X$ be a vector field on $M$ such that $X(\gamma(0))=0$. Show that $$\nabla_{\gamma'}(R(\gamma',X)\gamma')(0) = (R(\gamma', X')\gamma')(0)$$ where $X' = \frac{DX}{dt}$.
The solution is simple by using derivative of tensors:for all $Z$ on $M$ and at $t=0$, we have \begin{align} 0 &= (\nabla_{\gamma'}R)(\gamma', X, \gamma',Z) \\ &=\frac{d}{dt}\langle R(\gamma',X)\gamma', Z \rangle - \langle R(\gamma', X')\gamma',Z\rangle - \langle R(\gamma',X)\gamma', Z'\rangle \\ &=\langle\nabla_{\gamma'}(R(\gamma',X)\gamma'),Z\rangle - \langle R(\gamma',X') \gamma', Z \rangle. \end{align}
It is a trivial question, but I want to confirm some details :
Question 1: Is it true for $\nabla_{\gamma'}(R(\gamma',X)\gamma')(0)$, the order of operation is that you first calculate $\nabla_{\gamma'}(R(\gamma',X)\gamma')$ and then put $0$ into the formula ?
Question 2 Is it true that $\langle R(\gamma',X)\gamma', Z'\rangle(0)= \langle R(\gamma'(0),X(0),\gamma'(0), Z'(0)\rangle$ ?
Appreciate your feedback.