Two ways to differentiate a section of vector bundle Let $\pi:E\rightarrow M$ be a vector bundle, and $D$ a connection on it. Suppose $\sigma_1,\sigma_2\in\Gamma(E)$, $p\in M$, $V\in T_pM$ such that $\sigma_1(p)=\sigma_2(p)$. Are the following two conditions equivalent?


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*$d(\sigma_1)_p(V)=d(\sigma_2)_p(V)$

*$D_V\sigma_1=D_V\sigma_2$

 A: Yes: just write it out in a local basis: $\sigma_i = \sum f_{ij} E_j$ in some local choice of basis $E_j$, and $\sigma_1(p)=\sigma_2(p)$ means that $f_{1j}(p)=f_{2j}(p)$. Then $d(\sigma_1-\sigma_2)$ is defined at $p$ and represented by $d(f_{1j}-f_{2j}) \otimes E_j$, while $D(\sigma_1-\sigma_2)=\sum d(f_{1j}-f_{2j}) \otimes E_j + (f_{1j}-f_{2j}) \otimes D E_j$ so that at $p$, $D(\sigma_1-\sigma_2)(p)=\sum d(f_{1j}-f_{2j}) \otimes E_j$ represented by the same stuff.
To be careful, $d\sigma_1$ and $d\sigma_2$ are not really defined, even at $p$, because $d\sigma(p)$ is defined for a section $\sigma$ just at the points where $\sigma=0$.
A: The two ways are connected by $D_V\sigma = (K\circ d\sigma)(V)$ where $K:TE\to E$ is the connector for the covariant derivative. See 19.12 of 


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*Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008. (pdf)
A: Let $D:V(M)\times \Gamma(E)\rightarrow\Gamma(E)$ be the connection on the vector bundle $E$. Now we consider the local representation of $D$.
Let $V|_U=\sum V_ie_i\in V(M)$, $\sigma|_U=\sum \sigma_is_i\in\Gamma(E)$ and  $\{e_i\}$, $\{s_i\}$ be local vector basis on $U$. Then we have
$$D_V\sigma|_U$$
$$=\sum_k(\sum_i\sum_jV_i\sigma_j\Gamma_{ij}^k+V(\sigma_k))s_k$$

First, from $\sigma_1=\sigma_2$, we have $\sigma_{1k}=\sigma_{2k}$.
Second,
  $$d(\sigma_1)(V)=d(\sigma_2)(V)$$ 
  $$\Longleftrightarrow V(\sigma_{1k})=V(\sigma_{2k})$$ 
  $$\Longleftrightarrow D_V(\sigma_1)=D_V(\sigma_2)$$

So they are equal.
