# Covariant and partial derivative of a vector field (not component)

Is the covariant derivative of a vector field (not the components of a vector) same as the partial derivative? I am adding a screenshot from page 69 from General Relativity: An introduction for physicists by Hobson where they used the partial derivative operator $$\partial_bv$$ for vector field derivatives, where in other books the covariant derivative operator $$\nabla_bv$$ is used.

Relevant: is a vector operating on another like $$u(v)$$ (as in Lie Bracket) also mean covariant derivative of $$v$$ in the direction of $$u$$?

It is indeed a notational issue. The left-hand side represents the tensor that results from the covariant derivative of $$\boldsymbol{v}$$ along the direction of the coordinate vector $$\boldsymbol{e}_b$$. Personally I would put it slightly different as: $$\nabla_{\boldsymbol{e}_b}\boldsymbol{v} = \nabla_b v^a \boldsymbol{e}_a$$ where $$\nabla_b v^a$$ is the quantity given in your eq. (3.32) of your image.
Notice that $$\nabla_b v^a$$ is just notation for the components of this covariant derivative (maybe some people prefer something like $$(\nabla v)_b{}^a$$). What I mean is that $$\nabla_b$$ is not an operator here, although, in practice, it can be treated like that in component calculations if you understand its rules. On the other hand $$\nabla_{\boldsymbol{e}_b}$$ is a true operator. In fact, (3.32) comes from the Leibniz rule,
$$\nabla_{\boldsymbol{e}_b}\boldsymbol{v} = \nabla_{\boldsymbol{e}_b} (v^a \boldsymbol{e}_a) = (\nabla_{\boldsymbol{e}_b} v^a )\boldsymbol{e}_a +v^a (\nabla_{\boldsymbol{e}_b} \boldsymbol{e}_a)$$
$$\nabla_{\boldsymbol{e}_b} v^a = \partial_b v^a \qquad \nabla_{\boldsymbol{e}_b} \boldsymbol{e}_a= \Gamma^c{}_{ab} \boldsymbol{e}_c$$