I'm not really sure where to put this, so I thought it post it here!

I'm reading through my GR lecture notes, and have come across a comment that has confused me. I quote

Now, I don't really see how this is true. For example, consider a scalar field f. The covariant derivative of this is
[tex]\nabla_af\equiv\partial_af=\frac{\partial f}{\partial x^a}[/tex]

But, aren't [itex]\partial_af[/itex] the components of a covector?

I can only assume that they mean an isolated component of a covariant derivative eg [tex]\nabla_0f[/tex] is a scalar if f is scalar, vector if f is vector etc. Which would not be true for a partial derivative.

He defines the covariant derivative by setting up the situation of two vector fields u and v ona manifold. Then, picking some point P, u determines a flow line passing through P; parametrised by some path parameter s, such that [tex]\bold{u}_P=\left.\frac{d}{ds}\right|_{s=s_P}[/tex]
Suppose Q[itex]\neq[/itex]P is another point on the flowline corresponding to [itex]s_Q\neq s_P[/itex]. Then we write
[tex]\bold{v}(s_Q)\equiv\bold{v}_Q, \hspace{2cm} \tilde{\bold{v}}(s_Q)\equiv \tilde{\bold{v}}_{QP}[/tex] [he actually uses a symbol with two parallel lines over v, like one side of a "norm" sign-- anyway, it means the vector field parallely transported from P, evaluated at Q]
and define the covariant derivative as
[tex]\nabla_{\bold{u}}\bold{v}_P=\lim_{s_Q\rightarrow s_P}\frac{\tilde{\bold{v}}(s_Q)-\bold{v}(s_Q)}{s_Q-s_P}[/tex]

Ok, so would I be correct in thinking that whilst [tex]\nabla_{\bold{u}}v^b=u^a\nabla_av^b[/tex] is a vector, [tex]\nabla_bv^a[/tex] is a not a vector, but is a type (1,1) tensor?

I think this makes sense, since clearly u^a is a vector; if [tex]\nabla_bv^a[/tex] is a type (1,1) tensor, then the product of the two will be a vector.