Confusion with partial derivatives as basis vectors So I have seen that the directional derivative can be written as 
$$ \frac{df}{d\lambda} = \frac{dx^i}{d\lambda}\frac{df}{dx^i} $$
And we can identify $ \frac{d}{dx^i} $ as basis vectors and $  \frac{dx^i}{d\lambda} $ as components. What I don't understand is why is $\frac{df}{d\lambda} $ considered a vector? It's a derivative of a function w.r.t. a parameter and surely that's not a vector?
I.e. In vector notation the directional derivative is given by a dot product
$$ \frac{df}{d\lambda} = \hat{n} \cdot \nabla f $$ which is a scalar but in tensor notation that seems to not be the case?
 A: I think the physicspages author is just confused. $df/d\lambda$ is a scalar, not a vector. It's the scalar product of the covector $\nabla f$ with the vector $d\mathbf{x}/d\lambda$. They say, "Regarding the partial derivatives as basis vectors, ..." and go on as if $\partial f/\partial x$ and $\partial f/\partial y$ were the basis vectors. This is wrong. In the notational convention they have in mind, it's the operators $\partial/\partial x$ and $\partial/\partial y$ that are used as basis vectors.
In this notational convention, the partial derivative operators are never actually applied to anything. They never have anything written to the right of them. The convention is a notational trick that exploits an isomorphism between vectors and derivative operators, but it doesn't involve actually taking the derivative of anything.
The thought that they're probably trying to express is that in their example, their paraboloid is embedded in a higher-dimensional space (which would not normally be the case in general relativity). They're being sloppy/confused with their notation, because they're using the symbol $f$ to mean a scalar field defined on the $(x,y)$ plane, but they're also treating $f$ as if it were a position vector in $(x,y,z)$ space.
A: The directional derivative $\frac{df}{d\lambda}$ is not actually a vector in the space spanned by the $x^i$. What the source was trying to say was that in the abstract vector space spanned by the partial derivative operators, $\frac{d}{d\lambda}$ can be thought of as a vector.
*http://www.physicspages.com/2013/02/10/tangent-space-partial-derivatives-as-basis-vectors/
