Understanding tangent space basis Consider our manifold to be $\mathbb{R}^n$ with the Euclidean metric.
In several texts that I've been reading, $\{\partial/\partial x_i\}$ evaluated at $p\in U \subset \mathbb{R}^n$ is given as the basis set for the tangent space at p so that any $v\in T_pM$ can be written is terms of them. The texts further state that a $\partial/\partial x_k$ is a unit basis vector.
My question is why are the $\partial/\partial x_i$ unit vectors? I can see how they would give the canonical basis and be unit vectors if they acted on the standard coordinate curves $(x_1, ..., x_n)$, but don't they act on arbitrary functions? Or do they only act on the coordinate curves?
I have a feeling that I'm missing something fundamental in my understanding.
 A: It depends on the notation you are using. Let $(e_1,...,e_n)$ be the natural basis for $\mathbb{R}^n$.
One way to understand $\partial/\partial x_i$ is to imagine it as a derivation on the direction $e_i$ calculated at a point $p$. In this case, you can interpret $\partial/\partial x_i$ as the unit vector $e_i$.
To understand it better you should show that any given derivation on $C^\infty(V,\mathbb{R})$ (i.e. an element of $T_p \mathbb{R}^n$) is actually a linear combination of $\partial/\partial x_1, ..., \partial/\partial x_n$.
A: If your manifold $M$ is embedded within $\mathbb{R}^n$ (say a surface within $\mathbb{R}^3$), it has a dimension, say $d$. So the tangent spaces have dimension less than $n$. Every (suppose it $C^\infty$ to begin with) manifold comes with the algebra $C^\infty(M,\mathbb{R})$ and it can be shown that every derivation $d$ of $C^\infty(M,\mathbb{R})$ is of the form "evaluation at a tangent vector", i.e., for $f\in C^\infty(M,\mathbb{R})$ 
$$
d(f)[x]=f'(X_x)[x]
$$
where $X$ is in the tangent space. On the other hand $f'(X_x)=\sum_{i\leq n}a_i\frac{\partial}{\partial_i}(f)[x]$. That's why tangent vectors are operators and $X=\sum_{i\leq n}a_i\frac{\partial}{\partial_i}$. Why unit vectors ? in fact your manifold, as embedded comes with the metric inherited from that of $\mathbb{R}^n$. Note that it may happen that none of $\frac{\partial}{\partial_i}$ be in any tangent space (think of $M$ of eq. $x=y$ in the plane) but those who are, are unit vectors. Now, if your manifold is $\mathbb{R}^n$ itself ($M=\mathbb{R}^n$), the $\frac{\partial}{\partial_i}$ are a basis of $\mathfrak{Der}(C^\infty(\mathbb{R}^n,\mathbb{R}))$, the space of derivations of $C^\infty(\mathbb{R}^n,\mathbb{R}))$ hence 


*

* Each $d\in\mathfrak{Der}(C^\infty(\mathbb{R}^n,\mathbb{R}))$ is, in fact, a vector field and its evaluation at $p$ is a vector of the tangent space

* If the space is given a (Riemannian) metric (here it is the Euclidean structure) it provides a scalar product on every tangent space and here $(\frac{\partial}{\partial_i})_{1\leq i\leq n}$ is an orthonormal basis of it.  

* As every derivation in a vector field, the intuition is that it is, at each point, the "speed vector" of an integral curve and the norm is the (scalar) speed. You, again, find a coherent result.   

