Tangent Spaces and Tangent Vectors The Wikipedia page here gives one of the definitions of the tangent space at $x\in M$ as the set of all derivations at $x$. I'm guessing a tangent vector is just an element of the tangent space. This definition seems more abstract compared to the intuitive notion of a tangent vector at $x$ being the derivative of a curve in $M$ passing through $x$. To help elucidate the first definition, could I get some examples (including calculations) of elements in the tangent space of $\left(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}}\right)\in\mathbb{S}^2$ where $\mathbb{S}^2$ is the unit sphere? Thanks.
 A: There are something like five equivalent notions of "a tangent vector" at a point $P$ of a  manifold $M$ in $\mathbb R^n$: 


*

*the derivative of a curve passing through $P$ and lying on $M$

*A derivation at $P$

*A "collection of coordinates that transforms according to the rule ...." 

*An expression $a_1 \frac{\partial}{\partial x_1} + \ldots + a_n \frac{\partial}{\partial x_n}$ which, when considered as a differential operator on functions on $\mathbb R^n$, is zero (at $P$) on every function $f : \mathbb R^n \to R$ that's locally constant at $P$. 
... others I cannot remember. 
Michael Spivak's Comprehensive Introduction to Differential Geometry has a nice chapter showing that all of these are the same thing...but it is a whole chapter of a book. 
To answer your question (very slightly), for a smooth function $f$ on the 2-sphere, there's some smooth function $g$ defined on a neighborhood of the 2-sphere in 3-space that agrees with $f$ on the sphere itself (to show this probably requires the inverse function theorem or something); such a $g$ is called an extension of f. 
Now I'll define for you a derivation
$$
f \mapsto H(f)
$$
by saying that $H(f)$ is the restriction of $K(g)$ to the 2-sphere, where
$$
K(g) = \frac{\partial g}{\partial x} - \frac{\partial g}{\partial y}.
$$
Now you might ask "How do you know you get the same result no matter which smooth extension $g$ you pick?" Clearly $K(g)$ will be different for different $g$s. But for all of these, the restriction of $K(g)$ to the 2-sphere will be the same. Why? Well, because thinking of 
$$
\frac{\partial }{\partial x} - \frac{\partial }{\partial y}.
$$
as a vector (where the "partials" are the basis!), you have
$$
[1, -1, 0]
$$
which is orthogonal to the gradient $[2x, 2y, 2z]$ of $x^2 + y^2 + z^2 - 1$, which is the function whose zero-set is the sphere, at the point you chose. Critical to the proof of "independence of the choice of $g$" is something involving the chain rule, $f$, and $g$ which I cannot reconstruct in my head right this moment. :(
Now that's skipping a lot of steps...approximately one whole Spivak-chapter worth. But at least it's a concrete example for you. 
