Let $\mathbb{S}^2 = {(x, y, z) \in \mathbb{R}^3 |x^2 + y^2 + z^2 = 1}$ be the unit sphere, and let $p \in \mathbb{S}^2$. Let also $D$ be a 1’st order differential operator on $\mathbb{S}^2$ at $p$, i.e. a rule that assigns a real number to any function which is defined and smooth in some neighborhood of p and has the following properties:
(1) $D(cf) = cD(f)$ for any smooth function $f$ and any $c \in \mathbb{R}$.
(2) $D(f + g) = D(f) + D(g)$ in the intersection of domains of $f$ and $g$.
(3) $D(fg) = f(p)D(g) + D(f)g(p)$.
Show that there exists a vector $v \in \mathbb{R}^3$ tangent to $\mathbb{S}^2$ at $p$ such that $D$ is the differentiation in direction $v$.
I'm working out of Lee's Introduction to Smooth Manifolds and have machinery from about 3 chapters into the book.
I'm having a difficult time interpreting what to do on this problem. I know that $\mathbb{S}^2$ is to be treated like an arbitrary manifold, which means I don't really know how to do calculus on it since it won't have coordinates. For one I don't have a function to do calculus. I do have the stereographic projections which I probably use somehow. Something that seems sensible to me is given a point $p$, if we want to find the tangent plane, directional derivative, etc. -do calculus on $\mathbb{S}^2$ somehow, we could take two steriographic projections with $U := \mathbb{S}^2 - N$ and $V := \mathbb{S}^2 - S$ so that $\phi :U \rightarrow \mathbb{R}^2$ and $\psi :V \rightarrow \mathbb{R}^2$ ($N$ and $S$ are north and south poles respectively) and look at $\phi \circ \psi^{-1}$ so that we have a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ and take the preimage of a point in $\mathbb{R}^2$ to get our point in $\mathbb{R}^3$, but I can't do this at the north and south poles since they're not in $V \cap U$. For that, would I look at something else in the atlas that covers these and to the same composition as before to take the derivative? Is this just completely wrong? Any insight into how to do this would help.