I'm taking a first course on Smooth Manifolds and I'm studying from L. Tu's book. I have a little problem understanding something that Tu seems to think it's obvious: It says (specifically Introduction to Manifolds, 2nd Edition, page 88 example 8.4) that the vectors $$\frac{\partial}{\partial x^1}\bigg{\vert}_p,\dots,\frac{\partial}{\partial x^n}\bigg{\vert}_p$$ form a basis for the tangent space $T_p\mathbb{R}^n$, with no further explanation. Why is this so obvious? The definition of $T_pM$ is the set of all derivations at $C_p^\infty(M)$, but I can't see why a derivation at $C_p^\infty(\mathbb{R}^n)$ need to be a linear combination of the vectors above.

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    $\begingroup$ If one defines tangent vectors as derivations, that really isn't obvious, and requires proof. In effect, you consider the ring $R$ of germs of smooth functions at the origin, and show that $R$ is local and whose maximal ideal $I$ satisfies $I/I^2$ is an $n$-dimensional $R/I\cong\Bbb R$-module. $\endgroup$ – Lord Shark the Unknown Dec 2 '17 at 10:56
  • $\begingroup$ @LordSharktheUnknown Okay, I see, but It becomes quite obvious if you consider $T_pM$ as the tanent vectors of smooth curves that pass through p, right? $\endgroup$ – JustDroppedIn Dec 2 '17 at 11:09
  • $\begingroup$ "Obvious" is a judgment call, but if these derivations didn't span the entire space, then $(x^1, ..., x^n)$ would not be a full coordinate system about $p$. $\endgroup$ – Paul Sinclair Dec 2 '17 at 15:50

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