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In the definition of the Tangent space under definition via derivations

Pick a point $x$ in $M$. A derivation at $x$ is a linear map $D : C^{\infty}(M) \to \mathbb{R} $ which has the property that for all $ƒ, g$ in $C^{\infty}(M)$: $D(fg) = D(f)\cdot g(x) + f(x)\cdot D(g)$ modeled on the product rule of calculus. These derivations form a real vector space in a natural manner; this is the tangent space $T_xM$

My question is, it is given that $D : C^{\infty}(M) \to \mathbb{R}$ is real valued and gives out scalar and how does $D(fg)$ forms a vector space ?

Same is the case in the next sentence as well.

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Just to add that derivations are used when the manifold is not embedded in some ambient $\mathbb{R^n}$, since the tangent space as a tangent vector space only makes sense for embedded manifolds. –  gary Apr 10 '11 at 23:06
Please ignore my previous; I did not read carefully-enough, and the edit function seems disabled. Sorry. –  gary Apr 10 '11 at 23:13

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The tangent space $T_xM$ is the collection of all linear mappings $D:C^\infty(M)\to\mathbb R$ (not the scalars $D(fg)$ )that satisfy the property $$D(fg) = D(f)\cdot g(x) + f(x)\cdot D(g), \forall f,g\in C^\infty(M).$$ These mappings form a vector subspace of the space of all linear functionals on $C^\infty(M)$ (which is itself a vector space): for any two of such linear mappings $D_1$ and $D_2$ and any two real numbers $a$ and $b$, $D:=aD_1+bD_2$ is another linear mapping that satisfies the property. It's quite straightforward to check \begin{eqnarray} D(fg)&=&aD_1(fg)+bD_2(fg)\\ &=&a[(D_1f)g+f(D_1g)]+b[(D_2f)g+f(D_2g)]=(Df)g+f(Dg). \end{eqnarray}

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could you please explain how ? –  Rajesh D Apr 10 '11 at 22:37
I edited my answer. –  GWu Apr 10 '11 at 22:45
thank you. I was'nt aware of notion of vector space for linear functionals. –  Rajesh D Apr 10 '11 at 22:54
what kind of a vector space is $C^\infty(M)$ ? –  Rajesh D Apr 11 '11 at 6:19
please ignore my last question. –  Rajesh D Apr 11 '11 at 10:35

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