Prove the directional derivative operators at a point on manifold form a vector space

Question

One of the way to define tangent space is to use directional derivative. However, it's not clear at the first glance that the directional derivative operators form a vector space. Let $D$ be the set of all directional derivative operators at $p \in M$ where $M$ is a n dimensional manifold. So I was wondering, rigorously speaking, how to prove that $$a\frac{d}{d\lambda}+b \frac{d}{d\eta} \in D$$

Answer 1

Hint: Note that directional derivative can be written as $$\nabla_u:=u\cdot\nabla$$ So we take $\frac{\partial}{\partial x_i}$ as bases, $$\sum_iu_i\frac{\partial}{\partial x_i}$$ forms all directional derivatives.

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Addition: If you speak of Spacetime and Geometry, as you said, the author wants to show linear combination of directional derivatives is also directional derivative, in which case the set of directional derivatives is closed under scalar multiplication and addition and thus forms vector space. So to prove an operator is derivative operator, two conditions are required. They are linearity and following Leibniz rule. Usually when defining tangent space of smooth manifold where differential operators live, we don't want to adapt the classical definition that $\frac{\partial f(p)}{\partial x}=\lim_{h\to0}\frac{f(p+\hat xh)-f(x)}h$ because the point $p+\hat xh$ may not lie on the manifold. In other word, traditional definition of derivative needs ambient space which may not exists in smooth manifold. Hence abstractly, we call operators with those two properties differentials. Return back to what you ask, the author proves linear combination of two differentials satisfies those properties and concludes new operator is indeed a directional derivative.