I can't get a thing from "Advanced calculus" of Lynn H. Loomis, page 374.
Defyning the tangent space, it says that two curves through $x$, named $\varphi$ and $\psi$ are equivalent, i.e. $\varphi \sim \psi$, if for every differentiable real-valued function $f$ we have
$$\left. \frac{d f \circ\varphi}{dt} \right|_{t=0} = \left. \frac{d f \circ\psi}{dt} \right|_{t=0}$$
Then it says that for a chart $(U, \alpha)$ through $x$ we can write $F=f\circ\alpha^{-1}$ and $\Phi=\alpha \circ\varphi$ such that
$$\left. \frac{d f \circ\varphi}{dt} \right|_{t=0} = dF(\Phi'(0))$$
So it concludes that $\varphi \sim \psi$ iff $\Phi'(0) = \Psi'(0)$.
The $\Longleftarrow$ direction is obvious but for me it is not obvious the $\Longrightarrow$. I guess we must prove that if $\Phi'(0) \neq \Psi'(0)$ there is a function $f$ for witch the differential is different.
Any help? Thanks.
The question has already been done here Loomis and Sternberg: Tangent Space to a manifold, using equivalence classes; help justifying one step of an argument but not properly answered for me. I've already studied the difinition of the tangent spaces in other books without this problem but I would like to follow the passages of this book.