prove that this function is an immersion How I can show that $F: \mathbb{R} \rightarrow \mathbb{R}^2$ defined by $F(t)=(\cos(t),\sin(t))$ is an immersion? In my definition $F$ is an immersion if $\forall p\in\mathbb{R}$, $dF_p$ is injective. I have computed $DFp=(−\sin(t),\cos(t))$ and here we have already seen that it is injective.
But now I would like to work with the definition $dF_p:T_p\mathbb{R}\rightarrow T_{F(p)}\mathbb{R}^2$, $dF_p(v)(f)=v(f\circ F)$ where $f$ a function in the stalk at $p$, and $v$ is a tangent vector. 
How can I proceed?
 A: In your definition, $dF_t$ is a linear map
$$dF_t:T_t \mathbb{R}\rightarrow T_{F(t)}\mathbb{R}^2$$
where $T_t\mathbb{R}$ is the space of derivations of $C^\infty_t(\mathbb{R})$ at $t$. Here I have denoted by $C^\infty_t(\mathbb{R})$ the space of germs of smooth functions at $t$. Note that if $v:C^\infty_t(\mathbb{R})\rightarrow\mathbb{R}$ is a derivation, then there exists a $c\in\mathbb{R}$ such that $v(f)=c\cdot f^\prime(t)$ for all $f\in C^\infty_t(\mathbb{R})$. This establishes an isomorphism $\phi:T_t\mathbb{R}\rightarrow\mathbb{R}$, $v\mapsto c$.
Then $dF_t(v)(f)=v(f\circ F)=\phi(v)\cdot (f\circ F)^\prime(t)$.
Now notice that $f$ is a smooth function $\mathbb{R}^2\rightarrow\mathbb{R}$. So $f\circ F$ is a map $\mathbb{R}\overset{F}{\rightarrow}\mathbb{R}^2\overset{f}{\rightarrow}\mathbb{R}$. Therefore the chain rule from multivariable calculus tells us that
$$(f\circ F)^\prime(t) = Df_{F(t)}\cdot DF_t = -\frac{\partial f}{\partial x}|_t \sin(t)+\frac{\partial f}{\partial y}|_t \cos(t)$$
Note that a derivation $d:C^\infty_p(\mathbb{R}^2)\rightarrow\mathbb{R}$ at $p$ has the form $d(f)=c_1\frac{\partial f}{\partial x}|_p+c_2\frac{\partial f}{\partial y}|_p$. Therefore we obtain an isomorphism $\psi:T_p\mathbb{R}^2\rightarrow\mathbb{R}^2$, $d\mapsto (c_1,c_2)$.
By the above calculation we see that the map $\psi\circ dF_t\circ \phi^{-1}:\mathbb{R}\rightarrow\mathbb{R}^2$ is really given by $x\mapsto (-\sin(t)x, \cos(t)x)$.

Note:
In general, in the same way we can see that for $F:U\rightarrow V$ with $U\subset\mathbb{R}^n$, $V\subset\mathbb{R}^m$, we always have that $dF_p$ is "up to composition with isomorphisms" just given by the total differential $DF_p$ (i.e. the Jacobian) which we already know from multivariable calculus.
