Show there exists no diffeomorphism between positive real line and a spiral I'm given the following two subset of $\mathbb{R}^2$:
$$
M = \{(t,0) \in \mathbb{R}^2 \mid t>0\}, 
\quad 
N = \{(t \cos(1/t),\, t \sin(1/t)) \in \mathbb{R}^2 \mid t>0 \} 
$$
The goal of the exercise is to show that there exists no diffeomorphism
$\varphi: \mathbb{R}^2 \to \mathbb{R}^2 $ such that $\varphi (M) = N$.
In a previous exercise I showed that there exists a homeomorphism such that the same conditions are fullfiled.
I'm also given the hint that one should start by showing that the partial derivatives with respect to x of such a diffeomorphism
$\varphi: \mathbb{R}^2 \to \mathbb{R}^2$
(if he would exist) would vanish in $(0,0)$.
I know how to finish the exercise if I find a way of showing that the partial derivatives with respect to x vanish at $(0,0)$. Then the Jacobian determinant is zero at $(0,0)$ so the inverse function is not differentiable at $(0,0)$ and I find a contradiction to the existence of a differeomorphic $\varphi$.
So far I tried to use the difference quotient
$$
\lim_{h\to 0} \frac{\varphi_1(x+h,y) - \varphi_1(x,y)}{h}. 
$$
For this Limit to be zero the numerator has to decrease faster than the linear denominator. I figure that I will have to use that the line segment around $(0,0)$ in $N$ is infintely long while the length of the line segment of the preimage is finite but so far I'm not sure how to combine this with the difference quotient.
The concept of differentiating an unknown function is not something I encountered before so I would be thankful for any help.
 A: Assume there exists a diffeomorphism $\varphi$ on $\mathbb R^2$ such that $\varphi(M) = N$. Since $0$ is the only limit point of $M$ not belonging to $M$, also $\varphi(0)$ must be the only limit point of $N$ not belonging to $N$. We conclude $\varphi(0) = 0$.
The curve $\gamma : [0,1] \to \mathbb R^2, \gamma(t) = (t,0)$, is a smooth embedding with derivative $\gamma'(t) = (1,0) \ne 0$. Hence the curve $\varphi \circ \gamma : [0,1] \to \mathbb R^2$ is a smooth embedding with continuous derivative $\bar \gamma'(t) \ne 0$. There is a reparameterization $\bar \gamma : [0,b] \to \mathbb R^2$  of $\varphi \circ \gamma$ such that $\lVert \bar \gamma'(t) \rVert = 1$. By construction $\bar \gamma(0) = 0$ and  $\bar \gamma((0,b]) \subset N$. Moreover $\bar \gamma(t) + \bar \gamma'(t)$ is contained in the tangent at $N$ in the point $\bar \gamma(t)$. But in any neigborhood of $0$ the spiral $N$ has both horizontal and vertical tangents which means that $\lim_{n \to \infty} \bar \gamma'(t)$ does not exist (recall $\lVert \bar \gamma'(t) \rVert = 1$). This is a contradiction because $\bar \gamma'$ is continuous and thus $\lim_{n \to \infty} \bar \gamma'(t) = \bar \gamma'(0)$.
