How to compute the derivative for my particular line? Let me give a short definition then make a question.
Let $L$ be any smooth directional line on the XY plane. That is ,for any point $p$ on $L$ , just one tangent line exists. Let the tangent spin speed at $p$ on $L$ be defined as the limit of $function1$ at $p$ when a point $q$ approaches to $p$ along $L$ , against the direction of $L$.
Where $function1$ is the ratio =
$( tangent(q) - tangent(p) ) / (q.x - p.x)$ .
My question is this. Let $sL$ be an enough short segment of $L$. Where the tangent spin speed on $sL$ is known as constant $c$ ; and the initial tangent at the first end of $sL$ is known as $t$. Can you compute the derivative for $sL$? If yes , how?
Thank you in advance.
Please read the answer by user7530,it almost answers what I want. The remaining question follows.
I said that the point $q$ moves along $L$ , but the tangent spin speed concerns the speed of $q$ only along the x-axis, not along $L$.
 A: I'm having a lot of trouble understanding your question, but I will do the best I can.
If I understand correctly, what you are trying to capture with "tangent spin speed" is the curvature of the curve: the rate at which the unit tangent vector changes as you move at unit speed along the curve.
Let $\gamma(s)$ be your curve. We'll suppose $\gamma$ is parameterized by arc length, which means that the "speed" of the point $\gamma(s)$ that sweeps out the curve is constant and equal to one: $\|\gamma'(s)\| = 1$. Curvature $\kappa$ is then defined as the magnitude of the second derivative of $\gamma$: 
$$\kappa = \|\gamma''(s)\| = \left\|\frac{d}{ds} \gamma'(s)\right\|.$$
Curvature captures a notion of "tangent spin speed": straight line segments have 0 curvature, and the more sharply $\gamma$ bends, the higher the curvature at that point. (Curvature, or at least this definition of it, is always positive; you can extend this to signed curvature by e.g. defining curvature as positive if the curve bends clockwise and negative if it bends counterclockwise.) We use arc-length parametrization to guarantee that two curves that look the same have the same curvature. 
Your question, then, would be equivalent to the following: given a curve of constant curvature $c$ and initial tangent vector $\gamma'(0) = t$, what is the tangent vector $\gamma'(s)$ at arbitrary $s$?
This question has a particularly simple solution: curves in the plane of constant curvature $c$ are precisely circles of radius $\frac{1}{c}$. Thus $\gamma'(s) = R(\pm cs) t$, where $R(\theta)$ is rotation by $\theta$ radians, e.g.
$$R(\theta) = \left[\begin{array}{cc} \cos\theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right].$$
The precise sign will depend on your convention.
