Computing second order tangential derivative at a point on circle. This is probably a very basic question in the geometry but I have not been able to figure it out. Let P(x,y) be a point on a unit circle that is centered at (0,0). How to compute exactly the function $\frac{\partial^2x}{\partial s^2}$
where $x$ is the x-coordinate of the point P(x,y) and $s$ is the tangent at point P(x,y). Clearly,
$\frac{\partial x}{\partial s} = t_x = −n_y$
where $t_x$ is the x-component of the tangent at point P(x,y) and $n_y$ is the y-component of the normal to circle boundary at point P(x,y). I could verify $\frac{\partial x}{\partial s} = t_x = −n_y$ with finite difference. Now how do I obtain an exact expression for 
$\frac{\partial}{\partial s}\left(\frac{\partial x}{\partial s}\right)$
to get $\frac{\partial^2x}{\partial s^2}$?
 A: I'd suggest that what you want to do can most clearly be accomplished by parameterizing the unit circle with $s$ taken to be the arc length around the circle, i.e.
$$ (x,y) = (\cos s, \sin s) $$
In this framework you can express a unit normal vector and a unit tangent vector at $(x,y)$.
The outward unit normal direction is simply $(x,y)=(\cos s,\sin s)$, i.e. the point $(x,y)$ considered as a vector pointing away from the origin along the radial ray through $(x,y)$.  The inward unit normal is the opposite, $(-x,-y)=(-\cos s,-\sin s)$.
The counterclockwise unit tangent direction is $(-y,x)=(-\sin s, \cos s)$, which is easily verified to be perpendicular to both the outward and inward unit normals at each point on the unit circle.  Similarly the clockwise unit tangent is the opposite, $(y,-x)=(\sin s, -\cos s)$, and it too is perpendicular to both the outward and inward unit normals.
Happily as $s$ measures arc length around the unit circle, it also corresponds in the usual "infinitesimal" way with distance along the tangent in the counterclockwise direction.  Hereafter I will fix the choices of counterclockwise unit tangent vector and outward unit normal vector, largely because these agree with axis vectors at point $(1,0)$.
So we should be (and are) able to recover the $x$-component of the unit tangent vector, resp. $y$-component of the unit normal vector, or as you have denoted them:
$$ t_x = -y = -\sin s = -n_y  $$
Perform the easy calculus operation, differentiating $x$ with respect to $s$ (since $x$ is now a function of $s$, namely $\cos s$):
$$ \frac{dx}{ds} = -\sin s $$
Taking the next step, the second derivative is also easily seen to be:
$$ \frac{d^2 x}{ds^2} = -\cos s = -n_x = -t_y $$
That is, the second derivative of $x$ with respect to $s$ is the $x$-component of the negative outward unit normal vector (aka, of the inward normal unit vector) and the $y$-component of the negative counterclockwise unit tangent vector (aka, clockwise unit tangent vector).
