This is a simple question I asked in math.SE last month but unfortunately no one gives any comment. So I decided to try some luck here.
**You can skip examples below and read from "General setting" at the bottom**.

 
Given a smooth arc (part of an ellipse actually) on the complex plane by 
   > $z=\cos t + 0.5 i \sin t,\; t\in[\pi/10,\pi/5] $ , 

and a non-analytic function 
$f(z) = \text{Re } z $  defined on the arc.
Obviously, $f(z) = g(t) := \cos t.$  
Suppose we compute the "derivatives" of $f$ on the arc recursively by   
$f'(z) = g'(t)/z'(t),\quad$
$f''(z) = \dfrac{df'(z)}{dt}\dfrac{1}{z'(t)},\quad$
$f'''(z) = \dfrac{df''(z)}{dt}\dfrac{1}{z'(t)},\quad \dots$

Is there an estimate on the upper bound of magnitude of $n^\text{th}$ order derivative of  $f$ ? 
For example, can we show something like 

> $|f^{(n)}(z)|\leq C n! r^n $, where $C$ and $r$ are positive constants independent of $n$  ?

Note that in the case above the form of $f$ is really simple. 
If $f$ is more complicated, for  example, $f\circ z(t) := \frac{|z'(t)|}{z'(t)}$,
what can we say about $|f^{(n)}(z)|$ ?

**Update:** According To Fedor's answer, function $f$ in first example actually coincides with an analytic function on the arc. I need to modify the curve  so that it is not easy to find an analytic function that coincides with $f$ on the curve.

**New curve:**
Suppose the smooth arc is given by 

> $z(t) = [1+0.5\cos(4t)]\cos t + i [1+0.5\sin(4t)]\sin t,\quad t\in[\frac{\pi}{8},\frac{3\pi}{8}],$

and the function $f$ defined on the arc is given by  
    > $f\circ z(t) := \frac{|z'(t)|}{z'(t)}$.

With derivatives for $f$ defined recursively as before, can we derive an upper bound for $\lvert f^{(n)}(z) \rvert$ as above ?

**General setting:**    
Given a smooth Jordan arc parametrized by $z(t)$ 
on complex plane with $z'(t)\neq 0,\; t\in [0,1]$,
and a smooth function $f$ defined on the arc in the sense that $f\circ z(t) \in C^\infty$. 
Define derivatives of $f$ recursively as above, namely, let $g(t):=f\circ z(t)$,
>  $f'(z):= g'(t)/z'(t), \quad f''(z):= \dfrac{df'(z(t))}{dt}\dfrac{1}{z'(t)},\quad f'''(z) = \dfrac{df''(z(t))}{dt}\dfrac{1}{z'(t)},\quad \dots$

 and we ask if there is an estimate
> $||f^{(n)}(z)||_\infty \leq C n! r^n $, where $C$ and $r$ are positive constants independent of $n$  ?