Are there any functions that are continuously differentiable over a range with no known exact derivative? Defining the term "continuously differentiable" over a continuous range $X$, is to say that for any $x \in X$ there exists an $f'(x)$ and throughout $X$, $f'(x)$ is continuous. 
Using this term, can there exist some function $f(x)$ that is continuously differentiable over the range $X$ where $f'(x)$ cannot be expressed as an exact function over that entire range. A piecewise function is acceptable as long as for any of the values $x_p$ -- the piecewise break -- (meaning that $\lim_{x \to x_p^+} f'(x) = \lim_{x \to x_p^-} f'(x)$). 
Can and does such a function exist?
 A: The Fabius function might be an example of what you look for. It is an infinitely differentiable function (thus very smooth) that however is not locally analytic. Since all elementary functions are locally analytic, this function can not be expressed as an elementary function.
The Fabius function is defined on the unit interval, and is given by the cumulative distribution function of
$$\sum_{n=1}^\infty2^{-n}\xi_n,$$
where the $\xi_n$ are independent uniformly distributed random variables on the unit interval.
It can also be written as the Fourier transform of
$$\hat{f}(z) = \prod_{m=1}^\infty \left(\cos\frac{\pi z}{2^m}\right)^m$$
(Text has been copied from the article on Wikipedia.)
A: Any function $f$ is defined "exactly" at all points of its domain, and so is the derivative $f'$ of such a function on a possibly smaller domain. What you have in mind is something different, and can be studied by means of the following nice and simple example:
$$f(x):=\left\{\eqalign{{\sin x\over x}&\qquad(x\ne0)\cr 1\quad&\qquad(x=0)\ .\cr}\right.\ $$
This function is not only continuous at $0$, but even (arbitrarily often) continuously differentiable there. This is not evident from the given presentation of $f$, since $0$ will be an exceptional point at all stages of an iterated differentiation process. Fortunately there is the following way of writing $f$, from which it is immediately evident that $f$ is $C^\infty$ over all of ${\mathbb R}$:
$$f(x)=\int_0^1\cos(t\,x)\>dt\qquad(x\in{\mathbb R})\ .$$
