when does derivative of a function coincide with derivative of Fourier series? Example:
for function $$f(x)=x^{3}(1-x)^{3}=\sum f_{s}\exp(2\pi isx)$$ Fourier series of its fourth  derivative are different from derivative of its Fourier series
$$f^{(4)}(x)=-360x^{2}+360x-72=\sum g_{s}\exp(2\pi i s x)$$
with $g_{s}\neq(2\pi s)^{4}f_{s}$ 
Related question:
When  $$\sum\left|f_{s}\right|^{2}j^{2p}<\infty$$ is equivalent to $f\in C^{(p)}[0,1]$?  (for periodic f)

An advertisement for the utility and aptness of Sobolev theory is the perfect connection between $L^2$ "growth conditions" on Fourier coefficients, and $L^2$ notions of differentiability, mediated by Sobolev's lemma that says ${1\over 2}+k+\epsilon$ $L^2$ differentiability of a function on the circle implies $C^k$-ness. Yes, there is a "loss". However, the basis of this computation is very robust, and generalizes to many other interesting situations.
That is, rather than asking directly for a comparison of $C^k$ properties and convergence of Fourier series (etc.), I'd recommend seizing $L^2$ convergence, extending this to Sobolev theory for both differentiable and not-so-differentiable functions, and to many distributions (at least compactly-supported), and only returning to the "classical" notions of differentiability when strictly necessary.
I know this is a bit avante-garde, but all my experience recommends it. A supposedly readable account of the issue in the simplest possible case, the circle, is at  functions on circles .
