Question about Fourier series of derivative of $f$

Question

We have function $f(x)$ and we knows its fourier series. What can we say about fourier series of $f'$ ?

For example, fourier series of $f$ is $\sum ^{\infty }_{n=0}\dfrac{1}{7n^{6}}\cos nx$

What can I say for fourier series of derivative of $f(f')$?

Actually, I know that $\sum ^{\infty }_{n=0}\dfrac{1}{7n^{6}}\cos nx$ converges uniform by Weierstrass M-test with $$M_n= \sum \dfrac{1}{7n^6}$$

Edited: I guess answer is wrong because choose $f(x)=x$ and $f'(x)$ would be $1$ but their fourier series does not satisfy below. Besides fourier series of $x$ and $1$ is the following and they are not equal.

$$\sum _{n=1}^{\infty \:}-\frac{2\left(-1\right)^n\sin \left(nx\right)}{n} \neq 1$$

Answer 1

If the functional series is uniform convergent, then the limit and derivative can be interchanged, more precisely.

If $f_n:[a,b]\to \mathbb{R}$ is continuously differentiable, $f:[a,b]\to \mathbb{R}$, $g:[a,b]\to \mathbb{R}$, and \begin{align*} f_n &\to f \ \ \text{pointwise}, \\ f_n' &\to g \ \ \text{uniformly}, \end{align*} then $f$ is continuously differentiable and $g = f'$.

In your example put $$f_n(x) = \sum\limits^{n}_{k=1}\dfrac{1}{7k^{6}}\cos kx$$ and you get that $$f_n' \to f'.$$ That is $$f' = \sum\limits^{\infty}_{k=1}\left(\dfrac{1}{7k^{6}}\cos kx\right) = \sum\limits^{\infty}_{k=1}\dfrac{1}{7k^{5}}(-\sin kx).$$