# Question about Fourier series of derivative of $f$

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.

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

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 $$f' = \sum\limits^{\infty}_{k=1}\left(\dfrac{1}{7k^{6}}\cos kx\right) = \sum\limits^{\infty}_{k=1}\dfrac{1}{7k^{5}}(-\sin kx).$$