Period of derivative is the period of the original function Let $f:I\to\mathbb R$  be a differentiable and periodic function with prime/minimum period $T$ (it is $T$-periodic) that is, $f(x+T) = f(x)$ for all $x\in I$. It is clear that
$$
f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{f(x+T+h) - f(x+T)}{h} = f'(x+T),
$$
but how to prove that $f'$ has the same prime/minimum period $T$? I suppose that there exist $\tilde T < T$ such that $f'(x+\tilde T) = f'(x)$ for all $x\in I$ but can't find the way to get a contradiction.
 A: To see what happens, simply integrate (I will use $\tilde{T}$ instead of $T'$, since prime is being used for derivatives):
\begin{align*}
    f'(x+\tilde{T}) &= f'(x)\\
    \Rightarrow \int^y f'(x+\tilde{T})\,dx &= \int^y f'(x)\,dx \\
    \Rightarrow \int^{y+\tilde{T}} f'(\tilde x)\,d\tilde x &= \int^y f'(x)\,dx \\
\end{align*}
where we have substituted $\tilde x = x+\tilde{T}$.  So we get
$$
    f(y+\tilde{T}) = f(y) + C
$$
for some constant $C$.  But we already know $f$ is periodic, so we must have $C = 0$.  Hence $f(y+\tilde{T}) = f(y)$, so $\tilde{T}$ is some integer multiple of $T$ (since by assumption, $T$ is the prime period of $f$).
A: One solution is to note that $f(x)$ has an associated Fourier series, and since the derivative of a sinusoid of any frequency is another sinusoid of the same frequency, we deduce that the Fourier series of the derivative will have all the same sinusoidal terms as the original.  
Thus, the derivative must have the same frequency as the original function.
