Is the derivative of a periodic function always periodic?

If $f$ is periodic with period $T$, then $f'$ is also periodic with period $T$, because, if $f$ is differentiable at $x$,\begin{align}f'(x+T)&=\lim_{h\to0}\frac{f(x+T+h)-f(x+T)}h\\&=\lim_{h\to0}\frac{f(x+h)-f(x)}h\\&=f'(x).\end{align}And $\{x\}'$ is periodic with period $1$ (although its domain is not $\Bbb R$).

The answer is actually "true"; at least, the derivative of any differentiable periodic functions is periodic. This is easy to prove:

Suppose $f : \mathbb{R} \to \mathbb{R}$ is a differentiable periodic function with period $P$. Let $g : \mathbb{R} \to \mathbb{R}$ be the function $g(x) = x + P$. Then $f = f \circ g$ so $f' = (f \circ g)'$. By the chain rule, for all $x \in \mathbb{R}$ we have $$f'(x) = (f \circ g)'(x) = f'(g(x))g'(x) = f'(x+P).$$ In other words, $f'$ is periodic with period $P$.