Is the derivative of a periodic function always periodic?

Question

True or False : The derivative of a periodic functions is always periodic.

I thought it to be true , as everything about a periodic function repeats itself at regular intervals, and so should it's derivative . But , to my surprise it is given false , which suggests that it might be true most of the time but not always , I have given all my thoughts to finding a counter example but I just can't seem to find even one counter example .

One possibility was $\{x\}$ , which is not differentiable at every integer , and I am confused about whether I should call it periodic or not , because it's graph will be a straight line with holes at every integer , so in a sense it is periodically not defined , just like $\tan x$ wich is not defined at every odd multiple of $\pi\over 2$ but still it is said to be periodic .

Could someone please help me find a counter example and clarify about periodicity of derivative of $\{x\}$.

Thanks !

$\{x\}$ is fractional part of x .

Answer 1

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 $\Bbb R$).

Answer 2

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 $P$.

Answer 3

Just a shortcut to what @diracdeltafunk said above.

You could directly apply the chain rule on $f(x)=f(x+T):$ \begin{eqnarray*}f(x)=f(x+T),~\forall x\in\mathbb{R}&\Rightarrow&f'(x)=(x+T)'\cdot f'(x+T),~\forall x\in\mathbb{R}\\ &\Rightarrow&f'(x)=f'(x+T),~\forall x\in\mathbb{R} \end{eqnarray*}