Derivative of Periodic Real Function

Theorem
Let $f: \R \to \R$ be a real function.

Let $f$ be differentiable on all of $\R$.

Then if $f$ is periodic with period $L$, then its derivative is also periodic with period $L$.

Proof
Let $f$ be differentiable on all of $\R$.

Let $f$ be periodic with period $L$.

Then taking the derivative of both sides using the Chain Rule yields:
 * $f \left({x}\right) = f \left({x+L}\right) \implies f^\prime \left({x}\right) = f^\prime \left({x+L}\right)$

Let $L'$ be the period of $f^\prime$.

Suppose that $\lvert L' \rvert \lt \lvert L \rvert$.

$f$ is differentiable and therefore continuous, by Differentiable Function is Continuous.

From Image of Closed Real Interval is Bounded, it follows that $f$ is bounded on $\left[{0 \,. \, . \, {\lvert L \rvert}}\right]$.

But from the General Periodicity Property, it follows that $f$ is bounded on all of $\R$.

Then from Primitive of Periodic Function, it follows that $L'$ is the period of $f$.

But we had previously established that $L$ was the period of $f$.

This is a contradiction, therefore $L' = L$.

Hence the result.

Also see

 * Primitive of Periodic Function