Derivative of Periodic 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 for Derivatives yields:
 * $\map f x = \map f {x + L} \implies \map {f'} x = \map {f'} {x + L}$

Let $L'$ be the period of $f'$.

Suppose that $\size {L'} < \size L$.

$f$ 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 $\closedint 0 {\size L}$.

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