Prove that if f is differentiable and periodic
Prove that if $f$ is differentiable and periodic with period $\omega$, so that $f(x+\omega) = f(x)$ for all $x$, then $f'$ is periodic with the same period $\omega$.
Explanation of Method
Differentiating the periodicity relation leaves it unchanged.
Step 1: State the hypothesis
\[ f(x+\omega) = f(x) \qquad \text{for all } x \in \mathbb{R}. \tag{1} \]
Step 2: Differentiate both sides
On the left, the chain rule applies with $u = x+\omega$; since $\omega$ is a constant, $u' = 1$ and no extra factor appears: \[\frac{d}{dx}f(x+\omega) = f'(x+\omega)\cdot 1 = f'(x+\omega), \qquad \frac{d}{dx}f(x) = f'(x).\]
Step 3: Conclude
Since (1) holds identically, the derivatives agree: $f'(x+\omega) = f'(x) \qquad \text{for all } x \in \mathbb{R}$.
which is the definition of $f'$ being periodic with period $\omega$.
Step 4: Verification
\[(\sin x)' = \cos x:\quad \text{both have period } 2\pi \;\;\checkmark\]
\[(\tan x)' = \frac{1}{\cos^{2}x}:\quad \text{both have period } \pi \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,f(x+\omega) = f(x) \;\Longrightarrow\; f'(x+\omega) = f'(x)\,}\]
Remark
Remark. The statement is about a period, not the fundamental period: the derivative can have a shorter one. For instance $f(x) = \sin^{2}x$ has fundamental period $\pi$ while $\sin x$ has $2\pi$; and a constant function, periodic with every $\omega$, differentiates to $0$, which has no fundamental period at all. The converse also fails: $f(x) = x + \sin x$ has periodic derivative $1+\cos x$ but is not itself periodic.