$< Derivative of a periodic function.$

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$< Derivative of a periodic function.$

Show extensively that the derivative of a periodic function is always periodic.

Calculus Derivatives

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Suppose $f$ is a periodic function with period $T$, i.e.
\[f(x+T)=f(x) \text{for all} x.\]
Then
\[f'(x+T)=\lim_{\epsilon \rightarrow 0} \frac{f(x+T+\epsilon)-f(x+T)}{\epsilon}\]
\[=\lim_{\epsilon \rightarrow 0} \frac{f(x+\epsilon)-f(x)}{\epsilon}=f'(x)\]
for all $x$. Hence
\[f'(x+T)=f'(x) \text{for all} x.\]
Hence $f'$ is a periodic function.

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$< Derivative of a periodic function.$