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The simplest example of this is $e^x$ which we could say has period 1 (it is its own derivative). $e^{-x}$ would have period 2.

Using similar constructions, I can get a function that has a derivative of period $n$ by doing $e^{x\cdot (1)^{1/n}}$

Is this the only way to get periodic derivatives?

Note: I am treating sin and cos as special cases of this when $n=4$

Is there a proof to this effect?

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    The other thing to note is that a solution to a differential equation of order $n$ is determined by initial conditions on $y, y', \ldots, y^{(n-1)}$, and therefore the space of solutions has dimension $n$. Thus once you have solutions of the form $e^{\lambda x}$ for $n$ different $\lambda$ (and these are easily seen to be linearly independent), all solutions are linear combinations of these.2012-09-09

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Assume $f^{(n)}(x)=f(x)$ for all $x$. Let $g(x) = \sum_{k=0}^{n-1}\xi^{-k} f^{(k)}(x)$, where $\xi^n=1$. Then $g'(x)=\sum_{k=0}^{n-1}\xi^{-k} f^{(k+1)}(x) = \xi\sum_{k=1}^{n}\xi^{-k} f^{(k)}(x)=\xi g(x)$. We already know all solutions of $y'=c y$ and conclude that $g(x)=a e^{\xi x}$.

If we additionally assume that $\xi$ is primitive, we find for $0\le m that $ \sum_{k=0}^{n-1}\xi^{-mk} f^{(k)}(x)= a_m e^{\xi^m x}$ Adding all equations leads to $n f(x) = \sum_{m=0}^{n-1} a_m e^{\xi^m x}.$