The idea of the proof is captured by considering a 2-cycle _{1},
x_{2}} |

First, note _{1}) = x_{2}_{2}) = x_{1},_{1} and x_{2} are
fixed points of ^{2}(x): |

f^{2}(x_{1}) = f(f(x_{1})) = f(x_{2}) = x_{1} |

and |

f^{2}(x_{2}) = f(f(x_{2})) = f(x_{1}) = x_{2} |

Now to test the stability of x_{1} as a fixed point of f^{2}(x),
we compute the derivative
^{2})'(x_{1}). |

By the chain rule |

(f^{2})'(x_{1}) = f'(f(x_{1}))⋅f'(x_{1}) =
f'(x_{2})⋅f'(x_{1}) |

Similarly, |

(f^{2})'(x_{2}) = f'(f(x_{2}))⋅f'(x_{2}) =
f'(x_{1})⋅f'(x_{2}) |

That is, ^{2})'(x_{1}) = (f^{2})'(x_{2}),_{1}^{2}(x)_{2} |

Return to stability of cycles.