(Multiplier a Conjugacy Invariant) Let f : R - R and g : R - R be smooth (i.e. derivatives of all order exist everywhere). They are (smoothly) conjugate if there exists a smooth h : R - R with smooth inverse so that goh = hof. Let a ER be a point of period k for f. Then use the chain rule to prove that (g* ) (h(x)) = (fky'(z). Note: this proves that multipliers of corresponding cycles under conjugacy are equal