I'm working on a homework problem which seems obvious, but I am having a hard time proving/completing. The problem can be stated as follows:
Let $f,g:$ $\mathbb R$ $\rightarrow$ $\mathbb R$ be $\mathcal C^1$ functions and suppose they are topologically conjugated by a diffeomorphism $h$. Let $x_0$ be a periodic point of period $n$ of $f$. Prove that $y_0=h(x_0)$ is a periodic point of period $n$ of $g$, and that $(f^n)'(x_0)=(g^n)'(h(x_0))$
The first part seems pretty obvious from the definition of topological conjugacy - it preserves all the dynamics, so the periodic point is preserved as well. Anything beyond this seems to be lost on me, and I can't figure out how to start the second part.
Since you have $h(f^n) = g^n(h)$, just differentiate and use the chain rule: $$(h(f^n))' = (g^n(h))'$$ $$h'(f^n)\cdot(f^n)' = (g^n)'(h)\cdot h'$$ Evaluating at $x_0$, and recalling that $f^n(x_0)=x_0$ and $g^n(h(x_0))=h(x_0)$, $$h'(\;\underbrace{f^n(x_0)}_{x_0}\;)\cdot(f^n)'(x_0) = (g^n)'(h(x_0))\cdot h'(x_0)$$ $$\require{cancel}\cancel{h'(x_0)}\cdot(f^n)'(x_0) = (g^n)'(h(x_0))\cdot \cancel{h'(x_0)}$$ Cancelling $h'(x_0)$ (since it isn't zero), $$\boxed{(f^n)'(x_0) = (g^n)'(h(x_0))}$$ as desired.