I have a homework problem I'm working on about the discrete logistic equation:
$f(x)=rx(1-x)$
So far, through some experimentation and polynomial division I've dtermined that the fixed poits of $f(f(x))$ are:
$x=\frac{r-1}{r}$ and $x=\frac{r+1\pm\sqrt{(r-3)(r+1)}}{2r}$
As a result, there is only a non-trivial 2 cycle where $x_0\rightarrow f(x_0)=x_1\rightarrow f(x_1)=x_0$ when $r\gt3$. However, between what range $3<r<a$ is this non-trivial 2 cycle stable? To calculate a, I assume that need to determine the first order derivative of $f(f(x))$ and determine what value of r makes $\frac{d}{dx}(f(f(x_0))=1$.
However, it's been a couple years since I took differential equations, I've forgotten how to evaluate the derivative of a second-iterate map, and my professor's office hours aren't until Wednesday.
Is it just the chain rule such that $\frac{d}{dx}f(f(x))=\frac{d}{dx}(f(x_1))*\frac{d}{dx}(f(x_0))$?
We should first note that $x=0$ is another trivial fixed point.
Now, in general the chain rule says that $\frac{d}{dx}g(f(x))=g'(f(x))f'(x)$, and in the case where $f=g$, we have $\frac{d}{dx}f(f(x))=f'(f(x))f'(x)$.
So for your example, we have $f(x)=rx-rx^2$, so $$\begin{align}\frac{d}{dx}f(f(x))&=f'(f(x))f'(x) \\&=\left(r-2r(rx-rx^2)\right)(r-2rx) \\&=r^2\left(1-2rx-2rx^2\right)(1-2x) \\&=-4 r^3 x^3+6 r^3 x^2-2 r^3 x-2 r^2 x+r^2\end{align}$$
However, in this case it might be easier just to avoid the chain rule altogether by directly evaluating the derivative of $$\begin{align}f(f(x))&=f(rx(1-x)) \\&=r(rx(1-x))(1-rx(1-x)) \\&=r^2x(1-x)(1-rx+rx^2) \\&=-r^3 x^4+2 r^3 x^3-r^3 x^2-r^2 x^2+r^2 x\end{align}$$