I have been looking into fixed stable and unstable points in the iteration $x_{n+1}=f(x_{n})$. My textbook says that in order to determine whether a fixed point $x^*$ is stable, we consider $x^* + \eta_{n}$ and see whether $\eta_{n}$ grows or decays. Fair enough.
What I do not understand is the next step:
$$x^* + \eta_{n+1}= x_{n+1} =f(x^* + \eta_{n}) = f(x^*)+f'(x^*)\eta_{n} + O(\eta_{n}^2)$$
Where does $O(\eta_{n}^2)$ come from? I thought that the rule for differentiation was
$$f'(x^*)=\frac{f(x^* + \eta_{n})-f(x^*)}{\eta_{n}}$$
The problem is that the derivative is defined as the limit of $\frac{f(x+h)-f(x)}{h}$ as $h\to 0$, so your formula won't match it exactly (although it would be 'close' provided $\eta$ is 'small enough'). What's used is Taylor's theorem (assuming the function is at least twice differentiable), so that for $h \in \mathbb{R}$:
$$f(x+h) = f(x) + hf^\prime(x)+\tfrac{1}{2}h^2f''(\xi),$$
for some $\xi$ between $x$ and $x+h$. Then $\frac{1}{2}h^2 f''(\xi) = O(h^2)$, and the result follows.