Taylor series: Epsilon not differentiated? Why isn't epsilon differentiated with respect to time? (see my question on the right)

 A: It seems, given the fact that $\mathcal{O}(\epsilon^2)$ is written, that we are taking a Taylor series in $\epsilon$, which would involve differentiation with respect to $\epsilon$, rather than its arguments. Recall the expansion,
$$\frac{1}{(\epsilon+a)^3}= \frac{1}{a^3}-3\frac{\epsilon}{a^4} + \mathcal{O}(\epsilon^2)$$
where we think of the function as $f(\epsilon)$, rather than $f(t)$ or $f(a)$. Inserting the expansion yields,
$$2\ddot{\epsilon} +g -ga^3 \left(\frac{1}{a^3}-3\frac{\epsilon}{a^4} \right) + \mathcal{O}(\epsilon^2)$$
$$=2\ddot{\epsilon} +3g \frac{\epsilon}{a} + \mathcal{O}(\epsilon^2)$$
A: The Taylor expansion in your question is carried out as follows:
$$\frac{ga^3}{(a+\epsilon)^3}=\frac{ga^3}{a^3\left(1+\frac{\epsilon}{a}\right)^3} $$
Now, we want to expand 
$$f(\epsilon)=\frac{1}{\left(1+\frac{\epsilon}{a}\right)^3} \hspace{2cm} \text{around}\ \epsilon=0$$
This Taylor expansion contains derivatives of $f(\epsilon)$, evaluated at $\epsilon=0$, but no derivatives of $\epsilon$ as this is the variable rather than the function.
