# Taking the nth derivative as a function as n becomes very large (Taylor series)

Suppose I have a function:

$F(x) = \frac{1-x^2}{1-x^2-2x^3}$

How would I go about approximating the $n-th$ derivative of $F(x)$ when $n$ becomes very large?

Edit: Motivation: I ask because I want to approximate the n-th term of the Taylor series expansion at $x=0$ when n becomes very large.

Perhaps you could try the following. For $|x|$ small we have \begin{align*} F(x) &= \frac{1-x^2}{1-(x^2+2x^3)} = \frac{1-x^2}{1-x^2(1+2x)} = (1-x^2)\sum_{k=0}^\infty \big(x^2(1+2x)\big)^k \\ &= (1-x^2)\sum_{k=0}^\infty x^{2k}(1+2x)^k. \end{align*} If you specify a value of $n$, it won't be too bad to give a formula for the coefficient of $x^n$, hence $f^{(n)}(0)/n!$.
Define a recursive approximation of the $n$th derivative of $f(x)$ as:
$$f^{(n)}(0) \approx \frac{f^{(n - 1)}(0.0001) - f^{(n - 1)}(0)}{0.0001}$$
You can then compute $f'(0), f''(0), \dots, f^{(n)}(0)$ in an iterative manner to derive the required Maclaurin series up to the $n$th term. You can change $0.0001$ to something smaller for more accuracy.
However, as noted in this similar question, the loss of accuracy builds up with increasing $n$. Do check the linked question for better answers.