Find the Nth derivative using Maclaurin series

Here's a hint.

Factor the denominator, and expand via "partial fractions". Then find the series for each term, and sum.

$$f (x)=\frac {1}{1-x}+\frac {2}{1+2x} $$

$$=\Bigl(1+x+x^2+...x^8\Bigr)+$$ $$2 \Bigl (1-(2x)+(2x)^2-....+(2x)^8\Bigr)+x^8\epsilon (x) $$

thus

$$f^{(8)}(0)=8!(1+2^9)=8!.513 $$

put $t=-2x $ then $1/(1+2x)=1/(1-t)=1+t+t^2+... $.

Suppose we want to do this without expanding in partial fractions. Then we can just perform a long division of the fraction to obtain the coefficient of $x^8$. But a faster division algorithm is Newton-Raphson division, $n$ steps will yield the coefficients up to $x^{2^n -1}$ This method can be generalized such that it also works for arbitrary closed form functions, it is used in some computer algebra programs.

Newton-Raphson division for polynomials works in an analogous way as it does for real numbers. In the latter case this boils down to finding the reciprocal of some number $y$ by solving the equation $$\frac{1}{x} - y = 0$$ using the Newton-Raphson method, which yields a sequence of approximants $x_n$ satisfying the recurrence equation:

$$x_{n+1} = 2 x_{n} - y x_{n}^2\tag{1}$$

Then if $y$ is a polynomial and we take $x_0$ equal to the reciprocal of the constant term of that polynomial then iterating with Eq. (1) will yield successive approximations for the series expansion of the reciprocal of the polynomial, such that the number of correct terms doubles after each iteration.

If we put $p(x) = 1 + x - 2 x^2$ then the successive approximants $q_n(x)$ of the reciprocal of $p(x)$ satisfy the recurrence equation

$$q_{n+1}(x) = 2 q_n(x) - p(x) q_n(x)^2 \bmod x^{2^{n+1}}$$

where $q_0(x) = 1$. Here we compute modulo $x^{2^{n+1}}$ because only the coefficients of $x^r$ for $r<2^{n+1}$ will be correct. We then find:

$$ \begin{split} q_1(x) &= &1-x\\ q_2(x) &= -&5 x^3+3 x^2-x+1\\ q_3(x) &= -&85 x^7+43 x^6-21 x^5+11 x^4-5 x^3+3 x^2-x+1 \\ q_4(x) &= -&21845 x^{15}+10923 x^{14}-5461 x^{13}+2731 x^{12}-1365 x^{11}+\\ & &683 x^{10}-341 x^9+171 x^8-85 x^7+43 x^6-21 x^5+11 x^4-5 x^3+3 x^2-x+1 \end{split} $$

The 8th derivative of $\dfrac{3}{p(x)}$ is thus equal to $8!\times 3\times 171$