What is the $n^{th}$ term derivative of $f(x) = (x^2-x-1)(\ln(1-x))$? I have the first three terms but am struggling with finding the $n^{th}$ term derivative of the function. Here is my work:
$$\\$$
$$f(x) = (x^2-x-1)(\ln(1-x)) $$
$$f'(x) = (2x-1)(\ln(1-x))-\left(\dfrac{x^2-x-1}{1-x}\right)$$
$$f''(x) = \dfrac{3x^2-5x+2(x-1)^2 \ln(1-x)+3}{(1-x)^2}$$
$$f'''(x) = \dfrac{2x^2-5x+1}{(x-1)^3}$$
$$f^{(n)}(x) = \ ?$$

As you noticed,
$$
f^{(3)}(x)=\frac{2x^{2}-5x+1}{\left(x-1\right)^{3}}.
$$
Wolfram claims that
$$
f^{(3+n)}(x)=n!\frac{n^{2}+n\left(x+2\right)-2x^{2}+5x-1}{\left(x-1\right)^{3+n}}.
$$
To convince yourself that this is indeed the case, fix $n\geq0$ and differentiate the expression for $f^{(3+n)}$ to get the expression for $f^{(3+n+1)}$.
