TITLE: Find the general form of $n$th derivative $f(x) = \ln(1+x)$
QUESTION [5 upvotes]: I think I am doing it the correct way but I am not sure.
Is it $$(-1)^{n+1}n!(1+x)^{-n} ?$$
Thank you guys.

REPLY [5 votes]: This is how I would do it
$$f(x) = \ln(1+x)$$
$$f'(x) = \frac{1}{x+1}$$
$$f''(x) = -\frac{1}{(x+1)^2}$$
$$f'''(x) = (-1)(-2)\frac{1}{(x+1)^3}$$
$$f'^v(x) = (-1)(-2)(-3)\frac{1}{(x+1)^4}$$
$$f'^v(x) = (-1)^3\cdot \frac{3!}{(x+1)^4}$$
$$.....................$$
$$f'^n(x) = (-1)^{n-1}\cdot \frac{(n-1)!}{(x+1)^{n}}\forall n\ge 1$$