f(x) = 1/(1+2x) = (1+2x) ^ (-1)
f'(x) = (-1) (2) (1+2x) ^ (-2)
f''(x) = (-1)(-2) (2)^2 (1+2x) ^ (-3)
Do you see the pattern here?
The first term is the sign going from negative to positive to negative, etc. That means (-1)^n.
The second term is going to be 1*2*3*...*n coming from that negative exponent getting more negative each time = n!
The third term is that 2^n term. It comes from the derivative of (1+2x) repeated over and over.
Finally the fourth term is (1+2x) ^ (-n+1) because it started at one not zero.
f^n(x) = (-1)^n (2^n) (n!) (1+2n)^-(n+1)