>>64
f(x) = log (1 + 1/(1 + x)) = log ((x + 2)/(x + 1)) = log (x + 2) - log (x + 1)

だから

f'(x)　= 1/(x + 2) - 1/(x + 1)
f''(x) = -1/(x + 2)^2 + 1/(x + 1)^2

などなど。n 次導関数 (n ≧ 1)は

f^(n) (x) = (-1)^(n-1) (n-1)! {1/(x + 2)^n - 1/(x + 1)^n}