Question

fx=ln(1-x)/(x-1)+求f(0)的n阶导数

Answer 1

f(x)=ln(1-x)/(x-1)
(x-1)f(x)=ln(1-x)……①
令y=ln(1-x)
y'=-1/(1-x)=1/(x-1)
y''=-1/(x-1)^2
y'''=(-1)*(-2)*1/(x-1)^3
y^(n)=(-1)^(n-1)*(n-1)!/(x-1)^n
根据莱布尼兹公式，对①式两边求n阶导
C(n,0)*(x-1)*f^n(x)+C(n,1)*f^(n-1)(x)=(-1)^(n-1)*(n-1)!/(x-1)^n
(x-1)*f^n(x)+n*f^(n-1)(x)=(-1)^(n-1)*(n-1)!/(x-1)^n
将x=0代入上式，得：
-f^(n)(0)+n*f^(n-1)(0)=-(n-1)!
f^(n)(0)=n*f^(n-1)(0)+(n-1)!
两边同除以n!
f^(n)(0)/n!=f^(n-1)(0)/(n-1)!+1/n
f^(n)(0)/n!=f(0)/0!+1/1+1/2+1/3+...+1/n
f^(n)(0)/n!=1+1/2+1/3+...+1/n
f^(n)(0)=n!*(1+1/2+1/3+...+1/n)