f(x)=(x+1)(2x+1)(3x+1).(nx+1),求f'(0)

f(x)=y=(x+1)(2x+1)(3x+1).(nx+1),求f'(0)两边取自然对数得：lny=ln(x+1)+ln(2x+1)+ln(3x+1)+.+ln(nx+1)两边对x取导数得y'/y=1/(x+1)+2/(2x+1)+3/(3x+1)+.+n/(nx+1)故f'(x)=f(x)[1/(x+1)+2/(2x+1)+3/(3x+1)+.+n/(nx+...