f(x)=e^[ln(x+1)+ln(x+2)+...+ln(x+2018]f'(x)=e^[ln(x+1)+ln(x+2)+...+ln(x+2018]*[1/(1+x)+1/(x+2)+...+1/(x+2018)]=(x+1)(x+2)"""(x+2018)*[1/(1+x)+1/(x+2)+...+1/(x+2018)]f'(0)=2018!*(1+1/2+1/3+...+1/2018)
