Question

设f(x)＝e^(-x) *ln(2-x) + (1+3x^2)^(1/2),求f'(1)

Answer 1

对,先求导再代入 [e^(-x) *ln(2-x)]' =[e^(-x)]'*ln(2-x)+e^(-x)*[ln(2-x)]' =[e^(-x)*(-x)']*ln(2-x)+e^(-x)*[1/(2-x)*(2-x)'] =-e^(-x)*ln(2-x)-e^(-x)/(2-x) [(1+3x^2)^(1/2)]' =(1/2)(1+3x^2)^(1/2-1)*(1+3x^2)' =6x*(1/2)(1+3x^2)^(1/2-1) =3x(1+3x^2)^(-1/2) 所以f'(x)=-e^(-x)*ln(2-x)-e^(-x)/(2-x)+3x(1+3x^2)^(-1/2) f'(1)=(3/2)-(1/e) 查看原帖