(5x^3-4/x^4+3*(x^2)^(1/3)+7/(x^2)^(1/7)+11/x-13)'

Question

Помогите с решением (5x^3-4/x^4+3*(x^2)^(1/3)+7/(x^2)^(1/7)+11/x-13)' тут в ответе 11/x^2

=

(4x^2-2/x^2+4*(x^2)^(1/2)+4/(x^2)^(1/4)+11/x-13)' (5x^3-4/x^4+3*(x^2)^(1/3)+7/(x^2)^(1/7)+11/x-13)'

равно случайно поставил 2 разных примера

Мария Лютина · Accepted Answer

= (4х^2 +4х - 2/х^2 + 4/√х + 11/х - 13)' = 
(4х^2)' = 8х
(4х) ' = 4
(2/х^2)' = (2х^(-2))' = -4х^(-3) = 4/х^3
((4/√х) ' = 4х^(-1/2)) = -2х^(-3/2) = -2/х^3/2
(11/х) ' = (11х^(-1))' = -11х^(-2) = -11/х^2
13' = 0

= 8х + 4 +4/х^3 - 2/х^3/2 -11/х^2 = 
= 8х +4 +4/х^3 -2/(х√х) - 11/х^2