Question

Найти производные функций в точке х0

Answer 1

а) y = 6x^3 ; y' = 18x^2 ; y'(x0) = y'(-2) = 72

б) y = 2sinx ; y' = 2cosx ; y'(x0) = y'(-π/6) = sqrt(3)

в) y = 4cosx ; y' = -4sinx ; y'(x0) = y'(-π/4) = 2 × sqrt(2)

г) y = ctg(4x - 2π/3) ; y' = -4/sin^2 (4x - 2π/3) ; y'(x0) = y'(π/3) = -4/sin^2 (2π/3) = -16/3