Найти значение производной в точке x0 f (x) = (2x) (sin5x), x0=П/2 с подробным решением

F (x) = 2x * Sin5x

f ' (x) = 2 (x' * Sin5x + x (Sin5x) ') = 2 (Sin5x + x * Cos5x * (5x) ') = 2 (Sin5x +

+ 5x*Cos5x)

f ' (π/2) = 2 (Sin 5π/2 + 5π/2*Cos 5π/2) = 2[Sin (2π+π/2) + 5π/2 * Cos (2π+π/2) ]=

= 2 (Sin π/2 + 5π/2 * Cos π/2) = 2 (1 + 5π/2 * 0) = 2

У'=2 (1+5 кос (5 х)). если х0=π/2, то:

У' (х0) = 2 (1+0), у' (х0) = 2.