y=(3x2−3x+1)4
Найдите y''(x).
В ответ введите значение y''(1).
y' = 4(3x2 – 3x + 1)3·(6x – 3) = 12(3x2 – 3x + 1)3·(2x – 1)
y'' = 12·(3(3x2 – 3x + 1)2·(2x – 1) + (3x2 – 3x + 1)3·2)
y'' = 12(3x2 – 3x + 1)2·(3(2x–1) + 2(3x2 – 3x + 1)) = 12(3x2 – 3x + 1)2·(6x – 3 + 6x2 – 6x + 2)
y'' = 12(3x2 – 3x + 1)2·(6x2 – 1)
y''(1) = 12(3·12 – 3·1 + 1)(6·12 – 1) = 12·1·(6 – 1) = 60