Ответ Автор - vladbaranov201p67bwg 2af'(x) = (3x - 4)5^(3x - 5) = 3*5^(3x - 5)2bf'(x) = (4x - 7)'cos(4x - 7) = 4cos(4x - 7)2cf'(x) = (3x + 2)' / 2√(3x + 2) = 1,5 /√(3x + 2)2df'(x) = (x³ + 5x)' / (x³ + 5x) = (3x² + 5)/(x³ + 5x)