y=1/(x^2-2x+4)二阶导数怎么求

y = 1/(x² - 2x + 4) dy/dx = -(x² - 2x + 4)'/(x² - 2x + 4)² = -(2x - 2)/(x² - 2x + 4)² = 2(1 - x)/(x² - 2x + 4)² d²y/dx² = 2 • [(x² - 2x + 4)²(1 - x)' - (1 - x)(x² - 2x + 4)²']/(x² - 2x + 4)⁴ = 2 • [(x² - 2x + 4)²(-1) - (1 - x) • 2(x² - 2x + 4) • (2x - 2)]/(x² - 2x + 4)⁴ = 2 • (- x² + 2x - 4 + 4x² - 8x + 4)/(x² - 2x + 4)³ = 2 • (3x² - 6x)/(x² - 2x + 4)³

= 6x(x - 2)/(x² - 2x + 4)³ 4)³