Question

求隐函数x十y++2xy=1的导数

Answer 1

问：求隐函数x十y++2xy=1的导数

答：假设 y=f(x)，则有：x·f(x) + 2x·f'(x) + 2f(x) = 1对x求导：f(x) + 2f'(x) + 2x·f''(x) + 2f'(x) = 0即：f''(x) + 3f'(x) + 2f(x) = 0令f'(x)=p，则有：p' + 3p + 2f(x) = 0即：p' + 3p = -2f(x)解得：p=C·e^(-3x) - 2f(x)/3再求f(x)：f'(x) = C·e^(-3x) - 2f(x)/3令f(x)=y，则有：y' + 3y = -2C·e^(-3x)解得：y=C1·e^(-3x) + C2·e^(x) + 2C/3综上所述，可得：f(x)=C1·e^(-3x) + C2·e^(x) + 2C/3而f'(x)=C·e^(-3x) - 2f(x)/3综上，得出隐函数x·y++2xy=1的导数为：f'(x)=C·e^(-3x) - 2f(x)/3