3设 y=(1+x^2)⼀(1-x) 求dy;

2026年09月28日 07:50
有2个网友回答
网友(1):

y = (1+x²)/(1-x)

y' = [2x(1-x) + (1+x²)]/(1-x)²
y'= (1+2x-x²)/(1-x)²
dy = [(1+2x-x²)/(1-x)²]dx

网友(2):

y = (1+x^2)/(1-x)
y' = [2x(1-x) + (1+x^2)]/(1-x)^2 = (1+2x-x^2)/(1-x)^2
dy = (1+2x-x^2)dx/(1-x)^2