已知由方程xy+lny=1确定隐函数y=y(x),求dy⼀dx

2025-03-13 23:51:45
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回答1:

方程两边对x求导:
y+xy'+y'/y=0
(x+1/y)y'=-y
y'=-y/(x+1/y)=-y²/(xy+1)
即dy/dx=-y²/(xy+1)