一道简单的隐函数求导数 求详细过程,求导我会的,就那一步看不懂 怎么变成y✀=那一步,求详细解答,

2025-05-16 00:25:59
推荐回答(4个)
回答1:

就是将y'当成未知数,解一元一次方程而已:
-sin(x+y) * (1+y')+y'=0
-sin(x+y)-y'sin(x+y)+y'=0
[1-sin(x+y)]y'=sin(x+y)
y'=sin(x+y)/[1-sin(x+y)]

回答2:

回答3:

回答4:

cos(x+y)+y=1
-sin(x+y)*(x+y)'+y'=0
-sin(x+y)*(x'+y')+y'=0
-sin(x+y)*(1+y')+y'=0
-sin(x+y)-y'sin(x+y)+y'=0
[1-sin(x+y)]y'=sin(x+y)
y'=sin(x+y)/[1-sin(x+y)]