Question

x-y+1/2siny=0所确认的隐函数的二阶导数 y'=2/(2-cosy)

求由方程x-y+1/2siny=0所确认的隐函数的二阶导数
y'=2/(2-cosy)
到了二阶求导就不明白了
(y')'=(2/(2-cosy))'
然后我算出了y''=(2-cosy-2siny)/(2-cosy)^2
我高数方面很差,求详解下.

Answer 1

简单计算一下即可，答案如图所示

Answer 2

x-y+1/2siny=0 两边对x求导得
1-y'+1/2cosy*y'=0
y'=2/(2-cosy)
y''=dy'/dx
=(dy'/dy)*(dy/dx)
=[-2/(2-cosy)²]*siny*2/(2-cosy)
=-4siny/(2-cosy)³