y=sin(x+y)的隐函数的二阶导数。要详细的过程

发布网友发布时间：2022-06-07 22:04

共2个回答

热心网友时间：2023-10-28 12:47

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

y''=-sin(x+y)*(1+y')+y''cos(x+y)+y'*(-sin(x+y)*(1+y'))
∴[cos(x+y)-1]y''=sin(x+y)*(1+y')²，即y''=sin(x+y)/[cos(x+y)-1]³

热心网友时间：2023-10-28 12:48

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

y''=[cos(x+y]'(x+y)'+y''cos(x+y)+y'[cos(x+y)]'
=-(1+y')sin(x+y)+y''cos(x+y)-y'(1+y')sin(x+y)
y''[cos(x+y)-1]=(1+y')^2sin(x+y)
={1+cos(x+y)/[(1-cos(x+y)]}^2sin(x+y)
=sin(x+y)/[cos(x+y)-1]^2
y''=sin(x+y)/[cos(x+y)-1]^3

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