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y=sin(x+y)的隐函数的二阶导数。要详细的过程

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3 | 2011-11-9 21:17:07 | 显示全部楼层 |阅读模式

y'=cos(x+y)*(1+y')y'=-1+1/[1-cos(x+y)]，及1+y'=1/[1-cos(x+y)]y"= -1/[1-cos(x+y)]^2*sin(x+y)*(1+y') =-1/[1-cos(x+y)]^2*sin(x+y)*{1/[1-cos(x+y)]}
=-sin(x+y)/[1-cos(x+y)]^3...

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千问 | 2011-11-9 21:17:07 | 显示全部楼层

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

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