x-y+siny=2,求dy/dx
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x-y+siny=2,求dy/dxx-y+siny=2,求dy/dxx-y+siny=2,求dy/dx两边对x求导有1-y''+y''cosy=0所以y''=1/(cosy-1)1/(cosy-1)隐函数求
x-y+siny=2,求dy/dx
x-y+siny=2,求dy/dx
x-y+siny=2,求dy/dx
两边对x求导有
1-y'+ y'cosy =0
所以y'=1/(cosy-1)
1/(cosy-1)
隐函数求导即可
两边关于x求导:1-y'+cosy*y'=0
于是dy/dx = y'=1/(1-cosy)
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