y=sin(x+y),一阶隐导数y'=cos(x+y)/[1-cos(x+y)]我懂,但如何求二阶隐导数~
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一阶导数你求对了
令 t = dy/dx = cos(x+y)/[1-cos(x+y)]
dy = cos(x+y)·dx/[1-cos(x+y)]
dx+dy = cos(x+y)·dx/[1-cos(x+y)] + dx = dx/[1-cos(x+y)]
t+1 = 1+ cos(x+y)/[1-cos(x+y)] = 1/[1-cos(x+y)]
那么原函数的二阶导数即是 dt/dx
t[1-cos(x+y)] = cos(x+y)
[1-cos(x+y)]dt + tsin(x+y)(dx+dy) = -sin(x+y)(dx+dy)
[1-cos(x+y)]dt = -sin(x+y)(dx+dy)(t+1)
代入上面dx+dy和t+1的结论可得
[1-cos(x+y)]dt = -sin(x+y)·{dx/[1-cos(x+y)]}·{1/[1-cos(x+y)]}
[1-cos(x+y)]dt = -sin(x+y)·dx/[1-cos(x+y)]^2
移项即可得结果
