Question

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~

Answer 1

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