y=cos(x+y)隐函数求二次导y"能用反函数做吗

y=cos(x+y)隐函数求二次导y"能用反函数做吗
求一阶导数时可以用反函数作,待一阶求出后再求二阶时就不能再用反函数来求.
我用两种方法求一阶,可看出结果是一致的.
用“隐函数求导法”：
F(x,y)=y-cos(x+y)=0
dy/dx=-(∂F/∂x)/(∂F/∂y)=sin(x+y)/[1+sin(x+y)]
用“反函数求导法”：
x+y=arccosy,故x=arccosy-y,
dx/dy=-1/√(1-y²) - 1=-1/√(1-cos²(x+y)-1=-1/sin(x+y)-1=-[1+sin(x+y)]/sin(x+y)
∴dy/dx=y′=1/(dx/dy)=-sin(x+y)/[1+sin(x+y)]
显然,两种方法,结果相同.
但求二阶时,已无反函数可用,因此不能再用“反函数求导法”.可直接对x求导,但要记住：
要把y看作中间变量,遇到y时要用复合函数求导法.
d²y/dx²=dy′/dx=-{[1+sin(x+y)][cos(x+y)](1+y′)-[sin(x+y)cos(x+y)](1+y′)}/[1+sin(x+y)]²
=-(1+y′){[1+sin(x+y)][cos(x+y)]-[sin(x+y)cos(x+y)]}/[1+sin(x+y)]²
=-(1+y′)cos(x+y)/[1+sin(x+y)]²
再将y′=-sin(x+y)/[1+sin(x+y)]代入,化简,即得：
y″=-cos(x+y)/[1+sin(x+y)]³.