【题目】求由方程 x-y+1/2siny=0 所确定的隐函数的二阶导数(d^2y)/(dx^2)

【解析】解方法一把方程两边分别对x求导数,注意y=y(x).1-(dy)/(dx)+1/2cosy+(dy)/(dx)=0. 于是(dy)/(dx)=2/(2-cosy) 上式两边再对x求导,得(d^2y)/(dx^2)=(-2siny)/((2-cosy)^2)=(-4siny)/((2-cosy)^2) 方法二等式两边求微分,得d(x-y+1/2siny)=d(0)=0 ,dx-dy+1/2cosydy=0 dy=2/(2-cosy)dx 所以(dy)/(dx)=2/(2-cosy) 上式两边再求微分得d((dy)/(dx))=d(2/(2-cosy)) (2-cosy)2(2-cosy)2(2-cosy)3(d^2y)/(dx^2)=(-4siny)/((2-cosy)^3) 方法三把方程两边分别对x求导数,注意y=y(x)1-(dy)/(dx)+1/2cosy+(dy)/(dx)=0. 再对x求导得-(d^2y)/(dx^2)-1/2siny⋅((dy)/(dx))^2+1/2cosy⋅(d^2y)/(dx^2)=0. 于是(dy)/(dx)=2/(2-cosy) (dy)/(dx^2)=rac(rac12⋅rac(dx)(rac(dx)(ctany))=rac(dx)(cosy)=rac(-4siny)((2-cosy)^2) 方法四等式两边求微分,得d(x-y+1/2siny)=d(0)=0 dx-dy+1/2cosydy=0 1-(dy)/(dx)+1/2cosy(dy)/(dx)=0 再求微分,得(d^2y)/(dx^2)dx-1/2sinydy(dy)/(dx)+1/2cosy(d^2y)/(dx^2)dx=0 所以(dy)/(dx)=2/(2-cosy). (∂_y)/(dx^2)=1//(1/2cos(2x-1))=rac(sin(rac(4y+(((2-cosy^2))(cosy-2))^2