Question

y^2-2xy+x^2-2x=1隐函数的二阶导数

Answer 1

两边对x求导，把y看成x的复合函数，用复合函数的求导法则来求导：
1-y'+1/2y'cosy=0
得y'=1/(1-1/2cosy)=2/(2-cosy)
再继续对y'求导，注意的是仍然是对x的求导，cosy仍然是复合函数，其导数为-y'siny
故y"=-2/(2-cosy)^2*
(y'siny)
然后得再代入y'的式子，得y"=-2/(2-cosy)^2*2siny/(2-cosy)=-4siny/(2-cosy)^3

Answer 2

等式两边对 x 求导， 2yy'-2y-2xy'+2x-2=0,
即 yy'-xy'-y+x-1=0 ①
解得 y'=(y-x+1)/(y-x) ②
① 再对 x 求导，(y')^2+yy''-y'-xy''-y'+1=0,
解得 y''=[-(y')^2+2y'-1]/(y-x) =-(y'-1)^2/(y-x)
② 代入，得 y''= -(y-x+1-y+x)^2/(y-x)^3 = 1/(x-y)^3.
或由②对 x 求导，得 y''=(1-y')/(y-x)^2=1/(x-y)^3.