Question

求x^2-xy y^2=1这个方程所确定的隐函数的二阶导数

Answer 1

x^2-xy+ y^2=1
2x- ( xy' + y) +2y.y' =0
(x-2y)y' = 2x-y
y' = (2x-y)/(x-2y)
y''
= [(x-2y)(2-y') - (2x-y)(1-2y') ]/(x-2y)^2
= [(x-2y)( 2-(2x-y)/(x-2y) ) - (2x-y)( 1-2(2x-y)/(x-2y) ) ]/(x-2y)^2
= [(x-2y)( 2(x-2y)-(2x-y) ) - (2x-y)( (x-2y)-2(2x-y) ) ]/(x-2y)^3
= [(x-2y)( -3y ) - (2x-y)( -3x ) ]/(x-2y)^3
=-3( xy-6y^2 - 2x^2+xy) /(x-2y)^3
=6( x^2-xy+3y^2 ) /(x-2y)^3

追问

教材上给出答案是6/（x-2y）^3这是怎么算出来的呢？

追答

x^2-xy+ y^2=1 有没有错？

y''
= [(x-2y)(2-y') - (2x-y)(1-2y') ]/(x-2y)^2
= [(x-2y)( 2-(2x-y)/(x-2y) ) - (2x-y)( 1-2(2x-y)/(x-2y) ) ]/(x-2y)^2
= [(x-2y)( 2(x-2y)-(2x-y) ) - (2x-y)( (x-2y)-2(2x-y) ) ]/(x-2y)^3
= [(x-2y)( -3y ) - (2x-y)( -3x ) ]/(x-2y)^3
=-3( xy-2y^2 - 2x^2+xy) /(x-2y)^3
=6( x^2-xy+y^2 ) /(x-2y)^3

追问

我也算出来的是这个答案，但教材答案就是上面那个，所以我才会提问😂

总之多谢您的帮助