Question

隐函数求二阶导数问题.

Answer 1

对x求导得到
3x^2 + 3y^2y' -3ay -3axy' =0
y' = (3x^2 -3ay)/(3ax-3y^2) = (x^2-ay)/(ax-y^2)
dy'/dx = [(2x-ay')(ax-y^2) - (x^2-ay)(a-2yy')]/(ax-y^2)^2
然后把y' = (x^2-ay)/(ax-y^2)带入即可

Answer 2

x^3+y^3-3axy = 0, 两边对 x 求导， 得
3x^2+3y'y^2-3ay-3axy' = 0 (1)
y' = (ay-x^2)/(y^2-ax)
式 (1) 即 x^2+y'y^2-ay-axy' = 0，两边对 x 再求导，得
2x+2y(y')^2+y''y^2-2ay'-axy'' = 0
则 y'' = 2[ay'-x-y(y')^2]/(y^2-ax)
= 2[a(ay-x^2)/(y^2-ax)-x-y(ay-x^2)^2/(y^2-ax)^2]/(y^2-ax)
= 2[a(ay-x^2)(y^2-ax)-x(y^2-ax)^2-y(ay-x^2)^2]/(y^2-ax)^3

= 2(3ax^2y^2-a^3xy-xy^4-yx^4)/(y^2-ax)^3