Question

已知方程组 {x+y+z=0 x·y·z=1 求d2y/dx2?

d2y/dx2 即d[(yz-xy)/(xy-xz)]/dx,求计算及结果，谢谢！

Answer 1

x＋y＋z=0 ①
xyz=1 ②
由①②可以确定y=y(x)，z=z(x)
①式两边对x求导得1＋dy/dx＋dz/dx=0 ③
②式两边对x求导得yz＋x(ydz/dx＋zdy/dx)=0
即yz＋xydz/dx＋xzdy/dx=0 ④
由③④得
dz/dx=(xz－yz)/(xy－xz) ⑤
dy/dx=(xy－yz)/(xz－xy) ⑥
d²y/dx²=[(y＋xdy/dx－ydz/dx－zdy/dx)(xz－xy)－(xy－yz)(z＋xdz/dx－y－xdy/dx)]/(xz－xy)² ⑦
将⑤⑥代入⑦再结合①式，化简得
d²y/dx²=[y(xz－xy)(x²＋y²＋z²)－x(xy－yz)(x²＋y²＋z²)]/(xz－xy)³
=(xyz－xy²－x²y＋xyz)(x²＋y²＋z²)/(xz－xy)³
=[2xyz－xy(y＋x)](x²＋y²＋z²)/(xz－xy)³
=3xyz(x²＋y²＋z²)/(xz－xy)³
=3(x²＋y²＋z²)/(xz－xy)³

附
(y＋xdy/dx－ydz/dx－zdy/dx)(xz－xy)
=y(xz－xy)＋x(xy－yz)－y(yz－xz)－z(xy－yz)
=xyz－xy²＋x²y－xyz－y²z＋xyz－xyz＋yz²
=－xy²＋x²y－y²z＋yz²
=y(x²＋z²)－y²(x＋z)
=y(x²＋z²)＋y³
=y(x²＋y²＋z²)
(xy－yz)(z＋xdz/dx－y－xdy/dx)
=(xy－yz)[z＋x(xz－yz)/(xy－xz)－y－x(xy－yz)/(xz－xy)]
=(xy－yz)[z(xz－xy)－x(xz－yz)－y(xz－xy)－x(xy－yz)]/(xz－xy)
=(xy－yz)(xz²－xyz－x²z＋xyz－xyz＋xy²－x²y＋xyz)/(xz－xy)
=(xy－yz)(xz²－x²z＋xy²－x²y)/(xz－xy)
=(xy－yz)[x(y²＋z²)－x²(y＋z)]/(xz－xy)
=(xy－yz)[x(y²＋z²)＋x³]/(xz－xy)
=(xy－yz)[x(x²＋y²＋z²)]/(xz－xy)
=x(xy－yz)(x²＋y²＋z²)/(xz－xy)