終極微分, find dy/dx

Sum [(aij)(x^i)(y^j)] = 0 (i = 0, 1, ...n; j = 0, 1, ... n)

Take a general term,
(aij)(x^i)(y^j)
d[(aij)(x^i)(y^j)]/dx = aij(ix^(i-1)y^j + jx^n*y^(j-1)dy/dx)
Take the derivative w.r.t. x on both sides.
Sum [aij(ix^(i-1)y^j + aij*jx^i*y^(j-1)dy/dx)] = 0
dy/dx * Sum [aij*jx^i*y^(j-1)] = - Sum [aij(ix^(i-1)y^j]
dy/dx = - Sum [aij(ix^(i-1)y^j]/Sum [aij*jx^i*y^(j-1)]
= - {(a10)(x^0)(y^0) + (a11)(x^0)(y^1) + (a12)(x^0)(y^2)... + (a1n)(x^0)(y^n) +
2[(a20)(x^1)(y^0) + (a21)(x^1)(y^1) + (a22)(x^1)(y^2)... + (a2n)(x^1)(y^n)] + ... +
n[(an0)(x^(n-1))(y^0) + (an1)(x^(n-1))(y^1) + (an2)(x^(n-1))(y^2) + ... + (ann)(x^(n-1))(y^n)} /
{(a01)(x^0)(y^0) + 2(a02)(x^0)(y^1) + ... + (n)(a0n)(x^0)(y^(n-1)) +
(a11)(x^1)(y^0) + 2(a12)(x^1)(y^1) + ... + (n)(a1n)(x^0)(y^(n-1)) + ... +
(an1)(x^n)(y^0) + 2(an2)(x^n)(y^1) + ... + (n)(ann)(x^n)(y^(n-1))}

d/dx the both side,0+a01dy/dx+a02dy^2/dx+....+a0ndy^n/dx+a10+a11y+a11(x)dy/dx
+a12y^2+a12(x)dy^2/dx+.....+a1ny^n+a1n(x)dy^n/dx+a20(2x)+a21(2x)y+a21(x^2)dy/dx
+a22(2x)y^2+a22(x^2)dy^2/dx+...+a2n(2x)y^n+a2n(x^2)dy^n/dx+.....+ann(nx^n-1)y^n
+ann(x^n)dy^n/dx=0

so dy/dx=-(a10+a11y+a12y^2+.....+a1ny^n+a20(2x)+a21(2x)y+a22(2x)y^2+....+
a2n(2x)y^n+.......ann(dx^n/dx)y^n)/(a01+a02(2y)+....+a0n(ny^n-1)+a11x+a12x(2y)
+....+a12(x)(ny^n-1)+......ann(x^n)(ny^n-1))

唔知o岩唔o岩，同埋第一段寫個d野其實係廢既，sor.

2006-12-17 15:27:34 補充：
簡而言之，就係-(ann(dx^n/dx)y^n既總和)/(ann(x^n)(ny^n-1)既總和)

2006-12-17 15:41:05 補充：
x既次方要睇第1個n，y既要睇第2個