DIfferentiate the equations (12.69) with respect to yto obtain
Fy+Fuuy+Fvvy+Fwwy=0
Gy+Guuy+Gvvy+Gwwy=0
Hy+Huuy+Hvvy+Hwwy=0
(12 .74)
This produces three equations in the three unknowns uy, v y, w ywhich can be solved
using Cramers rule. If the Jacobian determinant of F, G, H is different from zero,
then the unique solution is given by
uy=−
∂(F,G,H )
∂(y,v,w )
∂(F,G,H )
∂(u,v,w )
, vy=−
∂(F,G,H )
∂(u,y,w )
∂(F,G,H )
∂(u,v,w )
, w y=−
∂(F,G,H )
∂(u,v,y)
∂(F,G,H )
∂(u,v,w )
(12 .75)
Generalization
The system of equations
F 1 (x 1 , x 2 ,... , x n, y 1 , y 2 ,... , ym) =0
F 2 (x 1 , x 2 ,... , x n, y 1 , y 2 ,... , ym) =0
..
.
..
.
Fm(x 1 , x 2 ,... , x n, y 1 , y 2 ,... , ym) =0
(12 .76)
in m+nunknowns implicitly defines the functions
y 1 =y 1 (x 1 , x 2 ,... , x n)
y 2 =y 2 (x 1 , x 2 ,... , x n)
..
.
..
.
ym=ym(x 1 , x 2 ,... , x n)
(12 .77)
If the Jacobian determinant
∂(F 1 , F 2 ,... , F m)
∂(y 1 , y 2 ,... , y m)
=
∣∣
∣∣
∣∣
∣∣
∣∣
∂F 1
∂y 1
∂F 1
∂y 2 ···
∂F 1
∂ym
∂F 2
∂y 1
∂F 2
∂y 2 ···
∂F 2
∂ym
..
.
..
. ...
..
.
∂F m
∂y 1
∂Fm
∂y 2 ···
∂Fm
∂ym
∣∣
∣∣
∣∣
∣∣
∣∣
is different from zero at a point (x^01 , x^02 ,... , x^0 n, y^01 , y 20 ,... , y^0 m), then one can calculate
the partial derivatives
∂y i
∂x j at this point for any combination of integer values for i, j
satisfying 1 ≤i≤mand 1 ≤j≤n. The above is true because if one calculates the
