Originally Posted by
ynj I do not exactly know what you are confusing about....
you may understand J(F(G(x,y)))=J(F(u0,v0))J(G(x,y)) like
(fg)'(x)=f'(g(x))g'(x)
the function G actually put (x,y) to (x^2+1,y^2)
and F put (u,v) to (u+v,v^2)
For the format reason, we may define f(x,y)'x to be the partial deriviation of x, say the lim (f(x+dx,y)-f(x,y))/dx.
if you want to know (FG)(x,y)'x or (FG)(x,y)'y, you may first know F(u,v)'v or F(u,v)'v or G(x,y)'x or G(x,y)'y because it will be easier.
The Jacobian's theorem actually provides you a perfect way to solve the problem.
Let we define a series of functions
f1(x1,x2,...xn)
f2(x1,x2....xn)
...
fm(x1,x2...,xn)
this set of functions will be written as F,where
F(x1,...,xn)=(y1,...ym)
yi=fi(x1,...xn)
then J(F) will be
(f1'x1,f1'x2.....f1'xn
f2'x1,f2'x2.....f2'xn
....
fm'x1,fm'x2,......fm'xn)
it is a n*m matrix
I hope you can understand the definition of Jacobian, although it will be a little abstract..
To get the answer, you may first write (FG)(x,y) as F(g1(x,y),g2(x,y)), where g1(x,y)=x^2+1,g2(x,y)=y^2
In this case, we can let u0=g1(x,y),v0=g2(x,y).
Then we can use our equation J(F(G(x,y)))=J(F(u0,v0))J(G(x,y)).
The first part J(F(u0,v0)) tells you that you can ignore the existence of x,y, just looking at u0 and v0.After calculating, you just let u0=g1(x,y),v0=g2(x,y), then everything is completed.
so J(F(u0,v0))=
((u0+v0)'u0,(u0+v0)'v0
(v0^2)'u0,(v0^2)'v0)
=(1,1
0,2v0)
The second part J(G(x,y)) tells you where the x,y will appear.
so J(G(x,y))=
((x^2+1)'x,(x^2+1)'y
(y^2)'x,(y^2 )'y)
=(2x,0
0,2y)
Then we calculate J(F(u0,v0))J(G(x,y))
it will be
(2x,2y
0,4v0y)
then v0=g2(x,y)=y^2
so the answer will be
(2x,2y
0,4y^3)
=(2,2
0,4)