quick question on multivariable calc chain rule problem

• Sep 28th 2009, 06:40 PM
holly123
quick question on multivariable calc chain rule problem
find g of f at the indicated point a.
g(x,y)= (x^2 * 3y, 3x-y^2)
f(x1, x2, x3)= (x1x3, (x2)/(x1x3) )
a= (3,1,-1)

so the matrix for D(g) is
2xy^3 3x^2 y^2
3 -2y

D(g(f(x1,x2,x3) is
(2(x2)^3)/(x1x3)^2 3(x2)^2
3 (-2x2)/(x1x3)

what is D(f) ?
i started with
x3 0 x1
but i dont know about the next row or if i even did the first row right.

any help? thanks so much
• Sep 28th 2009, 07:29 PM
holly123
okay so i think i got it. my final answer was the matrix:
-5/9 3 -7/3
-83/27 2/3 25/3

due to these weird fractions i get the feeling i made an error somewhere. anyone have any ideas??(Headbang)
• Sep 28th 2009, 08:24 PM
Calculus26
the problem starts with your calculation of D(g)

Quote:

g(x,y)= (x^2 * 3y, 3x-y^2)

so the matrix for D(g) is
2xy^3 3x^2 y^2
3 -2y
Were you using g(x,y) = (x^2*y^3,3x-y2)

or the formula for g(x,y) in your original post?

I've included attachments for both --check my work I still disagree somewhat.
• Sep 29th 2009, 07:10 AM
holly123
oh! it is y^3

i'm sorry for the typo, don't worry about the problem, i found a mistake, fixed, it, and still got weird fractions in my answer, so i'm just going to hope it's right!
• Sep 29th 2009, 02:52 PM
Calculus26
The attachment on the right was worked with y^3 do you get the same result?
• Sep 29th 2009, 03:24 PM
holly123
ah wow i actually got that!!! thank you for the reassurance(Happy)