Multivariable Chain Rule problem

[AskingForAlyssa]

Hey guys, I've got a problem on a final review, and I cannot for the life of me figure this thing out. If you could help me out, it would be awesome!

let z=f(x,y) where x=uv and y=u-v. suppose that: Fx(1,1)=1 Fx(1,0)=2, Fy(1,1)=-1 Fy(1,0)=4.

Then, <(Zu(1,1),Zv(1,1)> = ?

I came up with a solution <6,-2> which I know to be true as my professor agreed with my answer, but not my work (I tried to come up with formulas for Fx and Fy based on their equations. Can anyone help me figure out the method of solving for an answer without solving for Fx and Fy?

 

[HallsofIvy]

If your answer is correct but you are unsure of your work, then you should show your work so we can explicitely comment on it. You say "I tried to come up with formulas for Fx and Fy based on their equations". What do you mean by that? You are given the derivatives of F with respect to x and y only at a single point. That is not enough to determine what F is in general or what Fx and Fy are at other points.

You need to know the general formulas, which I am sure are in your textbook,
\(\displaystyle \frac{\partial F}{\partial u}= \frac{\partial F}{\partial x}\frac{\partial x}{\partial u}+ \frac{\partial F}{\partial y}
\frac{\partial y}{\partial u}\) and
\(\displaystyle \frac{\partial F}{\partial v}= \frac{\partial F}{\partial x}\frac{\partial x}{\partial v}+ \frac{\partial F}{\partial y}
\frac{\partial y}{\partial v}\)

Since you are asked for the derivatives at (1, 1), the derivatives of F at (1, 0) are irrelevant.

(Please do not use "F" and "f" and "Z" and "z" interchangeably!)

 

[AskingForAlyssa]

Well, let me put down what I did. I will have to replace the partial sign with d as I'm not sure how to type them, please forgive my ignorance.

Given Z=F(x,y); x=uv and y=u-v dF/dx(1,1)=1 dF/dx(1,0)=2 I came up with dF/dx=2x-y (not the only solution that I could come up with, but hang in there.
dF/dy(1,1)=-1 dF/dy(1,0)=4 I came up with dF/dy=4x-5y

so if dZ/du = dF/dx(dx/du) +dF/dy(dy/du) => (2x-y)v + (4x-5y)1; plug in the values of x and y => (2uv^2 -uv+v^2) + (4uv -5u +5v)
dZ/du(1,1) = 2-1+1 + 4 -5 +5 = 2 + 4 = 6

for simplicity sake, I'll leave out dz/dy but leave you with my answer of dZ/dy(1,1) = -2

so <dZ/du(1,1),dZ/dy(1,1)> = <6,-2>

if I only use the formula: dZ/du = dF/dx(dx/du) + dF/dy(dy/du); I get dF/dx(v) + dF/dy(1)
dZ/dv = dF/dx(dx/dv) + dF/dy(dy/dv); I get dF/dx(u) + dF/dy(-1)

This is where I'm getting stuck..

 

[AskingForAlyssa]

Figured it out. Thanks for the help though. But, in reply to HallsofIvy, Fx(1,0) and Fy(1,0) are in fact needed for the solution. It's just a bit confusing as to which it's referring to (either x,y , or u,v)

so for a quick condensed version of my final solution;

dZ/du = dF/dx(x(u,v),y(u,v))dx/du + dF/dy(x(u,v),y(u,v))dy/du
= dF/dx(1,1)(v) + dF/dy(1,0)(1)
= 2(v) + 4
dZ/du(1,1) = 6

again, for dZ/dv(1,1) we get;
dZ/du(1,1) = -2