# Homework Help: Calc III Chain rule - which vars to put in?

1. Feb 29, 2012

### 1MileCrash

1. The problem statement, all variables and given/known data

z = cos(x^2 + 3y^2)
x = ucosv
y=usinv

find dz/dv

2. Relevant equations

3. The attempt at a solution

I think I can do these fairly well, but I'm a little unsure of the "protocol" for which variables to put back in. Sometimes (in this case) I can't really put everything as v. So I choose to put everything as u and v since they are on the same "level."

for example, I get that

(dv/dv) = (dy/dv)(dz/dy) + (dx/dv)(dz/dx)

When taking those partials and multiplying out ( leaving variables as they are.. )

= - ucos(v)sin(x^2 + 3y^2)6y + usin(v)sin(x^2 + 3y^2)2x

So what is the technique? Replace all x and y with ucosv and usinv respectively, and leave u in the partial derivative?

2. Feb 29, 2012

### BruceW

This is not quite right. The partial derivatives you have defined are keeping x constant (to get dy/dv) and y constant to get dx/dv. So u is not necessarily kept constant, but you have taken the partial derivative as if it were constant.

EDIT: So sorry! I got myself confused! you have done it correctly! Ignore Everything I was saying please!

3. Feb 29, 2012

### 1MileCrash

I have no idea what you mean.

My partials are
dy/dv = ucos(v)
dz/dy = -sin(x^2 + 3y^2)6y
dx/dv = -usinv
dz/dx = -sin(x^2 + 3y^2)2x

Then I multiplied them and added as such:

(dy/dv)(dz/dy) + (dx/dv)(dz/dx)

What part of this is wrong?

EDIT:

Saw your edit, no apologies needed. Is it correct to use u and v in the final partial and phase out x and y?

4. Feb 29, 2012

### BruceW

OK, so now I've looked at it again, I see what you are saying:
$$\frac{dz}{dv} = \frac{\partial z}{\partial x} \LARGE | \small _y \normalsize \ \ \frac{dx}{dv} + \frac{\partial z}{\partial y} \LARGE | \small _x \normalsize \ \ \frac{dy}{dx}$$
Right? (I hope I did the latex correctly). Anyway, your evaluation of the partial derivatives are correct, but you derivatives (i.e. dx/dv and dy/dv) are not correct, because they are full derivatives, yet you have treated u as constant.

Edit: my latex skills aren't amazing, but the line and small x (or y) show what is being held constant when the partial derivative is taken.

Last edited: Feb 29, 2012
5. Feb 29, 2012

### 1MileCrash

I'm sorry, it's my fault for not being clear (latex stopped working on my browser) but, EVERY derivative I've written here is intended a partial, including dz/dv. My goal is the partial derivative of function z with respect to variable v. That's why I hold u constant.

6. Feb 29, 2012

### BruceW

Oh, I get you. So the goal is the partial derivative of z with respect to v, while holding u constant. Yes, I think you've got the right answer then. So in the final answer you can have u, v, x and y in there, it doesn't matter, you can rearrange it whichever way you like really.