# GRE Partial Derivatives

1. Aug 21, 2007

### domhal

I am reading "Cracking the GRE Math Subject Test - Princeton Review, 3rd Ed." and and confused by the section on the chain rule for partial derivatives. The method in the book is as follows:

1) Draw a diagram to show how the variables depend on each other, with an arrow meaning "depends on"

2) To find a derivative, find all paths from the dependent variable to the independent variable. Each path gives a product of partial derivatives.

I thought I understood this (and I think I have paraphrased it correctly!) but they give the following example of a complicated situation:

Let z = F(u, v, y), where u = f(v, x) and v = g(x, y). We are to find $$\frac{\partial z}{\partial y}$$. They provide the following diagram, which I have reproduced (in glorious code-o-vision):

Code (Text):

---->u------->x
|   ^ \     /
|   || \   /
|   ||  \ /
--z   ||   \
| |   ||  / \
| |   || /   \
| |   |v/     \
| ---->v------->y
|               ^
----------------|

Arrows from z to u, v and y
Arrows from u to x, y and v
Arrows from v to x, y and u

$$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial y} + \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial u}\frac{\partial u}{\partial y}$$

My problems:

1) Does their diagram have two too many arrows? (I think the arrows from v to u and from u to y are wrong.) If their arrows are correct, why are they there?

2) Using their method, there are paths from z to y they have not considered.

3) I get the answer (using my diagram)

$$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial z}{\partial u}\frac{\partial u}{\partial v}\frac{\partial v}{\partial y}$$

are these the same? (I don't think so.)

Thanks. I hope this is clear.

2. Aug 21, 2007

### proton

Your answer seems correct to me. They probably made a typo. I mean if u = f(v,x) has no y terms in it, then it makes no sense to take the partial derviative of u with respect to y