Homework Help: Manipulation of partial differential operators.

1. Dec 19, 2012

jj364

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

Given that u(x,y) and y(x,z) are both continuous, differentiable functions show that

($\frac{\partial u}{\partial z}$)x=($\frac{\partial u}{\partial y}$)x($\frac{\partial y}{\partial z}$)x

2. Relevant equations

Only equations given above

3. The attempt at a solution

I tried to write out the total differential for du and dy and then holding x constant so dx=0 I rearranged the equations to get

$\frac{du}{dz}$=($\frac{\partial u}{\partial y}$)x($\frac{\partial y}{\partial z}$)x

I then tried to get $\frac{du}{dz}$ in another form but couldn't see a way of getting ($\frac{\partial u}{\partial z}$)x from what I have.

Any help would be greatly appreciated!

2. Dec 19, 2012

pasmith

The chain rule gives
$$\left( \frac{\partial u}{\partial z} \right)_x = \left( \frac{\partial u}{\partial x} \right)_y \left( \frac{\partial x}{\partial z} \right)_x + \left( \frac{\partial u}{\partial y} \right)_x \left( \frac{\partial y}{\partial z} \right)_x$$
So why is it that the first term on the right always vanishes?

3. Dec 19, 2012

jj364

Sorry mis read that!

Last edited: Dec 19, 2012
4. Dec 19, 2012

jj364

I'm sorry I don't know. I thought that the chain rule for partial derivatives was;

$\frac{du}{dz}$ = ($\frac{\partial u}{\partial x}$)y($\frac{dx}{dz}$) + ($\frac{\partial u}{\partial y}$)x($\frac{dy}{dz}$)

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