Manipulation of partial differential operators.

[jj364]

Homework Statement

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

Homework Equations

Only equations given above

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!

 

[pasmith]

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

 

[jj364]

Sorry mis read that!

 

[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})