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
$$\left( \frac{\partial u}{\partial z} \right)_x<br /> = \left( \frac{\partial u}{\partial x} \right)_y<br /> \left( \frac{\partial x}{\partial z} \right)_x<br /> + \left( \frac{\partial u}{\partial y} \right)_x<br /> \left( \frac{\partial y}{\partial z} \right)_x$$
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($\frac{dy}{dz}$)