# Partial Derivative Relations

1. Sep 2, 2008

### Jacobpm64

Show that: $$\left(\frac{\partial z}{\partial y}\right)_{u} = \left(\frac{\partial z}{\partial x}\right)_{y} \left[ \left(\frac{\partial x}{\partial y}\right)_{u} - \left(\frac{\partial x}{\partial y}\right)_{z} \right]$$

I have Euler's chain rule and "the splitter." Also the property, called the "inverter" where you can reciprocate a partial derivative.

If I write Euler's chain rule, I only know how to write it when there are 3 variables, I usually write it in the form:
$$\left(\frac{\partial x}{\partial y}\right)_{z} \left(\frac{\partial y}{\partial z}\right)_{x} \left(\frac{\partial z}{\partial x}\right)_{y} = -1$$

Where I can write x,y,z in any order as long as each variable is used in every spot. However, I do not know how to work this chain rule if I have an extra variable (u in this case).

I also tried using the "splitter" to do something like writing:
$$\left(\frac{\partial z}{\partial y} \right)_{u} = \left(\frac{\partial z}{\partial x} \right)_{u} \left(\frac{\partial x}{\partial y}\right)_{u}$$

However, I do not know what to do with this because I have the term
$$\left(\frac{\partial z}{\partial x} \right)_{u}$$ , which doesn't appear in the original problem.

Any help would be appreciated.

(This is for a thermodynamics course, but we are still in the mathematics introduction.)

2. Sep 2, 2008

### Mapes

You can write dz as

$$dz=\left(\frac{\partial z}{\partial x}\right)_y dx + \left(\frac{\partial z}{\partial y}\right)_x dy$$

Now differentiate both sides with respect to y at constant u, and continue trying out Euler's chain rule in various places, and you should have it.

3. Sep 2, 2008

### Jacobpm64

Woot I got it.

Thanks a lot.

Last edited: Sep 2, 2008
4. Sep 3, 2008

### Mapes

Cool. That trick is useful if you want to compare relationships at different constant conditions. For example, which is larger, heat capacity at constant pressure or constant volume? Expand entropy the same way:

$$dS=\left(\frac{\partial S}{\partial T}\right)_V dT + \left(\frac{\partial S}{\partial V}\right)_T dV$$

$$\left(\frac{\partial S}{\partial T}\right)_p=\left(\frac{\partial S}{\partial T}\right)_V + \left(\frac{\partial S}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_p$$

$$T \left(\frac{\partial S}{\partial T}\right)_p-T\left(\frac{\partial S}{\partial T}\right)_V =C_p-C_V=T \left(\frac{\partial S}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_p=T \left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial T}\right)_p=-T\left(\frac{\partial T}{\partial V}\right)_p^{-1}\left(\frac{\partial V}{\partial p}\right)_T^{-1}\left(\frac{\partial V}{\partial T}\right)_p=\frac{\alpha^2 V T}{\beta}$$

Since all these terms are positive (thermal expansion can be negative, but it's squared), constant pressure specific heat is always higher.