 So we've seen how useful the chain rule and the reciprocal rule are, and those are both partial derivative identities that had very close correspondence with something you knew already about ordinary derivatives. There's some additional partial derivative identities like the cyclic rule that don't have a corresponding identity for ordinary derivatives, so that makes them a little less familiar, but they're certainly useful nonetheless. So to derive the cyclic rule, let's start out talking about any arbitrary function that depends on two other variables. So I'll write it as z as a function of x and y, but we should be thinking about this as some thermodynamic variable, maybe the Gibbs free energy or the pressure or something else that depends on volume and temperature or any two other thermodynamic variables. So if I know z depends on x and y, I can write the differential of z, how much z changes in response to a change in x and y, there's going to be some change in x because, I'm sorry, some change in z because x is changing, that's given by the rate of change of z as x changes without changing y, and there's also dz dy at constant x times dy, that's the response when I change y, that tells you how much z changes in response. That's the differential of z. So what I'm going to do next now is I'm going to take both sides of this equation and take the derivative with respect to y at constant z. So on the left side, that dz, I'll just put it over dy and it becomes dz dy at constant z. I need to do the same thing on the left and the right sides of the equation. So this coefficient dz dx at constant y stays just the way it is, and the differentials, the dx and the dy and the dz, those are the things I'm going to divide by dy or take the derivative with respect to y. So this leaves me with a dx dy at constant z, and the second term looks like the dz dy at constant x that I already had, now multiplied by dy dy at constant z. So some of those new derivatives that I just wrote down may have caught your ear as being a little bit unusual. dy dy at constant z doesn't look like the derivative you're used to seeing, but if we think about what that derivative means, the rate of change of y with respect to y while holding z constant. In other words, when I change y, how much does y change? Well, of course, y changes exactly as much as y changes. So the value of this derivative doesn't matter what we're holding constant. dy with respect to dy is just 1. When y changes, y changes the exact same amount. So that derivative is equal to 1. The other one that seems a little odd is this one, dz dy at constant z. When I change y, how much does z change when we hold z constant? Well, we've just answered our own question. When we hold z constant, z is not allowed to change, so the change in z must be 0. So if we change z dy when holding z constant, that's equal to 0. So with those two derivatives converted to numbers, we can rewrite this equation. 0 is equal to dz dx at constant y dx dy at constant z plus dz dy at constant x. That's a perfectly good partial derivative identity. We can clean it up, make it look a little more intuitive. If I take this term, the dz dy at constant x, and put it on the other side of the equal sign. So I'll leave the dz dx at constant y dx dy at constant z. I'll put the equal sign over here, and I'll move this dz dy over to the other side, so it requires a negative sign dz dy at constant x. So that's the same result as this one. Just rearranged a little bit. What I want to do next is do the same thing on both sides of this equal sign. And what I'm going to put on both sides of this equal sign is a dy dz at constant x. Why did I do that? I did that because dy dz at constant x is the reciprocal of this one. So after I multiply these two together, they're going to cancel themselves out. But I have to make sure and do the same thing on both sides. So the dy dz at constant x I put on the right, I have to also include one on the left. And then that guarantees the left is still equal to the right. Now, the reciprocal rule tells us dz dy multiplied by dy dz. Those are going to cancel and leave me with one, since I'm taking both derivatives at the same constraint, holding the same thing constant. And I still have a negative one, a negative sign out front. And if I rewrite the left side, I've got dz dx at constant y, dx dy at constant z, dy dz at constant x. So that's our final result. I'll put that one in a box. That is the identity we call the cyclic rule, sometimes also called the triple product rule. It's obvious why it's called the triple product rule. It's because it's a product of three partial derivatives. This particular product of three partial derivatives gives me a minus one. The reason it's called the cyclic rule may also be clear. We have three different partial derivatives, each one involving an x and a y and a z. And all I do is the first one looks like z with respect to x at constant y. If I move that x up to the numerator and the y to the denominator and the z around to being held constant, if I move them in a cycle like that, then I'll get dx dy at constant z. If I move them in a cycle again so that x moves down, y moves up, and z moves to the denominator, I'll get dy dz at constant x. If I were to do the cycle again, I'd get back to the first one I started with. So if I keep cycling the variables around, then the three different partial derivatives I can get by cycling those variables, multiply them all together, and what they come out to be equal to is negative one. So that's a very useful partial derivative identity. To show you why it's useful, what sort of circumstances it can be useful in, let's ask ourselves a question like, what is dp dt at constant v? So if we think about what that means physically, so if I change the temperature of some object, I have an object, I'm going to heat it up, change its temperature, while holding its volume constant. I'm not going to allow it to expand. Normally when I change the temperature of something, if I heat it up, it would tend to expand. If I prevent it from expanding, then the pressure it exerts on its container that's keeping it from expanding, that pressure is going to increase. So we might want to know how much does the pressure increase as I increase the temperature isochlorically at constant volume. So that quantity is not something we know the value of, or we know a relationship for at the moment, not just yet. But it looks familiar. That combination of variables, p's and t's and v's, is reminiscent of some things we've seen before. In particular, the isothermal compression, isothermal compressibility, is another collection of v's and t's and p's. Minus v times kappa, the isothermal compressibility is dv dt at constant p. That's a v and a t and a p just in a different relationship than these. Likewise, we know alpha, the thermal expansion coefficient. That's dv dt at constant p. A different combination of v's and t's and p's than either, not actually that's wrong, that's, this one's correct. The isothermal compressibility is, as I change the pressure isothermally, how does the volume change? That's the isothermal compressibility. So we have three different combinations of v's and p's and t's. If the one we're interested in is dp dt at constant v, what I'd really like is to find out how it's related to these other permutations of those variables. I can use the cyclic rule. So if I just take these three variables and I cycle them around, cyclic rule tells me dp dt at constant v multiplied by dt dv at constant p, cycling the variables, multiplied by dv dp at constant t, cycling them around one more time. That triple product is going to be negative one. This is the one we're interested in. So let's leave that on the left hand side and write dp dt at constant v is equal to the negative one that's on the right divided by these two. When I bring them over to the right side, they end up in the denominator. dt dv constant p and a dv dp at constant t, both in the denominator. So if we look back at the things we know the names of already, dv dp at constant t, that's very closely related to the isothermal compressibility. This one, dt dv at constant p, that looks like this one, but upside down. So we're going to need to use the reciprocal rule to understand this one. So I could rewrite this fraction. If I move the dt dv at constant p up to the numerator using the reciprocal rule, I'll have negative dv dt at constant p in the numerator, dv dp at constant t in the denominator. And now we can say that that's equal to dv dt at constant p. That's this term v alpha. Denominator dv dp at constant t, that's negative v kappa. There we go. So now the same cancellation happens. The volumes cancel, the negative signs cancel, but now I have an alpha over kappa. So what we've learned is the derivative we're interested in. If we want to know how quickly the pressure changes when I change the temperature of an object at constant volume, that's alpha over kappa. I've managed to relate that now to these two named quantities, that have been tabulated for many materials that we can go look up in a table. If I can look up alpha, look up kappa, I just divide them, take their ratio, and that tells me the value of this quantity we're interested in. So again, that's an example of the cyclic rule. We still have one more partial derivative identity to go that's equally useful. So that's what's coming up next.