 So the biggest question is, if I have a function, what do I need to look at? Do I need to look at the function, first derivative, second derivative, later on the anti-derivative, and so on? And one of the ways that we can answer this question is through dimensional analysis, and by looking at the units of the derivative. So one way to read this differential notation, Df, is it's the change in whatever our function is. And the important observation here, as far as the dimensions are concerned, is that the change in a function is measured in the same units as the function itself. So if I'm measuring a change in distance, the change is going to be measured in the same units I measure distance. If I'm going to measure change in mass, the units will be the same as the units of mass, and so on. Now our differential notation is particularly useful here, because if you read the differential notation as the change in F over the change in X, again it's not a fraction, don't read this as a fraction, but it's helpful to think about it as one, then that tells me that the units of the derivative, Df, dx, will be the units of F divided by the units of X. And there's a couple of ways that we can read it, we can read it as units of F over units of X, but more conventionally we read this as something per unit. For example, if H is something that's going to measure our height, and we're measuring that in meters, and T is our time measured in seconds, the units of the derivative, Dhdt, are going to be units of H meters over units of T seconds, that will be meters per second, and we'll read that as meters per second. Likewise, if P is our population, and we may measure a population in millions of persons, and A is an area measured in, I don't know, square miles, then the units of the derivative, Dp over Da will be millions of persons over square miles, and again we can read this millions of persons per square mile. Let's say I have a grade function, G, and G is our grade measured in points, H is the time spent studying in hours, the units of the derivative, G, D, H are going to be points over hour, and we'd read that as points per hour. And we can continue to differentiate, we can talk about second, third, fourth, fifth derivatives, and all these emerge from repeated applications of the differential operator, and each of these introduces another per unit. So if height is measured in feet and T in time, then our first derivative, D, DT, so there's our derivative operator, H, that's going to be measured in feet per second. If I differentiate again, again applying the derivative operator to the derivative, that's going to be measured in feet per second per second. So that's going to be the change in height over the change in time twice. So there's our feet change in height per second per second from each of the DT's. Then similar example, if C is money, and I'm going to measure that in dollars, and V is the number of visitors measured in persons, then the first derivative, D, V, D, C, is going to be measured in dollars per visitor. My second derivative is going to be measured in dollars per visitor, per visitor introduced by that next derivative. Identifying the units is going to be particularly helpful because, again, the problem we generally have is not so much finding the derivative, but finding what is relevant to answering the question. So here we have an example, H of T represents the height of an object in meters, T seconds after we fire it from a cannon. And so the question is, well, I know how to differentiate a function. The question is, do I need to? And suppose I want to find the object's velocity. So I have a choice, H of T, H prime of T, and again, later on, we may also be looking at the anti-derivative or the second derivative. So we have a lot of choices, and so the question is, which one is relevant? So remember, the prime symbol is equivalent to the derivative operator. So when I write H prime of T, well, that's the same as D H over D T. And I can just find the units, H of T are going to be measured in meters, object in meters. So that is not a unit of velocity. So I know that H of T is not going to be relevant. On the other hand, the units of H prime of T, which is the same as D H D T, again, using our differential notation, that's going to be units of H over units of T. H is measured in meters, T is measured in seconds. So D H D T has units of meters per second. And this is a unit of velocity. Velocity is distance over time, and that really is a distance over time. So whatever H prime of T is, I can expect that it's going to have something to do with velocity. Now an important note, do not get the association derivative means velocity. In this case it happens to be true, but that's because the units of the derivative are the same as the units of velocity. And that's the thing to check. Do the units match what we want? So for example, let's change the question. Suppose I have the velocity of a falling object given by V of T, some function meters per second, T seconds after it's dropped. And again, same question, V or V prime gives the rate of change of the object's height. And here's a quick wrong answer. Since the question asks about a rate of change, it must involve the derivative. Because we are used to the idea that derivative is rate of change. We've memorized that mantra and rate of change is derivative, except it's a quick wrong answer. A correct answer, let's check out what the units are. So in terms of the units given, the height, I want a rate of change of height, well height is going to be a length, and the only length unit we have on the board is meters. So we're going to measure our height in meters. And a rate of change is going to be a ratio of a change in height to something. And in this particular case, the only possible something is our other set of units, seconds. So our rate of change of height, presumably, is going to be something that's measured in meters over seconds, meters per second. So I note that V of T is already measured in meters per second. So V of T gives us a rate of change of the object's height. So while I could find V prime of T, the derivative, that's going to give me something else. That's going to give me a rate of change of velocity, that'll be meters per second, that's our change in velocity, per second. So V prime of T is going to be something that is measured in meters per second, per second, which is definitely not a rate of change of height.