 Remember, we introduced the derivative in connection with finding the instantaneous rate of change. So let's go back and see what we can do with it. So according to the definition of the derivative, f' of x will be the change in f, whatever our function is, divided by the change in x. And so that means that f' of x will be a ratio of changes, and equivalently we say it gives us a rate of change. Now importantly, the units of f' of x are going to be the units of f divided by the units of x. And this is actually easiest to remember when we write it in differential notation dy over dx. For example, let h of t be the height of a balloon in meters t minutes after release. Let's find the units of h' of t and determine the significance of h' of t. So first the units of h' of t will be the units of h over the units of t, and that will be meters over minutes. And this is a rate of change of height with respect to time. Now that's a literal interpretation of the units of h' of t and the significance of h' of t. And if the universe were a kind and gentle place, the literal interpretation would always be sufficient. Unfortunately the universe is not so kind and not so gentle, and one of the important things is that we often have to impose an additional interpretation on our results before we can get them to make sense to somebody who isn't a mathematician. And so we're looking at something that corresponds to a change in height with respect to time. And we have many different ways we can state this. For example we might describe this as how rapidly height is changing with respect to time, or we might connect it to the fact that we're talking about a balloon and say this is how quickly the balloon is rising. We could even go a little farther and say this is the eccentric rate of the balloon, or we can come up with yet other answers. Now it's actually not necessary to come up with these extra answers when looking for the significance of the derivative. We could simply leave this as the rate of change of height with respect to time. The reason that it is useful to come up with these other answers when working this problem is that you will almost never be asked for the rate of change of f with respect to x. In fact it's more likely that someone who wants to know an instantaneous rate of change will ask a question that does not have the word instantaneous, rate, or change. For example we might be asked something like how rapidly is the population increasing, and this is a rate of change of population. Or we might be asked how fast is the rocket moving, and this is a rate of change of distance. Or we might be asked how quickly is the balloon deflating, and this is a rate of change of volume. And there are many more possibilities, and it's worth keeping in mind the English language has over a hundred thousand commonly used words, and many of these words describe in one way or another a rate of change. So how can we find what we need? Well let's consider a problem like this. Now one extremely useful thing to do is to look for the units. So we might start out by looking for the units of h prime of t, and those units will be the units of h over the units of t. So that'll be meters per second, and the thing to notice is that in our question there is a reference to something that is measured in meters per second, and that suggests that h prime of t will have something to do with our problem. So let's find the derivative, and we know that this will be the rate of change of the object's height. And let's consider this information. The object's height when it was falling at 20 meters per second. And let's consider what that means. First, the important thing to recognize here is that the formula we have is for the height of the object, and so if the object is falling at 20 meters per second, what we want to do is we want to translate this information into some statement about the height of the object. And so one possibility is that the object's height is decreasing by 20 meters per second. However, an even more useful form is to keep in mind that we use the words like increase or decrease because we don't want to use signs like plus or minus. The problem is increase or decrease is good for English but bad for math. So let's rewrite this as a statement that does use the plus or minus. And so that means we'll use the is changing by format. The object's height is changing by minus 20 meters per second. Well, this is useful because we have the rate of change of the object's height. And so we want this rate of change of the object's height, h prime of t, to be minus 20. And so we'll set down that equation and solve. And we find t equals 6. So we can summarize our results. The object was falling at 20 meters per second, six seconds after it was thrown. This is a good answer. The problem is it's not actually the answer to our question. However, that shouldn't stop us from writing it down. Any information at all is useful. And there's nothing wrong with providing more information than we need. In this particular case, t equals 6 is relevant to our problem because it tells us when the object was falling at 20 meters per second. However, t equals 6 itself is not the actual answer to the question. We want the object's height. Fortunately, we have a way of finding that. So we'll let t equals 6 and substitute that into our equation for the height and get 318, which is an answer measured in meters.