 When we compute a rate of change, we should also compute the units. And if we're looking for a quantity, we can use dimensional analysis to help us decide what we want to find. For example, suppose the height of an object is given by some function, and this is in meters t seconds after an experiment begins, and we want to find out how rapidly was the object moving after 6 seconds. And so we find h of 6. Now, since h of t is measured in meters, then h of 6 will also be measured in meters. Now, before we box in this answer and turn in our work, let's remember something useful. Read your answer out loud with the question. And so here the question, how rapidly was the object moving after 6 seconds? The answer, 138 meters? And it's important to recognize that this answer of 138 meters doesn't really answer the question, how rapidly. So even if this number is correct, it's not an answer to our question. Well, this is calculus, and we don't have too many other things we can do. So what if we find the derivative at 6? Now, here's where differential notation is useful. This h prime of t, well, that's dh dt. And remember, the change in a quantity has the same units as the quantity. So dh, well, that represents a change in h, but h is measured in meters. So dh is also measured in meters. dt represents a change in t, our time, and since our time is measured in seconds, dt is also measured in seconds. And so while it might not be clear what the units of h prime of t are, the units of dh dt will be meters over seconds. And so let's say we go to the trouble and find h prime of 6. We find it's equal to... numerically, it's 4, and we know the units are meters per second. And so h prime of 6 is 4 meters per second. And again, we can gain some useful insight if we read our answer out loud with the question, how rapidly is the object moving after 6 seconds, 4 meters per second? And the important thing here is that 4 meters per second does sound like it could be an answer to our question. Now in the previous problem, it happened to be that our derivative was the velocity of the object. But keep in mind that this is only because our function gave us distance. Do not make the mistake of assuming that first derivative is velocity, second derivative is acceleration. That is almost never true. Outside of cases where the problem is constructed to make it true, it's always better to compute the units and figure things out for yourself. So here, v of t is a velocity in meters per second. t seconds after the experiment begins. Let's find the units of v prime of t and decide what v prime of t represents. Again, our differential notation is useful. v prime of t is dv dt. And so our units... dv is a change in velocity. So its unit will be the same as velocity, meters per second. dt is a change in time, so its unit will be seconds. And if we clean up the algebra, the units of our derivative will be meters per second squared. So v prime of t is an acceleration. So for example, suppose the distance an object travels is given by some function in meters. t seconds after an experiment begins. What's the acceleration of the object after three seconds? So we could find s of three, which will give us an answer in meters, but this doesn't answer the question. Acceleration is not answered in meters. We could find s prime of three. So again, we find s prime of t. That's really ds dt. And so our units will be... And so s prime of three will give us an answer in meters per second, and this is not an acceleration, so this still doesn't answer the question. However, s prime of t will be a velocity. Now we'll call it v of t, and we can actually find that derivative. And so now we have a function for our velocity. And notice that v prime of t has units... Well again, it's dv dt. v is measured in meters per second, and so v prime will have units of meters per second squared, which will be an acceleration. So let's find v prime of t, and so v prime of three is two meters per second squared.