 The real power of implicit differentiation comes with related rates problems. Now in order to work with related rates problems, it's helpful to distinguish between variables and constants. A constant never changes its value in the story of a problem. On the other hand, a variable will change during the story. And as a general rule, we want to make sure that we keep our variables as variables until we're ready to answer the question. And so, in general, we'll proceed as follows. First, we'll write down any relationship at all between some or all of the variables. And then, we'll identify the appropriate differential operator, the derivative with respect to something. This tends to be the hardest part of the problem. But once we have it, we can use implicit differentiation to find the required rate of change. For example, here's a standard related rates problems. We have a spherical balloon being inflated at a certain rate. And we want to find how rapidly the balloon's radius is increasing after five minutes. And so, first, we want to identify the variables and constants. Because we're being inflated, it seems that the volume is changing. And because we're interested in the rate of change of the radius, it seems that the radius is also increasing. So, radius and volume are both variables, and we'd like to write down a relationship between the two of them. And so, one possible relationship is that the volume is four thirds pi times the radius cubed. Next, we want to identify the differential operator. We know we want to look at the derivative because the question asks us how rapidly something is going on. To decide on the differential operator, we can use the units given in the problem. And we might start by noting that we're given this quantity of 100 pi cubic feet per minute. And that sounds like a rate of change of volume measured in cubic feet with respect to time measured in minutes. So, it appears we're looking for a rate of change of radius with respect to time. So, the differential operator is the derivative with respect to time. So, we have our relationship. Let's differentiate with respect to time. And now we have our relationship between the rate of change of volume with respect to time and our rate of change of radius with respect to time. So, how do we find how rapidly the balloon's radius is increasing? And so, we want to find dr dt, so we need to know dv dt and r. Again, to find these values, we can look at the units. dv dt is something that should be measured in cubic feet per minute. So, we look around and see that we have a quantity in cubic feet per minute. This must be dv dt. And so, dv dt is 100. We also need r. r is a quantity in feet. Since it's not given to us, we should set up and solve an equation to find it. To find r, we note that we're interested in finding how rapidly the balloon's radius is increasing after five minutes. And so, we need to ask ourselves, self, what can we do with this information? We do have a relationship between v and r, so if we know v, we can find r. And if we know that after five minutes, the volume will be 5 times 100 pi 500 pi cubic feet. So, if we substitute that into our equation, we can solve for r. So, r is 7.2112, which should be in feet. And now we can substitute and find dr dt. And now we can substitute our value for r into our equation and solve for dr dt. And we get our final value 0.4808 feet per minute. Or consider this problem. Here we have two cars approaching an intersection. One car is moving at 20 meters per second southbound, and the other at 15 meters per second eastbound. And we'd like to know how rapidly the cars are approaching each other when they are both 50 meters from the intersection. So to begin with, we have at least three variables. First, the distance from the intersection of the eastbound car, which we'll call x. Next, the distance from the intersection of the southbound car, which we'll call y. And finally, the distance between the cars, which we'll call s. So let x be the distance from the intersection of the eastbound car, y the distance from the intersection of the southbound car, and s the distance between the cars. We'd like to express a relationship between the variables. And if we look at our picture, it seems that x, y, and s form two sides and a hypotenuse of a right triangle. And so we can use the Pythagorean relationship between them. Now let's try and determine our differential operator. Now we do want to find the rate of change of the distance between the cars. So maybe we're looking for ds dx or ds dy. Well, let's look at the units. If we find ds dx or ds dy, the units will be meters per meter. And these will cancel and leave us with a pure number. This cannot answer a how rapidly question. What else can we do? Since we also have a rate of change of meters per second, we might be looking at the derivative with respect to time. And if we find ds dt, the units will be meters per second. And this could answer a how rapidly question. So a differential operator should be the derivative with respect to time. So differentiating. In order to find ds dt, we need to know s, x, y, dx dt, and dy dt. So we know we're looking at a time when both cars are 50 meters from the intersection. And so we know x and y are both equal to 50. We also know one car is moving southbound at 20 meters per second and the other eastbound at 15 meters per second. Since x was the distance from the intersection of the eastbound car, then we know that dx dt is minus 15 because the eastbound car is getting closer to the intersection. Similarly, since y is the distance of the southbound car from the intersection and that distance is decreasing at 20 meters per second, then dy dt is minus 20. But we don't have s, so we can't solve the problem. Or can we? We know x, y, and we do have a relationship with s, so we can substitute these in and solve for s. And now we can substitute these in to find the value of ds dt. So we'll start with our relationship between the rates and substitute in our known values. Then we can solve for ds dt. And so we find ds dt is about negative 24.7487 meters per second.