 So one of the nice things about the derivative is that it's what's called a linear operator, and that doesn't really mean anything until you've taken a few more advanced math classes, but what it translates into is that we can turn some very complicated problems into much easier ones. So here's a hard and painful problem. I suppose I have a snowball that maintains a spherical shape as it melts at a rate of 12 cubic centimeters per hour. We might want to know how rapidly its radius is decreasing when its radius is 6 centimeters. Now let's take this problem apart. So we're given a rate of change of volume with respect to time. So we know it's a rate because it says it's a rate, but it's also volume cubic centimeters with respect to time per hour. So what we have is a rate of change of volume with respect to time. We're looking for a rate of change of radius, presumably also with respect to time. In a sense we have set the independent variable here as time, so any further questions about rates are going to be presumably with respect to time unless we are specifically asking otherwise. So here we want a rate of change of radius with respect to time, and so one of the possibilities is maybe we can do some algebra and find the relationship between radius and time, and then because it's a rate of change I can look for the derivative. So I can do this. After some rather painful and complicated geometric and algebraic arguments, we can eventually construct the function. The radius as a function of time is going to be something like this, and t equals zero is going to correspond to the time when the radius is 6 centimeters, and what I'm interested in finding is when r prime of zero, the rate at which the radius is changing at the time when the radius is 6 centimeters. Now the challenge here is constructing this function. The derivative, barely straightforward derivative of a cube root function, well that involves fractions and it's a chain rule problem. Maybe this isn't the easiest way. So let's see if there's an easier way of doing this. Mathematicians like to look for easier ways. Sometimes they'll spend hours looking for a way to save a few minutes on a problem. So let's approach this problem from a different point of view. So again we have a rate of change of volume with respect to time, and I can at least write a symbol for that. This is the rate of change of volume with respect to time. It's a derivative that's dv over dt is minus 12. I'm going to drop the units there, but keep in mind that this is a minus 12 because it's a decrease, it's melting, and the units are cubic centimeters per hour. What am I looking for? Well I'm looking for how rapidly the radius is decreasing. So I want to find the rate of change of the radius, d, the derivative of r with respect to time, dr dt. So my first step is to look for any sort of relationship at all that I can find between our quantities. Notably this derivative, this volume, and this radius. So what do we know? Well it's a, the snowball has a spherical shape, and there's some formula that relates the volume of a sphere to the radius of the sphere. It's going to be four-thirds pi r cubed gives us the volume, and well that's the relationship between v and r. I actually want to say something about the derivatives. So let's differentiate both of these with respect to time. So this is what I'm about to do. I'm about to differentiate with respect to time, but I haven't done it yet, and there's two reasons for that. First is that this is going to take a couple of steps to work our way through. The other thing is to remember that we are not differentiating with respect to r or v, which means that these expressions, v and r, should be considered to be functions in and of themselves. And what that means is we're going to have to apply the chain rule. Again, the rule of thumb is you can never go wrong by applying the chain rule. You can always apply the chain rule, and the worst that'll happen is, if you apply it correctly, you'll end up with a string of ones. So let's go ahead and differentiate. The left-hand side's actually pretty easy. We can't really do very much. This is just the derivative of v with respect to t. I don't know what it is, but I can write down its symbol. This four-thirds pi is a constant, so I have the derivative of constant times functions, so the constant can be pulled out front. And the function is r cubed. What are we doing here? We are cubing, raising it to the third power. So when I differentiate to the third power, what I get is three times whatever squared times, don't forget the chain rule, derivative with respect to time of whatever was being raised to the third power. And once again, I can't really do anything with this other than to write down what it is in notation. So there's a little bit of a simplification there. That four-thirds pi and the three, simplify it to four pi, are squared. And that gives me this wonderful relationship between dv dt and dr dt. And the reason that that's really helpful is I want to find dr dt, and I know what dv dt actually is. So at this point, I can just substitute in the values that I know for dv dt minus 12. I need to know what r is. Well, I am interested in the point where radius is six centimeters, so I know what r is as well. I'll substitute in those values. Four and pi are just numbers, so I can combine them. And again, keeping in mind I'm actually trying to find dr dt, I can solve the equation for it. So let's see, six squared is 36, and combine the numbers, and solve for dr dt by dividing. And maybe we'll do a little bit of simplification there. One last thing we should check out. The units of dr dt are going to be units of r. What's the radius measured in? Well, evidently the radius is measured in centimeters. Radius is six centimeters, so the units of r should be centimeters. And t time, again, the only time unit we have up here are hours. So presumably if we want to find dr dt, we are getting an answer that is going to be measured in centimeters over hours. We should indicate that as part of our answer. Now, one thing to be careful with here. The proper expression is the snowball's radius is changing at a rate of minus one over 12 centimeters per hour. And that gives us a correct rate of change. Now, you might want to rephrase that. The question is asking how rapidly the radius is decreasing. The radius is not decreasing at negative one over 12 pi centimeters per hour. That negative tells you that the radius is decreasing, and the amount is one over 12 pi centimeters. So if you want to answer the question, how rapidly is the radius decreasing? The actual answer to that question is just one over 12 pi centimeters per hour. If you want to phrase an answer so that you cover all bets, you can just give the rate of change and not worry about the interpretation of whether the amount indicates a decrease or an increase. Well, about another problem. So here we have a coffee shop selling coffee at $3 a cup to 5,000 customers, cheaper than Starbucks. It determines that it will gain or lose 10 customers for every .01 dollar increase of changing the price of a cup of coffee. And of course, their big question is, well, how rapidly is our revenue going to change? And should it raise prices or lower prices? Big important question if you're an owner of the coffee shop. And again, some of the important pieces of analysis here were given a rate of change. So gain or lose the number of customers per dollar. So this is a rate of change in terms of customers with respect to dollars prices, probably what we would interpret that to measure. So this is a rate of change of customers with respect to price. So presumably when we try to answer this question, how rapidly is revenue changing? We're also looking for a rate of change of revenue also with respect to price. So we have to think about what this is going to be. So let's throw down some variables. What I have is information about a number of customers. So I'll let my variable C represent the number of customers. I also have information about price $3 per cup. So maybe I want to let price be another variable. And I'm interested in finding revenue. And I need to know the relationship between revenue and anything else. Well, revenue is just the number of cups that we sell, the number of customers, times how much we can persuade each of them to pay for a cup of coffee. So revenue is price per cup. That's x times the number of cups I've sold, that's c. And I'm interested in the rate of change of revenue with respect to price. So I'm to differentiate with respect to x. So there's my original function. I'll differentiate everything with respect to x. Again, a little bit of analysis goes a long way. This is a product x times c. So when I differentiate, I'm going to differentiate a product. First times derivative second plus second times derivative of the first. And I can simplify that a little bit. The derivative of x with respect to x. Here's where the differential notation is particularly nice. If I were to try and write this, this is dx over dx. And again, the derivative, the differential notation does not indicate a fraction. But you can't look at dx over dx without thinking that that's equal to one. And I can drop it or the dx just will cancel out. But here I do get the relationship between revenue, cost, number of customers, and all of the given rates of change. So what are those rates of change? Well, so x is the price of a cup of coffee. C is the number of customers. So I want to know x, c, and dx to find dr dx. So one of the pieces of information we gain or lose customers for every change in the price of cup of coffee. So again, here's where our differential notation is very useful. I know that dc, the change in the number of customers, I'm going to lose 10 customers when dx, I change the price by 0.01. So dc over dx, again, not a fraction. But it's convenient to think about it as one. That's minus 10 over 0.01, or after all the dust settles, minus 1,000. So what else do I know? I'm interested at a price of $3. I do have 5,000 customers. I have dc dx equals minus 1,000. I can substitute those in and compute dr dx works out to be 2,000. And so note that dr dx, the units of this expression are, revenue is measured in dollars, x price of a cup of coffee also measured in dollars. So we can read this as having units of dollars of revenue per dollar of price. And that is going to be positive 2,000. So I can take that as an increase. So revenue is increasing by $2,000 of revenue per dollar of price. And since our revenue is increasing as we increase x, we should probably raise prices. And again, this explains Starbucks.