Review of Calculus Tools. c 007 Je rey A. Miron Outline 1. Derivatives. Optimization 3. Partial Derivatives. Optimization again 5. Optimization subject to constraints 1 Derivatives The basic tool we need to review is derivatives. The basic, intuititive de nition of a derivative is that it is the rate of change of a function in response to a change in its argument. Let s take an eample and look at it more slowly. Say we have some variable y that is a function of another variable, e.g., y = f() For eample, we could have y = or y = 7 + 3 or y = ln Graphically, I am just assuming that we have something that looks like the following: 1
Graph: A Standard Di erentiable Function with a Maimum y = ( 3) + y 9 7 5 3 1 0 0 1 3 5 7 9 Now say that we are interested in knowing how y will change if we change. Let s say that y is test scores, and is hours of studying. Assume we are initially at some amount of, e.g., you have been in the habit of studying 0 hours per week. You want to know how much higher your test scores would be at some other amount of, + h. One thing you could do, if you know the formula, is take this alternate + h, and compute f() as well as f( + h). You could then look at the di erence: f( + h) f() This would be the change in y. you care about. For some purposes, that might be eactly what In other instances, however, you might care about not just how much of a change there would be, but how much per amount of change in, i.e., per h: That is also easy to calculate:
f( + h) f() h Now look at this graphically: 3
Graph: Calculating the Rate of Change in f Over a Discrete Interval y = 1= y 9 f(+h) 7 f(+h) f() f() 5 3 h 1 1 0 1 3 5 7 9 +h As you can see, we are just calculating the ratio of two legs of a triangle; that ratio is the slope of the line that connects the two points, as seen above. The problem is that this calculation would have a di erent answer if we calculated it at a di erent point:
Graph: Calculating the Rate of Change at a Di erent Point y = 1= f(+h) y f() 9 7 5 3 1 h' f(+h') f() 1 0 1 3 5 7 9 +h' So, what if we calculated all this, but for a smaller h? Then, we would get a di erent rate of change/slope: 5
Graph: Rate of Change as h Shrinks y = 1= y 9 7 5 3 h 1 0 0 1 3 5 7 9 So, let s think about the limiting case of this. Say we eamine lim h!0 f( + h) h f() At one level, this thing might seem a bit confusing or ill-de ned. The numerator obviously goes to zero as h gets small. The denominator also goes to zero. So, why should we epect the limit to converge to anything? The proof is outside this course. But, looking at the graph, we can see that it seems plausible that as we let h go to zero, the ratio should approach the slope of the line that is tangent to the function. This is indeed the case, and it can be proven, but we will just accept it as reasonable. To summarize, we have shown that the rate of change of a function at a given point (assuming it has a well-de ned rate of change) is equal to the slope of a line that is tangent to the curve at that point.
So, we simply want to de ne the derivative as dy=d = f 0 () = lim h!0 f( + h) f() h The key thing to keep in your head is that the derivative is both: 1) the rate of change of the function at that point, and ) the slope of the tangent line at that point. Here are a few additional things to consider: 1. The derivative is usually di erent at di erent points.. Some functions do not have derivatives at various points: 7
Graph: Functions with Non-Di erentiabilities y 9 7 5 3 1 0 0 1 3 5 7 9 ' y 9 7 5 3 1 0 0 1 3 5 7 9 ' y = ( + 3)= +
y ' = 0 0 0.5 1.0 1.5.0.5 3.0 3.5.0 3. We know the formula for the derivatives of a lot of functions: constant linear polynomial to any power ln e and many more, but we will only need the ones above.. We also know some rules about combinations of functions. The product rule: if f() = g()h() Then 9
f 0 () = g()h 0 () + h()g 0 () The chain rule: if f() = g(h()) Then f 0 () = g 0 (h())h 0 () Optimization of a Function of a Single Variable So far, we have talked about the idea that the change in a variable y that depends on a variable, per unit of, might be a useful thing to measure in some settings. And, we have seen that the derivative we have de ned the change in y per unit of, for small changes in seems to measure that concept. But we have not been that eplicit about why derivatives are useful in economics. We ll take a step in that direction now. So, imagine that we have some y that depends on, and we control. We know that di erent values of y lead to di erent values of, and we want to choose the that gives us the highest y. For eample, assume y is a measure of happiness, and is the number of pints of Ben and Jerry s that a consumer eats each night. You might think that for small values of, y increases with. But at some point, as increases, happiness decreases (because you can feel your arteries clogging as you eat your th pint that night). Graphically, we have
Graph: A Single-Peaked Function y = ( 3) + y 9 7 5 3 1 0 0 1 3 5 7 9 So, graphically, it s easy to pick the right point. The key thing about this point, of course, other than the fact that it seems to be where y is highest, is that the slope at that point, i.e., the derivative, is zero. So, this suggests a strategy for nding the that leads to the maiumum y: take the derivative, set it equal to zero, and then solve. That is, compute f 0 () set this to zero f 0 () = 0 and solve for. This kind of equation is known as the rst-order condition (FOC) for a maimum. 11
The phrase rst-order is important; it suggests that this is not the whole story, and that there may be second-order things we have to worry about. Let s leave that aside for a second. Intuitively, it seems clear (and one can prove rigorously under some assumptions) that the that satis es this condition is the at which the maimum y occurs. There are some caveats, but ignore them for a moment and look at an eample. Let s say y = f() = + + : Then the problem we want to solve can be written as ma + + We therefore compute the derivative + set this to zero + = 0 and solve for ; turns out to be 3..1 Caveats: The graph that I drew, and the eample I considered, had nice features: 1. eactly one peak. de nitely had a ma 3. everywhere di erentiable This is not true for all functions: 1
Graph: Functions Without Well-De ned Maima y no ma or min 3 1 1 3 5 7 9 y infinitely many ma = min 3 1 1 3 5 7 9 13
y 9 7 5 min but no ma 3 1 0 0 1 3 5 7 9 So, the condition we have stated, the FOC, is not su cient for a point to be a maimum. Indeed, it is not even necessary, if we allow for functions that are not di erentiable. There is a standard approach to dealing with this that handles these weird cases for di erentiable functions. This method is known as the second-order conditions. It basically says that the second derivative has to be negative for a maimum. What is a second derivative? It s just a derivative of a derivative. And you probably remember, or can at least see intuitively, why this makes sense: If the second derivative is negative, the derivative is getting smaller. Don t worry about this for now. it is relevant later. I will review it again in a few eamples where Most, although not all, of the problems we eamine are nice. For now, I want you to be aware of the issue, and we will see some eamples where it is relevant later. 1
But, it s not the key thing to focus on now - just be sure to understand the intuition and mechanics of the FOC. To be clear, it is very important that you be aware that the FOC is not a su cient condition; there are special cases where the point that satis es the FOC is not the maimizing point. But we re not going to worry about the details yet, and not to a signi cant degree in this course overall. NB: everything I ve said is applicable for nding minima instead of maima. That is one reason we have to check the SOC s. But again, in most applications that we will consider, this will take care of itself. 3 Partial Derivatives The net, and basically last, calculus topic that we need is partial derivatives. The reason is that many interesting economics eamples relate one variable, say y, to two (or more) other variables, say k and l. A common eample can be found in a production function: y = f(k; l) or, in a utility function, u = u( 1; ) So, the standard calculus of one variable is not su cient. Imagine that we have a function of two variables, e.g., y = f(; z) Now, this is a bit more of a pain graphically. But, in principle, we can draw this: 15
Graph: A Function of Two Variables z = ( ) = (y ) = + z 5 0 0 0 5 5 y So, y changes in responses to both and z. If we held one variable constant that is, looked at a particular slice of this picture in either the or z direction we would have a univariate function and picture. If we were only working with that, then we might just apply the standard approach from before. So, we might consider the rate of change of y with respect to either one of those variables. It is therefore natural to de ne what are called partial derivatives: @y @ = lim f( + h; z) f(; z) h!0 h Now, this might look messy. But it simply treats z as a constant, and then takes a standard derivative. 1
Then This is easiest to see by considering eamples. Assume y = z @y @ = z: Why? Because if we treat z as a constant, then y equals just a constant times, and we know how to take that derivative. What eactly is this partial telling us? It is telling us the rate at which y changes as we change, holding z constant. Furthermore, it makes sense that this depends on the value of z. Take z = 0 - then changing has no e ect on y. Of course, we could also think about the e ect of z on y. take the derivative of y wrt z, treating as a constant: To calculate that, we @y @z = : So, if we have a function y = f( 1; ; : : : n ) i.e., a function of n variables, there will be eactly n partial derivatives. More eamples: Let y = a + bz + cq Then @y @ = a @y @z = b: 17
Now say @y @z = c: y = z 3 Then Or, let @y @ = z3 @y @ = 3 z : u( 1; ) = 1 Then @u( 1 ; ) @ 1 = 1 1 @u( 1 ; ) = 1 1 @ 3.1 Discussion: You need to know two things about partials. First, given a general function or some speci c function, you should know how to calculate them. That should be pretty straightforward, since once you understand the approach treat all other variables as constants, and then apply standard rules from univariate calculus it s a totally mechanical application of univariate calculus. Second, you need to know how to interpret partials. 1
This again should not be hard; it is just a tad di erent than the univariate case, but in a way that matters. In words, the partial of a function with respect to one argument is the rate of change in the function in response to a small change in the argument, holding the other arguments ed. This is di erent than adjusting both arguments. For eample, increasing a consumer s consumption of goods 1 and is normally going to have a di erent e ect on utility than just increasing, say, good 1. As a second eample, increasing both K and L will have a di erent e ect than, say, increasing L and holding K constant. We ll see this in practice soon. Optimization with of Functions of Several Variables The last topic we need to consider is how to nd the maimizing values for functions of several variables. Indeed, this is the case of real interest, since key eamples in economics are of this variety. That is what creates all the tension about how much math to use in intermediate courses. Everyone agrees that it s nice to be able to use calculus. we need just a little bit of multivariate calculus. But it turns out that Virtually all basic calculus courses, however, focus only on univariate, rather than multivariate, calculus; in particular, they do not teach partial derivatives. Thus, in most sequences, you do linear algebra, and then multivariate calculus. This makes 19
sense, since you need linear algebra (but only a tiny amount) for some parts of multivariate calculus. This standard approach makes life di cult. So, the key tool we need to do micro theory with calculus is partial derivatives. That means that if we cannot use partials, the bene ts of using calculus are not large; that s why most books put it in an appendi, or skip it entirely. That s also why many departments do not require calculus for an econ major; Harvard did not until or 15 years ago. But, this seems nutty to me: for good students who have had some introduction to basic calculus, learning a partial derivative is not big deal; it s really just a baby step away from what you already know. Indeed, if you think about it the right way, you already know what a partial is, as we have seen. Now we can see why it is useful. Let s rst consider an abstract eample, because there s one small wrinkle that I want to leave aside for the moment that comes in when we get to the economics eamples. Say we have y = f( 1; ) We know that if f is a smooth function, it could look something like: 0
Graph: A Smooth Function of Two Variables z = ( ) = (y ) = + z 5 0 0 0 5 5 y We also know we could think about this in only one of two dimensions. Then this would look like: 1
Graph: A Slice of the Graph Above z = ( ) = ( ) = + z 1 1 1 1 0 So, intuitively, we want to make sure we re at a peak from either angle. Well, looking from either angle is like holding one of 1 or ed. So, say we do the following: Calculate @y=@ 1 and @y=@ set both to zero, and nd the combination of 1 and that simultaneously solves the two equations. I am assuming you can see intuitively that this is analogous to the univariate case. In words, if we are at the combination of 1 and that produces the maimum y, then two things must be true:
1) A small change in 1 leads to a decrease in y, whichever direction we go in. ) A small change in leads to a decrease in y, whichever direction we go in. Thus we have two conditions involving two unknowns, and we can solve these. These two equations need not be linear. But, under some regularity assumptions, there will be an 1 and an that works. Take for an eample y = 3 1 + 5 1 + 1 + Then, the FOC s are @y=@ 1 = 1 + 5 + = 0 @y=@ = + 5 1 + = 0 This is just two linear equations in two unknowns.. We can easily solve for 1 and Of course, these are just rst-order conditions. As with the univariate case, we have to worry about whether we re getting a ma or min; we also have to worry about kinks and boundaries, etc. Ignore all this for the time being, but be aware that it could be an issue. We ll look more carefully in some particular cases as needed. 5 Constraints So far, we have not talked about the multivariate case under the assumption that there might be constraints. We are going to nesse that issue for the most part, in ways you will see shortly. So, it is again something we will have to worry about a bit, but it s best handled case-by-case with speci c eamples, rather than with general theory. 3