Lecture 16: The chain rule

Transcription

1 Lecture 6: The chain rule Nathan Pflueger 6 October 03 Introuction Toay we will a one more rule to our toolbo. This rule concerns functions that are epresse as compositions of functions. The iea of a composition is: you can sometimes interpret one function as a sequence of two steps. The chain rule allows you to ifferentiate the function be ifferentiating the two steps iniviually an multiplying the results. This rule will allow us to compute a great eal more erivatives, especially when it is use in conjunction with other rules. The reference for toay is Stewart 3.5. The chain rule The basic iea that unerlies the chain rule is: the faster the inputs of a function change, the faster its outputs will change. So for eample, if f() is one function, an f() is another, then the inputs to f in the secon function are moving twice as fast as the inputs to f in the first. So it s erivative is magnifie by a factor of : f() f(). The chain rule generalizes this principle. There are two stanar ways to write it, which are name after the two mathematicians usually creite with inventing calculus. The chain rule (Newton notation) (f g) () f (g()) g () The chain rule (Leibniz notation) z z y y Here, the symbol means composition (NOT multiplication). It means: fee the outputs from one function into the other. So the function f g() is just the same thing as f(g()). In the Leibniz notation, the symbol y shoul refer to something which is a function of, an the symbol z shoul refer to something that is a function of y (an therefore also a function of ). At first glance, it is not at all obvious how these two statements are relate. To show how they both work, I will illustrate them both to compute the erivative of sin(). Newton notation Leibniz notation Let f() sin an g(). Let z sin() an let y. Then f g() sin(). Then z sin(y). So (sin()) f (g()) g () So sin(y) () sin() y cos() cos(y) cos() cos() cos() The iea is the same in both cases: when you have a composite function (that is, a function forme by plugging the output of one function into the input of another), you can preten the inner function is a variable an ifferentiate with respect to it. Then you must multiply the result by the rate of change of the inner function. The iea is that the term f (g()) (in Newton notation) or the term z y (in Leibniz notation)

2 tells how quickly the output changes per unit change in the input to the outer function, an then the terms g () an y tell how quickly the inputs to the outer function change per unit change in. I think you will probably fin the Newton notation easier to apply initially, but I fin the Leibniz notation more intuitively helpful in the long term. In fact, for the first century or so after calculus was invente, the British preferre Newton s notation while the French an Germans preferre Leibniz s notation; it tune out that Leibniz s notation was more practical in leaing to further avances, an French scientific knowlege avance somewhat faster uring this time. Now of course, we can set patriotism asie an use the two notations interchangeably, accoring to which is more useful at any given time. As an eample of how to use the chain rule (in Newton notation this time), consier the following problem. Eample.. Suppose that you know the following information about two functions f() an g(). Determine (g f) (). f() g() f () g () Solution. By the chain rule, (g f) () g (f()) f (). By the value in the table, f(), so this is the same as g () f (). By the values in the table, this is 7 ( 6) 4. 3 First eamples I will illustrate the chain rule by ifferentiating the following eight functions.. ( + ) 7. sin(5) ( + ) e + 7. sin(e ) 8. (e + ) 6 These can be ifferentiate as follows. I will use Leibniz notation in this section, since I personally prefer it. Note that in homework an eams, you o not nee to show as many steps as I o here over time you will get use to skipping some of the more obvious parts (I will also begin to omit some steps in my notes as well). ( + )7 [ ] ( + ) ( + )7 ( + ) 7( + ) 6 4( + ) 6 For a iscussion, see Philip E. B. Jourain s The Nature of Mathematics, chapter 5.

3 sin(5) sin(5) (5) cos(5) 5 5 cos(5) (5) (7 + ) (7 + ) ( + ) 7 ( + ) 7 ( + ) ( + ) 7( + ) 6 () 4( + ) 6 ( ) ( ) ( ) e + e + (e + ) e + e e e + (e + ) sin(e ) (sin(e )) e e cos(e ) e e cos(e ) (e + ) 6 (e + ) 6 (e + ) (e + ) 6(e + ) 5 e 6e (e + ) 5 3

4 4 Differentiating eponential functions The chain rule gives us the necessary tool to ifferentiate arbitrary eponential functions. Remember that we chose the number e specifically to be the number such that e e. This one fact, plus the chain rule, allows us to ifferentiate any eponential function. For eample, take f(). Then using the laws of eponential functions, this can also be rewritten f() (e ln ) e ln. This is a composition of two function: the chain rule says that its erivative will be e ln ln ln e ln ln. The same iea works for all eponentials to give the following fact. b ln(b) b Here, b is a constant value. One thing that this fact reveals is that the number e is totally inescapable in calculus: even if you on t want to write your eponential functions in terms of the base e, you still must introuce the iea of a natural logarithm (an therefore the iea of the number e) to ifferentiate eponential functions. 5 Eamples with multiple rules In the following eamples, we can ifferentiate the given functions with the help of the chain rule, but the chain rule must be use in conjunction with some of the other rules we have seen in class. In this section, I will start to be a little more terse when applying the chain rule, rather than spelling all steps out in full as in the last section. Eample 5.. Differentiate f() tan( ). Solution. This function is a composite of two functions: tan an. The first can be ifferentiate using the quotient rule, as we ve seen: the result is sec. The secon can be ifferentiate using the prouct rule (an the fact mentione in the previous section). The result is: tan( ) sec ( ) ( ) (chain rule) ( ) sec ( ) + (prouct rule) sec ( ) ( + ln ) (previous section) sec ( ) ( + ln ) ( ) Eample 5.. Differentiate f() sin +. Solution. Here we nee to apply the chain rule an the quotient rule in sequence. ( ) ( ) ( ) sin cos (chain rule) ( ) cos ( + ) ( + ) + ( + ) (quotient rule) ( ) ( + ) cos + ( + ) ( ) cos + ( + ) 4

5 Eample 5.3. Differentiate f() cos( ). Solution. This problem requires the chain rule to be applie twice in sequence. cos( ) cos( ) cos( ) (chain rule) cos( ) ( sin( )) (chain rule again) cos( ) ( sin( )) () sin( ) cos( ) 6 Optimization problems Here are a couple eamples where the chain rule can be use in the solution of an optimization problem. Eample 6.. Suppose that a horse begins at the point mile north of a river, which runs east to west. She wishes to walk to the river to rink, an then to walk to the stable, which is 3 miles east an miles north of where she stans. What is the shortest possible istance that she coul walk to o this? Solution. The situation is epicte visually in the following iagram. (0, ) y 0 The ashe line shows two possible paths that the horse coul take. She gets to choose where she will come to the river, but then shoul walk in a straight line to that point an then back to the stable.to fin the best possible point to reach the river, introuce the variable to be the -coorinate of the place she arrives at the river. Then her path can be broken into the sum of the hypotenuses of two right triangles, as shown. (0, ) 3 3 Total istance (3 ) There are two etreme cases: 0 (when she goes straight to the river) an 3 (when she comes straight back from the river) shown below. Clearly no negative value of coul be better, since both legs of 5

6 the trip woul be longer; nor coul any > 3 be better than 3 for the same reason. So the best value of must be in the interval [0, 3]. 0 3 (0, ) (0, ) So let s now optimize the function f() (3 ) on the interval [0, 3]. The first step is to ifferentiate it to fin the critical points. We can o this with the chain rule. f () + ( + ) (3 ) (9 + (3 ) ) + + (3 ) (3 ) 9 + (3 ) (3 ) To fin the critical numbers, set this equal to 0 an solve (3 ) (3 ) + (3 ) 9 + (3 ) (by squaring both sies) + (3 ) ( + ) (9 + (3 ) ) (cross-multiplying) (3 ) + (3 ) 9 + (3 ) (istributing) (3 ) 9 (canceling like terms on both sies) (3 ) 3 (square root of both sies) 3 4 3/4 So there is one critical number: 3/4. Now check the critical number an the two enpoints to fin the absolute minimum. f(0) f(3/4) 5 f(3)

7 So the absolute minimum is 5 miles, an occurs at 3/4. Clever solution. This problem also has a classic, much more clever solution. Of course, a solution like this oesn t generalize well to more complicate situations, but you might enjoy seeing it. The iea that you simply preten that the stable is on the other sie of the river. Then walking to the river an back to the stable is just like walking across the river; just reflect the path across the river. (0, ) (3, 3) But now it is obvious what the faster path is: she shoul just walk in a straight line! So the shortest possible istance is the istance from (0, ) to (3, 3), which is (0, ) (3, 3) Remark. Although this problem is obviously contrive an silly, it is mathematically ientical to an important problem in physics: that of a photon bouncing off a mirror. One of the basic principles of physics (iscovere by Lagrange, I believe) is that light always takes the shortest possible time to go between two points. So when it must go from one point to another by way of a mirror, it takes the path that minimizes istance. In a sense, it solves the problem above (of course, the photon oesn t solve any equations; it just obeys mathematical laws as if it i). The same sort of problem becomes very important in stuying 7

8 the properties of lenses: in that setting, the relevant fact is that light travels more slowly through glass than through air. So to etermine how light will ben as it passes through a lens is like asking the best path for a horse to take through a lens-shape patch of tar, if it wants to get to the other sie as quickly as possible. Eample 6.. A cone-shape coffee filter is 5cm tall, an has raius 5cm at the top. Suppose that it is filling up with liqui, an that height of the liqui in the filter is given by h(t) 5 +t, where t is measure in secons. What is the rate of change of the volume at time t 4 secons? Solution. The liqui occupies a space shape like a cone, with volume 3 πr h, where r is the raius at the top an h is the height. In this case, the proportions of this cone are the same as the proportions of the filter as a whole, so the raius will be equal to the height: r h. So the volume of liqui will be 3 πh3. Now, h is in fact a function of t: h(t) 5 +t, so we can write the volume also as a function of t: V (t) 3 πh(t)3. So we can fin the rate of change of the volume using the chain rule. V (t) V h h t 3 π 3h(t) h t π h(t) 5 t + t ( π h(t) 5 ) ( + t) 5π ( + t) h(t) So in the specific case we are intereste in, V (4) 5π 5 h(4) 5π π 5. 7 Appeni: The chain rule an linear approimation As usual, this appeni is not part of the course material; it s inclue just in case of interest. An alternative way to formulate the chain rule is: the linear approimation of a composition is the composition of the linear approimations. This formulation turns out to be the one that generalizes best to other situations (especially in multivariable calculus). To my min, it is also he most intuitive way to think about it, although this may not be apparent the first time you learn the topic. 8

9 This formulation also happens to be the right strategy to use if you actually want to write own a proof of the chain rule. To see why this is so, consier the composite function f g(). Then its linear approimation aroun a given constant c is given as follows. Now, the linear approimation of g aroun c is: (f g)() f g(c) + (f g) (c) ( c) () g() g(c) + g (c)( c) () Now, consier the linear approimation of f(), not aroun the input c, but rather aroun the input g(c) (that is the input that actually gets plugge into the function f): f() f(g(c)) + f (g(c))( g(c)) (3) Now look what happens when you combine these last two approimations. They say that: f(g()) f(g(c)) + f (g(c) (g() g(c)) f(g(c)) + f (g(c)) (g(c) + g (c)( c) g(c)) f(g(c)) + f (g(c))g (c)( c) The fact that this is the same as the linear approimation of f g() is just the same thing as (f g) () f (g(c))g (c). 9