Differentiation Rules c 2002 Donald Kreider and Dwight Lahr

Transcription

1 Dierentiation Rules c 00 Donal Kreier an Dwigt Lar Te Power Rule is an example o a ierentiation rule. For unctions o te orm x r, were r is a constant real number, we can simply write own te erivative rater tan go troug a long computation o te it o a ierence quotient. Developing a repertoire o suc basic rules, an gaining skill in using tem, is a large part o wat calculus is about. Inee, calculus may be escribe as te stuy o elementary unctions, a ew o teir basic properties suc as continuity an ierentiability, an a toolbox o computational tecniques or computing erivatives (an later integrals). It is te latter computational ingreient tat most stuents recall wit suc pleasure as tey relect back on learning calculus. An it is skill wit tose tecniques tat enables one to apply calculus to a large variety o applications. Builing te Toolbox We begin wit an observation. It is possible or a unction to be continuous at a point a an not be ierentiable at a. Our avorite example is te absolute value unction x wic is continuous at x = 0 but wose erivative oes not exist tere. Te grap o x as a sarp corner at te point (0, 0) (c. Example 9 in Section.3). Continuity says only tat te grap as no gaps or jumps, wereas ierentiability says someting more te grap not only is not broken at te point but it is in act smoot. It as no corner. Te ollowing teorem ormalizes tis important act: Teorem : I (a) exists, ten is continuous at a. Te proo is not iicult. To sow tat is continuous at a we must sow tat x a (x) = (a), or tat 0 (a + ) = (a). We will accomplis tis by sowing tat 0 ((a + ) (a)) = 0: (a + ) (a) ((a + ) (a)) 0 (a + ) (a) = (a) 0 = 0 0 We ave use in te proo only elementary properties o its concerning sums or proucts o two unctions. Notice tat te crucial step was in isolating te ierence quotient ((a + ) (a))/, wose it, (a), exists by our assumption. Te teorem conirms our intuition tat ierentiability is a stronger notion tan continuity. A unction can be continuous witout being ierentiable, but it cannot be ierentiable witout also being continuous. A unction wose erivative exists at every point o an interval is not only continuous, it is smoot, i.e. it as no sarp corners. We now procee to evelop ierentiation rules. Cognizant o te way unctions are built rom a small number o simple unctions using algebraic operations an composition, we examine ow ierentiation regars tese operations. In te teorems tat ollow we assume tat an g are unctions wose erivatives an g exist. Teorem : Suppose y = (x) is a unction tat as erivative. Ten, (c) = c, were c is a constant. Or in Leibniz s notation x (c(x)) = c x (x). A proo simply uses te corresponing property o its: c(x + ) c(x) (x + ) (x) (c(x)) = c = c (x). x Just as constants can be move outsie a it, so tey can be move outsie te operation o ierentiation. Example : (3x ) = 3(x ) = 3 x = 6x. In Leibniz s notation x (3x ) = 3 x (x ) = 3 x = 6x. Teorem 3: I an g are unctions wit erivatives an g, respectively, ten ( + g) = + g. In wors, te erivative o a sum is te sum o te erivatives. Again, tis ollows immeiately rom te corresponing property o its: [(x + ) + g(x + )] [(x) + g(x)] ((x) + g(x)) x 0 (x + ) (x) g(x + ) g(x) + = (x) + g (x)

2 In act it ollows tat te erivative o any number o terms is te sum o te erivatives o eac term. For example ( + g + ) = (( + g) + ) = ( + g) + = + g +. Example : Teorems an 3 taken togeter enable us to ierentiate any polynomial. For example An, similarly x (3x + x + 7) = x (3x ) + x (x) + x (7) = = 3 x (x ) + x (x) + x (7) = 3 x = 6x +. x (x + x) = + x x/ = + (/)x / = + x. Teorem 4 (Te Prouct Rule): I an g are unctions wit erivatives an g, respectively, ten (g) = g + g. In wors, te erivative o a prouct is te irst actor times te erivative o te secon, plus te secon actor times te erivative o te irst. It is, in act, useul to learn to state te teorem in wors. Comparing a given example to te matematical statement is prone to error, wereas carrying out te necessary computations wile reciting te rule is a convenient skill to learn. We look at several examples o te rule in action an ten provie a proo. Example 3: Fin (x) in two ways, given (x) = (5x + 3)(x + ). Te irst way, o course, migt be to multiply out te given expression an ten ierentiate te resulting polynomial: [(5x + 3)(x + )] = (5x + 3x + 6) = 0x + 3. Using te prouct rule we get [(5x + 3)(x + )] = (5x + 3)(x + ) + (x + )(5x + 3) = (5x + 3) + (x + ) 5 = 5x x + 0 = 0x + 3. Example 4: I y = x(x +), in y x. Using te prouct rule, tis time carrying out te computations as we recite te rule: y x = x x + (x + ) x. Stuents always ask te question Must I simpliy tis? Te answer is yes i you know wat you want to o wit te erivative or wat oter results it nees to be compare wit. Tis is normally te case. However, note tat tere is no unique simplest orm. It einitely oes epen on wat use you inten to make o te result. A reasonable simpliication in tis case migt be y = x 3/ + (/)x 3/ + (/)x / = (5/)x 3/ + (/)x /. To give a proo o te prouct rule, we start start wit te it we must evaluate an ten subtract an a (x + )g(x) to pull te proucts apart: (g) (x) 0 (x + )g(x + ) (x)g(x) (x + )g(x + ) (x + )g(x) + (x + )g(x) (x)g(x) (x + )(g(x + ) g(x)) + g(x)((x + ) (x)) 0 0 ( (x + ) = (x)g (x) + g(x) (x) g(x + ) g(x) + g(x) ) (x + ) (x) Note tat we ave use te act tat a ierentiable unction is continuous to conclue tat 0 (x+) = (x).

3 3 Teorem 5 (Te Reciprocal Rule): Suppose as erivative. Ten or any x suc tat (x) 0, ( ) ( ) (x) = (x). Tat is, = (x). Again, tis ollows rom te it einition: ( ) ( ) ( ) (x + ) (x) (x) 0 (x+) (x) (x) (x + ) (x)(x + ) (x + ) (x) = (x + ) (x) (x + ) (x) = 0 = (x) (x) (x) = (x) (x) (x + ) 0 In evaluating te tree its, we recognize te irst as te einition o (x). In te secon we use te continuity o at x (Teorem ). An te tir was inepenent o. Example 5: Let (x) = x +. Ten (x) = x (x +). Again we wrote tis own wile we recite te rule minus te erivative o te enominator ivie by te square o te enominator. Teorem 6 (Te Quotient Rule): Suppose an g ave erivatives an g, respectively. Ten or ( ) any x suc tat g(x) 0, (x) = g(x)(x) (x)g(x) ( ) g g(x). Tat is, = g g g g. In wors, te erivative o a quotient is te enominator times te erivative o te numerator minus te numerator times te erivative o te enominator all ivie by te enominator square. Te quotient rule is really just te prouct an reciprocal rules combine, or ( ) ( = ) ) = ( g g g g + g = g g g. Example 6: (x) = x+ x+. Ten, writing as we recite te rule: Example 7: (x) = (x) = (x) (x + )() (x + )() (x + ) = x + x (x + ) = (x + ). + x x +3x+. Ten (x) = (x + 3x + ) x ( + x)(x + 3) (x + 3x + ). It woul try one s patience to obtain tis result using te it einition instea o te quotient rule. Soul we simpliy it? No, unless we know wat use is to be mae o it. Example 8: For (x) = x = x, in te erivative tree ways, using te power rule, te reciprocal rule, an te quotient rule: Power Rule: (x) = ( )x = x Reciprocal Rule: (x) = x Quotient Rule: (x) = x 0 x = x

4 4 Te collection o rules tat we now ave enable us to write own te erivatives o a remarkable variety o unctions, knowning only te erivatives o a ew basic unctions. Tere is one situation not covere by our rules, owever, namely ow o we eal wit te composition o unctions? How woul we ierentiate x +, or example? We will a one inal rule to our arsenal to anle unctions tat are built up by te operation o composition, te so-calle cain rule. It is peraps te most important ierentiation rule o all. Teorem 7 (Te Cain Rule): Let ( g)(x) = (g(x)) be te unction eine rom an g by composition. Assume tat g is ierentiable at te point x an tat is ierentiable at te point g(x). Ten te composite unction g is ierentiable at te point x, an ( g) (x) = [(g(x))] = (g(x))g (x). Leibniz notation gives us a useul alternative way to write te cain rule. I we eine u = g(x), we can write te composition as te cain o unctions y = (u), were u = g(x). Ten te cain rule takes te orm y x = (u) u x = y u u u=g(x) x, y or simply x = y u u x i we remember tat te irst actor y u is evaluate at u = g(x). Example 9: For te unction x + we woul ave, applying te irst statement o te cain rule, [ ] x + = x + x To apply te secon orm o te rule we write y = u, were u = x + ; ten we ave y x = y u u x = u u x = u x = x + x = x x +. Leibniz notation really comes into ull bloom in writing te cain rule in te secon orm, above. Remembering tat erivatives measure rate o cange, we interpret y u as measuring ow muc aster y canges tan u, an u x as measuring ow muc aster u canges tan x. Tus it seems perectly natural tat y x soul be te prouct o tese two erivatives, measuring ow muc aster y canges tan x. (I y canges twice as ast as u, an u canges tree times as ast as x, ten y is canging six times as ast as x.) Beore coming back to a proo o te cain rule we consier a ew more examples tat illustrate te ease o its use in practice. Example 0: Dierentiate y = (x + ) 0. Writing tis as y = u 0, were u = x +, we ave y x = y u u x = 0u9 x = 0x(x + ) 9. As wit nearly any rule, ater gaining some acility wit its use one can in sortcuts. Instea o explicitly substituting u = g(x) one simply tinks it instea. To ierent (g(x)), ten, one tinks o te insie unction g(x) as an inivisible glob, an recites take te erivative o wit respect to glob an ten multiply by te erivative o glob wit respect to x. In tis way te erivatives o many composite unctions may be written own irectly as one recites te rule. Example : Dierentiate (x) = ( + 3 x) 35. Tinking o + 3 x as te glob in tis case, we tink, an write, immeiately (x) = 35(glob) 34 x (glob) = 35( + 3 x) 34 3 x (O course, we normally on t reveal te glob part outsie te amily.) ( 3, x+ Example : For (x) = x +) we ave ( ) x + (x) = 3 x (x + )() (x + )(x) + (x + ).

5 5 Simpliy? Sure, go aea. Let us prove te Cain Rule: Assume tat y = (g(x)), tat g is ierentiable at x 0, an is ierentiable at g(x 0 ). Ten we must sow tat (g(x)) is ierentiable at x 0 an tat ( g) (x 0 ) = (g(x 0 ))g (x 0 ). As usual we begin wit te it einition o te erivative at x 0 : ( g) (g(x)) (g(x 0 )) (x 0 ), were we must sow tat te it exists an as te given value. Can we rewrite te ierence quotient in a more transparent orm? A naive (an not completely correct) irst step migt be to multiply an ivie by g(x) g(x 0 ) as ollows: (g(x)) (g(x 0 )) = (g(x)) (g(x 0)) g(x) g(x 0) x x 0 g(x) g(x 0 ) x x 0 Te secon actor we immeiately recognize to be te ierence quotient or g, wose it as x x 0 is g (x 0 ). An, eeling on a roll, we notice tat te irst actor is also a ierence quotient o sorts it is te slope o a secant line to te grap o at te point g(x 0 ). As suc, wen x x 0, g(x) g(x 0 ) (g is continuous at x 0 ), an te quotient soul approac te slope (g(x 0 )) o te grap o. Tis woul yiel exactly te expression (g(x 0 ))g (x 0 ) tat we want. Wat, i anyting, is wrong wit te above proo? Te one sticky point is tat we multiplie an ivie by g(x) g(x 0 ), an tis woul be a problem i we were iviing by zero. Coul g(x) g(x 0 ) = 0 or values o x ierent rom x 0? An coul tis appen or values o x arbitrarily close to x 0? Te ba news is tat it can, even toug suc circumstances are rare. But it takes only one exception to rener our proo invali. Te goo news is tat we can ix te problem by taking a sligtly closer look at te argument we gave above. We begin wit our assumption tat is ierentiable at te point u 0 = g(x 0 ), i.e. tat u u0 ((u) (u 0 ))/(u u 0 ) exists an is equal to (u 0 ). Let us introuce te unction Q(u) = (u) (u 0) u u 0 i u 0 (u 0 ) i u = u 0 an notice tat it is simply te continuous extension o te ierence quotient to te point u 0. I.e. u u0 Q(u) = (u 0 ) = Q(u 0 ). Notice also tat (u)(u u 0 ) = (u) (u 0 ). (I u u 0 tis is immeiate rom te einition o Q, an i u = u 0 it is obvious since bot sies o te equation are zero.) In particular, (g(x)) (g(x 0 )) = (u) (u 0 ) = Q(u)(u u 0 ) = Q(g(x))(g(x) g(x 0 )), an we can return to our initial argument: ( g) (x 0 ) = (g(x)) (g(x 0 )) = Q(g(x))(g(x) g(x 0 )) = g(x) g(x 0 ) Q(g(x)) x x 0 = Q(g(x 0 ))g (x 0 ) = Q(u 0 )g (x 0 ) = (u 0 )g (x 0 ) = (g(x 0 ))g (x 0 ) In concluing tat x x0 Q(g(x)) = Q(g(x 0 )) we mae key use o te continuity o g at x 0 an Q at u 0 = g(x 0 ). Exercises: Problems Ceck wat you ave learne! Vieos: Tutorial Solutions See problems worke out!