4.4 The Derivative. 51. Disprove the claim: If lim f (x) = L, then either lim f (x) = L or. 52. If lim x a. f (x) = and lim x a. g(x) =, then lim x a

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1 Capter 4 Real Analysis Disprove te claim: If lim f () = L, ten eiter lim f () = L or a a lim f () = L. a 52. If lim a f () = an lim a g() =, ten lim a f + g =. 53. If lim f () = an lim g() = L R f (), ten lim a a a g() =. 54. If lim f () = L, ten lim(f () L) = 0. a a 55. Disprove te two claims: (a) If lim f () = L, ten f (a) = L. a (b) If f (a) = L, ten lim f () = L. a 56. Te squeeze teorem (teorem 4.3.4). In eercises 57 68, prove eac matematical statement about continuity. 57. Te constant function is continuous (from teorem 4.3.5). 58. Te sum of two continuous functions is continuous (from teorem 4.3.5). 59. Te ifference of two continuous functions is continuous (from teorem 4.3.5). 60. Te prouct of two continuous functions is continuous (from teorem 4.3.5). 61. If f is continuous at = a, ten f is continuous at = a. 62. If f is continuous at = a, ten f is continuous at = a. 63. Prove tat f () = n is continuous for all n N via inuction (an teorem 4.3.5). 64. If lim f (a + ) = f (a), ten f is continuous at = a Disprove te claim: If f an g are not continuous at = a, ten f + g is not continuous at = a. 66. Disprove te claim: If f is continuous at = a, ten f is continuous at = a. 67. Disprove te claim: If te composite function f (g()) is continuous, ten f () an g() are bot continuous. 68. Te following variation on te caracteristic function of Q is continuous at = 0: { if Q f () = 0 if Q In eercises 69 70, state bot an intuitive escription an a efinition of eac limit. 69. lim a f () = 70. lim f () = L 4.4 Te Derivative Calculus is te stuy of cange. Wile Sir Isaac Newton an Gottfrie Leibniz are bot creite for inepenently eveloping calculus in te late 1600s, matematicians a alreay been working wit erivatives for nearly a alf century. Te stuy of cange as epresse by te erivative was motivate by a siteent an seventeent century European reflection on an ultimate rejection of ancient Greek astronomy an pysics. Te European astronomers Nicolaus Copernicus, Tyco Brae, an Joannes Kepler

2 282 A Transition to Avance Matematics eac a insigts tat callenge te teories of te ancient Greeks, setting te stage for te groun-breaking work of te Italian scientist Galileo Galilei in te early 1600s. Many of te questions about a moving object (tat is, an object canging position an velocity) tat tese scientists were stuying are reaily answere by consiering lines tangent to curves. A number of matematicians from many ifferent countries mae important contributions to te question of fining te equation of a tangent line. Pierre e Fermat stuie maima an minima of curves via tangent lines, essentially using te approac stuie in contemporary calculus courses. Tis work prompte fellow Frenc matematician Josep-Louis Lagrange to assert tat Fermat soul be creite wit te evelopment of calculus! Te Englis matematician Isaac Barrow, wo was Newton s teacer an mentor, correspone regularly wit Leibniz on tese matematical ieas. As we ave mentione, neiter Newton nor Leibniz tougt of te erivative as a measure of cange in terms of our contemporary efinition involving limits. Our stuy of te erivative follows more closely te work of Fermat an Barrow from te early 1600s, in wic we tink of a tangent line as a limit of secant lines. Naturally, te contemporary presentation is informe by an unerstaning of Caucy s notion of te limit from te early 1800s. Te erivative enables te etermination of te equation of a line tangent to a given curve at a given point. Given a function y = f (), te slope of a secant line joining two points (c, f (c)) an (c +, f (c + )) is m = rise run = y = f (c + ) f (c) (c + ) c = f (c + ) f (c). In tis contet, te symbol m is te first letter in te Frenc wor montrer wic translates as to climb. To fin te slope of te line tangent to a function f at te point (c, f (c)), we take a limit of te slopes of secant lines, letting approac 0. Figure 4.13 illustrates wy tis limit process makes intuitive sense; you can see tat te slopes of te secant lines get closer an closer to te slope of te tangent line as te point (c +, f (c + )) gets closer an closer to (c, f (c)). Te efinition of te erivative reflects tese ieas. Te following efinition epresses te real number c as a variable quantity to ientify a general formula for te erivative, enabling us to etermine te slope of te line tangent to f () wenever tis slope is efine. Definition4.4.1 Let f () be a function wit omain D. Ten te erivative of f () is f f ( + ) f () (), 0 wenever tis limit eists. We say tat f () is ifferentiable at = c wen f (c) eists for c D, an tat f () is ifferentiable wen f () eists for all D. Te ratio f ( + ) f () is calle te ifference quotient of te erivative.

3 Capter 4 Real Analysis Figure4.13 A tangent line at = 2 as a limit of secant lines Recall from calculus tat many ifferent notations are use for te erivative of a function y = f (), incluing f () = f = (f ) = y = y = D (y) = ẏ. Various prases also refer to te erivative, incluing f (or y) prime, te erivative of f (or y) wit respect to, te letters f spoken iniviually, an y spoken iniviually. Most of tis notation for te erivative is attributable to Leibniz, wo gave consierable tougt to carefully ientifying a useful symbolism an is recognize as a genius in eveloping notation to make subtle concepts unerstanable. Te alternate efinition of te erivative is sometimes elpful; if te limit eists, ten f f (t) f () (). t t Te proof of te equivalence of tis alternate efinition an te one given in efinition is left for eercise 50 at te en of tis section. Eample We use te two efinitions of te erivative to etermine te equation of a line tangent to f () = 2 at (2, 4). Applying te efinition, f (+) () 2+ = Hence te slope of te tangent line at (2, 4) is f (2) = 2 2 = 4, an te equation of te line tangent to f () = 2 at (2, 4) is given by y 4 = 4( 2). Applying te alternate efinition prouces te same result: f t 2 2 () t t (t + )(t ) t + = 2. t t t Wen using te efinition (as in eample 4.4.1), we often algebraically manipulate te ifference quotient so tat appears as a factor in te numerator. Tis factor

4 284 A Transition to Avance Matematics ten cancels te enominator, simplifying te ifference quotient so te limit can be evaluate. If te original function f () is a rational function, ten fining a common enominator will simplify te ifference quotient in tis way. If f () contains a square root, multiplying by te conjugate square root function will simplify te ifference quotient. Eample4.4.2 We use te efinition of te erivative to fin te erivative of f () = Multiplying bot te numerator an te enominator of te ifference quotient by te conjugate square root function an ten simplifying yiels te following calculation. f 5 () [( + + 1) ( + 1)] 5( ) 0 = ( ) Question4.4.1 Using te efinition of te erivative, ifferentiate eac function. (a) f () = (b) g() = 7 3 (c) s() = () t() = 1 3 Wile we can use te formal efinition of te erivative to compute erivatives of a given function, teoretical applications of te efinition are more important. Using te efinition, we can prove general teorems tat ol for all erivatives, making it easy to ifferentiate many familiar functions witout eplicitly applying te efinition one function at at time. Many functions are so complicate in structure tat irectly using te ifference quotient becomes unwiely or impossible. Te net teorem states analytic properties of erivatives to facilitate suc computations. Using tese results is a common eercise in calculus courses, but you may not ave consiere te unerlying proofs tat justify tem. Tese proofs are te focus of te remainer of tis section. Teorem4.4.1 If c R an bot f an g are ifferentiable functions, ten te following ol. Te constant rule: Te scalar multiple rule: [ c ] = 0 [ c f () ] = c f ()

5 Capter 4 Real Analysis 285 Te sum rule: Te ifference rule: Te power rule: Te prouct rule: Te quotient rule: Te cain rule: [ f + g ] = f + g [ f g ] = f g [ n ] = n n 1, for n R [ f g ] = g f + f g [ ] f = g f f g g g 2, provie tat g() = 0 [ f (g()) ] = f (g()) g () A stanar goal of a calculus course is to evelop a mastery in using tese ifferentiation rules. Before iving into te proofs of various parts of tis teorem, te net eample provies te opportunity to revisit te skills you learne in calculus. Question4.4.2 Using teorem 4.4.1, ifferentiate eac function. (a) f () = (b) g() = cos (3 + 1) (c) () = () p() = ( 5 + ) tan(2) (e) q() = ln( ) sin 2 (5 + 3) (f) r() = ( ) 3 4e + 6 Te net tree eamples give te proofs of some of tese ifferentiation rules. As in te stuy of limits an continuity, we first consier te scalar multiple an sum rules, an ten iscuss a couple of ifferent approaces to proving te power rule. Eample4.4.3 We prove te scalar multiple rule from teorem 4.4.1: For any constant c R an ifferentiable function f, [ c f () ] = c f (). Proof Apply te efinition of te erivative an te limit of a scalar multiple rule. c f ( + ) c f () [ c f () ] 0 = c lim 0 f ( + ) f () 0 c [ f ( + ) f ()] = c f () Eample4.4.4 We prove te sum rule: If f an g are ifferentiable functions, ten [ f + g ] = f () + g ().

6 286 A Transition to Avance Matematics Proof Apply te efinition of te erivative an te limit of a sum rule. [ f + g ] [ f ( + ) + g( + )] [ f () + g()] 0 0 [ f ( + ) f ()] + [ g( + ) g()] f ( + ) f () g( + ) g() + lim 0 0 = f () + g () Eample4.4.5 We prove te power rule: If n R, ten [ n ] = n n 1. Proof We prove te power rule in te case of te positive integers n N by using te binomial teorem to epan te term f ( + ) = ( + ) n in te ifference quotient as follows: ( + ) n = n + n n 1 n(n 1) + n n n 1 + n. 2 Applying te efinition of te erivative, [ n ] (+) n n 0 n(n 1) [n +n n 1 + n n n 1 + n ] n 2 0 n(n 1) n n 1 + n n n 1 + n 2 0 n(n 1) [n n 1 + n 2 + +n n 2 + n 1 ] 2 0 n n 1 + n(n 1) n 2 + +n n 2 + n = n n 1. Alternatively, te power rule for n N follows by inuction (see eercise 67 in section 3.6). Te efinition of te erivative proves te base case [ ] = 1 0 = 1, an te prouct rule applies in te inuctive step (for n+1 = n ). A complete proof of te power rule must consier arbitrary real numbers n R, not just positive integers n N. Te power rule etens to te negative integers via te quotient rule, to rational powers via implicit ifferentiation, an to all real numbers via logaritmic ifferentiation. Te etails of suc a complete proof are left for your later stuies.

7 Capter 4 Real Analysis 287 Question Te following steps outline a proof of te quotient rule: If f (), g() are ifferentiable functions wit g() = 0, ten [ ] f () = g() f () f () g () g() g() 2. (a) Wat is te ifference quotient for te function f () g()? (b) Using te common enominator g() g( + ), simplify te ifference quotient from part (a). (c) In te numerator from part (b), subtract an a te term g() f (). Now split te fraction into a ifference of two ifferences, gatering togeter te two terms wit g() as a common factor an te two terms wit f () as a common factor. () Wat is te limit of te ifference of ifference quotients from part (c) as approaces 0? (e) Base on parts (a) (), craft a complete proof of te quotient rule as moele in eamples 4.4.3, 4.4.4, an Question Te following steps outline a proof of te cain rule: If f (), g() are ifferentiable functions, ten [ f [g()] ] = f [g()] g (). (a) Wat is te ifference quotient (t) () t (from te alternate efinition of te erivative) for te function () = f [ g() ]? (b) Assuming tere are no values for wic g() = g(t), multiply bot te numerator an te enominator of te ifference quotient from part (a) by g(t) g(). Factor out te resulting ifference quotient for g(). (c) Take te limit of te prouct of ifference quotients from part (b) as t approaces to obtain te cain rule formula. () Base on parts (a) (), craft a proof of te cain rule uner te assumption tat g() = g(t) as moele in eamples 4.4.3, 4.4.4, an Te assumption tat tere are no values for wic g() equals g(t) may be unreasonable; a complete proof of te cain rule tat oes not use tis assumption is outline in eercises at te en of tis section. We en tis section by consiering te relationsip between two of te most significant properties of functions stuie in tis capter: continuity an ifferentiability. Some properties of functions are completely inepenent of one anoter, as we saw in our iscussion of one-to-one an onto functions; some functions are bot, some are neiter, wile still oters ave just one of tese properties. Tis observation leas us

8 288 A Transition to Avance Matematics to ask if continuity an ifferentiability are inepenent of one anoter, or is tere a connection between tese two properties? As you may recall, every ifferentiable function is continuous, but not every continuous function is ifferentiable. We consier te teorem an its proof, along wit a countereample tat togeter justify tese assertions. Teorem4.4.2 If a function f wit omain D is ifferentiable at a (b, c) D, ten f is continuous at a. Proof By te alternate efinition of te erivative, given any ε > 0, tere eists a value δ > 0 so tat f () f (a) f (a) a < ε wenever 0 < a < δ. Multiplying bot sies by a, we see tat f () f (a) f (a)( a) < ε a. Applying te secon inequality ( y y ) from teorem in section 4.3, we ave f () f (a) f (a)( a) f () f (a) f (a)( a). Tis fact implies f () f (a) < f (a)( a) + ε a, an so f () f (a) < ( f (a) + ε) a. Te term on te rigt can be mae arbitrarily small: we restrict values of in tat term so tat a is smaller tan bot δ (so tat te first inequality ols) an ε/( f (a) + ε). Ten f () f (a) < ε, wic proves te result. Teorem asserts tat every ifferentiable function is continuous. Are tere continuous functions tat are not ifferentiable? Peraps you can recall from calculus eamples of continuous functions tat are not ifferentiable. Te net eample provies one suc countereample. Eample4.4.6 We iscuss te continuity an ifferentiability of f () = at = 0. We can sow tat y = is continuous at = 0, using te efinition. Let ε > 0 an coose δ = ε. For any suc tat 0 < δ, te following string of relations ols: f () f (0) = 0 = = < ε. By te efinition of continuity, is continuous at = 0. On te oter an, we can sow tat is not ifferentiable at = 0, using te alternate efinition of te erivative. Te ifference quotient for f () at = 0 is f () f (0) 0 = 0 =. Taking te limit of tis ifference quotient as approaces 0, lim 0 = 1 an lim 0 + = 1.

9 Capter 4 Real Analysis 289 Terefore te limit f () f (0) lim 0 0 oes not eist, an f () = is not ifferentiable at = 0. Question Give an eample of a continuous function tat is not ifferentiable at te following points: (a) = 1 (b) bot = 1 an = 1 (c) = n π, for every n Z () = 2n, for every n Z Tese results sow tat (intuitively speaking) it is more ifficult for a function to be ifferentiable tan continuous. From an informal, grapical perspective, tis fact is quite natural; at a point of iscontinuity for a grap, we cannot raw a unique tangent line. Te results also provie anoter reason for te importance of stuying continuity: te functions tat are te most well beave from te perspective of ifferential calculus are continuous. Section 4.6 will ientify an important connection between continuity an Riemann integrability. Te erivative as transforme te way matematicians tink about functions. Many questions about matematical objects an our real-worl can be prase in terms of te erivative s measure of cange. In tis way, te evelopment of te erivative set te stage for muc of te last tree centuries of investigations into function teory. From your calculus courses, you know tat tese investigations inclue fining maima an minima, an etermining increasing an ecreasing sections of curves, concavity, an points of inflection, as well as te construction of power series. In summary, te erivative flows troug function teory in a useful an meaningful way Reaing Questions for Section Define an give an eample of te slope of a line. 2. Describe an intuitive motivation for te efinition of te erivative in terms of secant lines an tangent lines to a curve. 3. State te efinition of te erivative f (). 4. State te alternative efinition of te erivative f (). 5. Give an eample of a ifferentiable function. 6. Wat is te istinction between a function being ifferentiable at a point = c an a function being ifferentiable? 7. State teorem How is tis result elpful wen stuying erivatives? 8. Give an eample of eac ifferentiation rule state in teorem Define an give an eample of a conjugate square root function. 10. State te binomial teorem. How is tis result elpful wen stuying erivatives? 11. Discuss te relationsip between continuity an ifferentiability. 12. Give two eamples of functions tat are continuous, but not ifferentiable.

10 290 A Transition to Avance Matematics Eercises for Section 4.4 In eercises 1 6, epress te slope of a secant line to eac function for te esignate -coorinates as a ifference quotient, an sketc te corresponing grap. 1. f () = at = 3 an = 4 2. f () = at = 3 an = f () = at = 3 an = f () = 3 at = 0 an = 1 5. f () = 3 at = 0 an = f () = 3 at = 0 an = In eercises 7 18, use te efinition of te erivative to compute te erivative (if it eists) of eac function. 7. f () = g() = () = j() = p() = 1/ 12. q() = r() = s() = 15. t() = u() = + 7 { 4 if v() = 2 2 if > 2 { if w() = 2 2 if > 2 In eercises 19 28, compute te erivative of eac function using te analytic ifferentiation rules from teorem 4.4.1, along wit your recollection of te erivatives of functions from calculus. 19. f () = ( ) f () = ( + 1 ) f () = ( ) (2 + 1/) 22. f () = ( 2 +1) 3 ( ) f () = sin 5 ( 3 + 2) 24. f () = ln() cos(2 + 7) 25. f () = log 3 (cot(2)) 26. f () = ln( 2 + 2) log 5 (csc() + 2) 27. f () = (k 5 + 2) 3, were k R ( f () = ) 3n, k were k, n R In eercises 29 34, etermine te eact value of (3π/4) an state te equation of te line tangent to () at = 3π/4 using te information in te following table. f () f () g() g () = 3π/ () = 7 f () sec() + π () = g() cos() 31. () = g() + f () () = tan() + π cot 2 (g()) 33. () = sin[π f ()]+cos[π g()] 34. () = f () 2 g()

11 Capter 4 Real Analysis 291 In eercises 35 38, answer eac question about f () =. 35. Using te efinition of te erivative, fin f (). 36. Using te power rule, fin f (). 37. Determine te equation of te tangent line to f () = at (9, 3). 38. Determine te equation of te tangent line to f () = tat is perpenicular to te line etermine by 2y + 8 = 16. Eercises evelop a proof tat te erivative of sin θ is cos θ. 39. Prove tat sin θ cos θ < θ < tan θ. Hint: Compare te areas of te tree neste regions in figure 4.14 an use te fact tat a pie-sape sector of te unit circle wit central angle θ (in raians) as an area of θ/ Ientify upper an lower bouns on sin θ/θ using te inequalities from eercise 39. Hint: Divie by sin θ an take reciprocals. 41. Prove tat lim θ 0 sin θ/θ = 1. Hint: Apply te squeeze teorem (see teorem from section 4.3) to te inequalities from eercise Prove tat lim θ 0 (1 cos θ)/θ = 0. Hint: Multiply bot te numerator an te enominator by 1 + cos θ an ten use bot te Pytagorean ientity sin 2 θ + cos 2 θ = 1 an te limit from eercises Prove tat te erivative of sin θ is cos θ. Hint: Working wit te efinition of te erivative, simplify te resulting ifference quotient using te limits from eercises 41 an 42 along wit te trigonometric ientity sin(u + v) = sin u cos v + sin v cos u. In eercises 44 48, erive te formulas for te erivative of te oter trigonometric functions; all but eercise 44 use te quotient rule. 44. Prove tat te erivative of cos θ is sin θ. Hint: Use te cofunction ientity cos = sin(π/2 ) an te erivative from eercises 43. (cos q, sin q) (1, tan q) Figure 4.14 Figure for eercise 39 q (0,0) (1,0)

12 292 A Transition to Avance Matematics 45. Prove tat te erivative of tan θ is sec 2 θ. 46. Prove tat te erivative of cot θ is csc 2 θ. 47. Prove tat te erivative of sec θ is sec θ tan θ. 48. Prove tat te erivative of csc θ is csc θ cot θ. In eercises 51 66, prove eac matematical statement about erivatives. 49. Te erivative of a ifferentiable function is unique. Hint: See te unique limit teorem (teorem from section 4.3). 50. Te two efinitions of te erivative are equivalent. Hint: Let = a. 51. Te constant rule from teorem Te ifference rule from teorem Te prouct rule from teorem Hint: A an subtract f ( + ) g() in te numerator of te ifference quotient for f () g(). 54. Te quotient rule from teorem Hint: See question Te cain rule from teorem Hint: See question Every polynomial is ifferentiable. 57. Te erivative of a polynomial of egree n is a polynomial of egree n Te erivative of an even function is o; tat is, if f () = f ( ), ten f () = f ( ). 59. If f is a ifferentiable function on an interval (, + ) for some R, ten te erivative f () equals lim 0 f ( + ). Hint: Apply L Hôpital s rule from calculus to te limit of te ifference quotient. 60. Appling te alternative efinition of te erivative at = 0, te following function is not ifferentiable at = 0. [ ] 1 sin if = 0 f () = 0 if = Te function f () efine as follows as erivative f (0) = 0. [ ] 1 2 sin if = 0 f () = 0 if = For every k R, te function f () efine as follows as erivative f (0) = 0. f () = { k 2 if Q 0 if Q 63. If a function f is ifferentiable on (b, c) an f (a) = 0 for a (b, c), ten it is not necessarily true tat eiter a relative maimum or relative minimum for f occurs at = a. 64. If f an g are ifferentiable functions on (a, b) wit te same erivative, ten f () g() is a constant for any (a, b).

13 Capter 4 Real Analysis If f an g are ifferentiable functions on (a, b) wit f g a constant, ten f an g ave te same erivative at any (a, b). 66. Define a function f tat is nowere ifferentiable, wile f 2 is everywere ifferentiable. Hint: Consier a variation on te caracteristic function of Q. Eercises evelop a proof of te cain rule in a fuller generality tan was iscusse in question Trougout tese eercises assume tat g() is ifferentiable at a point = a an tat f () is ifferentiable at g(a). 67. Prove tat te following function F is continuous at = 0; intuitively, we tink of F as te erivative of f wit respect to t = g(a). f [g(a) + ] f [g(a)] if = 0 f () = f [g(a)] if = Prove tat f [g(a) + ] = f [g(a)] + F() for sufficiently small values of by taking te limit of tese two epressions as approaces In a parallel way, we can efine a function G so tat G(0) = g (a) an g(a + k) = g(a) + k G(k) for sufficiently small values of k. Use tis fact, te result from eercise 68, an te coice of = g(a + k) g(a) = k G(k) to prove tat: f [g(a) + ] = f [g(a + k)] an F() = k G(k) F(k G(k)). 70. Using te two equations obtaine in eercise 69, substitute te first equation into te secon to prove tat f [g(a + k)] = f [g(a)] + k G(k) F(k G(k)). Te last term on te rigt is continuous at 0 base on te efinitions of F an G. Subtract f [g(a)] on bot sies of tis equation, ivie bot sies by k an take te limit as k approaces 0 to obtain te cain rule. 4.5 Unerstaning Infinity Te notion of infinity as been an important element in many cultures attempts to unerstan life: people refer to eternal time; an eternal spiritual afterlife; a bounless universe; an all-powerful eity. Matematics as a unique an important perspective on infinity; te insigts arising from matematics rigorous, logical approac to infinity ave a an important influence on Western society s view of te worl. But many avance matematical results on infinity (especially tose tat grew out of Georg Cantor s work in te late 1860s) are not wiely known. In tis section, we eplore a matematical unerstaning of te infinite. We ave alreay taken te first steps in tis irection in our stuy of limits. One major breaktroug in te evelopment of calculus is te arnessing of infinity in te very specific an powerful way epresse by te notion of limit to obtain te erivative (an te integral as iscusse in section 4.6). As matematicians evelope an refine teir unerstaning of limits, erivatives, an integrals in te eigteent an nineteent