The Derivative. Chapter 3

Transcription

1 59957_CH0a_-90q 9/5/09 7:7 PM Page Capter Te Derivative cos (0, ) e e In Tis Capter Te wor calculus is a iminutive form of te Latin wor cal, wic means stone In ancient civilizations small stones or pebbles were often use as a means of reckoning Consequentl, te wor calculus can refer to an sstematic meto of computation However, over te last several unre ears te connotation of te wor calculus as evolve to mean tat branc of matematics concerne wit te calculation an application of entities known as erivatives an integrals Tus, te subject known as calculus as been ivie into two rater broa but relate areas: ifferential calculus an integral calculus In tis capter we will begin our stu of ifferential calculus Te Derivative Power an Sum Rules Prouct an Quotient Rules Trigonometric Functions Cain Rule Implicit Differentiation Derivatives of Inverse Functions Eponential Functions Logaritmic Functions Hperbolic Functions Capter in Review Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

2 59957_CH0a_-90q 9/5/09 7:7 PM Page CHAPTER Te Derivative Te Derivative Introuction In te last section of Capter we saw tat te tangent line to a grap of a function f () is te line troug a point (a, f (a)) wit slope given b mtan lim Recall, mtan is also calle te slope of te curve at (a, f(a)) S0 f (a ) f (a) wenever te limit eists For man functions it is usuall possible to obtain a general formula tat gives te value of te slope of a tangent line Tis is accomplise b computing lim S0 f ( ) f () () for an (for wic te limit eists) We ten substitute a value of after te limit as been foun A Definition Te limit of te ifference quotient in () efines a function a function tat is erive from te original function f () Tis new function is calle te erivative function, or simpl te erivative, of f an is enote b f Definition Derivative Te erivative of a function f () at is given b f () lim S0 f ( ) f () () wenever te limit eists Let us now reconsier Eamples an of Section 7 A Derivative Fin te erivative of f () EXAMPLE Solution As in te calculation of mtan in Section 7, te process of fining te erivative f () consists of four steps: (i) f ( ) ( ) (ii) f ( ) f () [ ] ( ) f ( ) f () ( ) cancel s f ( ) f () (iv) lim lim [ ] S0 S 0 (iii) From step (iv) we see tat te erivative of f () is f () Observe tat te result mtan obtaine in Eample of Section 7 is obtaine b evaluating te erivative f () at, tat is, f () Value of te Derivative For f (), fin f ( ), f (0), f A B, an f () Interpret EXAMPLE Solution From Eample we know tat te erivative is f () Hence, at, at 0, e f ( ) 6 f ( ) f (0) e f (0) 0 point of tangenc is (, 6) slope of tangent line at (, 6) is m point of tangenc is (0, ) slope of tangent line at (0, ) is m 0 Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

3 59957_CH0a_-90q 9/5/09 7:7 PM Page Te Derivative at, e at, e f A B 9 f A B f () f () point of tangenc is (, 9 ) slope of tangent line at (, 9 ) is m point of tangenc is (, ) slope of tangent line at (, ) is m Recall tat a orizontal line as 0 slope So te fact tat f (0) 0 means tat te tangent line is orizontal at (0, ) B te wa, if ou trace back troug te four-step process in Eample, ou will fin tat te erivative of g() is also g () f () Tis makes intuitive sense; since te grap of f () is a rigi vertical translation or sift of te grap of g() for a given value of, te points of tangenc cange but not te slope of te tangent line at te points For eample, at, g () 6 f () but te points of tangenc are (, g()) (, 9) an (, f ()) (, ) A Derivative Fin te erivative of f () EXAMPLE Solution To calculate f ( ), we use te Binomial Teorem (i) f ( ) ( ) (ii) f ( ) f () [ ] ( ) (iii) Recall from algebra tat (a b) a ab ab b Now replace a b an b b f ( ) f () [ ] (iv) lim S0 f ( ) f () lim [ ] S0 Te erivative of f () is f () Tangent Line Fin an equation of te tangent line to te grap of f () at EXAMPLE Solution From Eample we ave two functions f () an f () As we saw in Eample, wen evaluate at te same number tese functions give ifferent information: fq RQ R 8 f Q R Q R slope of tangent at A, 8 B is Tus, b te point slope form of a line, an equation of te tangent line is given b Q R 8 or Te grap of te function an te tangent line are given in FIGURE A Derivative Fin te erivative of f () EXAMPLE 5 Solution In tis case ou soul be able to sow tat te ifference is f ( ) f () Terefore, point of tangenc is A, 8 B ( ) fractions b using a a common enominator f ( ) f () lim S0 S0 ( ) lim S0 ( ) lim Te erivative of f () is f () Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION 冢, 8冣 FIGURE Tangent line in Eample

4 59957_CH0a_-90q 0/6/09 5:6 PM Page CHAPTER Te Derivative Notation Te following is a list of some of te common notations use trougout matematical literature to enote te erivative of a function: f (),,, D, D For a function suc as f (), we write f () ; if te same function is written, we ten utilize Ⲑ,, or D We will use te first tree notations trougout tis tet Of course oter smbols are use in various applications Tus, if z t, ten z t t or z t Te Ⲑ notation as its origin in te erivative form of () of Section 7 Replacing b an enoting te ifference f ( ) f () b in (), te erivative is often efine as f ( ) f () lim lim S0 S0 () A Derivative Using () Use () to fin te erivative of EXAMPLE 6 Solution In te four-step proceure te important algebra takes place in te tir step: (i) f ( ) (ii) f ( ) f () (iii) f ( ) f () (iv) lim S0 ( ) ( ) of rationalization numerator lim S0 Te erivative of is Ⲑ >AB Value of a Derivative Te value of te erivative at a number a is enote b te smbols f (a), `, a (a), D ` a A Derivative From Eample 6, te value of te erivative of at, sa, 9 is written EXAMPLE 7 ` ` Alternativel, to avoi te clums vertical bar we can simpl write (9) 6 Differentiation Operators Te process of fining or calculating a erivative is calle ifferentiation Tus ifferentiation is an operation tat is performe on a function f () Te Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

5 59957_CH0a_-90q 9/5/09 7:7 PM Page 5 Te Derivative 5 operation of ifferentiation of a function wit respect to te variable is represente b te smbols an D Tese smbols are calle ifferentiation operators For instance, te results in Eamples,, an 6 can be epresse, in turn, as ( ),, Te smbol ten means Differentiabilit If te limit in () eists for a given number in te omain of f, te function f is sai to be ifferentiable at If a function f is ifferentiable at ever number in te open intervals (a, b), ( q, b), an (a, q ), ten f is ifferentiable on te open interval If f is ifferentiable on ( q, q ), ten f is sai to be ifferentiable everwere A function f is ifferentiable on a close interval [ a, b] wen f is ifferentiable on te open interval (a, b), an f (a) lim S0 f (a ) f (a) f (b ) f (b) f (b) lim S0 () bot eist Te limits in () are calle rigt-an an left-an erivatives, respectivel A function is ifferentiable on [a, q ) wen it is ifferentiable on (a, q ) an as a rigt-an erivative at a A similar efinition in terms of a left-an erivative ols for ifferentiabilit on ( q, b] Moreover, it can be sown: A function f is ifferentiable at a number c in an interval (a, b) if an onl if f (c) f (c) (5) Horizontal Tangents If f () is continuous at a number a an f (a) 0, ten te tangent line at (a, f (a)) is orizontal In Eamples an we saw tat te value of erivative f () of te function f () at 0 is f (0) 0 Tus, te tangent line to te grap is orizontal at (0, f (0)) or (0, 0) It is left as an eercise (see Problem 7 in Eercises ) to verif b Definition tat te erivative of te continuous function f () is f () Observe in tis latter case tat f () 0 wen 0 or Tere is a orizontal tangent at te point (, f ()) (, 5) Were f Fails to be Differentiable A function f fails to ave a erivative at a if (i) te function f is iscontinuous at a, or (ii) te grap of f as a corner at (a, f (a)) In aition, since te erivative gives slope, f will fail to be ifferentiable (iii) at a point (a, f (a)) at wic te tangent line to te grap is vertical Te omain of te erivative f, efine b (), is te set of numbers for wic te limit eists Tus te omain of f is necessaril a subset of te omain of f Differentiabilit (a) Te function f () is ifferentiable for all real numbers, tat is, te omain of f () is ( q, q ) (b) Because f () > is iscontinuous at 0, f is not ifferentiable at 0 an consequentl not ifferentiable on an interval containing 0 EXAMPLE 8 Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

6 59957_CH0a_-90q 6 9/5/09 7:7 PM Page 6 CHAPTER Te Derivative (a) Absolute-value function ƒ Eample 7 of Section 7 Revisite In Eample 7 of Section 7 we saw tat te grap of f () 0 0 possesses no tangent at te origin (0, 0) Tus f () 0 0 is not ifferentiable at 0 But f () 0 0 is ifferentiable on te open intervals (0, q ) an ( q, 0) In Eample 5 of Section 7, we prove tat te erivative of a linear function f () m b is f () m Hence, for 7 0 we ave f () 0 0 an so f () Also, for 6 0, f () 0 0 an so f () Since te erivative of f is a piecewise-efine function, EXAMPLE 9 ƒ() ƒ (), > 0 f () e,, , we can grap it as we woul an function We see in at 0 ƒ (), < 0 FIGURE (b) tat f is iscontinuous In ifferent smbols, wat we ave sown in Eample 9 is tat f (0) an f (0) Since f (0) f (0) it follows from (5) tat f is not ifferentiable at 0 (b) Grap of te erivative ƒ FIGURE Graps of f an f in Eample 9 Vertical Tangents Let f () be continuous at a number a If lim 0 f () 0 q, ten te Sa grap of f is sai to ave a vertical tangent at (a, f (a)) Te graps of man functions wit rational eponents possess vertical tangents In Eample 6 of Section 7 we mentione tat te grap of > possesses a vertical tangent line at (0, 0) We verif tis assertion in te net eample Vertical Tangent It is left as an eercise to prove tat te erivative of f () > is given b EXAMPLE 0 f () / (See Problem 55 in Eercises ) Altoug f is continuous at 0, it is clear tat f is not efine at tat number In oter wors, f is not ifferentiable at 0 Moreover, because lim f () q S0 FIGURE Tangent lines to te grap of te function in Eample 0 an lim f () q S 0 we ave 0 f () 0 S q as S 0 Tis is sufficient to sa tat tere is a tangent line at (0, f (0)) or (0, 0) an tat it is vertical FIGURE sows tat te tangent lines to te grap on eiter sie of te origin become steeper an steeper as S 0 Te grap of a function f can also ave a vertical tangent at a point (a, f (a)) if f is ifferentiable onl on one sie of a, is continuous from te left (rigt) at a, an eiter 0 f () 0 S q as S a or 0 f () 0 S q as S a -ais is tangent to te grap at (0, 0) > One-Sie Vertical Tangent Te function f () is not ifferentiable on te interval [0, q ) because it is seen from te erivative f () >A B tat f (0) oes not eist Te function f () is continuous on [0, q ) but ifferentiable on (0, q ) In aition, because f is continuous at 0 an lim f () q, tere is a one-sie vertical tangent at te origin (0, 0) We see in FIGURE S0 tat te vertical tangent is te -ais EXAMPLE FIGURE Vertical tangent in Eample Important Te functions f () 0 0 an f () > are continuous everwere In particular, bot functions are continuous at 0 but neiter are ifferentiable at tat number In oter wors, continuit at a number a is not sufficient to guarantee tat a function is ifferentiable at a However, if a function f is ifferentiable at a, ten f must be continuous at tat number We summarize tis last fact in te net teorem Teorem Differentiabilit Implies Continuit If f is ifferentiable at a number a, ten f is continuous at a Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

7 59957_CH0a_-90q 9/5/09 7:7 PM Page 7 Te Derivative 7 PROOF To prove continuit of f at a number a it is sufficient to prove tat lim f () f (a) Sa or equivalentl tat lim [ f () f (a)] 0 Te potesis is tat Sa f (a) lim S0 f (a ) f (a) eists If we let a, ten as S 0 we ave S a Tus, te foregoing limit is equivalent to f (a) lim Sa f () f (a) a Ten we can write f () f (a) ( a) Sa a f () f (a) lim lim ( a) Sa Sa a f (a) 0 0 lim [ f () f (a)] lim Sa multiplication b a a bot limits eist Postscript A Bit of Histor It is acknowlege tat Isaac Newton (6 77), an Englis matematician an psicist, was te first to set fort man of te basic principles of calculus in unpublise manuscripts on te meto of fluions, ate 665 Te wor fluion originate from te concept of quantities tat flow tat is, quanti ties tat cange at a certain rate Newton use te ot notation to represent a fluion, or as we now know it: te erivative of a function Te smbol never acieve overwelming popularit among matematicians an is use toa primaril b psicists For tpograpical reasons, te so-calle fl-speck Newton notation as been supersee b te prime notation Newton attaine everlasting fame wit te publication of is law of universal gravitation in is monumental treatise Pilosopiae Naturalis Principia Matematica in 687 Newton was also te first to prove, using te calculus an is law of gravitation, Joannes Kepler s tree empirical laws of planetar motion an was te first to prove tat wite ligt is compose of all colors Newton was electe to Parliament, was appointe waren of te Roal Mint, an was knigte in 705 Sir Isaac Newton sai about is man accomplisments: If I ave seen farter tan oters, it is b staning on te soulers of giants Te German matematician, lawer, an pilosoper Gottfrie Wilelm Leibniz (66 76) publise a sort version of is calculus in an article in a perioical journal in 68 Te > notation for a erivative of a function is ue to Leibniz In fact, it was Leibniz wo introuce te wor function into matematical literature But, since it was well known tat Newton s manuscripts on te meto of fluions ate from 665, Leibniz was accuse of approprileibniz ating is ieas from tese unpublise works Fuele b nationalistic pries, a controvers about wo was te first to invent calculus rage for man ears Historians now agree tat bot Leibniz an Newton arrive at man of te major premises of calculus inepenent of eac oter Leibniz an Newton are consiere te co-inventors of te subject NOTES FROM THE CLASSROOM (i) In te preceing iscussion, we saw tat te erivative of a function is itself a function tat gives te slope of a tangent line Te erivative is, owever, not an equation of a tangent line Also, to sa tat 0 f () ( 0) is an equation of te tangent at (0, 0) is incorrect Remember tat f () must be evaluate at 0 before it is use in te point slope form If f is ifferentiable at 0, ten an equation of te tangent line at (0, 0) is 0 f (0) ( 0) Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

8 59957_CH0a_-90q 8 9/5/09 7:7 PM Page 8 CHAPTER Te Derivative (ii) Altoug we ave empasize slopes in tis section, o not forget te iscussion on average rates of cange an instantaneous rates of cange in Section 7 Te erivative f () is also te instantaneous rate of cange of te function f () wit respect to te variable More will be sai about rates in subsequent sections (iii) Matematicians from te seventeent to te nineteent centuries believe tat a continuous function usuall possesse a erivative (We ave note eceptions in tis section) In 87 te German matematician Karl Weierstrass conclusivel estroe tis tenet b publising an eample of a function tat was everwere continuous but nowere ifferentiable Eercises Answers to selecte o-numbere problems begin on page ANS-000 Funamentals In Problems 0, use () of Definition to fin te erivative of te given function f () 0 f () f () 5 f () p 5 f () 6 f () 7 f () 9 ( ) f () f () 6 7 f () ( 5) f () f () 8 f () f () In Problems, use () of Definition to fin te erivative of te given function Fin an equation of te tangent line to te grap of te function at te inicate value of f () 7; f () ; 0 6 ; ; In Problems 5 8, use () of Definition to fin te erivative of te given function Fin point(s) on te grap of te given function were te tangent line is orizontal 5 f () f () 6 f () ( 5) 8 f () In Problems 9, use () of Definition to fin te erivative of te given function Fin point(s) on te grap of te given function were te tangent line is parallel to te given line 9 f () ; 0 f () ; 0 f () ; f () 6 ; In Problems an, sow tat te given function is not ifferentiable at te inicate value of, ; f () e, 7 f () e 6 0 ; 0 0,, In te proof of Teorem we saw tat an alternative formulation of te erivative of a function f at a is given b f (a) lim Sa f () f (a), a (6) wenever te limit eists In Problems 5 0, use (6) to compute f (a) 5 f () 0 6 f () 7 f () 9 f () 8 f () 0 f () Fin an equation of te tangent line sown in re in FIGURE 5 Wat are f ( ) an f ( )? ƒ() FIGURE 5 Grap for Problem Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

9 59957_CH0a_-90q 9/5/09 7:7 PM Page 9 Te Derivative 9 Fin an equation of te tangent line sown in re in FIGURE 6 Wat is f ()? Wat is te -intercept of te tangent line? (c) () ƒ () ƒ () 9, 0 ƒ() (e) (f) ƒ () FIGURE 6 Grap for Problem ƒ () In Problems 8, sketc te grap of f from te grap of f ƒ() (, ) ƒ() 9 50 ƒ() ƒ() FIGURE 8 Grap for Problem FIGURE 7 Grap for Problem 5 6 ƒ() ƒ() 5 5 FIGURE 9 Grap for Problem 5 FIGURE Grap for Problem a b 5 FIGURE 5 Grap for Problem 5 5 ƒ() a ƒ() ƒ() FIGURE Grap for Problem 7 8 FIGURE 6 Grap for Problem 5 a 5 ƒ() FIGURE 0 Grap for Problem 6 7 FIGURE Grap for Problem 50 ƒ() FIGURE 7 Grap for Problem 5 (, ) FIGURE 8 Grap for Problem 5 ƒ() Tink About It (, ) FIGURE Grap for Problem 8 In Problems 9 5, matc te grap of f wit a grap of f from (a) (f) ƒ () (a) (b) ƒ () 55 Use te alternative efinition of te erivative (6) to fin te erivative of f () > [Hint: Note tat a ( >) (a >) ] 56 In Eamples 0 an, we saw, respectivel, tat te functions f () > an f () possesse vertical tangents at te origin (0, 0) Conjecture were te graps of ( ) > an ma ave vertical tangents 57 Suppose f is ifferentiable everwere an as te tree properties: (i) f ( ) f () f (), (ii) f (0), (iii) f (0) Use () of Definition to sow tat f () f () for all Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

10 59957_CH0a_-90q 0 9/5/09 7:7 PM Page 0 CHAPTER Te Derivative 58 (a) Suppose f is an even ifferentiable function on ( q, q ) On geometric grouns, eplain w f ( ) f (); tat is, f is an o function (b) Suppose f is an o ifferentiable function on ( q, q ) On geometric grouns, eplain w f ( ) f (); tat is, f is an even function 59 Suppose f is a ifferentiable function on [ a, b] suc tat f (a) 0 an f (b) 0 B eperimenting wit graps iscern weter te following statement is true or false: Tere is some number c in (a, b) suc tat f (c) 0 60 Sketc graps of various functions f tat ave te propert f () 7 0 for all in [a, b] Wat o tese functions ave in common? Calculator/CAS Problem 6 Consier te function f () n 0 0, were n is a positive integer Use a calculator or CAS to obtain te grap of f for n,,,, an 5 Ten use () to sow tat f is not ifferentiable at 0 for n,,,, an 5 Can ou prove tis for an positive integer n? Wat is f (0) an f (0) for n 7? Power an Sum Rules Introuction Te efinition of a erivative f () lim S0 f ( ) f () () as te obvious rawback of being rater clums an tiresome to appl To fin te erivative of te polnomial function f () using te above efinition we woul onl ave to juggle 7 terms in te binomial epansions of ( )00 an ( )5 Tere are more efficient was of computing erivatives of a function tan using te efinition eac time In tis section, an te sections tat follow, we will see tat tere are sortcuts or general rules wereb erivatives of functions suc as f () can be obtaine, literall, wit just a flick of a pencil In te last section we saw tat te erivatives of te power functions f (), f (), f (), f () > were, in turn, See Eamples, 5, an 6 in Section f (), f (), f (), f () > If te rigt-an sies of tese four erivatives are written,, ( ),, we observe tat eac coefficient (inicate in re) correspons wit te original eponent of in f an te new eponent of in f can be obtaine from te ol eponent (also inicate in re) b subtracting from it In oter wors, te pattern for te erivative of te general power function f () n appears to be bring own eponent as a multiple ( T) (c) ecrease eponent b () Derivative of te Power Function Te pattern illustrate in () oes inee ol for an real-number eponent n, an we will state it as a formal teorem, but at tis point in te course we o not possess te necessar matematical tools to prove its complete valiit We can, owever, reail prove a special case of tis power rule; te remaining parts of te proof will be given in te appropriate sections aea Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

11 59957_CH0a_-90q 9/5/09 7:8 PM Page Power an Sum Rules Teorem Power Rule For an real number n, n n n () PROOF We present te proof onl in te case wen n is a positive integer To compute () for f () n we use te four-step meto: general Binomial Teorem (i) f ( ) ( ) n n n n (ii) f ( ) f () n n n n(n ) n n n n! See te Resource Pages for a review of te Binomial Teorem n(n ) n n n n n! n(n ) n n n n! n(n ) n n n n c n n! n n (iii) f ( ) f () c n n n n n(n ) n n n n! n(n ) n n n n! f ( ) f () S0 n(n ) n lim c n n n n n n n S0! (iv) f () lim tese terms S 0 as S 0 EXAMPLE Power Rule Differentiate (a) 7 (b) (c) > () Solution B te Power Rule (), , 0, (b) wit n : Q R ( >) 5> 5>, (c) wit n : () wit n : (a) wit n 7: m Observe in part (b) of Eample tat te result is consistent wit te fact tat te slope of te line is m See FIGURE Teorem Constant Function Rule If f () c is a constant function, ten f () 0 () Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION FIGURE Slope of line m is consistent wit >

12 59957_CH0a_-90q 9/5/09 7:8 PM Page CHAPTER Te Derivative ƒ() c (, c) (, c) PROOF If f () c were c is an real number, ten it follows tat te ifference is f ( ) f () c c 0 Hence from (), f () lim S 0 FIGURE Slope of a orizontal line is 0 c c lim 0 0 S 0 Teorem as an obvious geometric interpretation As sown in FIGURE, te slope of te orizontal line c is, of course, zero Moreover, Teorem agrees wit () in te case wen 0 an n 0 Teorem Constant Multiple Rule If c is an constant an f is ifferentiable at, ten cf is ifferentiable at, an c f () c f () (5) PROOF Let G() c f () Ten G () lim S0 G( ) G() c f ( ) c f () lim S0 lim c c S0 c lim S0 f ( ) f () f ( ) f () c f () EXAMPLE A Constant Multiple Differentiate 5 Solution From () an (5), 5 5( ) 0 Teorem Sum an Difference Rules If f an g are functions ifferentiable at, ten f g an f g are ifferentiable at, an [ f () g()] f () g (), (6) [ f () g()] f () g () (7) PROOF OF (6) Let G() f () g() Ten G () lim S0 G( ) G() [ f ( ) g( )] [ f () g()] lim S0 lim S0 since limits eist, limit of a sum is S te sum of te limits f ( ) f () g( ) g() regrouping terms f ( ) f () g( ) g() lim S0 S0 lim f () g () Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

13 59957_CH0a_-90q 9/5/09 7:8 PM Page Power an Sum Rules Teorem ols for an finite sum of ifferentiable functions For eample, if f, g, an are functions tat are ifferentiable at, ten [ f () g() ()] f () g () () Since f g can be written as a sum, f ( g), tere is no nee to prove (7) since te result follows from a combination of (6) an (5) Hence, we can epress Teorem in wors as: Te erivative of a sum is te sum of te erivatives Derivative of a Polnomial Because we now know ow to ifferentiate powers of an constant multiples of tose powers we can easil ifferentiate sums of tose constant multiples Te erivative of a polnomial function is particularl eas to obtain For eample, te erivative of te polnomial function f () , mentione in te introuction to tis section, is now reail seen to be f () EXAMPLE Polnomial wit Si Terms Differentiate Solution Using (), (5), an (6), we obtain Since 6 0 b (), we obtain (5 ) ( ) 9( ) 0() () Tangent Line Fin an equation of a tangent line to te grap of f () 7 at te point corresponing to EXAMPLE Solution From te Sum Rule, f () ( ) ( ) 7() 6 7 Wen evaluate at te same number te functions f an f give: f ( ) 8 f ( ) point of tangenc is (, 8) slope of tangent at (, 8) is Te point slope form gives an equation of te tangent line 8 ( ( )) or 5 Rewriting a Function In some circumstances, in orer to appl a rule of ifferentiation efficientl it ma be necessar to rewrite an epression in an alternative form Tis alternative form is often te result of some algebraic manipulation or an application of te laws of eponents For eample, we can use () to ifferentiate te following epressions, but first we rewrite tem using te laws of eponents, 0, S rewrite square roots as powers S ten rewrite using negative eponents S 0,, ( )>, >, 0 >, >, te erivative of eac term using () S 8, 5 >, > Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION Tis iscussion is wort remembering

14 59957_CH0a_-90q 9/5/09 7:8 PM Page CHAPTER Te Derivative A function suc as f () (5 )> can be rewritten as two fractions f () From te last form of f it is now apparent tat te erivative f is f () 5( ) ( ) EXAMPLE 5 5 Rewriting te Terms of a Function Differentiate Solution Before ifferentiating we rewrite te first tree terms as powers of : > 8 6 > 0 Ten > 8 6 > 0 B te Power Rule () an (), we obtain > 8 ( ) 6 Q R > 0 8 > (, 6) 6 5 Horizontal Tangents Fin te points on te grap of f () were te tangent line is orizontal EXAMPLE 6 (0, ) FIGURE Grap of function in Eample 6 Solution At a point (, f ()) on te grap of f were te tangent is orizontal we must ave f () 0 Te erivative of f is f () 6 an te solutions of f () 6 0 or ( ) 0 are 0 an Te corresponing points are ten (0, f (0)) (0, ) an (, f ()) (, 6) See FIGURE Normal Line gent line at P tangent normal FIGURE Normal line in Eample 7 EXAMPLE 7 Equation of a Normal Line Fin an equation of te normal line to te grap of at Solution Since >, we know tat m tan at (, ) Tus te slope of te normal line sown in green in FIGURE is te negative reciprocal of te slope of te tangent line, tat is, m B te point-slope form of a line, an equation of te normal line is ten (, ) A normal line at a point P on a grap is one tat is perpenicular to te tan- / ( ) FIGURE 5 Grap of function in Eample 8 EXAMPLE 8 Vertical Tangent For te power function f () > te erivative is f () or > > Observe tat lim f () q wereas lim f () q Since f is continuous at 0 an S0 S0 冟 f ()冟 S q as S 0, we conclue tat te -ais is a vertical tangent at (0, 0) Tis fact is apparent from te grap in FIGURE 5 Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

15 59957_CH0a_-90q 9/5/09 7:9 PM Page 5 Power an Sum Rules 5 Cusp Te grap of f () > in Eample 8 is sai to ave a cusp at te origin In general, te grap of a function f () as a cusp at a point (a, f (a)) if f is continuous at a, f () as opposite signs on eiter sie of a, an 冟 f ()冟 S q as S a Higer-Orer Derivatives We ave seen tat te erivative f () is a function erive from f () B ifferentiation of te first erivative, we obtain et anoter function calle te secon erivative, wic is enote b f () In terms of te operation smbol >, we efine te secon erivative wit respect to as te function obtaine b ifferentiating f () twice in succession: Q R Te secon erivative is commonl enote b te smbols f (),, EXAMPLE 9 f (), D,, D Secon Derivative Fin te secon erivative of Solution We first simplif te function b rewriting it as Ten b te Power Rule (), we ave Te secon erivative follows from ifferentiating te first erivative ( ) ( 5) 5 5 Assuming tat all erivatives eist, we can ifferentiate a function f () as man times as we want Te tir erivative is te erivative of te secon erivative; te fourt erivative is te erivative of te tir erivative; an so on We enote te tir an fourt erivatives b > an > an efine tem b Q R an Q R In general, if n is a positive integer, ten te nt erivative is efine b n n a n b n Oter notations for te first n erivatives are f (), f (), f (), f ()(), p, f (n)(),,,, (), p, (n), f (), f (), f (), f (), p, n f (), n D, D, D, D, p, D n, D, D, D, D, p, D n Note tat te prime notation is use to enote onl te first tree erivatives; after tat we use te superscript (), (5), an so on Te value of te nt erivative of a function f () at a number a is enote b f (n)(a), (n)(a), an n ` n a Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

16 59957_CH0a_-90q 6 0/6/09 5:6 PM Page 6 CHAPTER Te Derivative Fift Derivative Fin te first five erivatives of f () EXAMPLE 0 Solution We ave f () f () 6 f () 8 6 f ()() 8 f (5)() 0 After reflecting a moment, ou soul be convince tat te (n )st erivative of an nt-egree polnomial function is zero NOTES FROM THE CLASSROOM (i) In te ifferent contets of science, engineering, an business, functions are often epresse in variables oter tan an Corresponingl we must aapt te erivative notation to te new smbols For eample, Function Derivative (t) t t A A (r) pr r r r (u) 8u u D D (p) 9 p p (t) A(r) pr r(u) u u D(p) 800 9p p (ii) You ma be wonering wat interpretation can be given to te iger-orer erivatives If we tink in terms of graps, ten f gives te slope of tangent lines to te grap of te function f ; f gives te slope of te tangent lines to te grap of f, an so on In aition, if f is ifferentiable, ten te first-erivative f gives te instantaneous rate of cange of f Similarl, if f is ifferentiable, ten f gives te instantaneous rate of cange of f Eercises Answers to selecte o-numbere problems begin on page ANS-000 Funamentals In Problems 8, fin > p In Problems 9 6, fin f () Simplif 9 f () f () f () ( 5 6) 5 f () Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

17 59957_CH0a_-90q 0/6/09 5:6 PM Page 7 Power an Sum Rules 7 f () ( 5) 5 f () A B f () ( ) 6 f () (9 )(9 ) In Problems 7 0, fin te erivative of te given function 7 (u) (u) 8 p(t) (t) (t ) t 5 t 9 g(r) 0 Q(t) r 6 r r r In Problems, fin an equation of te tangent line to te grap of te given function at te inicate value of ; f () ; 8 ; f () 6 ; In Problems 5 8, fin te point(s) on te grap of te given function at wic te tangent line is orizontal f () 9 8 f () In Problems 9, fin an equation of te normal line to te grap of te given function at te inicate value of 9 ; 0 ; f () ; f () ; In Problems 8, fin te secon erivative of te given function ( 9) f () 0 8 f () Q R In Problems 9 an 0, fin te inicate iger erivative 9 f () 6 5 ; f ()() 0 0 ; 5> 5 In Problems an, etermine intervals for wic f () 7 0 an intervals for wic f () 6 0 f () 8 f () 9 In Problems an, fin te point(s) on te grap of f at wic f () 0 f () 0 f () In Problems 5 an 6, etermine intervals for wic f () 7 0 an intervals for wic f () f () ( ) 6 f () 50 Fin te point on te grap of f () at wic te tangent line is Fin te point on te grap of f () at wic te slope of te normal line is 5 Fin te point on te grap of f () at wic te tangent line is parallel to te line 0 5 Fin an equation of te tangent line to te grap of at te point were te value of te secon erivative is zero 5 Fin an equation of te tangent line to te grap of at te point were te value of te tir erivative is Applications 55 Te volume V of a spere of raius r is V pr Fin te surface area S of te spere if S is te instantaneous rate of cange of te volume wit respect to te raius 56 Accoring to te Frenc psician Jean Louis Poiseuille ( ) te velocit of bloo in an arter wit a constant circular cross-section raius R is (r) (P Ⲑnl)(R r ), were P, n, an l are constants Wat is te velocit of bloo at te value of r for wic (r) 0? 57 Te potential energ of a spring-mass sstem wen te spring is stretce a istance of units is U() k, were k is te spring constant Te force eerte on te mass is F UⲐ Fin te force if te spring constant is 0 N/m an te amount of stretc is m 58 Te eigt s above groun of a projectile at time t is given b gt 0 t s0, were g, 0, an s0 are constants Fin te instantaneous rate of cange of s wit respect to t at t s(t) Tink About It In Problems 59 an 60, te smbol n represents a positive integer Fin a formula for te given erivative n n n 60 n n 6 From te graps of f an g in FIGURE 6, etermine wic function is te erivative of te oter Eplain our coice in wors 59 An equation containing one or more erivatives of an unknown function () is calle a ifferential equation In Problems 7 an 8, sow tat te function satisfies te given ifferential equation 7 ; 0 8 ; 9 Fin te point on te grap of f () 6 at wic te slope of te tangent line is 5 g () ƒ() FIGURE 6 Graps for Problem 6 Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

18 59957_CH0a_-90q 8 9/5/09 7:9 PM Page 8 CHAPTER Te Derivative 6 From te grap of te function f () given in FIGURE 7, sketc te grap of f 75 Fin conitions on te coefficients a, b, an c so tat te grap of te polnomial function f () a b c ƒ() FIGURE 7 Grap for Problem 6 6 Fin a quaratic function f () a b c suc tat f ( ), f ( ) 7, an f ( ) 6 Te graps of f () an g() are sai to be ortogonal if te tangent lines to eac grap are perpenicular at eac point of intersection Sow tat te graps of 8 an are ortogonal 65 Fin te values of b an c so tat te grap of f () b possesses te tangent line c at 66 Fin an equation of te line(s) tat passes troug (, ) an is tangent to te grap of f () 67 Fin te point(s) on te grap of f () 5 suc tat te tangent line at te point(s) as -intercept (, 0) 68 Fin te point(s) on te grap of f () suc tat te tangent line at te point(s) as -intercept (0, ) 69 Eplain w te grap of f () 5 5 as no tangent line wit slope 70 Fin coefficients A an B so tat te function A B satisfies te ifferential equation 7 Fin values of a an b suc tat te slope of te tangent to te grap of f () a b at (, ) is 5 7 Fin te slopes of all te normal lines to te grap of f () tat pass troug te point (, ) [Hint: Draw a figure an note tat at (, ) tere is onl one normal line] 7 Fin a point on te grap of f () an a point on te grap of g() at wic te tangent lines are parallel 7 Fin a point on te grap of f () 5 5 at wic te tangent as te least possible slope as eactl one orizontal tangent Eactl two orizontal tangents No orizontal tangents 76 Let f be a ifferentiable function If f () 7 0 for all in te interval (a, b), sketc possible graps of f on te interval Describe in wors te beavior of te grap of f on te interval Repeat if f () 6 0 for all in te interval (a, b) 77 Suppose f is a ifferentiable function suc tat f () f () 0 Fin f (00)() 78 Te graps of an given in FIGURE 8 sow tat tere are two lines L an L tat are simultaneousl tangent to bot graps Fin te points of tangenc on bot graps Fin an equation of eac tangent line L L FIGURE 8 Graps for Problem 78 Calculator/CAS Problems 79 (a) Use a calculator or CAS to obtain te grap of f () (b) Evaluate f () at,, 0,,,, an (c) From te ata in part (b), o ou see an relationsip between te sape of te grap of f an te algebraic signs of f? 80 Use a calculator or CAS to obtain te grap of te given functions B inspection of te graps inicate were eac function ma not be ifferentiable Fin f () at all points were f is ifferentiable (a) f () 0 0 (b) f () 0 0 Prouct an Quotient Rules Introuction in turn: So far we know tat te erivative of a constant function an a power of are, c0 an n n n Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION ()

19 59957_CH0a_-90q 9/5/09 7:50 PM Page 9 Prouct an Quotient Rules 9 We also know tat for ifferentiable functions f an g: cf () c f () an [ f () g()] f () g () () Altoug te results in () an () allow us to quickl ifferentiate man algebraic functions (suc as polnomials) neiter () nor () are of immeiate elp in fining erivatives of functions suc as or >( ) We nee aitional rules for ifferentiating proucts fg an quotients f>g Prouct Rule Te rules of ifferentiation an te erivatives of functions ultimatel stem from te efinition of te erivative Te Sum Rule in (), erive in te preceing section, follows from tis efinition an te fact tat te limit of a sum is te sum of te limits wenever te limits eist We also know tat wen te limits eist, te limit of a prouct is te prouct of te limits Arguing b analog, it woul ten seem plausible tat te erivative of a prouct of two functions is te prouct of te erivatives Regrettabl, te Prouct Rule state net is not tat simple Teorem Prouct Rule If f an g are functions ifferentiable at, ten fg is ifferentiable at, an [ f ()g()] f ()g () g()f () () PROOF Let G() f ()g() Ten b te efinition of te erivative along wit some algebraic manipulation: G () lim S0 G( ) G() f ( )g( ) f ()g() lim S0 zero f ( )g( ) f ( )g() f ( )g() f ()g() g( ) g() f ( ) f () lim c f ( ) g() S0 g( ) g() f ( ) f () lim f ( ) lim lim g() lim S0 S0 S0 S0 lim S0 Because f is ifferentiable at, it is continuous tere an so lim f ( ) f () Furtermore, S0 lim g() g() Hence te last equation becomes S0 G () f ()g () g()f () Te Prouct Rule is best memorize in wors: Te first function times te erivative of te secon plus te secon function times te erivative of te first Prouct Rule Differentiate ( )(7 ) EXAMPLE erivative of secon secon erivative of first first Solution From te Prouct Rule (), ( ) (7 ) (7 ) ( ) ( )( ) (7 )( ) 5 7 Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

20 59957_CH0a_-90q 7:50 PM Page 0 CHAPTER Te Derivative Alternative Solution Te two terms in te given function coul be multiplie out to obtain a fift-egree polnomial Te erivative can ten be gotten using te Sum Rule Tangent Line Fin an equation of te tangent line to te grap of ( )( ) at EXAMPLE Solution Before taking te erivative we rewrite as > Ten from te Prouct Rule () ( >) ( ) ( ) ( >) ( >) ( ) > Evaluating te given function an its erivative at gives: () A B( ) 6 7 ` point of tangenc is (, 6) slope of te tangent at (, 6) is 7 B te point slope form, te tangent line is 7 6 ( ) or 7 8 Altoug () is state for onl te prouct of two functions, it can be applie to functions wit a greater number of factors Te iea is to group two (or more) functions an treat tis grouping as one function Te net eample illustrates te tecnique EXAMPLE Prouct of Tree Functions Differentiate ( )( )( 8) Solution We ientif te first two factors as te first function : secon erivative of first erivative of secon first ( )( ) ( 8) ( 8) ( )( ) Notice tat to fin te erivative of te first function, we must appl te Prouct Rule a secon time: Prouct Rule again ( )( ) ( 8) ( 8) [( )( ) ( ) ] ( )( )( 8) ( 8)(6 ) ( 8)( ) 0 9/5/09 Quotient Rule Teorem Te erivative of te quotient of two functions f an g is given net Quotient Rule If f an g are functions ifferentiable at an g() 0, ten f>g is ifferentiable at, an g()f () f ()g () f () c g() [g()] Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION ()

21 59957_CH0a_-90q 9/5/09 7:50 PM Page Prouct an Quotient Rules PROOF Let G() f ()>g() Ten f ( ) f () G( ) G() g( ) g() G () lim lim S0 S0 lim S0 g()f ( ) f ()g( ) g( )g() zero lim S0 g()f ( ) g()f () g()f () f ()g( ) g( )g() g() lim S0 f ( ) f () g( ) g() f () g( )g() f ( ) f () g( ) g() lim f () lim S0 S0 S0 lim g( ) lim g() lim g() lim S0 S0 S0 Since all limits are assume to eist, te last line is te same as G () g()f () f ()g () [g()] In wors, te Quotient Rule starts wit te enominator: Te enominator times te erivative of te numerator minus te numerator times te erivative of te enominator all ivie b te enominator square Quotient Rule Differentiate 5 7 EXAMPLE enominator erivative of numerator numerator erivative of enominator Solution From te Quotient Rule (), ( 5 7) ( ) ( ) ( 5 7) ( 5 7) square of enominator EXAMPLE 5 ( 5 7) 6 ( ) (6 0) multipl out numerator ( 5 7) ( 5 7) Quotient an Prouct Rule Fin te points on te grap of ( )( ) were te tangent line is orizontal Solution We begin wit te Quotient Rule an ten use te Prouct Rule wen ifferentiating te numerator: Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

22 59957_CH0a_-90q 9/5/09 7:5 PM Page CHAPTER Te Derivative Prouct Rule ere ( ) [( )( )] ( )( ) ( ) ( ) ( )[( ) ( ) ] ( )( )6 multipl out numerator ( ) 8 ( ) 5 At a point were te tangent line is orizontal we must ave > 0 Te erivative just foun can onl be 0 wen te numerator satisfies Of course, values of tat make te numerator zero must not simultaneousl make te enominator zero or ( 8) 0 (5) In (5) because 8 0 for all real numbers, we must ave 0 Substituting tis number into te function gives (0) Te tangent line is orizontal at te -intercept (0, ) Postscript Power Rule Revisite Remember in Section we state tat te Power Rule, (>) n n n, is vali for all real number eponents n We are now in a position to prove te rule wen te eponent is a negative integer m Since, b efinition, m > m, were m is a positive integer, we can obtain te erivative of m b te Quotient Rule an te laws of eponents: subtract eponents m Q mr m T m m m m m m ( m) NOTES FROM THE CLASSROOM (i) Te Prouct an Quotient Rules will usuall lea to epressions tat eman simplification If our answer to a problem oes not look like te one in te tet answer section, ou ma not ave performe sufficient simplifications Do not be content to simpl carr troug te mecanics of te various rules of ifferentiation; it is alwas a goo iea to practice our algebraic skills (ii) Te Quotient Rule is sometimes use wen it is not require Altoug we coul use te Quotient Rule to ifferentiate functions suc as 5 6 an 0, it is simpler (an faster) to rewrite te functions as 6 5 an 0 an ten use te Constant Multiple an Power Rules: Eercises an 0 0 Answers to selecte o-numbere problems begin on page ANS-000 Funamentals In Problems 0, fin > ( 7)( ) (7 )( 9) 6 a b a b a b a b Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

23 59957_CH0a_-90q 9/5/09 7:5 PM Page Prouct an Quotient Rules (6 ) ( 5) 5 6 f () ( ) a 0 b f () f () 7 F() g() f () 9 F() a f ()b g() In Problems 0, fin f () f () a b ( 5 ) In Problems 5 0, f an g are ifferentiable functions Fin F () if f (), f (), an g() 6, g () 5 F() f ()g() 6 F() f ()g() 0 ( ) 5 f () ( )( )( ) 6 f () ( )( )( ) ( )( 5) 5 7 f () 8 f () ( )( ) 9 f () ( ) a b 0 f () ( ) a b In Problems, fin an equation of te tangent line to te grap of te given function at te inicate value of 5 ; ; ( )( 5 ); ( )( 5 ); 0 In Problems 5 8, fin te point(s) on te grap of te given function at wic te tangent line is orizontal 5 ( )( 6) 6 ( ) In Problems 9 an 0, fin te point(s) on te grap of te given function at wic te tangent line as te inicate slope ; m ( )( 5); m In Problems an, fin te point(s) on te grap of te given function at wic te tangent line as te inicate propert ; perpenicular to 5 ; parallel to Fin te value of k suc tat te tangent line to te grap of f () (k )> as slope 5 at Sow tat te tangent to te grap of f () ( )>( 9) at is perpenicular to te tangent to te grap of g() ( )( ) at 8 F() f () g() 0 F() f () g() Suppose F() f (), were f is a ifferentiable function Fin F () if f () 6, f (), an f () Suppose F() f () g(), were f an g are ifferentiable functions Fin F (0) if f (0) an g (0) 6 Suppose F() f ()>, were f is a ifferentiable function Fin F () Suppose F() f (), were f is a ifferentiable function Fin F () In Problems 5 8, etermine intervals for wic f () 7 0 an intervals for wic f () f () 6 f () 7 f () ( 6)( 7) 8 f () ( )( 8 ) Applications 9 Te Law of Universal Gravitation states tat te force F between two boies of masses m an m separate b a istance r is F kmm>r, were k is constant Wat is te instantaneous rate of cange of F wit respect to r wen r km? 50 Te potential energ U between two atoms in a iatomic molecule is given b U() q> q> 6, were q an q are positive constants an is te istance between te atoms Te force between te atoms is efine as 6 F() U () Sow tat F q>q 0 5 Te van er Waals equation of state for an ieal gas is ( ap ) a b (V b) RT, V were P is pressure, V is volume per mole, R is te universal gas constant, T is temperature, an a an b are constants epening on te gas Fin P>V in te case were T is constant 5 For a conve lens, te focal lengt f is relate to te object istance p an te image istance q b te lens equation f p q Fin te instantaneous rate of cange of q wit respect to p in te case were f is constant Eplain te significance of te negative sign in our answer Wat appens to q as p increases? Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

24 59957_CH0a_-90q 9/5/09 7:5 PM Page CHAPTER Te Derivative (c) Conjecture a rule for fining te erivative of [f ()] n, were n is a positive integer () Use our conjecture in part (c) to fin te erivative of ( 6) Suppose () satisfies te ifferential equation P() 0, were P is a known function Sow tat u()() satisfies te ifferential equation P() f () wenever u() satisfies u> f ()>() Tink About It (b) Fin all te points on te grap of f suc tat te normal lines pass troug te origin 5 (a) Grap te rational function f () 5 Suppose f () is a ifferentiable function (a) Fin > for [ f ()] (b) Fin > for [ f ()] Trigonometric Functions Introuction In tis section we evelop te erivatives of te si trigonometric functions Once we ave foun te erivatives of sin an cos we can etermine te erivatives of tan, cot, sec, an csc using te Quotient Rule foun in te preceing section We will see immeiatel tat te erivative of sin utilizes te following two limit results lim S0 sin an lim S0 cos 0 () foun in Section Derivatives of Sine an Cosine inition of te erivative To fin te erivative of f () sin we use te basic ef f ( ) f () lim S0 () an te four-step process introuce in Sections 7 an In te first step we use te sum formula for te sine function, sin( ) sin cos cos sin, () but wit an plaing te parts of te smbols an (i) f ( ) sin( ) sin cos cos sin (ii) f ( ) f () sin cos cos sin sin sin (cos ) cos sin from () factor sin from first an tir terms As we see in te net line, we cannot cancel te s in te ifference quotient but we can rewrite te epression to make use of te limit results in () (iii) f ( ) f () sin (cos ) cos sin cos sin sin cos (iv) In tis line, te smbol plas te part of te smbol in (): f () lim S0 f ( ) f () cos sin sin lim cos lim S0 S0 From te limit results in (), te last line is te same as f () lim S0 Hence, f ( ) f () sin 0 cos cos sin cos Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION ()

25 59957_CH0a_-90q 9/5/09 7:5 PM Page 5 Trigonometric Functions 5 In a similar manner it can be sown tat cos sin (5) See Problem 50 in Eercises Equation of a Tangent Line Fin an equation of te tangent line to te grap of f () sin at p> EXAMPLE Solution From () te erivative of f () sin is f () cos Wen evaluate at te same number p> tese functions give: fq p p R sin p p f Q R cos point of tangenc is A p, B slope of tangent at A p, B is From te point slope form of a line, an equation of te tangent line is p Q R Te tangent line is sown in re in or p point of tangenc FIGURE cos sin sin cos (cos ) cos (cos ) sin ( sin ) (cos ) tis equals cos sin cos Using te funamental Ptagorean ientit sin cos an te fact tat cos ( cos ) sec, te last equation simplifies to tan sec (6) Te erivative formula for te cotangent cot csc (7) is obtaine in an analogous fasion an left as an eercise See Problem 5 in Eercises Now sec >cos Terefore, we can use te Quotient Rule again to fin te erivative of te secant function: cos B writing cos cos (cos ) 0 ( sin ) (cos ) (, ) slope is ( ) ƒ FIGURE Tangent line in Eample Oter Trigonometric Functions Te results in () an (5) can be use in conjunction wit te rules of ifferentiation to fin te erivatives of te tangent, cotangent, secant, an cosecant functions To ifferentiate tan sin >cos, we use te Quotient Rule: sin cos sin sin cos (8) sin sin sec tan cos cos cos Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

26 59957_CH0a_-90q 6 9/5/09 7:5 PM Page 6 CHAPTER Te Derivative we can epress (8) as sec sec tan (9) Te final result also follows immeiatel from te Quotient Rule: csc csc cot (0) See Problem 5 in Eercises EXAMPLE Prouct Rule Differentiate sin Solution Te Prouct Rule along wit () gives sin sin cos sin EXAMPLE Prouct Rule Differentiate cos Solution One wa of ifferentiating tis function is to recognize it as a prouct: (cos )(cos ) Ten b te Prouct Rule an (5), cos cos cos cos cos ( sin ) (cos )( sin ) sin cos In te net section we will see tat tere is an alternative proceure for ifferentiating a power of a function EXAMPLE Quotient Rule Differentiate sin sec Solution B te Quotient Rule, (), an (9), ( sec ) sin sin ( sec ) ( sec ) ( sec ) cos sin (sec tan ) sec cos an sin (sec tan ) sin >cos ( sec ) cos tan ( sec ) Secon Derivative Fin te secon erivative of f () sec EXAMPLE 5 Solution From (9) te first erivative is f () sec tan To obtain te secon erivative we must now use te Prouct Rule along wit (6) an (9): tan tan sec sec (sec ) tan (sec tan ) sec sec tan f () sec Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

27 59957_CH0a_-90q 9/5/09 7:5 PM Page 7 Trigonometric Functions 7 For future reference we summarize te erivative formulas introuce in tis section Teorem Derivatives of Trigonometric Functions Te erivatives of te si trigonometric functions are sin cos, cos sin, () tan sec, cot csc, () sec sec tan, csc csc cot () NOTES FROM THE CLASSROOM Wen working te problems in Eercises ou ma not get te same answer as given in te answer section in te back of tis book Tis is because tere are so man trigonometric ientities tat answers can often be epresse in a more compact form For eample, te answer in Eample : sin cos is te same as sin b te ouble-angle formula for te sine function Tr to resolve an ifferences between our answer an te given answer Eercises Answers to selecte o-numbere problems begin on page ANS-000 Funamentals In Problems, fin cos 5 sin 7sin tan cos 5 cot 5 sin 6 A B cos 7 ( ) tan 8 cos cot 9 ( sin ) sec 0 csc tan cos sin cos sin In Problems, fin f () f () (csc ) cot 7 f () tan 5 f () cos cot 6 6 f () cos sin 8 f () f () csc sec sin f () sin tan f () cos In Problems 6, fin an equation of te tangent line to te grap of te given function at te inicate value of f () cos ; p> f () tan ; p 5 f () sec ; p>6 6 f () csc ; p> 9 f () sin cos 0 f () In Problems 7 0, consier te grap of te given function on te interval [0, p] Fin te -coorinates of te point(s) on te grap of te function were te tangent line is orizontal sin 7 f () cos 8 f () cos 9 f () 0 f () sin cos cos In Problems, fin an equation of te normal line to te grap of te given function at te inicate value of f () sin ; p f () tan ; p f () cos ; p ; p f () sin In Problems 5 an 6, fin te erivative of te given function b first using an appropriate trigonometric ientit 5 f () sin 6 f () cos In Problems 7, fin te secon erivative of te given function 7 f () sin 8 f () cos sin 9 f () 0 f () cos csc tan Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

28 59957_CH0a_-90q 8 9/5/09 8: PM Page 8 CHAPTER Te Derivative In Problems an, C an C are arbitrar real constants Sow tat te function satisfies te given ifferential equation C cos C sin cos ; sin cos sin C C ; A B 0 Applications 5 Wen te angle of elevation of te sun is u, a telepone pole 0 ft ig casts a saow of lengt s as sown in FIGURE Fin te rate of cange of s wit respect to u wen u pⲑ raians Eplain te significance of te minus sign in te answer 0 ft S FIGURE Saow in Problem 5 6 Te two ens of a 0-ft boar are attace to perpenicular rails, as sown in FIGURE, so tat point P is free to move verticall an point R is free to move orizontall (a) Epress te area A of triangle PQR as a function of te inicate angle u (b) Fin te rate of cange of A wit respect to u (c) Initiall te boar rests flat on te orizontal rail Suppose point R is ten move in te irection of point Q, tereb forcing point P to move up te vertical rail Initiall te area of te triangle is 0 (u 0), but ten it increases for a wile as u increases an ten ecreases as R approaces Q Wen te boar is vertical, te area of te triangle is again 0 (u pⲑ) Grap te erivative AⲐu Interpret tis grap to fin values of u for wic A is increasing an values of u for wic A is ecreasing Now verif our interpretation of te grap of te erivative b graping A(u) () Use te graps in part (c) to fin te value of u for wic te area of te triangle is te greatest rail Tink About It 7 (a) Fin all positive integers n suc tat n n cos cos ; n sin sin ; n n n sin cos n cos sin ; n (b) Use te results in part (a) as an ai in fining sin, sin, cos, an cos Fin two istinct points P an P on te grap of cos so tat te tangent line at P is perpenicular to te tangent line at P 9 Fin two istinct points P an P on te grap of sin so tat te tangent line at Pl is parallel to te tangent line at P 50 Use (), (), an te sum formula for te cosine to sow tat cos sin 5 Use () an (5) an te Quotient Rule to sow tat cot csc 5 Use () an te Quotient Rule to sow tat csc csc cot Calculator/CAS Problems In Problems 5 an 5, use a calculator or CAS to obtain te grap of te given function B inspection of te grap inicate were te function ma not be ifferentiable 5 f () 05(sin 0 sin 0 ) 5 f () 0 sin 0 55 As sown in FIGURE, a bo pulls a sle on wic is little sister is seate If te sle an girl weig a total of 70 lb, an if te coefficient of sliing friction of snowcovere groun is 0, ten te magnitue F of te force (measure in pouns) require to move te sle is 70(0) F, 0 sin u cos u were u is te angle te tow rope makes wit te orizontal (a) Use a calculator or CAS to obtain te grap of F on te interval [, ] (b) Fin te erivative FⲐu (c) Fin te angle (in raians) for wic FⲐu 0 () Fin te value of F corresponing to te angle foun in part (c) (e) Use te grap in part (a) as an ai in interpreting te numbers foun in parts (c) an () P 0 ft F Q R rail FIGURE Boar in Problem 6 FIGURE Sle in Problem 55 Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

29 59957_CH0b_-90q 9/5/09 7:56 PM Page 9 5 Cain Rule 5 Cain Rule Introuction As iscusse in Section, te Power Rule n n n is vali for all real number eponents n In tis section we see tat a similar rule ols for te erivative of a power of a function [g()]n Before stating te formal result, let us consier an eample wen n is a positive integer Suppose we wis to ifferentiate ( 5 ) () B writing () as ( ) ( 5 ), we can fin te erivative using te Prouct Rule: 5 5 ( ) ( 5 ) ( 5 ) ( 5 ) ( 5 ) ( 5 ) 5 ( 5 ) 5 ( 5 ) 5 () Similarl, to ifferentiate te function ( 5 ), we can write it as ( 5 ) ( 5 ) an use te Prouct Rule an te result given in (): we know tis from () 5 ( ) ( 5 ) ( 5 ) 5 ( ) ( 5 ) ( 5 ) ( 5 ) 5 ( 5 ) ( 5 ) 5 ( 5 ) 5 ( 5 ) () In like manner, b writing ( 5 ) as ( 5 ) ( 5 ) we can reail sow b te Prouct Rule an () tat 5 ( ) ( 5 ) 5 () Power Rule for Functions Inspection of (), (), an () reveals a pattern for ifferentiating a power of a function g For eample, in () we see bring own eponent as a multiple T T erivative of function insie parenteses ( 5 ) 5 c ecrease eponent b For empasis, if we enote a ifferentiable function b [ ], it appears tat [ ] n n[ ] n [ ] Te foregoing iscussion suggests te result state in te net teorem Teorem 5 Power Rule for Functions If n is an real number an u g() is ifferentiable at, ten or equivalentl, [g()] n n[ g()] n g (), (5) u n u nu n (6) Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION 9

30 59957_CH0b_-90q 5:9 PM Page 50 CHAPTER Te Derivative Teorem 5 is itself a special case of a more general teorem, calle te Cain Rule, wic will be presente after we consier some eamples of tis new power rule Power Rule for Functions EXAMPLE Differentiate ( )7 Solution Wit te ientification tat u g(), we see from (6) tat u> u n { n 50 0/6/09 7( )6 ( ) 7( )6( ) Power Rule for Functions To ifferentiate >( ), we coul, of course, use te Quotient Rule However, b rewriting te function as ( ), it is also possible to use te Power Rule for Functions wit n : EXAMPLE ()( ) ( ) ()( ) ( ) EXAMPLE Power Rule for Functions Differentiate (7 5 )0 Solution Write te given function as (7 5 )0 Ientif u 7 5, n 0 an use te Power Rule (6): 0(5 ) 0(7 5 ) (7 5 ) (7 5 ) Power Rule for Functions Differentiate tan EXAMPLE Solution For empasis, we first rewrite te function as (tan ) an ten use (6) wit u tan an n : (tan ) tan Recall from (6) of Section tat (>)tan sec Hence, tan sec EXAMPLE 5 Quotient Rule ten Power Rule Differentiate ( ) (5 )8 Solution We start wit te Quotient Rule followe b two applications of te Power Rule for Functions: Power Rule for Functions T T (5 ) ( ) ( ) (5 )8 (5 )6 (5 )8 ( ) ( ) 8(5 )7 5 (5 )6 8 Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

31 59957_CH0b_-90q 9/5/09 7:57 PM Page 5 5 Cain Rule EXAMPLE 6 6(5 )8( ) 0(5 )7( ) (5 )6 ( )( 0 6 0) (5 )9 Power Rule ten Quotient Rule Differentiate A 8 Solution B rewriting te function as Q > R we can ientif u 8 8 an n Tus in orer to compute u> in (6) we must use te Quotient Rule: Q 8 Q 8 > R Q R 8 > (8 ) ( ) 8 R (8 ) > 6 Q R 8 (8 ) Finall, we simplif using te laws of eponents: > ( ) (8 )> Cain Rule A power of a function can be written as a composite function If we ientif f () n an u g(), ten f (u) f (g()) [ g()] n Te Cain Rule gives us a wa of ifferentiating an composition f ⴰ g of two ifferentiable functions f an g Teorem 5 Cain Rule If te function f is ifferentiable at u g(), an te function g is ifferentiable at, ten te composition ( f ⴰ g)() f (g()) is ifferentiable at an f (g()) f (g()) g () or equivalentl, u u (7) (8) PROOF FOR u ⴝ 0 In tis partial proof it is convenient to use te form of te efinition of te erivative given in () of Section For 0, u g( ) g() (9) or g( ) g() u u u In aition, f (u u) f (u) f (g( )) f (g()) Wen an are in some open interval for wic u 0, we can write u u Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION 5

32 59957_CH0b_-90q 5 9/5/09 7:57 PM Page 5 CHAPTER Te Derivative Since g is assume to be ifferentiable, it is continuous Consequentl, as S 0, g( ) S g(), an so from (9) we see tat u S 0 Tus, lim S0 Q lim u R Q S0 lim R Q lim u R Q S0 lim R S0 u us0 u note tat u S 0 in te first term From te efinition of te erivative, () of Section, it follows tat u u Te assumption tat u 0 on some interval oes not ol true for ever ifferentiable function g Altoug te result given in (7) remains vali wen u 0, te preceing proof oes not It migt elp in te unerstaning of te erivative of a composition f (g()) to tink of f as te outsie function an u g() as te insie function Te erivative of f (g()) f (u) is ten te prouct of te erivative of te outsie function (evaluate at te insie function u) an te erivative of te insie function (evaluate at ): erivative of outsie function T f (u) f (u) u (0) c erivative of insie function Te result in (0) is written in various was Since f (u), we ave f (u) >u, an of course u u> Te prouct of te erivatives in (0) is te same as (8) On te oter an, if we replace te smbols u an u in (0) b g() an g () we obtain (7) Proof of te Power Rule for Functions As note previousl, a power of a function can be written as a composition of ( f ⴰ g)() were te outsie function is f () n an te insie function is u g() Te erivative of te insie function f (u) u n is nu n u an te erivative of te outsie function is Te prouct of tese erivatives is ten u n u nu n[g()]n g () u Tis is te Power Rule for Functions given in (5) an (6) Trigonometric Functions We obtain te erivatives of te trigonometric functions compose wit a ifferentiable function g as anoter irect consequence of te Cain Rule For eample, if sin u, were u g(), ten te erivative of wit respect to te variable u is cos u u Hence, (8) gives u u cos u u or equivalentl, sin[ ] cos[ ] [ ] Similarl, if tan u were u g(), ten >u sec u an so u u sec u u We summarize te Cain Rule results for te si trigonometric functions Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

33 59957_CH0b_-90q 0/6/09 5:9 PM Page 5 5 Cain Rule Teorem 5 Derivatives of Trigonometric Functions If u g() is a ifferentiable function, ten u sin u cos u, u tan u secu, u secu sec u tan u, EXAMPLE 7 u cos u sin u, u cot u cscu, u csc u csc u cot u () () () Cain Rule Differentiate cos Solution Te function is cos u wit u From te secon formula in () of Teorem 5 te erivative is u u sin sin EXAMPLE 8 Cain Rule Differentiate tan(6 ) sec u u Solution Te function is tan u wit u 6 From te first formula in () of Teorem 5 te erivative is sec(6 ) (6 ) sec (6 ) EXAMPLE 9 Prouct, Power, an Cain Rule Differentiate (9 ) sin 5 Solution We first use te Prouct Rule: (9 ) sin 5 sin 5 (9 ) followe b te Power Rule (6) an te first formula in () of Teorem 5, from () T from (6) T (9 ) cos 5 5 sin 5 (9 ) (9 ) (9 ) 5 cos 5 sin 5 (9 ) 7 (9 )(5 cos 5 5 cos 5 5 sin 5) In Sections an we saw tat even toug te Sum an Prouct Rules were state in terms of two functions f an g, te were applicable to an finite number of ifferentiable functions So too, te Cain Rule is state for te composition of two functions f an g but we can appl it to te composition of tree (or more) ifferentiable functions In te case of tree functions f, g, an, (7) becomes f (g(())) f (g(())) g(()) f (g(())) g (()) () Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION 5

34 59957_CH0b_-90q 5 9/5/09 7:59 PM Page 5 CHAPTER Te Derivative Repeate Use of te Cain Rule Differentiate cos (7 6 ) EXAMPLE 0 Solution For empasis we first rewrite te given function as [cos(7 6 )] Observe tat tis function is te composition ( f ⴰ g ⴰ )() f (g(())) were f (), g() cos, an () 7 6 We first appl te Cain Rule in te form of te Power Rule (6) followe b te secon formula in (): [cos (7 6 )] cos (7 6 ) cos(7 6 ) c sin(7 6 ) (7 6 ) ( 6) cos(7 6 ) sin (7 6 ) Cain Rule: first ifferentiate te power Cain Rule: secon ifferentiate te cosine In te final eample, te given function is a composition of four functions EXAMPLE Repeate Use of te Cain Rule Differentiate sin (tan ) Solution Te function is f (g((k()))), were f () sin, g() tan, (), an k() In tis case we appl te Cain Rule tree times in succession cos A tan B tan cos A tan B sec > cos A tan B sec ( ) Cain Rule: first ifferentiate te sine Cain Rule: secon ifferentiate te tangent rewrite power tir Cain Rule: cos Atan B sec ( ) > ( ) ifferentiate te power cos Atan B sec ( ) > 6 simplif B sec cos Atan You soul, of course, become so aept at appling te Cain Rule tat ou will not ave to give a moment s tougt as to te number of functions involve in te actual composition NOTES FROM THE CLASSROOM (i) Probabl te most common mistake is to forget to carr out te secon alf of te Cain Rule, namel te erivative of te insie function Tis is te u> part in u u For instance, te erivative of ( ) 57 is not > 57( ) 56 since 57( ) 56 is onl te >u part It migt elp to consistentl use te operation smbol >: ( ) 57 57( ) 56 ( ) 57( ) 56 ( ) Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

35 59957_CH0b_-90q 9/5/09 8:00 PM Page 55 5 Cain Rule 55 (ii) A less common but probabl a worse mistake tan te first is to ifferentiate insie te given function A stuent wrote on an eamination paper tat te erivative of cos ( ) was > sin (); tat is, te erivative of te cosine is te negative of te sine an te erivative of is Bot observations are correct, but ow te are put togeter is incorrect Bear in min tat te erivative of te insie function is a multiple of te erivative of te outsie function Again, it migt elp to use te operation smbol > Te correct erivative of cos ( ) is te prouct of two erivatives sin ( ) ( ) sin ( ) Eercises 5 Answers to selecte o-numbere problems begin on page ANS-000 Funamentals In Problems 9, fin te slope of te tangent line to te grap of te given function at te inicate value of ; 0 9 ( ); 0 ( ) In Problems 0, fin > ( 5)0 (>) 5 Q R ( )00 5 ( 7) 6 7 ( ) ( 9) 9 sin 5 A sin cos 5; p 50 tan ; p>6 0 8 ( ) 0 sec 6 (5 ) [ ( ) ] c ( ) 5 ( ) 7 sin(p ) 9 sin 5 6 ( ) 8 cos ( 7) 0 cos 0 In Problems 8, fin f () f () cos f () f () ( sin )0 f () 5 f () tan(>) 7 f () sin cos 9 f () (sec tan ) sin 5 cos 6 ( cos ) ( sin 5) 6 f () cot(5> ) 8 f () sin cos 0 f () csc csc f () tan Q cos R f () sin (sin ) f () cos A sin 5 B f () tan(tan ) 6 f () sec (tan ) f () sin( ) f () ( ( ( ))5)6 f () c a b In Problems 6, fin an equation of te tangent line to te grap of te given function at te inicate value of b; a ( ); 5 tan ; p> 6 ( cos ); p>8 In Problems 7 an 8, fin an equation of te normal line to te grap of te given function at te inicate value of p 7 sin Q R cos (p ); 6 8 sin ; p In Problems 9 5, fin te inicate erivative 9 f () sin p; f () 50 cos( ); 5> 5 5 sin 5; > 5 f () cos ; f () 5 Fin te point(s) on te grap of f () >( ) were te tangent line is orizontal Does te grap of f ave an vertical tangents? 5 Determine te values of t at wic te instantaneous rate of cange of g(t) sin t cos t is zero 55 If f () cos(>), wat is te slope of te tangent line to te grap of f at p? 56 If f () ( ), wat is te slope of te tangent line to te grap of f at? Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

36 59957_CH0b_-90q 56 0/6/09 5:0 PM Page 56 CHAPTER Te Derivative Applications (a) Verif tat (t) satisfies te ifferential equation A 0>gB sin u 57 Te function R gives te range of a projectile fire at an angle u from te orizontal wit an initial velocit 0 If 0 an g are constants, fin tose values of u at wic R>u 0 58 Te volume of a sperical balloon of raius r is V pr Te raius is a function of time t an increases at a constant rate of 5 in/min Wat is te instantaneous rate of cange of V wit respect to t? 59 Suppose a sperical balloon is being fille at a constant rate V>t 0 in/min At wat rate is its raius increasing wen r in? 60 Consier a mass on a spring sown in FIGURE 5 In te absence of amping forces, te isplacement (or irecte istance) of te mass measure from a position calle te equilibrium position is given b te function (t) 0 cos t 0 sin t, were k>m, k is te spring constant (an inicator of te stiffness of te spring), m is te mass (measure in slugs or kilograms), 0 is te initial isplacement of te mass (measure above or below te equilibrium position), 0 is te initial velocit of te mass, an t is time measure in secons Equilibrium 0 t (b) Verif tat (t) satisfies te initial conitions (0) 0 an (0) 0 Tink About It F()? 6 Let G be a ifferentiable function Wat is [G( )]? 6 Suppose f (u) Wat is f (0 7)? u u 6 Suppose f () Wat is f ( )? 6 Let F be a ifferentiable function Wat is In Problems 65 an 66, te smbol n represents a positive integer Fin a formula for te given erivative 65 n ( ) n 66 n n 67 Suppose g(t) ( f (t)), were f (), f () 6, an () Wat is g ()? 68 Suppose g(), g (), g (), f (), an f () Wat is f (g()) `? 69 Given tat f is an o ifferentiable function, use te Cain Rule to sow tat f is an even function 70 Given tat f is an even ifferentiable function, use te Cain Rule to sow tat f is an o function 0 0 FIGURE 5 Mass on a spring in Problem 60 6 Implicit Differentiation Introuction Te graps of man equations tat we stu in matematics are not te graps of functions For eample, te equation () escribes a circle of raius centere at te origin Equation () is not a function, since for an coice of satisfing 6 6 tere correspons two values of See FIGURE 6(a) Neverteless, graps of equations suc as () can possess tangent lines at various points (, ) Equation () efines at least two functions f an g on te interval [, ] Grapicall, te obvious functions are te top alf an te bottom alf of te circle To obtain formulas for tese functions we solve for in terms of : an f (), upper semicircle () g() lower semicircle () Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

37 59957_CH0b_-90q 9/5/09 8:0 PM Page 57 6 Implicit Differentiation 57 See Figures 6(b) an (c) We can now fin slopes of tangent lines for 6 6 b ifferentiating () an () b te Power Rule for Functions In tis section we will see ow te erivative > can be obtaine for (), as well as for more complicate equations F(, ) 0, witout te necessit of solving te equation for te variable (a) Not a function 6 (b) Function 6 (, ) are ientities on te interval [, ] Te grap of te equation sown in FIGURE 6(a) is a famous curve calle te Folium of Descartes Wit te ai of a CAS suc as Matematica or Maple, one of te implicit functions efine b is foun to be [ f ()] an [ g()] (, ) Eplicit an Implicit Functions A function in wic te epenent variable is epresse solel in terms of te inepenent variable, namel, f (), is sai to be an eplicit function For eample, is an eplicit function On te oter an, an equivalent equation 0 is sai to efine te function implicitl, or is an implicit function of We ave just seen tat te equation efines te two functions f () an g() implicitl In general, if an equation F(, ) 0 efines a function f implicitl on some interval, ten F(, f ()) 0 is an ientit on te interval Te grap of f is a portion or an arc (or all) of te grap of te equation F(, ) 0 In te case of te functions in () an (), note tat bot equations (c) Function FIGURE 6 Equation etermines at least two functions () Te grap of tis function is te re arc sown in Figure 6(b) Te grap of anoter implicit function efine b is given in Figure 6(c) (c) Function (b) Function (a) Folium FIGURE 6 Te portions of te grap in (a) tat are sown in re in (b) an (c) are graps of two implicit functions of Implicit Differentiation Do not jump to te conclusion from te preceing iscussion tat we can alwas solve an equation F(, ) 0 for an implicit function of as we i in (), (), an () For eample, solving an equation suc as 5 (5) for in terms of is more tan an eercise in callenging algebra or a lesson in te use of te correct snta of a CAS It is impossible! Yet (5) ma etermine several implicit functions on a suitabl restricte interval of te -ais Neverteless, we can etermine te erivative > b a process known as implicit ifferentiation Tis process consists of ifferentiating bot sies of an equation wit respect to, using te rules of ifferentiation, an ten solving for > Since we tink of as being etermine b te given equation as a ifferentiable function of, te Cain Rule, in te form of te Power Rule for Functions, gives te useful result n n n, (6) Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION Altoug we cannot solve certain equations for an eplicit function, it still ma be possible to grap te equation wit te ai of a CAS We can ten see te functions as we i in Figure 6

38 59957_CH0b_-90q 58 9/5/09 8:0 PM Page 58 CHAPTER Te Derivative were n is an real number For eample, wereas Similarl, if is a function of, ten b te Prouct Rule,, an b te Cain Rule, sin 5 cos cos 5 Guielines for Implicit Differentiation (i) Differentiate bot sies of te equation wit respect to Use te rules of ifferentiation an treat as a ifferentiable function of For powers of te smbol use (6) (ii) Collect all terms involving > on te left-an sie of te ifferentiate equation Move all oter terms to te rigt-an sie of te equation (iii) Factor > from all terms containing tis term Ten solve for > In te following eamples we sall assume tat te given equation etermines at least one ifferentiable implicit function EXAMPLE Using Implicit Differentiation Fin > if Solution We ifferentiate bot sies of te equation an ten utilize (6): use Power Rule (6) ere T 0 Solving for te erivative iels (7) As illustrate in (7) of Eample, implicit ifferentiation usuall iels a erivative tat epens on bot variables an In our introuctor iscussion we saw tat te equation efines two ifferentiable implicit functions on te open interval 6 6 Te smbolism > > represents te erivative of eiter function on te interval Note tat tis erivative clearl inicates tat functions () an () are not ifferentiable at an since 0 for tese values of In general, implicit ifferentiation iels te erivative of an ifferentiable implicit function efine b an equation F(, ) 0 Slope of a Tangent Line Fin te slopes of te tangent lines to te grap of at te points corresponing to EXAMPLE Solution Substituting into te given equation gives or Hence, tere are tangent lines at A, B an A, B Altoug A, B an A, B are points on te Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

39 59957_CH0b_-90q 0/6/09 5:0 PM Page 59 6 Implicit Differentiation 59 graps of two ifferent implicit functions, inicate b te ifferent colors in of Eample gives te correct slope at eac point for (, ) We ave ` A, B EXAMPLE FIGURE 6, an Prouct Rule ere T > from factor secon an fourt terms 5 Secon Derivative Fin > if Solution From Eample, we alrea know tat te first erivative is > > Te secon erivative is te erivative of >, an so b te Quotient Rule: substituting for > T T Q R Noting tat permits us to rewrite te secon erivative as EXAMPLE 5 (, ) FIGURE 6 Tangent lines in Eample are sown in green Higer Derivatives Troug implicit ifferentiation we etermine > B ifferentiating > wit respect to we obtain te secon erivative > If te first erivative contains, ten > will again contain te smbol >; we can eliminate tat quantit b substituting its known value Te net eample illustrates te meto Q R Power Rule (6) ere T 5 ⴢ 5 ( 5) Solution In tis case, we use (6) an te Prouct Rule: Using Implicit Differentiation (, ) ` A, B Fin > if 5 EXAMPLE (7) Cain an Prouct Rules Fin > if sin cos Solution From te Cain Rule an Prouct Rule we obtain sin cos (sin ) cos (cos cos ) sin cos sin cos cos Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

40 59957_CH0b_-90q 60 9/5/09 8:0 PM Page 60 CHAPTER Te Derivative Postscript Power Rule Revisite So far we ave prove te Power Rule (>) n n n for all integer eponents n Implicit ifferentiation provies a wa of proving tis rule wen te eponent is a rational number p>q, were p an q are integers an q 0 In te case n p>q, te function p>q gives q p iels q q Now for 0, implicit ifferentiation q p p p Solving te last equation for > an simplifing b te laws of eponents gives p p p p p p p p>q q p>q q p p>q q q ( ) q q Eamination of te last result sows tat it is () of Section wit n p>q Eercises 6 Answers to selecte o-numbere problems begin on page ANS-000 Funamentals In Problems, assume tat is a ifferentiable function of Fin te inicate erivative cos sin In Problems 5, assume tat te given equation efines at least one ifferentiable implicit function Use implicit ifferentiation to fin > ( ) ( ) 9 cos ( )6 ( ) ( ) ( ) 5 8 sin( ) sec 9 5 cos() sin cos 0 In Problems 5 an 6, use implicit ifferentiation to fin te inicate erivative 5 r sin u; r>u 6 pr 00; >r In Problems 7 an 8, fin > at te inicate point 7 0; (, ) 8 sin ; (p>, ) In Problems 9 an 0, fin > at te points tat correspon to te inicate number 9 0; 0 ; In Problems, fin an equation of te tangent line at te inicate point or number ; (, ) ; tan ; p> cos ; (, 0) In Problems 5 an 6, fin te point(s) on te grap of te given equation were te tangent line is orizontal Fin te point(s) on te grap of 5 at wic te slope of te tangent is 8 Fin te point were te tangent lines to te grap of 5 at (, ) an (, ) intersect 9 Fin te point(s) on te grap of at wic te tangent line is perpenicular to te line Fin te point(s) on te grap of 7 at wic te tangent line is parallel to te line 5 In Problems 8, fin > sin 6 tan In Problems 9 5, first use implicit ifferentiation to fin > Ten solve for eplicitl in terms of an ifferentiate Sow tat te two answers are equivalent sin Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

41 59957_CH0b_-90q 9/5/09 8:0 PM Page 6 6 Implicit Differentiation 6 In Problems 5 56, etermine an implicit function from te given equation suc tat its grap is te blue curve in te figure 5 ( ) 5 If all te curves of one famil of curves G(, ) c, c a constant, intersect ortogonall all te curves of anoter famil H(, ) c, c a constant, ten te families are sai to be ortogonal trajectories of eac oter In Problems 6 an 6, sow tat te families of curves are ortogonal trajectories of eac oter Sketc te two families of curves 6 c, c 6 c, c Applications FIGURE 6 Grap for Problem 5 65 A woman rives towar a freewa sign as sown in FIGURE 69 Let u be er viewing angle of te sign an let be er istance (measure in feet) to tat sign (a) If er ee level is ft from te surface of te roa, sow tat FIGURE 65 Grap for Problem ( ) tan u FIGURE 66 Grap for Problem 55 FIGURE 67 Grap for Problem 56 In Problems 57 an 58, assume tat bot an are ifferentiable functions of a variable t Fin >t in terms of,, an >t Te grap of te equation is te Folium of Descartes given in Figure 6(a) (a) Fin an equation of te tangent line at te point in te first quarant were te Folium intersects te grap of (b) Fin te point in te first quarant at wic te tangent line is orizontal 60 Te grap of ( ) ( ) sown in FIGURE 68 is calle a lemniscate (a) Fin te points on te grap tat correspon to (b) Fin an equation of te tangent line to te grap at eac point foun in part (a) (c) Fin te points on te grap at wic te tangent is orizontal 5 (b) Fin te rate at wic u canges wit respect to (c) At wat istance is te rate in part (b) equal to zero? ft 8 ft FIGURE 69 Car in Problem A jet figter loops te loop in a circle of raius km as sown in FIGURE 60 Suppose a rectangular coorinate sstem is cosen so tat te origin is at te center of te circular loop Te aircraft releases a missile tat flies on a straigt-line pat tat is tangent to te circle an its a target on te groun wose coorinates are (, ) (a) Determine te point on te circle were te missile was release ) on te (b) If a missile is release at te point (, circle, at wat point oes it it te groun? FIGURE 68 Lemniscate in Problem 60 In Problems 6 an 6, sow tat te graps of te given equations are ortogonal at te inicate point of intersection See Problem 6 in Eercises 6, 5; (, ) 6, ; (, ) Groun Target FIGURE 60 Jet figter in Problem 66 Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

42 59957_CH0b_-90q 6 9/5/09 8:0 PM Page 6 CHAPTER Te Derivative Tink About It 67 Te angle u (0 6 u 6 p) between two curves is efine to be te angle between teir tangent lines at te point P of intersection If m an m are te slopes of te tangent lines at P, it can be sown tat tan u (m m )>( m m ) Determine te angle between te graps of 6 an at (, ) 68 Sow tat an equation of te tangent line to te ellipse >a >b at te point (0, 0) is given b 0 a 0 b 69 Consier te equation Make up anoter implicit function () efine b tis equation for ifferent from te ones given in (), (), an Problem For 6 6 an p/ 6 6 p/, te equation sin efines a ifferentiable implicit function (a) Fin > in terms of (b) Fin > in terms of 7 Derivatives of Inverse Functions Introuction In Section 5 we saw tat te graps of a one-to-one function f an its inverse f are reflections of eac oter in te line As a consequence, if (a, b) is a point on te grap of f, ten (b, a) is a point on te grap of f In tis section we will also see tat te slopes of tangent lines to te grap of a ifferentiable function f are relate to te slopes of tangents to te grap of f We begin wit two teorems about te continuit of f an f Continuit of f Altoug we state te net two teorems witout proof, teir plausibilit follows from te fact tat te grap of f is a reflection of te grap of f in te line Teorem 7 Continuit of an Inverse Function Let f be a continuous one-to-one function on its omain X Ten f is continuous on its omain Increasing Decreasing Functions Suppose f () is a function efine on an interval I, an tat an are an two numbers in te interval suc tat 6 Ten from Section an Figure recall tat f is sai to be increasing on te interval if f () 6 f (), an ecreasing on te interval if f () 7 f () (ƒ(b), b) ƒ () () () Te net two teorems establis a link between te notions of increasing/ecreasing an te eistence of an inverse function (b, ƒ(b)) Teorem 7 (a, ƒ(a)) ƒ(a) Let f be a continuous function an increasing on an interval [a, b] Ten f eists an is continuous an increasing on [ f (a), f (b)] ƒ() (ƒ(a), a) a ƒ(b) b Eistence of an Inverse Function FIGURE 7 f (blue curve) an f (re curve) are continuous an increasing f increasing an ifferentiable means te tangent lines ave positive slope Teorem 7 also ols wen te wor increasing is replace wit te wor ecreasing an te interval in te conclusion is replace b [ f (b), f (a)] See FIGURE 7 In aition, we can conclue from Teorem 7 tat if f is continuous an increasing on an interval ( q, q ), ten f eists an is continuous an increasing on its omain Inspection of Figures an 7 also sows tat if f in Teorem 7 is a ifferentiable function on (a, b), ten: f is increasing on te interval [ a, b] if f () 7 0 on (a, b), an f is ecreasing on te interval [ a, b] if f () 6 0 on (a, b) We will prove tese statements in te net capter Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

43 59957_CH0b_-90q 9/5/09 8:05 PM Page 6 7 Derivatives of Inverse Functions 6 Teorem 7 Eistence of an Inverse Function Suppose f is a ifferentiable function on an open interval (a, b) If eiter f () 7 0 on te interval or f () 6 0 on te interval, ten f is one-to-one Moreover, f is ifferentiable for all in te range of f Eistence of an Inverse Prove tat f () as an inverse EXAMPLE Solution Since f is a polnomial function it is ifferentiable everwere, tat is, f is ifferentiable on te interval ( q, q ) Also, f () for all implies tat f is increasing on ( q, q ) It follows from Teorem 7 tat f is one-to-one an ence f eists Derivative of f If f is ifferentiable on an interval I an is one-to-one on tat interval, ten for a in I te point (a, b) on te grap of f an te point (b, a) on te grap of f are mirror images of eac oter in te line As we see net, te slopes of te tangent lines at (a, b) an (b, a) are also relate Derivative of an Inverse In Eample 5 of Section 5 we sowe tat te inverse of te one-to-one function f (), 0 is f () At, EXAMPLE f () 5 f (5) an 5 Now from f () ( f ) () an we see f () an ( f ) (5) Tis sows tat te slope of te tangent to te grap of f at (, 5) an te slope of te tangent to te grap of f at (5, ) are reciprocals: See f () or ( f ) (5) f ( f (5)) Derivative of an Inverse Function Suppose tat f is ifferentiable on an interval I an f () is never zero on I If f as an inverse f on I, ten f is ifferentiable at a number an f () f ( f ()) () PROOF As we ave seen in (5) of Section 5, f ( f ()) for ever in te omain of f B implicit ifferentiation an te Cain Rule, Solving te last equation for f or f ( f ()) ƒ () (ƒ ) (5) ƒ () (5, ) 5 6 FIGURE 7 Tangent lines in Eample Te net teorem sows tat te result in Eample is no coincience f ( f ()) (, 5) FIGURE 7 Teorem 7, ⱖ 0 ( f ) (5) 6 f () f () gives () Equation () clearl sows tat to fin te erivative function for f we must know () eplicitl For a one-to-one function f () solving te equation f () for is Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

44 59957_CH0b_-90q 6 9/5/09 8:05 PM Page 6 CHAPTER Te Derivative sometimes ifficult an often impossible In tis case it is convenient to rewrite () using ifferent notation Again b implicit ifferentiation f () gives f () Solving te last equation for > an writing > f () iels > () If (a, b) is a known point on te grap of f, te result in () enables us to evaluate te erivative of f at (b, a) witout an equation tat efines f () Derivative of an Inverse It was pointe out in Eample tat te polnomial function f () is ifferentiable on ( q, q ) an ence continuous on te interval Since te en beavior of f is tat of te single-term polnomial function 5 we can conclue tat te range of f is also ( q, q ) Moreover, since f () for all, f is increasing on its omain ( q, q ) Hence b Teorem 7, f as a ifferentiable inverse f wit omain ( q, q ) B intercanging an, te inverse is efine b te equation 5 8 9, but solving tis equation for in terms of is ifficult (it requires te cubic formula) Neverteless, using > 5 8, te erivative of te inverse function is given b (): EXAMPLE 5 8 (5) For eample, since f () we know tat f () Tus, te slope of te tangent line to te grap of f at (, ) is given b (5): ` ` 5 8 Rea tis paragrap a secon time In Eample, te erivative of te inverse function can also be obtaine irectl from using implicit ifferentiation: (5 8 9) gives 5 8 Solving te last equation for > gives (5) As a consequence of tis observation implicit ifferentiation can be use to fin te erivative of an inverse function wit minimum effort In te iscussion tat follows we will fin te erivatives of te inverse trigonometric functions Derivatives of Inverse Trigonometric Functions A review of Figures 55 an 57(a) reveals tat te inverse tangent an inverse cotangent are ifferentiable for all However, te remaining four inverse trigonometric functions are not ifferentiable at eiter or We sall confine our attention to te erivations of te erivative formulas for te inverse sine, inverse tangent, an inverse secant an leave te oters as eercises Inverse Sine: sin if an onl if sin, were an p> p> Terefore, implicit ifferentiation sin an so gives cos cos (6) For te given restriction on te variable, cos 0 an so cos sin B substituting tis quantit in (6), we ave sown tat sin Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION (7)

45 59957_CH0b_-90q 9/5/09 8:06 PM Page 65 7 Derivatives of Inverse Functions 65 As preicte, note tat (7) is not efine at an Te inverse sine or arcsine function is ifferentiable on te open interval (, ) Inverse Tangent: tan if an onl if tan, were q 6 6 q an p> 6 6 p> Tus, tan l sec gives sec or (8) In view of te ientit sec tan, (8) becomes tan Inverse Secant: (9) For an 0 6 p> or p> 6 p, sec sec if an onl if Differentiating te last equation implicitl gives sec tan (0) In view of te restrictions on, we ave tan sec, 冟冟 7 Hence, (0) becomes sec () We can get ri of te sign in () b observing in Figure 57(b) tat te slope of te tangent line to te grap of sec is positive for 6 an positive for 7 Tus, () is equivalent to sec μ,, 6 () 7 Te result in () can be rewritten in a compact form using te absolute value smbol: sec 0 0 () Te erivative of te composition of an inverse trigonometric function wit a ifferentiable function u g() is obtaine from te Cain Rule Teorem 75 Inverse Trigonometric Functions If u g() is a ifferentiable function, ten u sin u, u u cos u, u () u tan u, u u cot u, u (5) u sec u, 0 u 0 u u csc u 0 u 0 u (6) In te formulas in () we must ave 0 u 0 6, wereas in te formulas in (6) we must ave 0 u 0 7 Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

46 59957_CH0b_-90q 66 9/5/09 8:07 PM Page 66 CHAPTER Te Derivative Derivative of Inverse Sine Differentiate sin 5 EXAMPLE Solution Wit u 5, we ave from te first formula in (), 5 5 (5) 5 Derivative of Inverse Tangent Differentiate tan EXAMPLE 5 Solution Wit u, we ave from te first formula in (5), ( )> A B ( ) > ( ) ( ) Derivative of Inverse Secant Differentiate sec EXAMPLE 6 Solution For 7 7 0, we ave from te first formula in (6), 0 0 ( ) sec FIGURE 7 Grap of function in Eample 6 Wit te ai of a graping utilit we obtain te grap of sec given in Notice tat (7) gives positive slope for 7 an negative slope for 6 (7) FIGURE 7 Tangent Line Fin an equation of te tangent line to te grap of f () cos at EXAMPLE 7 Solution B te Prouct Rule an te secon formula in (), f () a b cos Since cos ( ) p>, te two functions f an f evaluate at give: p f Q R point of tangenc is (, p6 ) 6 p p f Q R slope of tangent at (, p6 ) is, 6 冢 冣 B te point slope form of a line, te unsimplifie equation of te tangent line is cos FIGURE 7 Tangent line in Eample 7 p p a ba b 6 Since te omain of cos is te interval [, ] te omain of f is [, ] Te corresponing range is [0, p] FIGURE 7 was obtaine wit te ai of a graping utilit Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

47 59957_CH0b_-90q 9/5/09 8:08 PM Page 67 8 Eponential Functions 67 Eercises 7 Answers to selecte o-numbere problems begin on page ANS-000 Funamentals cot tan In Problems, witout graping etermine weter te given function f as an inverse f () 0 8 f () f () f () In Problems 5 an 6, use () to fin te erivative of f at te inicate point 5 f () 8; A f A, BB 6 f () 7; ( f ( ), ) 8 f () (5 7) In Problems 9, witout fining te inverse, fin, at te inicate value of, te corresponing point on te grap of f Ten use () to fin an equation of te tangent line at tis point 9 7; 0 ; 0 ( 5 ); 8 6 ; cos a b 7 tan sin 9 cos 5 cot 6 0 sec 5 8 (tan )(cot ) sin 0 sin tan sec sin cos 8 5 a 9 tan b t 7 F(t) arctan a b t 9 f () arcsin (cos ) f () tan (sin ) 6 cos ( ) 8 g(t) arccost 0 f () arctan a sin b f () cos ( sin ) In Problems an, use implicit ifferentiation to fin > sin cos In Problems 5 an 6, sow tat f () 0 Interpret te result 5 f () sin cos 6 f () tan tan (>) In Problems 7 an 8, fin te slope of te tangent line to te grap of te given function at te inicate value of 7 sin ; 8 (cos ); > In Problems 9 an 0, fin an equation of te tangent line to te grap of te given function at te inicate value of 9 f () tan ; In Problems, fin te erivative of te given function sin (5 ) tan In Problems 7 an 8, fin f Use () to fin ( f ) an ten verif tis result b irect ifferentiation of f 7 f () Fin te points on te grap of f () 5 sin, 0 p, at wic te tangent line is parallel to te line Fin all tangent lines to te grap of f () arctan tat ave slope 0 f () sin ( ); Tink About It If f an ( f ) are ifferentiable, use () to fin a formula for ( f ) () Eponential Functions Introuction In Section 6 we saw tat te eponential function f () b, b 7 0, b, is efine for all real numbers, tat is, te omain of f is ( q, q ) Inspection of Figure 6 sows tat f is everwere continuous It turns out tat an eponential function is also ifferentiable everwere In tis section we evelop te erivative of f () b Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

48 59957_CH0c_-90q 68 9/5/09 7:6 PM Page 68 CHAPTER Te Derivative Derivative of an Eponential Function To fin te erivative of an eponential function f () b we will use te efinition of te erivative given in () of Definition We first compute te ifference quotient f ( ) f () () in tree steps For te eponential function f () b, we ave (i) f ( ) b b b (ii) f ( ) f () b b b b b b (b ) laws of eponents of eponents laws an factoring f ( ) f () b (b ) b b (iii) In te fourt step, te calculus step, we let S 0 but analogous to te erivatives of sin an cos in Section, tere is no apparent wa of canceling te in te ifference quotient (iii) Noneteless, te erivative of f () b is f () lim b S0 b () Because b oes not epen on te variable, we can rewrite () as b S0 f () b lim () Now ere are te amazing results Te limit in (), b, S0 lim can be sown to eist for ever positive base b However, as one migt epect, we will get a ifferent answer for eac base b So for convenience let us enote te epression in () b te smbol m(b) Te erivative of f () b is ten b f () b m (b) (0, ) () Slope at (0, ) is m(b) You are aske to approimate te value of m(b) in te four cases b 5,,, an 5 in Problems of Eercises 8 For eample, it can be sown tat m(0) 0585p an as a consequence if f () 0, ten f () (0585p )0 FIGURE 8 Fin a base b so tat te slope m(b) of te tangent line at (0, ) is (5) (6) We can get a better unerstaning of wat m(b) is b evaluating (5) at 0 Since b0, we ave f (0) m(b) In oter wors, m(b) is te slope of te tangent line to te grap of f () b at 0, tat is, at te -intercept (0, ) See FIGURE 8 Given tat we ave to calculate a ifferent m(b) for eac base b, an tat m(b) is likel to be an ugl number as in (6), over time te following question arose naturall: Is tere a base b for wic m(b)? (7) Derivative of te Natural Eponential Function To answer te question pose in (7), we must return to te efinitions of e given in Section 6 Specificall, () of Section 6, e lim ( )> S0 (8) provies te means for answering te question pose in (7) We know tat on an intuitive level, te equalit in (8) means tat as gets closer an closer to 0 ten ( )> can be mae arbitraril close to te number e Tus for values of near 0, we ave te approimation ( )> e an so it follows tat e Te last epression written in te form e Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION (9)

49 59957_CH0c_-90q 9/5/09 7:6 PM Page 69 8 Eponential Functions 69 suggests tat e S0 lim (0) Since te left-an sie of (0) is m(e) we ave te answer to te question pose in (7): Te base b for wic m(b) is b e () In aition, from () we ave iscovere a wonerfull simple result Te erivative of f () e is e In summar, e e () Te result in () is te same as f () f () Moreover, if c 0 is a constant, ten te onl oter nonzero function f in calculus wose erivative is equal to itself is ce since b te Constant Multiple Rule of Section ce c e ce Derivative of f () ⴝ b Revisite In te preceing iscussion we saw tat m(e), but left unanswere te question of weter m(b) as an eact value for eac b 7 0 It as From te ientit eln b b, b 7 0, we can write an eponential function f () b in terms of te e base: f () b (eln b) e (ln b) From te Cain Rule te erivative of b is f () (ln b) e e (ln b) (ln b) e (ln b) (ln b) Returning to b e (ln b), te preceing line sows tat b b (ln b) () Matcing te result in (5) wit tat in () we conclue tat m(b) ln b For eample, te erivative of f () 0 is f () 0 (ln 0) Because ln we see f () 0 (ln 0) is te same as te result in (6) Te Cain Rule forms of te results in () an () are given net Teorem 8 Derivatives of Eponential Functions If u g() is a ifferentiable function, ten an EXAMPLE u u e eu, () u u b bu(ln b) (5) Cain Rule Differentiate (a) e (b) e> (c) 85 Solution (a) Wit u we ave from (), e ( ) e ( ) e Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

50 59957_CH0c_-90q 70 9/5/09 7:6 PM Page 70 CHAPTER Te Derivative (b) B rewriting u > as u we ave from (), e> e> e> ( ) (c) Wit u 5 we ave from (5), 85 (ln 8) (ln 8) Prouct an Cain Rule Fin te points on te grap of e were te tangent line is orizontal EXAMPLE Solution We use te Prouct Rule along wit (): (, e ) e e ( e ) 6e e ( 6 6) (, e ) 0 wen Factoring te last equation gives ( )( ) 0 an so 0,, an Te corresponing points on te grap of te given function are ten (0, 0), (, e ), an (, e ) Te grap of e along wit te tree tangent lines (in re) are sown in FIGURE 8 Since e 0 for all real numbers, e (0, 0) FIGURE 8 Grap of function in Eample In te net eample we recall te fact tat an eponential statement can be written in an equivalent logaritmic form In particular, we use (9) of Section 6 in te form e (6) Tangent Line Parallel to a Line Fin te point on te grap of f () e at wic te tangent line is parallel to ( ln, ) Solution Let ( 0, f ( 0)) ( 0, e 0) be te unknown point on te grap of f () e were te tangent line is parallel to From te erivative f () e te slope of te tangent line at tis point is ten f (0) e 0 Since an te tangent line are parallel at tat point, te slopes are equal: ln EXAMPLE 5 if an onl if e f (0) FIGURE 8 Grap of function an lines in Eample or e 0 or e 0 From (6) te last equation gives 0 ln or 0 ln Hence, te point is ( ln, eln ) Since eln, te point is ( ln, ) In FIGURE 8 te given line is sown in green an te tangent line in re NOTES FROM THE CLASSROOM Te numbers e an p are transcenental as well as irrational numbers A transcenental number is one tat is not a root of a polnomial equation wit integer coefficients For eample, is irrational but is not transcenental, since it is a root of te polnomial equation 0 Te number e was prove to be transcenental b te Frenc matematician Carles Hermite (8 90) in 87, wereas p was prove to be transcenental nine ears later b te German matematician Ferinan Linemann (85 99) Te latter proof sowe conclusivel tat squaring a circle wit a rule an a compass was impossible Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

51 59957_CH0c_-90q 9/5/09 7:6 PM Page 7 8 Eponential Functions 7 Eercises 8 Answers to selecte o-numbere problems begin on page ANS-000 Funamentals 9 If C an k are real constants, sow tat te function Ce k satisfies te ifferential equation k 0 Use Problem 9 to fin a function tat satisfies te given conitions (a) 00 an (0) 00 P (b) 05P 0 an P(0) P0 t In Problems 6, fin te erivative of te given function e e e e sin e 8 e sin p e 9 f () 0 f () e 5 e e (e e )0 e e e e > e e > 5 e7 e 6 eee 8 a 7 (e) 00 b e 9 f () e (e )> 0 f () ( )e ( ) f () e tan e f () sec e f () e e > 5 ee Applications 6 e e e 7 Fin an equation of te tangent line to te grap of (e ) at 0 8 Fin te slope of te normal line to te grap of ( )e at 0 9 Fin te point on te grap of e at wic te tangent line is parallel to 7 5 sin e ; In Problems 7 an 8, C an C are arbitrar real constants Sow tat te function satisfies te given ifferential equation 7 Ce Ce ; 6 0 were a an b are positive constants, often serves as a matematical moel for an epaning but limite population (a) Sow tat P(t) satisfies te ifferential equation (t) t e6 099t ; e 6 e ; ap0, bp0 (a bp0) e at In Problems 6, fin te inicate iger erivative P(t) In Problems an, fin te point(s) on te grap of te given function at wic te tangent line is orizontal Use a graping utilit to obtain te grap of eac function f () e sin f () ( )e 9 Te logistic function P P(a bp) t (b) Te grap of P(t) is calle a logistic curve were P(0) P0 is te initial population Consier te case wen a, b, an P0 Fin orizontal asmptotes for te grap of P(t) b etermining te limits limq P(t) an limq P(t) ts ts (c) Grap P(t) () Fin te value(s) of t for wic P (t) 0 50 Te Jenss matematical moel (97) represents one of te most accurate empiricall evise formulas for preicting te eigt (in centimeters) in terms of age t (in ears) for prescool-age cilren ( monts to 6 ears): 0 Fin te point on te grap of 5 e at wic te tangent line is parallel to 6 e ; In Problems 6, use implicit ifferentiation to fin > e e cos e e( ) 5 e > 6 e e 7 (a) Sketc te grap of f () e 00 (b) Fin f () (c) Sketc te grap of f () Is te function ifferentiable at 0? 8 (a) Sow tat te function f () ecos is perioic wit perio p (b) Fin all points on te grap of f were te tangent is orizontal (c) Sketc te grap of f 8 Ce cos Ce sin ; 5 0 (a) Wat eigt oes tis moel preict for a -ear-ol? (b) How fast is a -ear-ol increasing in eigt? (c) Use a calculator or CAS to obtain te grap of on te interval [, 6] () Use te grap in part (c) to estimate te age of a prescool-age cil wo is 00 cm tall Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

52 59957_CH0c_-90q 7 9/5/09 7:6 PM Page 7 CHAPTER Te Derivative Tink About It 58 5 Sow tat te -intercept of te tangent line to te grap of e at 0 is one unit to te rigt of 0 5 How is te tangent line to te grap of e at 0 relate to te tangent line to te grap of e at 0? 5 Eplain w tere is no point on te grap of e at wic te tangent line is parallel to 5 Fin all tangent lines to te grap of f () e tat pass troug te origin 59 In Problems 55 an 56, te smbol n represents a positive integer Fin a formula for te given erivative n e n 56 n e n S S S Calculator/CAS Problems In Problems 57 60, use a calculator to estimate te value b m(b) lim for b 5, b, b, an b 5 b S0 filling out te given table 57 S Use a calculator or CAS to obtain te grap of f () e e >, 0, 0 0 Sow tat f is ifferentiable for all Compute f (0) using te efinition of te erivative (5) 9 Logaritmic Functions Introuction Because te inverse of te eponential function b is te logaritmic function log b we can fin te erivative of te latter function in tree ifferent was: () of Section 7, implicit ifferentiation, or from te funamental efinition () of Section We will emonstrate te last two metos Derivative of te Natural Logaritm We know from (9) of Section 6 tat ln is te same as e B implicit ifferentiation, te Cain Rule, an () of Section 8, e e gives e Terefore Replacing e b, we get te following result: Like te inverse trigonometric functions, te erivative of te inverse of te natural eponential function is an algebraic function ln () Derivative of f () ⴝ logb In precisel te same manner use to obtain (), te erivative of logb can be gotten b ifferentiating b implicitl: b Terefore gives b (ln b) b (ln b) Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

53 59957_CH0_-90q 9/5/09 8:5 PM Page 7 9 Logaritmic Functions 7 Replacing b b gives log b (ln b) () Because ln e, () becomes () wen b e Prouct Rule Differentiate f () ln EXAMPLE Solution B te Prouct Rule an () we ave f () ln (ln ) (ln ) f () ln or Slope of a Tangent Line Fin te slope of te tangent to te grap of log0 at EXAMPLE Solution B () te erivative of log0 is (ln 0) Wit te ai of a calculator, te slope of te tangent line at (, log0 ) is ` 07 ln 0 We summarize te results in () an () in teir Cain Rule forms Teorem 9 Derivatives of Logaritmic Functions If u g() is a ifferentiable function, ten u ln u, u u log b u u (ln b) an EXAMPLE () () Cain Rule Differentiate (a) f () ln(cos ) (b) ln(ln ) an Solution (a) B (), wit u cos we ave f () or cos ( sin ) cos cos f () tan (b) Using () again, tis time wit u ln, we get ln ln ln ln Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

54 59957_CH0_-90q 7 9/5/09 8:5 PM Page 7 CHAPTER Te Derivative Cain Rule Differentiate f () ln EXAMPLE Solution Because must be positive it is unerstoo tat 7 0 Hence b (), wit u we ave f () ( ) Alternative Solution: From (iii) of te laws of logaritms (Teorem 6), ln N c c ln N an so we can rewrite ln as ln an ten ifferentiate: f () ln Altoug te omain of te natural logaritm ln is te set (0, q ), te omain of ln 0 0 etens to te set ( q, 0) (0, q ) For te numbers in tis last omain, 00 e,, Terefore for 7 0, ln for 6 0, ln ( ) ( ) (5) Te erivatives in (5) prove tat for 0, ln 0 0 (6) Te result in (6) ten generalizes b te Cain Rule For a ifferentiable function u g(), u 0, u ln 0 u 0 (7) u Using (6) Fin te slope of te tangent line to te grap of ln 0 0 at an at EXAMPLE 5 ln (, ln ) (, ln ) Solution Since (6) gives > >, we ave ` FIGURE 9 Graps of tangent lines an function in Eample 5 an ` (8) Because ln 0 0 ln, (8) gives, respectivel, te slopes of te tangent lines at te points (, ln ) an (, ln ) Observe in FIGURE 9 tat te grap of ln 0 0 is smmetric wit respect to te -ais; te tangent lines are sown in re EXAMPLE 6 Using (7) Differentiate (a) ln( ) an (b) ln 0 0 Solution (a) For 7 0, or 7, we ave from (), ( ) (9) (b) For 0, or, we ave from (7), ( ) Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION (0)

55 59957_CH0_-90q 9/5/09 8:5 PM Page 75 9 Logaritmic Functions 75 Altoug (9) an (0) appear to be equal, te are efinitel not te same function Te ifference is simpl tat te omain of te erivative in (9) is te interval (, q ), wereas te omain of te erivative in (0) is te set of real numbers ecept A Distinction Te functions f () ln an g() ln are not te same Since 7 0 for all 0, te omain of f is te set of real numbers ecept 0 Te omain of g is te interval (0, q ) Tus, EXAMPLE 7 f (), EXAMPLE 8 0 g (), 7 0 wereas Simplifing Before Differentiating Differentiate ln >( 7) ( ) Solution Using te laws of logaritms given in Section 6, for 7 0 we can rewrite te rigt-an sie of te given function as ln >( 7) ln( ) ln > ln( 7) ln( ) ln ln ( 7) ln( ) so tat 6 7 or 8 7 ln (M>N) ln M ln N ln (MN) ln M ln N ln N c c ln N Logaritmic Differentiation Differentiation of a complicate function f () tat consists of proucts, quotients, an powers can be simplifie b a tecnique known as logaritmic ifferentiation Te proceure consists of tree steps Guielines for Logaritmic Differentiation (i) Take te natural logaritm of bot sies of f () Simplif te rigt-an sie of ln ln f () as muc as possible using te general properties of logaritms (ii) Differentiate te simplifie version of ln ln f () implicitl: ln ln f () (iii) Since te erivative of te left-an sie is, multipl bot sies b an replace b f () We know ow to ifferentiate an function of te tpe (constant)variable an (variable)constant For eample, p p (ln p) an p pp Tere are functions were bot te base an te eponent are variable: (variable)variable () Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

56 59957_CH0_-90q 76 9/5/09 8:5 PM Page 76 CHAPTER Te Derivative For eample, f () ( >) is a function of te tpe escribe in () Recall, in Section 6 we saw tat f () ( >) plae an important role in te efinition of te number e Altoug we will not evelop a general formula for te erivative of functions of te tpe given in (), we can noneteless obtain teir erivatives troug te process of logaritmic ifferentiation Te net eample illustrates te meto for fining > Logaritmic Differentiation Differentiate, 7 0 EXAMPLE 9 Solution Taking te natural logaritm of bot sies of te given equation an simplifing iels ln ln ln propert (iii) of te laws of logaritms, Section 6 Ten we ifferentiate implicitl: > ln ln c ( ln ) FIGURE 9 Grap of function in Eample 9 Prouct Rule now replace b common enominator an laws of eponents We obtaine te grap of in FIGURE 9 wit te ai of a graping utilit Note tat te grap as a orizontal tangent at te point at wic > 0 Tus, te -coorinate of te point of orizontal tangenc is etermine from ln 0 or ln Te last equation gives e EXAMPLE 0 Logaritmic Differentiation Fin te erivative of 6 (8 )5 ( 7)> Solution Notice tat te given function contains no logaritms As suc, we can fin > using te orinar application of te Quotient, Prouct, an Power Rules Tis proceure, wic is teious, can be avoie b first taking te logaritm of bot sies of te given equation, simplifing as we i in Eample 9 b te laws of logaritms, an ten ifferentiating implicitl We take te natural logaritm of bot sies of te given equation an simplif te rigt-an sie: ln ln 6 (8 )5 ( 7)> ln 6 ln (8 )5 ln ( 7)> ln ( 6 ) 5 ln (8 ) ln ( 7) Differentiating te last line wit respect to gives ( ) c multipl bot sies b 8 ( 6 ) ( 7) 6 (8 )5 replace b te 0 8 c original epression > 8 ( 7) ( 6 ) ( 7) Postscript Derivative of f () ⴝ logb Revisite As state in te introuction to tis section we can obtain te erivative of f () logb using te efinition of te erivative From () of Section, Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

57 59957_CH0_-90q 9/5/09 8:5 PM Page 77 9 Logaritmic Functions 77 logb ( ) logb lim logb algebra an te laws of logaritms S0 lim logb a b ivision of b S0 lim logb a b multiplication b > S0 > lim logb a b te laws of logaritms S0 > logb c lim a b S0 f () lim S0 () Te last step, taking te limit insie te logaritmic function, is justifie b invoking te continuit of te function on (0, q ) an assuming tat te limit insie te brackets eists If we let t > in te last equation, ten since is fie, S 0 implies t S 0 Consequentl, we see from () of Section 6 tat > lim a b lim ( t)>t e S0 ts0 Hence te result in () sows tat, log logb e b () Wen te natural coice of b e is mae, () becomes () since loge e ln e Postscript Power Rule Revisite We are finall in a position to prove te Power Rule (>) n n n, () of Section, for all real number eponents n Our emonstration uses te following fact: For 7 0, n is efine for all real numbers n Ten in view of te ientit eln we can write Tose wit sarp ees an long memories will ave notice tat () is not te same as () Te results are equivalent, since b te cange of base formula for logaritms logb e ln e>ln b >ln b n (eln )n e n ln n n e n ln en ln (n ln ) en ln Tus, Substituting en ln n in te last result completes te proof for 7 0, n n n n n Te last erivative formula is also vali for 6 0 wen n p>q is a rational number an q is an o integer Eercises 9 Answers to selecte o-numbere problems begin on page ANS-000 Funamentals In Problems, fin te erivative of te given function 0 ln ln 0 > ln (ln )> 5 ln ( ) 6 ln ( )0 7 ln 8 ln ln 9 0 (ln ) ln ln ln ln 0 cos 0 ln f () ln ( ln ) g() ln H(t) ln t (t 6) G(t) ln 5t (t ) f () ln ( )( ) Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION ln 0 sin 0 6 ln 8 f () ln (ln (ln )) 0 w(u) u sin (ln 5u) f () ln ( )5 B 7

58 59957_CH0_-90q 78 9/5/09 8:6 PM Page 78 CHAPTER Te Derivative 5 Fin an equation of te tangent line to te grap of ln at 6 Fin an equation of te tangent line to te grap of ln ( ) at 7 Fin te slope of te tangent to te grap of ln (e ) at 0 8 Fin te slope of te tangent to te grap of ln (e ) at 9 Fin te slope of te tangent to te grap of f at te point were te slope of te tangent to te grap of f () ln is 0 Determine te point on te grap of ln at wic te tangent line is perpenicular to In Problems an, fin te point(s) on te grap of te given function at wic te tangent line is orizontal ln f () f () ln In Problems 6, fin te inicate erivative an simplif as muc as possible ln A B ln a b 5 ln (sec tan ) 6 ln (csc cot ) In Problems 7 0, fin te inicate iger erivative 7 ln ; 8 ln ; 9 (ln 0 0 ); 0 ln (5 ); In Problems an, C an C are arbitrar real constants Sow tat te function satisfies te given ifferential equation for 7 0 C > C > ln ; 8 0 C cos A ln B C sin A ln B; 0 In Problems 8, use implicit ifferentiation to fin > ln ln ( ) 5 ln 6 ln 7 ln ( ) 8 ln ( ) 0 In Problems 9 56, use logaritmic ifferentiation to fin > 9 sin 50 (ln 0 0) 5 ( ) 5 ( )( ) 5 5 ( ) ( )5( ) 56 (7 5)9 57 Fin an equation of te tangent line to te grap of at 58 Fin an equation of te tangent line to te grap of (ln ) at e 55 In Problems 59 an 60, fin te point on te grap of te given function at wic te tangent line is orizontal Use a graping utilit to obtain te grap of eac function on te interval [00, ] Tink About It 6 Fin te erivatives of (a) tan (b) e (c) 6 Fin > for 6 Te function f () ln 0 0 is not ifferentiable onl at 0 Te function g() 0 ln 0 is not ifferentiable at 0 an at one oter value of 7 0 Wat is it? 6 Fin a wa to compute log e Calculator/CAS Problems 65 (a) Use a calculator or CAS to obtain te grap of (sin )ln on te interval (0, 5p) (b) Eplain w tere appears to be no grap on certain intervals Ientif te intervals 66 (a) Use a calculator or CAS to obtain te grap of 0 cos 0 cos on te interval [0, 5p] (b) Determine, at least approimatel, te values of in te interval [0, 5p ] for wic te tangent to te grap is orizontal 67 Use a calculator or CAS to obtain te grap of f () ln Ten fin te eact value of te least value of f () Hperbolic Functions Introuction If ou ave ever toure te 60-ft-ig Gatewa Arc in St Louis, Missouri, ou ma ave aske te question, Wat is te sape of te arc? an receive te rater crptic repl: te sape of an inverte catenar Te wor catenar stems from te Latin wor catena an literall means a anging cain (te Romans use te catena as a og leas) It Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION

59 59957_CH0_-90q 9/5/09 8:6 PM Page 79 0 Hperbolic Functions 79 can be emonstrate tat te sape assume b a long fleible wire, cain, cable, or rope anging uner its own weigt between two points is te sape of te grap of te function f () k c (e e c ) () for appropriate coices of te constants c an k Te grap of an function of te form given in () is calle a catenar Hperbolic Functions Combinations suc as () involving te eponential functions e an e occur so often in applie matematics tat te warrant special efinitions Te Gatewa Arc in St Louis, MO Definition 0 Hperbolic Sine an Cosine For an real number, te perbolic sine of is sin e e () cos e e () an te perbolic cosine of is Because te omain of eac of te eponential functions e an e is te set of real numbers ( q, q ), te omain of sin an of cos is ( q, q ) From () an () of Definition 0 it is also apparent tat sin 0 0 an cos 0 Analogous to te trigonometric functions tan, cot, sec, an csc tat are efine in terms of sin an cos, we efine four aitional perbolic functions in terms of sin an cos Te sape of te St Louis Gatewa Arc is base on te matematical moel A B cos (C>L), were A , B 68767, L 999, C 00, an an are measure in feet Wen 0, we get te approimate eigt of 60 ft Definition 0 Oter Hperbolic Functions sin For a real number, te perbolic tangent of is tan sin e e, cos e e () e te perbolic cotangent of, 0, is cos e e cot, sin e e (5), cos e e (6) sin e e (7) e te perbolic secant of is sec (a) sin cos te perbolic cosecant of, 0, is csc (0, ) e Graps of Hperbolic Functions Te graps of te perbolic sine an perbolic cosine are given in FIGURE 0 Note te similarit of te grap in Figure 0(b) an te sape of te Gatewa Arc in te poto at te beginning of tis section Te graps of te perbolic tangent, cotangent, secant, an cosecant are given in FIGURE 0 Note tat 0 is a vertical asmptote of te graps of cot an csc Jones an Bartlett Publisers, LLC NOT FOR SALE OR DISTRIBUTION e (b) cos FIGURE 0 Graps of perbolic sine an cosine