Derivatives. Brett Mulcahy / Shutterstock

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1 2 Derivatives For a roller coaster rie to be smoot, te straigt stretces of te track nee to be connecte to te curve segments so tat tere are no abrupt canges in irection. In te project on page 4 ou will see ow to esign te first ascent an rop of a new coaster for a smoot rie. Brett Mulca / Sutterstock In tis capter we begin our stu of ifferential calculus, wic is concerne wit ow one quantit canges in relation to anoter quantit. Te central concept of ifferential calculus is te erivative, wic is an outgrowt of te velocities an slopes of tangents tat we consiere in Capter. After learning ow to calculate erivatives, we use tem to solve problems involving rates of cange an te approimation of functions. 3

2 4 CHAPTER 2 DERIVATIVES 2. Derivatives an Rates of Cange Te problem of fining te tangent line to a curve an te problem of fining te velocit of an object bot involve fining te same tpe of limit, as we saw in Section.4. Tis special tpe of limit is calle a erivative an we will see tat it can be interprete as a rate of cange in an of te sciences or engineering. P{a, f(a)} Q{, ƒ} ƒ-f(a) Tangents If a curve C as equation f an we want to fin te tangent line to C at te point P a, f a, ten we consier a nearb point Q, f, were a, an compute te slope of te secant line PQ: -a m PQ f f a a a t Q Q P Q FIGURE Ten we let Q approac P along te curve C b letting approac a. If m PQ approaces a number m, ten we efine te tangent t to be te line troug P wit slope m. (Tis amounts to saing tat te tangent line is te limiting position of te secant line PQ as Q approaces P. See Figure.) Definition Te tangent line to te curve f at te point P a, f a is te line troug P wit slope provie tat tis limit eists. m l a f f a a In our first eample we confirm te guess we mae in Eample in Section.4. v EXAMPLE Fin an equation of te tangent line to te parabola 2 at te point P,. SOLUTION Here we ave a an f 2, so te slope is m l f f l l 2 2 l Point-slope form for a line troug te point, wit slope m: m Using te point-slope form of te equation of a line, we fin tat an equation of te tangent line at, is 2 or 2 We sometimes refer to te slope of te tangent line to a curve at a point as te slope of te curve at te point. Te iea is tat if we zoom in far enoug towar te point, te curve looks almost like a straigt line. Figure 2 illustrates tis proceure for te curve 2 in

3 SECTION 2. DERIVATIVES AND RATES OF CHANGE 5 TEC Visual 2. sows an animation of Figure 2. Eample. Te more we zoom in, te more te parabola looks like a line. In oter wors, te curve becomes almost inistinguisable from its tangent line (, ) (, ) (, ) FIGURE 2 Zooming in towar te point (, ) on te parabola = t Q{a+, f(a+)} Tere is anoter epression for te slope of a tangent line tat is sometimes easier to use. If a, ten a an so te slope of te secant line PQ is P{a, f(a)} f(a+)-f(a) a a+ FIGURE 3 m PQ f a f a (See Figure 3 were te case is illustrate an Q is to te rigt of P. If it appene tat, owever, Qwoul be to te left of P.) Notice tat as approaces a, approaces (because a) an so te epression for te slope of te tangent line in Definition becomes 2 m l f a f a +3-6= 3 = (3, ) EXAMPLE 2 Fin an equation of te tangent line to te perbola 3 at te point 3,. SOLUTION Let f 3. Ten te slope of te tangent at 3, is m l f 3 f 3 l 3 3 l 3 l 3 3 Terefore an equation of te tangent at te point 3, is l wic simplifies to 3 6 FIGURE 4 Te perbola an its tangent are sown in Figure 4. Velocities In Section.4 we investigate te motion of a ball roppe from te CN Tower an efine its velocit to be te limiting value of average velocities over sorter an sorter time perios.

4 6 CHAPTER 2 DERIVATIVES position at time t=a s f(a+)-f(a) f(a) f(a+) position at time t=a+ In general, suppose an object moves along a straigt line accoring to an equation of motion s f t, were s is te isplacement (irecte istance) of te object from te origin at time t. Te function f tat escribes te motion is calle te position function of te object. In te time interval from t a to t a te cange in position is f a f a. (See Figure 5.) Te average velocit over tis time interval is average velocit isplacement time f a f a FIGURE 5 s P{a, f(a)} Q{a+, f(a+)} wic is te same as te slope of te secant line PQ in Figure 6. Now suppose we compute te average velocities over sorter an sorter time intervals a, a. In oter wors, we let approac. As in te eample of te falling ball, we efine te velocit (or instantaneous velocit) v a at time t a to be te limit of tese average velocities: a a+ t f(a+)-f(a) m PQ = average velocit FIGURE 6 Recall from Section.4: Te is tance (in meters) fallen after t secons is 4.9t 2. 3 v a l f a f a Tis means tat te velocit at time t a is equal to te slope of te tangent line at P (compare Equations 2 an 3). Now tat we know ow to compute limits, let s reconsier te problem of te falling ball. v EXAMPLE 3 Suppose tat a ball is roppe from te upper observation eck of te CN Tower, 45 m above te groun. (a) Wat is te velocit of te ball after 5 secons? (b) How fast is te ball traveling wen it its te groun? SOLUTION We will nee to fin te velocit bot wen t 5 an wen te ball its te groun, so it s efficient to start b fining te velocit at a general time t a. Using te equation of motion s f t 4.9t 2, we ave v a l f a f a Te velocit of te ball as it its te groun is terefore l 4.9 a 2 4.9a a 2 2a 2 a a 2 l l l 4.9 2a 9.8a (a) Te velocit after 5 s is v m s. (b) Since te observation eck is 45 m above te groun, te ball will it te groun at te time t wen s t 45, tat is, 4.9t 2 45 Tis gives t t an 9.6 s v t 9.8t m s

5 Derivatives SECTION 2. DERIVATIVES AND RATES OF CHANGE 7 We ave seen tat te same tpe of limit arises in fining te slope of a tangent line (Equation 2) or te velocit of an object (Equation 3). In fact, limits of te form lim l f a f a arise wenever we calculate a rate of cange in an of te sciences or engineering, suc as a rate of reaction in cemistr or a marginal cost in economics. Since tis tpe of limit occurs so wiel, it is given a special name an notation. 4 Definition Te erivative of a function f at a number a, enote b f a, is f a is rea f prime of a. if tis limit eists. f a l f a f a If we write a, ten we ave a an approaces if an onl if approaces a. Terefore an equivalent wa of stating te efinition of te erivative, as we saw in fining tangent lines, is 5 f a l a f f a a v EXAMPLE 4 Fin te erivative of te function f at te number a. SOLUTION From Definition 4 we ave f a l f a f a l a 2 8 a 9 a 2 8a 9 l a 2 2a 2 8a 8 9 a 2 8a 9 2a 2 8 2a 8 l l 2a 8 We efine te tangent line to te curve f at te point P a, f a to be te line tat passes troug P an as slope m given b Equation or 2. Since, b Defini tion 4, tis is te same as te erivative f a, we can now sa te following. Te tangent line to f at a, f a is te line troug a, f a wose slope is equal to f a, te erivative of f at a.

6 8 CHAPTER 2 DERIVATIVES If we use te point-slope form of te equation of a line, we can write an equation of te tangent line to te curve f at te point a, f a : = -8+9 (3, _6) =_2 f a f a a v EXAMPLE 5 Fin an equation of te tangent line to te parabola at te point 3, 6. SOLUTION From Eample 4 we know tat te erivative of f at te number ais f a 2a 8. Terefore te slope of te tangent line at 3, 6 is f Tus an equation of te tangent line, sown in Figure 7, is FIGURE or 2 Q{, } Rates of Cange Suppose is a quantit tat epens on anoter quantit. Tus is a function of an we write f. If canges from to 2, ten te cange in (also calle te increment of ) is 2 P{, fl} Î an te corresponing cange in is average rate of cange m PQ instantaneous rate of cange slope of tangent at P FIGURE 8 Î Te ifference quotient f 2 f f 2 f 2 is calle te average rate of cange of wit respect to over te interval, 2 an can be interprete as te slope of te secant line PQ in Figure 8. B analog wit velocit, we consier te average rate of cange over smaller an smaller intervals b letting 2 approac an terefore letting approac. Te limit of tese average rates of cange is calle te (instantaneous) rate of cange of wit respect to at, wic is interprete as te slope of te tangent to te curve f at P, f : 6 instantaneous rate of cange l 2 l f 2 f 2 We recognize tis limit as being te erivative f. We know tat one interpretation of te erivative f a is as te slope of te tangent line to te curve f wen a. We now ave a secon interpretation: Te erivative f a is te instantaneous rate of cange of f wit respect to wen a.

7 SECTION 2. DERIVATIVES AND RATES OF CHANGE 9 P FIGURE 9 Te -values are canging rapil at P an slowl at Q. Q Te connection wit te first interpretation is tat if we sketc te curve f, ten te instantaneous rate of cange is te slope of te tangent to tis curve at te point were a. Tis means tat wen te erivative is large (an terefore te curve is steep, as at te point P in Figure 9), te -values cange rapil. Wen te erivative is small, te curve is relativel flat (as at point Q) an te -values cange slowl. In particular, if s f t is te position function of a particle tat moves along a straigt line, ten f a is te rate of cange of te isplacement s wit respect to te time t. In oter wors, f a is te velocit of te particle at time t a. Te spee of te particle is te absolute value of te velocit, tat is, f a. In te net eample we iscuss te meaning of te erivative of a function tat is efine verball. v EXAMPLE 6 A manufacturer prouces bolts of a fabric wit a fie wit. Te cost of proucing ars of tis fabric is C f ollars. (a) Wat is te meaning of te erivative f? Wat are its units? (b) In practical terms, wat oes it mean to sa tat f 9? (c) Wic o ou tink is greater, f 5 or f 5? Wat about f 5? SOLUTION (a) Te erivative f is te instantaneous rate of cange of C wit respect to ; tat is, f means te rate of cange of te prouction cost wit respect to te number of ars prouce. (Economists call tis rate of cange te marginal cost. Tis iea is iscusse in more etail in Sections 2.7 an 3.7.) Because C f l Here we are assuming tat te cost function is well beave; in oter wors, C oesn t oscillate rapil near. te units for f are te same as te units for te ifference quotient C. Since C is measure in ollars an in ars, it follows tat te units for f are ollars per ar. (b) Te statement tat f 9 means tat, after ars of fabric ave been manufacture, te rate at wic te prouction cost is increasing is $9 ar. (Wen, C is increasing 9 times as fast as.) Since is small compare wit, we coul use te approimation f C C C an sa tat te cost of manufacturing te t ar (or te st) is about $9. (c) Te rate at wic te prouction cost is increasing (per ar) is probabl lower wen 5 tan wen 5 (te cost of making te 5t ar is less tan te cost of te 5t ar) because of economies of scale. (Te manufacturer makes more efficient use of te fie costs of prouction.) So f 5 f 5 But, as prouction epans, te resulting large-scale operation migt become inefficient an tere migt be overtime costs. Tus it is possible tat te rate of increase of costs will eventuall start to rise. So it ma appen tat f 5 f 5

8 CHAPTER 2 DERIVATIVES t t D t D t D 99 t A Note on Units Te units for te average rate of cange D t are te units for D ivie b te units for t, namel, billions of ollars per ear. Te instantaneous rate of cange is te limit of te average rates of cange, so it is measure in te same units: billions of ollars per ear. In te following eample we estimate te rate of cange of te national ebt wit respect to time. Here te function is efine not b a formula but b a table of values. v EXAMPLE 7 Let D t be te US national ebt at time t. Te table in te margin gives approimate values of tis function b proviing en of ear estimates, in billions of ollars, from 98 to 25. Interpret an estimate te value of D 99. SOLUTION Te erivative D 99 means te rate of cange of D wit respect to t wen t 99, tat is, te rate of increase of te national ebt in 99. Accoring to Equation 5, D t D 99 D 99 t l99 t 99 So we compute an tabulate values of te ifference quotient (te average rates of cange) as sown in te table at te left. From tis table we see tat D 99 lies somewere between an billion ollars per ear. [Here we are making te reasonable assumption tat te ebt in t fluctuate will between 98 an 2.] We estimate tat te rate of increase of te national ebt of te Unite States in 99 was te average of tese two numbers, namel D billion ollars per ear Anoter meto woul be to plot te ebt function an estimate te slope of te tangent line wen t 99. In Eamples 3, 6, an 7 we saw tree specific eamples of rates of cange: te velocit of an object is te rate of cange of isplacement wit respect to time; marginal cost is te rate of cange of prouction cost wit respect to te number of items prouce; te rate of cange of te ebt wit respect to time is of interest in economics. Here is a small sample of oter rates of cange: In psics, te rate of cange of work wit respect to time is calle power. Cemists wo stu a cemical reaction are intereste in te rate of cange in te concentration of a reactant wit respect to time (calle te rate of reaction). A biologist is intereste in te rate of cange of te population of a colon of bacteria wit respect to time. In fact, te computation of rates of cange is important in all of te natural sciences, in engineering, an even in te social sciences. Furter eamples will be given in Section 2.7. All tese rates of cange are erivatives an can terefore be interprete as slopes of tangents. Tis gives ae significance to te solution of te tangent problem. Wenever we solve a problem involving tangent lines, we are not just solving a problem in geometr. We are also implicitl solving a great variet of problems involving rates of cange in science an engineering. 2. Eercises. A curve as equation f. (a) Write an epression for te slope of te secant line troug te points P 3, f 3 an Q, f. (b) Write an epression for te slope of te tangent line at P. ; 2. Grap te curve sin in te viewing rectangles 2, 2 b 2, 2,, b,, an.5,.5 b.5,.5. Wat o ou notice about te curve as ou zoom in towar te origin? 3. (a) Fin te slope of te tangent line to te parabola 4 2 at te point, 3 ( i) using Definition ( ii) using Equation 2 (b) Fin an equation of te tangent line in part (a). ; Graping calculator or computer require. Homework Hints available at stewartcalculus.com

9 SECTION 2. DERIVATIVES AND RATES OF CHANGE ; (c) Grap te parabola an te tangent line. As a ceck on our work, zoom in towar te point, 3 until te parabola an te tangent line are inistinguisable. 4. (a) Fin te slope of te tangent line to te curve 3 at te point, ( i) using Definition ( ii) using Equation 2 (b) Fin an equation of te tangent line in part (a). ; (c) Grap te curve an te tangent line in successivel smaller viewing rectangles centere at, until te curve an te line appear to coincie. 5 8 Fin an equation of te tangent line to te curve at te given point , 2, , 2, s, (, 8.,, 2 9. (a) Fin te slope of te tangent to te curve at te point were a. (b) Fin equations of te tangent lines at te points, 5 an 2, 3. ; (c) Grap te curve an bot tangents on a common screen.. (a) Fin te slope of te tangent to te curve s at te point were a. (b) Fin equations of te tangent lines at te points, an (4, 2 ). ; (c) Grap te curve an bot tangents on a common screen.. (a) A particle starts b moving to te rigt along a orizontal line; te grap of its position function is sown. Wen is te particle moving to te rigt? Moving to te left? Staning still? (b) Draw a grap of te velocit function. s (meters) 4 (b) At wat time is te istance between te runners te greatest? (c) At wat time o te ave te same velocit? 3. If a ball is trown into te air wit a velocit of 4 ft s, its eigt ( in feet) after t secons is given b 4t 6t 2. Fin te velocit wen t If a rock is trown upwar on te planet Mars wit a velocit of m s, its eigt ( in meters) after t secons is given b H t.86t 2. (a) Fin te velocit of te rock after one secon. (b) Fin te velocit of te rock wen t a. (c) Wen will te rock it te surface? () Wit wat velocit will te rock it te surface? 5. Te isplacement ( in meters) of a particle moving in a straigt line is given b te equation of motion s t 2, were t is measure in secons. Fin te velocit of te par - ticle at times t a, t, t 2, an t Te isplacement ( in meters) of a particle moving in a straigt line is given b s t 2 8t 8, were t is measure in secons. (a) Fin te average velocit over eac time interval: (i) 3, 4 (ii) 3.5, 4 ( iii) 4, 5 ( iv) 4, 4.5 (b) Fin te instantaneous velocit wen t 4. (c) Draw te grap of s as a function of t an raw te secant lines wose slopes are te average velocities in part (a) an te tangent line wose slope is te instantaneous velocit in part (b). 7. For te function t wose grap is given, arrange te following numbers in increasing orer an eplain our reasoning: t 2 t t 2 t 4 2 = t (secons) 2. Sown are graps of te position functions of two runners, A an B, wo run a -m race an finis in a tie. s (meters) _ A 4 B t (secons) (a) Describe an compare ow te runners run te race. 8. Fin an equation of te tangent line to te grap of t at 5 if t 5 3 an t If an equation of te tangent line to te curve f at te point were a 2 is 4 5, fin f 2 an f If te tangent line to f at (4, 3) passes troug te point (, 2), fin f 4 an f 4.

10 2 CHAPTER 2 DERIVATIVES 2. Sketc te grap of a function f for wic f, f 3, f, an f Sketc te grap of a function t for wic t t 2 t 4, t t 3, t t 4, t 2, lim l 5 t, an lim l t. 23. If f 3 2 3, fin f an use it to fin an equation of te tangent line to te curve at te point, If t 4 2, fin t an use it to fin an equation of te tangent line to te curve 4 2 at te point,. 25. (a) If F 5 2, fin F 2 an use it to fin an equation of te tangent line to te curve 5 2 at te point 2, 2. ; (b) Illustrate part (a) b graping te curve an te tangent line on te same screen. 26. (a) If G 4 2 3, fin G a an use it to fin equations of te tangent lines to te curve at te points 2, 8 an 3, 9. ; (b) Illustrate part (a) b graping te curve an te tangent lines on te same screen Fin f a. 27. f f t 2t 3 t 29. 2t f t t 3 3. f 2 3. f s f s Eac limit represents te erivative of some function f at some number a. State suc an f an a in eac case. s lim 34. lim l l lim 36. l5 5 cos 37. lim 38. l tan lim l 4 4 t 4 t 2 lim t l t 39 4 A particle moves along a straigt line wit equation of motion s f t, were s is measure in meters an t in secons. Fin te velocit an te spee wen t f t 5t 4.9t 2 4. f t t t 4. A warm can of soa is place in a col refrigerator. Sketc te grap of te temperature of te soa as a function of time. Is te initial rate of cange of temperature greater or less tan te rate of cange after an our? 42. A roast turke is taken from an oven wen its temperature as reace 85 F an is place on a table in a room were te temperature is 75 F. Te grap sows ow te temperature of te turke ecreases an eventuall approaces room temperature. B measuring te slope of te tangent, estimate te rate of cange of te temperature after an our. T ( F) 2 P t (min) 43. Te number N of US cellular pone subscribers ( in millions) is sown in te table. (Miear estimates are given.) t N (a) Fin te average rate of cell pone growt ( i) from 22 to 26 ( ii) from 22 to 24 ( iii) from 2 to 22 In eac case, inclue te units. (b) Estimate te instantaneous rate of growt in 22 b taking te average of two average rates of cange. Wat are its units? (c) Estimate te instantaneous rate of growt in 22 b measuring te slope of a tangent. 44. Te number N of locations of a popular coffeeouse cain is given in te table. (Te numbers of locations as of October are given.) Year N 8569,24 2,44 5, 6,68 (a) Fin te average rate of growt ( i) from 26 to 28 ( ii) from 26 to 27 ( iii) from 25 to 26 In eac case, inclue te units. (b) Estimate te instantaneous rate of growt in 26 b taking te average of two average rates of cange. Wat are its units? (c) Estimate te instantaneous rate of growt in 26 b measuring te slope of a tangent. () Estimate te intantaneous rate of growt in 27 an compare it wit te growt rate in 26. Wat o ou conclue? 45. Te cost ( in ollars) of proucing units of a certain commoit is C (a) Fin te average rate of cange of C wit respect to wen te prouction level is cange ( i) from to 5 ( ii) from to (b) Fin te instantaneous rate of cange of C wit respect to wen. (Tis is calle te marginal cost. Its significance will be eplaine in Section 2.7.)

11 SECTION 2. DERIVATIVES AND RATES OF CHANGE If a clinrical tank ols, gallons of water, wic can be raine from te bottom of te tank in an our, ten Torricelli s Law gives te volume V of water remaining in te tank after t minutes as V t,( 6 t) 2 t 6 Fin te rate at wic te water is flowing out of te tank (te instantaneous rate of cange of V wit respect to t) as a function of t. Wat are its units? For times t,, 2, 3, 4, 5, an 6 min, fin te flow rate an te amount of water remaining in te tank. Summarize our finings in a sentence or two. At wat time is te flow rate te greatest? Te least? 47. Te cost of proucing ounces of gol from a new gol mine is C f ollars. (a) Wat is te meaning of te erivative f? Wat are its units? (b) Wat oes te statement f 8 7 mean? (c) Do ou tink te values of f will increase or ecrease in te sort term? Wat about te long term? Eplain. 48. Te number of bacteria after t ours in a controlle laborator eperiment is n f t. (a) Wat is te meaning of te erivative f 5? Wat are its units? (b) Suppose tere is an unlimite amount of space an nutrients for te bacteria. Wic o ou tink is larger, f 5 or f? If te suppl of nutrients is limite, woul tat affect our conclusion? Eplain. 49. Let T t be te temperature ( in F) in Poeni t ours after minigt on September, 28. Te table sows values of tis function recore ever two ours. Wat is te meaning of T 8? Estimate its value. te ogen content of water.) Te grap sows ow ogen solubilit S varies as a function of te water temperature T. (a) Wat is te meaning of te erivative S T? Wat are its units? (b) Estimate te value of S 6 an interpret it. S (mg/ L) T ( C) Aapte from Environmental Science: Living Witin te Sstem of Nature, 2 e.; b Carles E. Kupcella, 989. Reprinte b permission of Prentice-Hall, Inc., Upper Sale River, NJ. 52. Te grap sows te influence of te temperature T on te maimum sustainable swimming spee S of Coo salmon. (a) Wat is te meaning of te erivative S T? Wat are its units? (b) Estimate te values of S 5 an S 25 an interpret tem. S (cm/s) 2 2 T ( C) t T Determine weter f eists. 5. Te quantit ( in pouns) of a gourmet groun coffee tat is sol b a coffee compan at a price of p ollars per poun is Q f p. (a) Wat is te meaning of te erivative f 8? Wat are its units? (b) Is f 8 positive or negative? Eplain. 5. Te quantit of ogen tat can issolve in water epens on te temperature of te water. (So termal pollution influences sin f if if f 2 sin if if

12 4 CHAPTER 2 DERIVATIVES WRITING PROJECT EARLY METHODS FOR FINDING TANGENTS Te first person to formulate eplicitl te ieas of limits an erivatives was Sir Isaac Newton in te 66s. But Newton acknowlege tat If I ave seen furter tan oter men, it is because I ave stoo on te soulers of giants. Two of tose giants were Pierre Fermat (6 665) an Newton s mentor at Cambrige, Isaac Barrow (63 677). Newton was familiar wit te metos tat tese men use to fin tangent lines, an teir metos plae a role in Newton s eventual formulation of calculus. Te following references contain eplanations of tese metos. Rea one or more of te references an write a report comparing te metos of eiter Fermat or Barrow to moern metos. In particular, use te meto of Section 2. to fin an equation of te tangent line to te curve 3 2 at te point (, 3) an sow ow eiter Fermat or Barrow woul ave solve te same problem. Altoug ou use erivatives an te i not, point out similarities between te metos.. Carl Boer an Uta Merzbac, A Histor of Matematics (New York: Wile, 989), pp. 389, C. H. Ewars, Te Historical Development of te Calculus (New York: Springer-Verlag, 979), pp. 24, Howar Eves, An Introuction to te Histor of Matematics, 6t e. (New York: Sauners, 99), pp. 39, Morris Kline, Matematical Tougt from Ancient to Moern Times (New York: Ofor Universit Press, 972), pp. 344, Te Derivative as a Function In te preceing section we consiere te erivative of a function f at a fie number a:. f a l f a f a Here we cange our point of view an let te number a var. If we replace a in Equation b a variable, we obtain 2 f l f f Given an number for wic tis limit eists, we assign to te number f. So we can regar f as a new function, calle te erivative of f an efine b Equation 2. We know tat te value of f at, f, can be interprete geometricall as te slope of te tangent line to te grap of f at te point, f. Te function f is calle te erivative of f because it as been erive from f b te limiting operation in Equation 2. Te omain of f is te set f eists an ma be smaller tan te omain of f.

13 SECTION 2.2 THE DERIVATIVE AS A FUNCTION 5 =ƒ v EXAMPLE Te grap of a function f is given in Figure. Use it to sketc te grap of te erivative f. FIGURE SOLUTION We can estimate te value of te erivative at an value of b rawing te tangent at te point, f an estimating its slope. For instance, for 5 we raw 3 te tangent at P in Figure 2(a) an estimate its slope to be about 2, so f 5.5. Tis allows us to plot te point P 5,.5 on te grap of f irectl beneat P. Repeating tis proceure at several points, we get te grap sown in Figure 2(b). Notice tat te tangents at A, B, an C are orizontal, so te erivative is tere an te grap of f crosses te -ais at te points A, B, an C, irectl beneat A, B, an C. Between A an B te tangents ave positive slope, so f is positive tere. But between B an C te tangents ave negative slope, so f is negative tere. B m= A m= =ƒ P 3 må 2 m= 5 C TEC Visual 2.2 sows an animation of Figure 2 for several functions. (a) =fª() Pª(5,.5) Aª Bª Cª 5 FIGURE 2 (b) v EXAMPLE 2 (a) If f 3, fin a formula for f. (b) Illustrate b comparing te graps of f an f.

14 6 CHAPTER 2 DERIVATIVES 2 f _2 2 _2 2 fª _2 2 SOLUTION (a) Wen using Equation 2 to compute a erivative, we must remember tat te variable is an tat is temporaril regare as a constant uring te calculation of te limit. f l f f l 3 3 l l l FIGURE 3 _2 (b) We use a graping evice to grap f an f in Figure 3. Notice tat f wen f as orizontal tangents an f is positive wen te tangents ave positive slope. So tese graps serve as a ceck on our work in part (a). EXAMPLE 3 If f s, fin te erivative of f. State te omain of f. Here we rationalize te numerator. SOLUTION f l f f s s l l (s s ) l s s 2s s s l s s s s s We see tat f eists if, so te omain of f is,. Tis is smaller tan te omain of f, wic is,. Let s ceck to see tat te result of Eample 3 is reasonable b looking at te graps of f an f in Figure 4. Wen is close to, s is also close to, so f (2s ) is ver large an tis correspons to te steep tangent lines near, in Figure 4(a) an te large values of f just to te rigt of in Figure 4(b). Wen is large, f is ver small an tis correspons to te flatter tangent lines at te far rigt of te grap of f an te orizontal asmptote of te grap of f. FIGURE 4 (a) ƒ=œ (b) fª()= 2œ

15 SECTION 2.2 THE DERIVATIVE AS A FUNCTION 7 EXAMPLE 4 Fin f if f. 2 SOLUTION f l f f l 2 2 a c - b e = a-bc b e l l l l Leibniz Gottfrie Wilelm Leibniz was born in Leipzig in 646 an stuie law, teolog, pilosop, an matematics at te universit tere, grauating wit a bacelor s egree at age 7. After earning is octorate in law at age 2, Leibniz entere te iplomatic service an spent most of is life traveling to te capitals of Europe on political missions. In particular, e worke to avert a Frenc militar treat against Ger man an attempte to reconcile te Catolic an Protestant curces. His serious stu of matematics i not begin until 672 wile e was on a iplomatic mission in Paris. Tere e built a calculating macine an met scientists, like Hugens, wo irecte is attention to te latest evelop - ments in matematics an science. Leibniz sougt to evelop a smbolic logic an sstem of notation tat woul simplif logical reasoning. In particular, te version of calculus tat e publise in 684 establise te notation an te rules for fining erivatives tat we use toa. Unfortunatel, a reaful priorit ispute arose in te 69s between te followers of Newton an tose of Leibniz as to wo a invente calculus first. Leibniz was even accuse of plagiarism b members of te Roal Societ in Englan. Te trut is tat eac man invente calculus inepenentl. Newton arrive at is version of calculus first but, because of is fear of controvers, i not publis it immeiatel. So Leibniz s 684 account of calculus was te first to be publise. Oter Notations If we use te traitional notation f to inicate tat te inepenent variable is an te epenent variable is, ten some common alternative notations for te erivative are as follows: f Te smbols D an are calle ifferentiation operators because te inicate te operation of ifferentiation, wic is te process of calculating a erivative. Te smbol, wic was introuce b Leibniz, soul not be regare as a ratio (for te time being); it is simpl a snonm for f. Noneteless, it is a ver useful an suggestive notation, especiall wen use in conjunction wit increment notation. Referring to Equation 2..6, we can rewrite te efinition of erivative in Leibniz notation in te form If we want to inicate te value of a erivative in Leibniz notation at a specific number a, we use te notation wic is a snonm for f a. f f Df D f a l or 3 Definition A function f is ifferentiable at a if f a eists. It is ifferentiable on an open interval a, b [or a, or, a or, ] if it is ifferentiable at ever number in te interval. a

16 8 CHAPTER 2 DERIVATIVES v EXAMPLE 5 Were is te function ifferentiable? SOLUTION If, ten an we can coose small enoug tat an ence. Terefore, for, we ave f l an so f is ifferentiable for an. Similarl, for we ave an can be cosen small enoug tat an so. Terefore, for, f l an so f is ifferentiable for an. For we ave to investigate f l l l l l l f l l f f Let s compute te left an rigt limits separatel: if it eists (a) =ƒ= an lim l lim l l l l l l l Since tese limits are ifferent, f oes not eist. Tus f is ifferentiable at all ecept. A formula for f is given b _ f if if FIGURE 5 (b) =fª() an its grap is sown in Figure 5(b). Te fact tat f oes not eist is reflecte geometricall in te fact tat te curve oes not ave a tangent line at,. [See Figure 5(a).] Bot continuit an ifferentiabilit are esirable properties for a function to ave. Te following teorem sows ow tese properties are relate.

17 SECTION 2.2 THE DERIVATIVE AS A FUNCTION 9 4 Teorem If f is ifferentiable at a, ten f is continuous at a. PROOF To prove tat f is continuous at a, we ave to sow tat lim l a f f a. We o tis b sowing tat te ifference f f a approaces. Te given information is tat f is ifferentiable at a, tat is, f a l a f f a a PS An important aspect of problem solving is tring to fin a connection between te given an te unknown. See Step 2 (Tink of a Plan) in Principles of Problem Solving on page 97. eists (see Equation 2..5). To connect te given an te unknown, we ivie an multipl f f a b a (wic we can o wen a): f f a f f a a a Tus, using te Prouct Law an (2..5), we can write lim f f a l a l a To use wat we ave just prove, we start wit f an a an subtract f a : Terefore f is continuous at a. l a f f a a f f a a f a a lim f f a f f a l a l a l a a l a f a l a f f a f a f a NOTE Te converse of Teorem 4 is false; tat is, tere are functions tat are continuous but not ifferentiable. For instance, te function is continuous at because f lim f f l l (See Eample 7 in Section.6.) But in Eample 5 we sowe tat at. f is not ifferentiable How Can a Function Fail to Be Differentiable? We saw tat te function in Eample 5 is not ifferentiable at an Figure 5(a) sows tat its grap canges irection abruptl wen. In general, if te grap of a function f as a corner or kink in it, ten te grap of f as no tangent at tis point an f is not ifferentiable tere. [In tring to compute f a, we fin tat te left an rigt limits are ifferent.]

18 2 CHAPTER 2 DERIVATIVES vertical tangent line Teorem 4 gives anoter wa for a function not to ave a erivative. It sas tat if f is not continuous at a, ten f is not ifferentiable at a. So at an iscontinuit ( for instance, a jump iscontinuit) f fails to be ifferentiable. A tir possibilit is tat te curve as a vertical tangent line wen a; tat is, f is continuous at a an lim f l a FIGURE 6 a Tis means tat te tangent lines become steeper an steeper as l a. Figure 6 sows one wa tat tis can appen; Figure 7(c) sows anoter. Figure 7 illustrates te tree possibilities tat we ave iscusse. a a a FIGURE 7 Tree was for ƒ not to be ifferentiable at a (a) A corner (b) A iscontinuit (c) A vertical tangent A graping calculator or computer provies anoter wa of looking at ifferentiabilit. If f is ifferentiable at a, ten wen we zoom in towar te point a, f a te grap straigtens out an appears more an more like a line. (See Figure 8. We saw a specific eample of tis in Figure 2 in Section 2..) But no matter ow muc we zoom in towar a point like te ones in Figures 6 an 7(a), we can t eliminate te sarp point or corner (see Figure 9). a a FIGURE 8 ƒ is ifferentiable at a. FIGURE 9 ƒ is not ifferentiable at a. Higer Derivatives If f is a ifferentiable function, ten its erivative f is also a function, so f ma ave a erivative of its own, enote b f f. Tis new function f is calle te secon erivative of f because it is te erivative of te erivative of f. Using Leibniz notation, we write te secon erivative of f as 2 2

19 SECTION 2.2 THE DERIVATIVE AS A FUNCTION 2 2 f fª f _.5.5 _2 FIGURE TEC In Moule 2.2 ou can see ow canging te coefficients of a polnomial f affects te appearance of te graps of f, f, an f. EXAMPLE 6 If f 3, fin an interpret f. SOLUTION In Eample 2 we foun tat te first erivative is f 3 2. So te secon erivative is f f l f f l l l Te graps of f, f, an f are sown in Figure. We can interpret f as te slope of te curve f at te point, f. In oter wors, it is te rate of cange of te slope of te original curve f. Notice from Figure tat f is negative wen f as negative slope an positive wen f as positive slope. So te graps serve as a ceck on our calculations. In general, we can interpret a secon erivative as a rate of cange of a rate of cange. Te most familiar eample of tis is acceleration, wic we efine as follows. If s s t is te position function of an object tat moves in a straigt line, we know tat its first erivative represents te velocit v t of te object as a function of time: v t s t s t Te instantaneous rate of cange of velocit wit respect to time is calle te acceleration a t of te object. Tus te acceleration function is te erivative of te velocit function an is terefore te secon erivative of te position function: a t v t s t or, in Leibniz notation, a v t 2 s t 2 Te tir erivative f is te erivative of te secon erivative: f f. So f can be interprete as te slope of te curve f or as te rate of cange of f. If f, ten alternative notations for te tir erivative are f Te process can be continue. Te fourt erivative f is usuall enote b f 4. In general, te nt erivative of f is enote b f n an is obtaine from f b ifferentiating n times. If f, we write n f n n n EXAMPLE 7 If f 3, fin f an f 4. SOLUTION In Eample 6 we foun tat f 6. Te grap of te secon erivative as equation 6 an so it is a straigt line wit slope 6. Since te erivative f is

20 22 CHAPTER 2 DERIVATIVES te slope of f, we ave f 6 for all values of. So f is a constant function an its grap is a orizontal line. Terefore, for all values of, f 4 We can also interpret te tir erivative psicall in te case were te function is te position function s s t of an object tat moves along a straigt line. Because s s a, te tir erivative of te position function is te erivative of te acceleration function an is calle te jerk: j a t 3 s t 3 Tus te jerk j is te rate of cange of acceleration. It is aptl name because a large jerk means a suen cange in acceleration, wic causes an abrupt movement in a veicle. We ave seen tat one application of secon an tir erivatives occurs in analzing te motion of objects using acceleration an jerk. We will investigate anoter application of secon erivatives in Section 3.3, were we sow ow knowlege of f gives us information about te sape of te grap of f. In Capter we will see ow secon an iger erivatives enable us to represent functions as sums of infinite series. 2.2 Eercises 2 Use te given grap to estimate te value of eac erivative. Ten sketc te grap of f.. (a) f 3 (b) f 2 (c) f () f (e) f (f) f 2 (g) f 3 3. Matc te grap of eac function in (a) () wit te grap of its erivative in I IV. Give reasons for our coices. (a) (b) (c) () 2. (a) f (b) f (c) f 2 () f 3 (e) f 4 (f) f 5 (g) f 6 () f 7 I II III IV ; Graping calculator or computer require. Homework Hints available at stewartcalculus.com

21 SECTION 2.2 THE DERIVATIVE AS A FUNCTION 23 4 Trace or cop te grap of te given function f. (Assume tat te aes ave equal scales.) Ten use te meto of Eample to sketc te grap of f below it A recargeable batter is plugge into a carger. Te grap sows C t, te percentage of full capacit tat te batter reaces as a function of time t elapse ( in ours). (a) Wat is te meaning of te erivative C t? (b) Sketc te grap of C t. Wat oes te grap tell ou? C percentage of full carge t (ours) Te grap (from te US Department of Energ) sows ow riving spee affects gas mileage. Fuel econom F is measure in miles per gallon an spee v is measure in miles per our. (a) Wat is te meaning of te erivative F v? (b) Sketc te grap of F v. (c) At wat spee soul ou rive if ou want to save on gas? 9.. F (mi/gal) (mi/) 5. Te grap sows ow te average age of first marriage of Japanese men varie in te last alf of te 2t centur. Sketc te grap of te erivative function M t. During wic ears was te erivative negative? M 2. Sown is te grap of te population function P t for east cells in a laborator culture. Use te meto of Eample to grap te erivative P t. Wat oes te grap of P tell us about te east population? 27 P 5 (east cells) t 5 5 t (ours) 6. Make a careful sketc of te grap of te sine function an below it sketc te grap of its erivative in te same manner as in Eercises 4. Can ou guess wat te erivative of te sine function is from its grap?

22 24 CHAPTER 2 DERIVATIVES ; 7. Let f 2. (a) Estimate te values of f, f ( 2 ), f, an f 2 b using a graping evice to zoom in on te grap of f. (b) Use smmetr to euce te values of f ( 2 ), f, an f 2. (c) Use te results from parts (a) an (b) to guess a formula for f. () Use te efinition of erivative to prove tat our guess in part (c) is correct. ; 8. Let f 3. (a) Estimate te values of f, f ( 2 ), f, f 2, an f 3 b using a graping evice to zoom in on te grap of f. (b) Use smmetr to euce te values of f ( 2 ), f, f 2, an f 3. (c) Use te values from parts (a) an (b) to grap f. () Guess a formula for f. (e) Use te efinition of erivative to prove tat our guess in part () is correct Fin te erivative of te function using te efinition of erivative. State te omain of te function an te omain of its erivative. 9. f f m b 2. f t 5t 9t f Te unemploment rate U t varies wit time. Te table (from te Bureau of Labor Statistics) gives te percentage of unemploe in te US labor force from 999 to 28. t (a) Wat is te meaning of U t? Wat are its units? (b) Construct a table of estimate values for U t. 34. Let P t be te percentage of Americans uner te age of 8 at time t. Te table gives values of tis function in census ears from 95 to 2. t U t P t (a) Wat is te meaning of P t? Wat are its units? (b) Construct a table of estimate values for P t. (c) Grap P an P. () How woul it be possible to get more accurate values for P t? t t U t P t f t s t G t 3 t f 4 t t st f f Te grap of f is given. State, wit reasons, te numbers at wic f is not ifferentiable _ (a) Sketc te grap of f s6 b starting wit te grap of s an using te transformations of Sec - tion.3. (b) Use te grap from part (a) to sketc te grap of f. (c) Use te efinition of a erivative to fin f. Wat are te omains of f an f? ; () Use a graping evice to grap f an compare wit our sketc in part (b). 3. (a) If f 4 2, fin f. ; (b) Ceck to see tat our answer to part (a) is reasonable b comparing te graps of f an f. 32. (a) If f, fin f. ; (b) Ceck to see tat our answer to part (a) is reasonable b comparing te graps of f an f _2 4 f s _2 2 ; 39. Grap te function. Zoom in repeatel, first towar te point (, ) an ten towar te origin. Wat is ifferent about te beavior of f in te vicinit of tese two points? Wat o ou conclue about te ifferentiabilit of f? ; 4. Zoom in towar te points (, ), (, ), an (, ) on te grap of te function t Wat o ou notice? Account for wat ou see in terms of te ifferentiabilit of t.

23 SECTION 2.2 THE DERIVATIVE AS A FUNCTION Te figure sows te graps of f, f, an f. Ientif eac curve, an eplain our coices. a ; Use te efinition of a erivative to fin f an f. Ten grap f, f, an f on a common screen an ceck to see if our answers are reasonable. 45. f b 46. f 3 3 c ; 47. If f 2 2 3, fin f, f, f, an f 4. Grap f, f, f, an f on a common screen. Are te graps consistent wit te geometric interpretations of tese erivatives? 42. Te figure sows graps of f, f, f, an f. Ientif eac curve, an eplain our coices. a b c 48. (a) Te grap of a position function of a car is sown, were s is measure in feet an t in secons. Use it to grap te velocit an acceleration of te car. Wat is te acceleration at t secons? s 43. Te figure sows te graps of tree functions. One is te position function of a car, one is te velocit of te car, an one is its acceleration. Ientif eac curve, an eplain our coices. a b c t 2 t (b) Use te acceleration curve from part (a) to estimate te jerk at t secons. Wat are te units for jerk? 49. Let f s 3. (a) If a, use Equation 2..5 to fin f a. (b) Sow tat f oes not eist. (c) Sow tat s 3 as a vertical tangent line at,. (Recall te sape of te grap of f. See Figure 3 in Section.2.) 44. Te figure sows te graps of four functions. One is te position function of a car, one is te velocit of te car, one is its acceleration, an one is its jerk. Ientif eac curve, an eplain our coices. a b t c 5. (a) If t 2 3, sow tat t oes not eist. (b) If a, fin t a. (c) Sow tat 2 3 as a vertical tangent line at,. ; () Illustrate part (c) b graping 2 3. f 6 5. Sow tat te function is not ifferentiable at 6. Fin a formula for f an sketc its grap. 52. Were is te greatest integer function f not ifferentiable? Fin a formula for f an sketc its grap. f 53. (a) Sketc te grap of te function. (b) For wat values of is f ifferentiable? (c) Fin a formula for f.

24 26 CHAPTER 2 DERIVATIVES 54. Te left-an an rigt-an erivatives of f at a are efine b an f a l f a l if tese limits eist. Ten f a eists if an onl if tese onesie erivatives eist an are equal. (a) Fin f 4 an f 4 for te function f (b) Sketc te grap of f. 5 5 f a f a f a f a if if 4 if 4 (c) Were is f iscontinuous? () Were is f not ifferentiable? 55. Recall tat a function f is calle even if f f for all in its omain an o if f f for all suc. Prove eac of te following. (a) Te erivative of an even function is an o function. (b) Te erivative of an o function is an even function. 56. Wen ou turn on a ot-water faucet, te temperature T of te water epens on ow long te water as been running. (a) Sketc a possible grap of T as a function of te time t tat as elapse since te faucet was turne on. (b) Describe ow te rate of cange of T wit respect to t varies as t increases. (c) Sketc a grap of te erivative of T. 57. Let be te tangent line to te parabola 2 at te point,. Te angle of inclination of is te angle tat makes wit te positive irection of te -ais. Calculate correct to te nearest egree. 2.3 Differentiation Formulas c =c slope= If it were alwas necessar to compute erivatives irectl from te efinition, as we i in te preceing section, suc computations woul be teious an te evaluation of some limits woul require ingenuit. Fortunatel, several rules ave been evelope for fining erivatives witout aving to use te efinition irectl. Tese formulas greatl simplif te task of ifferentiation. Let s start wit te simplest of all functions, te constant function f c. Te grap of tis function is te orizontal line c, wic as slope, so we must ave f. (See Figure.) A formal proof, from te efinition of a erivative, is also eas: f l f f c c l l FIGURE Te grap of ƒ=c is te line =c, so fª()=. In Leibniz notation, we write tis rule as follows. Derivative of a Constant Function c = slope= Power Functions We net look at te functions f n, were n is a positive integer. If n, te grap of f is te line, wic as slope. (See Figure 2.) So FIGURE 2 Te grap of ƒ= is te line =, so fª()=.

25 SECTION 2.3 DIFFERENTIATION FORMULAS 27 (You can also verif Equation from te efinition of a erivative.) We ave alrea investigate te cases n 2 an n 3. In fact, in Section 2.2 (Eercises 7 an 8) we foun tat For n 4 we fin te erivative of f 4 as follows: f l f f 4 4 l l l l Tus Comparing te equations in, 2, an 3, we see a pattern emerging. It seems to be a reasonable guess tat, wen n is a positive integer, n n n. Tis turns out to be true. We prove it in two was; te secon proof uses te Binomial Teorem. Te Power Rule If n is a positive integer, ten n n n FIRST PROOF Te formula n a n a n n 2 a a n 2 a n can be verifie simpl b multipling out te rigt-an sie (or b summing te secon factor as a geometric series). If f n, we can use Equation 2..5 for f a an te equation above to write f a l a f f a a n a n l a a l a n n 2 a a n 2 a n a n a n 2 a aa n 2 a n na n

26 28 CHAPTER 2 DERIVATIVES SECOND PROOF f l f f l n n Te Binomial Teorem is given on Reference Page. In fining te erivative of 4 we a to epan 4. Here we nee to epan n an we use te Binomial Teorem to o so: n n n n n n n 2 2 n n n 2 f l n n n n n 2 2 n n n 2 l n n n l n n n 2 n n 2 2 n n because ever term ecept te first as as a factor an terefore approaces. We illustrate te Power Rule using various notations in Eample. EXAMPLE (a) If f 6, ten f 6 5. (b) If, ten 999. (c) If t 4 3, ten 4t. () t r r 3 3r 2 New Derivatives from Ol Wen new functions are forme from ol functions b aition, subtraction, or multiplication b a constant, teir erivatives can be calculate in terms of erivatives of te ol functions. In particular, te following formula sas tat te erivative of a constant times a function is te constant times te erivative of te function. GEOMETRIC INTERPRETATION OF THE CONSTANT MULTIPLE RULE =2ƒ =ƒ Multipling b c 2 stretces te grap verticall b a factor of 2. All te rises ave been ouble but te runs sta te same. So te slopes are ouble too. Te Constant Multiple Rule If c is a constant an f is a ifferentiable function, ten PROOF Let t cf. Ten cf c f t t cf cf t l l l c c lim l cf f f f f (b Law 3 of limits)

27 SECTION 2.3 DIFFERENTIATION FORMULAS 29 EXAMPLE 2 (a) (b) Te net rule tells us tat te erivative of a sum of functions is te sum of te erivatives. Using prime notation, we can write te Sum Rule as f t f t Te Sum Rule If f an t are bot ifferentiable, ten f t f t PROOF Let F f t. Ten F l F F f t f t l l f f t t l f f l t t (b Law ) f t Te Sum Rule can be etene to te sum of an number of functions. For instance, using tis teorem twice, we get f t f t f t f t B writing f t as f t an appling te Sum Rule an te Constant Multiple Rule, we get te following formula. Te Difference Rule If f an t are bot ifferentiable, ten f t f t Te Constant Multiple Rule, te Sum Rule, an te Difference Rule can be combine wit te Power Rule to ifferentiate an polnomial, as te following eamples emonstrate.

28 3 CHAPTER 2 DERIVATIVES EXAMPLE {_œ 3, _5} (, 4) {œ 3, _5} v EXAMPLE 4 Fin te points on te curve were te tangent line is orizontal. SOLUTION Horizontal tangents occur were te erivative is zero. We ave FIGURE 3 Te curve =$-6@+4 an its orizontal tangents Tus if or 2 3, tat is, s3. So te given curve as orizontal tangents wen, s3, an s3. Te corresponing points are, 4, (s3, 5), an ( s3, 5). (See Figure 3.) EXAMPLE 5 Te equation of motion of a particle is s 2t 3 5t 2 3t 4, were s is measure in centimeters an t in secons. Fin te acceleration as a function of time. Wat is te acceleration after 2 secons? SOLUTION Te velocit an acceleration are v t s t 6t 2 t 3 a t v t Te acceleration after 2 s is a 2 4 cm s 2. 2t Net we nee a formula for te erivative of a prouct of two functions. B analog wit te Sum an Difference Rules, one migt be tempte to guess, as Leibniz i tree centuries ago, tat te erivative of a prouct is te prouct of te erivatives. We can see, owever, tat tis guess is wrong b looking at a particular eample. Let f an t 2. Ten te Power Rule gives f an t 2. But ft 3, so ft 3 2. Tus ft f t. Te correct formula was iscovere b Leibniz (soon after is false start) an is calle te Prouct Rule. We can write te Prouct Rule in prime notation as ft ft tf Te Prouct Rule If f an t are bot ifferentiable, ten f t f t t f

29 SECTION 2.3 DIFFERENTIATION FORMULAS 3 PROOF Let F f t. Ten F l F F l f t f t In orer to evaluate tis limit, we woul like to separate te functions f an t as in te proof of te Sum Rule. We can acieve tis separation b subtracting an aing te term f t in te numerator: F l f t f t f t f t t t t l f f f t t f t l l l l f f f t t f Note tat lim l t t because t is a constant wit respect to te variable. Also, since f is ifferentiable at, it is continuous at b Teorem 2.2.4, an so lim l f f. (See Eercise 59 in Section.8.) In wors, te Prouct Rule sas tat te erivative of a prouct of two functions is te first function times te erivative of te secon function plus te secon function times te erivative of te first function. EXAMPLE 6 Fin F if F SOLUTION B te Prouct Rule, we ave F Notice tat we coul verif te answer to Eample 6 irectl b first multipling te factors: F ? F But later we will meet functions, suc as 2 sin, for wic te Prouct Rule is te onl possible meto.

30 32 CHAPTER 2 DERIVATIVES v EXAMPLE 7 If t an it is known tat t 3 5 an t 3 2, fin 3. SOLUTION Appling te Prouct Rule, we get t t t t t Terefore 3 3t 3 t In prime notation we can write te Quotient Rule as f t t f ft t 2 Te Quotient Rule If f an t are ifferentiable, ten t f t f f t t 2 PROOF Let F f t. Ten F F F l l l f t f t t t f t f t We can separate f an t in tis epression b subtracting an aing te term f t in te numerator: F l t l f t f t f t f t t t lim t l l f f f t t t f f t t 2 f f lim l t t t t f l l t l t Again t is continuous b Teorem 2.2.4, so lim l t t. In wors, te Quotient Rule sas tat te erivative of a quotient is te enominator times te erivative of te numerator minus te numerator times te erivative of te enominator, all ivie b te square of te enominator. Te teorems of tis section sow tat an polnomial is ifferentiable on an an rational function is ifferentiable on its omain. Furtermore, te Quotient Rule an te

31 SECTION 2.3 DIFFERENTIATION FORMULAS 33 oter ifferentiation formulas enable us to compute te erivative of an rational function, as te net eample illustrates. We can use a graping evice to ceck tat te answer to Eample 8 is plausible. Figure 4 sows te graps of te function of Eample 8 an its erivative. Notice tat wen grows rapil (near 2), is large. An wen grows slowl, is near..5 _4 4 FIGURE 4 ª _.5 v EXAMPLE 8 Let Ten NOTE Don t use te Quotient Rule ever time ou see a quotient. Sometimes it s easier to rewrite a quotient first to put it in a form tat is simpler for te purpose of ifferentiation. For instance, altoug it is possible to ifferentiate te function using te Quotient Rule, it is muc easier to perform te ivision first an write te function as before ifferentiating. General Power Functions F 3 2 2s F Te Quotient Rule can be use to eten te Power Rule to te case were te eponent is a negative integer. If n is a positive integer, ten n n n PROOF n n n n n 2 n n n 2n nn 2n n n 2n n n

32 34 CHAPTER 2 DERIVATIVES EXAMPLE 9 (a) If, ten (b) So far we know tat te Power Rule ols if te eponent n is a positive or negative integer. If n, ten, wic we know as a erivative of. Tus te Power Rule ols for an integer n. Wat if te eponent is a fraction? In Eample 3 in Section 2.2 we foun tat wic can be written as t t t t t 4 8 t 4 s 2s Tis sows tat te Power Rule is true even wen n 2. In fact, it also ols for an real number n, as we will prove in Capter 6. (A proof for rational values of n is inicate in Eercise 48 in Section 2.6.) In te meantime we state te general version an use it in te eamples an eercises. Te Power Rule (General Version) If n is an real number, ten n n n EXAMPLE (a) If f, ten f. (b) Let s 3 2 Ten In Eample, a an b are constants. It is customar in matematics to use letters near te beginning of te alpabet to represent constants an letters near te en of te alpabet to represent variables. EXAMPLE Differentiate te function f t st a bt. SOLUTION Using te Prouct Rule, we ave f t st t a bt a bt t (st ) st b a bt 2 t 2 bst a bt 2st a 3bt 2st

33 SECTION 2.3 DIFFERENTIATION FORMULAS 35 SOLUTION 2 If we first use te laws of eponents to rewrite f t, ten we can procee irectl witout using te Prouct Rule. f t ast btst at 2 bt 3 2 f t 2at 2 3 2bt 2 wic is equivalent to te answer given in Solution. Te ifferentiation rules enable us to fin tangent lines witout aving to resort to te efinition of a erivative. Te also enable us to fin normal lines. Te normal line to a curve C at point P is te line troug P tat is perpenicular to te tangent line at P. (In te stu of optics, one nees to consier te angle between a ligt ra an te normal line to a lens.) EXAMPLE 2 Fin equations of te tangent line an normal line to te curve s 2 at te point (, 2 ). SOLUTION Accoring to te Quotient Rule, we ave 2 (s ) s s 2 2s s 2 2 So te slope of te tangent line at (, 2 ) is 3 2 2s 2 2 tangent normal 3 2 2s We use te point-slope form to write an equation of te tangent line at (, 2 ): 2 4 or FIGURE 5 Te slope of te normal line at (, 2 ) is te negative reciprocal of 4, namel 4, so an equation is 2 4 or Te curve an its tangent an normal lines are grape in Figure 5. EXAMPLE 3 At wat points on te perbola 2 is te tangent line parallel to te line 3? SOLUTION Since 2 can be written as 2, we ave

34 36 CHAPTER 2 DERIVATIVES (2, 6) =2 Let te -coorinate of one of te points in question be a. Ten te slope of te tangent line at tat point is 2 a 2. Tis tangent line will be parallel to te line 3, or 3, if it as te same slope, tat is, 3. Equating slopes, we get (_2, _6) FIGURE 6 3+= 2 a 2 3 or a 2 4 or a 2 Terefore te require points are 2, 6 an 2, 6. Te perbola an te tangents are sown in Figure 6. We summarize te ifferentiation formulas we ave learne so far as follows. Table of Differentiation Formulas c n n n cf cf f t f t f t f t ft ft tf f t tf ft t Eercises 22 Differentiate te function.. f f 2 3. f t t 4. F f f t 2 t 6 3t 4 t 7. t t t 2t 3 4. B c 6. A s s 5 3. S p sp p 4. s 23. Fin te erivative of f in two was: b using te Prouct Rule an b performing te multiplication first. Do our answers agree? 24. Fin te erivative of te function in two was: b using te Quotient Rule an b simplifing first. Sow tat our answers are equivalent. Wic meto o ou prefer? Differentiate. 25. V F s 2 5. R a 3a 2 6. S R 4 R s s 2 9. H 3 2. t u s2 u s3u 2. u s 5 t 4st v s 2 s L F J v v 3 2v v 4 v 2 2 t f 3 3 ; Graping calculator or computer require. Homework Hints available at stewartcalculus.com

35 SECTION 2.3 DIFFERENTIATION FORMULAS v3 2vsv 34. v t t t t 4 3t 2 t t 2 t st t Fin an equation of te tangent line to te curve at te given point. 5. 2,, ,, a 2 b c f t 2t 4. 2 st 4. s 3 t t 2 t t f 44. c A B C 2 c c u6 2u 3 5 u 2 f a b c 53. (a) Te curve 2 is calle a witc of Maria Agnesi. Fin an equation of te tangent line to tis curve at te point (, 2 ). ; (b) Illustrate part (a) b graping te curve an te tangent line on te same screen. 54. (a) Te curve 2 is calle a serpentine. Fin an equation of te tangent line to tis curve at te point 3,.3. ; (b) Illustrate part (a) b graping te curve an te tangent line on te same screen. 45. Te general polnomial of egree n as te form P a n n a n n a 2 2 a a were a n. Fin te erivative of P. ; Fin f. Compare te graps of f an f an use tem to eplain w our answer is reasonable. 46. f f f ; 49. (a) Use a graping calculator or computer to grap te function f in te viewing rectangle 3, 5 b, 5. (b) Using te grap in part (a) to estimate slopes, make a roug sketc, b an, of te grap of f. (See Eample in Section 2.2.) (c) Calculate f an use tis epression, wit a graping evice, to grap f. Compare wit our sketc in part (b). ; 5. (a) Use a graping calculator or computer to grap te function t 2 2 in te viewing rectangle 4, 4 b,.5. (b) Using te grap in part (a) to estimate slopes, make a roug sketc, b an, of te grap of t. (See Eample in Section 2.2.) (c) Calculate t an use tis epression, wit a graping evice, to grap t. Compare wit our sketc in part (b) Fin equations of te tangent line an normal line to te curve at te given point. 55. s,, , 3 57.,, s, Fin te first an secon erivatives of te function. 6. f Te equation of motion of a particle is s t 3 3t, were s is in meters an t is in secons. Fin (a) te velocit an acceleration as functions of t, (b) te acceleration after 2 s, an (c) te acceleration wen te velocit is. 64. Te equation of motion of a particle is, 9 4, f G r sr s 3 r s t 4 2t 3 t 2 t f 3 were s is in meters an t is in secons. (a) Fin te velocit an acceleration as functions of t. (b) Fin te acceleration after s. ; (c) Grap te position, velocit, an acceleration functions on te same screen.

36 38 CHAPTER 2 DERIVATIVES 65. Bole s Law states tat wen a sample of gas is compresse at a constant pressure, te pressure P of te gas is inversel proportional to te volume V of te gas. (a) Suppose tat te pressure of a sample of air tat occupies.6 m 3 at 25 C is 5 kpa. Write V as a function of P. (b) Calculate V P wen P 5 kpa. Wat is te meaning of te erivative? Wat are its units? ; 66. Car tires nee to be inflate properl because overinflation or unerinflation can cause premature treaware. Te ata in te table sow tire life L ( in tousans of miles) for a certain tpe of tire at various pressures P ( in lb in 2 ). (a) Use a graping calculator or computer to moel tire life wit a quaratic function of te pressure. (b) Use te moel to estimate L P wen P 3 an wen P 4. Wat is te meaning of te erivative? Wat are te units? Wat is te significance of te signs of te erivatives? 67. Suppose tat f 5, f 5 6, t 5 3, an t 5 2. Fin te following values. (a) ft 5 (b) f t 5 (c) t f Fin 2, given tat f 2 3, t 2 4, f 2 2, an t 2 7. (a) 5f 4t (b) f t (c) P L f t 69. If f s t, were t 4 8 an t 4 7, fin f If 2 4 an 2 3, fin 7. If f an t are te functions wose graps are sown, let u f t an v f t. (a) Fin u. (b) Fin v 5. f () 2 g t f 72. Let P F G an Q F G, were F an G are te functions wose graps are sown. (a) Fin P 2. (b) Fin Q If t is a ifferentiable function, fin an epression for te erivative of eac of te following functions. (a) t (b) (c) t t 74. If f is a ifferentiable function, fin an epression for te erivative of eac of te following functions. (a) (b) f 2 f 2 f (c) 2 () f s 75. Fin te points on te curve were te tangent is orizontal. 76. For wat values of oes te grap of f ave a orizontal tangent? 77. Sow tat te curve as no tangent line wit slope Fin an equation of te tangent line to te curve s tat is parallel to te line Fin equations of bot lines tat are tangent to te curve 3 an are parallel to te line Fin equations of te tangent lines to te curve tat are parallel to te line Fin an equation of te normal line to te parabola tat is parallel to te line Were oes te normal line to te parabola 2 at te point (, ) intersect te parabola a secon time? Illustrate wit a sketc. 83. Draw a iagram to sow tat tere are two tangent lines to te parabola 2 tat pass troug te point, 4. Fin te coorinates of te points were tese tangent lines intersect te parabola. 84. (a) Fin equations of bot lines troug te point 2, 3 tat are tangent to te parabola 2. F G

37 SECTION 2.3 DIFFERENTIATION FORMULAS 39 (b) Sow tat tere is no line troug te point 2, 7 tat is tangent to te parabola. Ten raw a iagram to see w. 85. (a) Use te Prouct Rule twice to prove tat if f, t, an are ifferentiable, ten ft f t ft ft. (b) Taking f t in part (a), sow tat (c) Use part (b) to ifferentiate Fin te nt erivative of eac function b calculating te first few erivatives an observing te pattern tat occurs. (a) f n (b) f 87. Fin a secon-egree polnomial P suc tat P 2 5, P 2 3, an P Te equation 2 2 is calle a ifferential equation because it involves an unknown function an its erivatives an. Fin constants A, B, an C suc tat te function A 2 B C satisfies tis equation. (Differential equations will be stuie in etail in Capter 9.) 89. Fin a cubic function a 3 b 2 c wose grap as orizontal tangents at te points 2, 6 an 2,. 9. Fin a parabola wit equation a 2 b c tat as slope 4 at, slope 8 at, an passes troug te point 2, In tis eercise we estimate te rate at wic te total personal income is rising in te Ricmon-Petersburg, Virginia, metropolitan area. In 999, te population of tis area was 96,4, an te population was increasing at rougl 92 people per ear. Te average annual income was $3,593 per capita, an tis average was increasing at about $4 per ear (a little above te national average of about $225 earl). Use te Prouct Rule an tese figures to estimate te rate at wic total personal income was rising in te Ricmon-Petersburg area in 999. Eplain te meaning of eac term in te Prouct Rule. 92. A manufacturer prouces bolts of a fabric wit a fie wit. Te quantit q of tis fabric (measure in ars) tat is sol is a function of te selling price p ( in ollars per ar), so we can write q f p. Ten te total revenue earne wit selling price p is R p pf p. (a) Wat oes it mean to sa tat f 2, an f 2 35? (b) Assuming te values in part (a), fin R 2 an interpret our answer. 93. Let f 3 3 f 2 f f 2 if if Is f ifferentiable at? Sketc te graps of f an f. 94. At wat numbers is te following function t ifferentiable? Give a formula for t an sketc te graps of t an t. 95. (a) For wat values of is te function ifferentiable? Fin a formula for f. (b) Sketc te graps of f an f. 96. Were is te function ifferenti - able? Give a formula for an sketc te graps of an. 97. For wat values of a an b is te line 2 b tangent to te parabola a 2 wen 2? 98. (a) If F f t, were f an t ave erivatives of all orers, sow tat F f t 2f t ft. (b) Fin similar formulas for F an F 4. (c) Guess a formula for. 99. Fin te value of c suc tat te line is tangent to te curve cs.. Let t f 2 Fin te values of m an b tat make f ifferentiable everwere.. An eas proof of te Quotient Rule can be given if we make te prior assumption tat F eists, were F f t. Write f Ft; ten ifferentiate using te Prouct Rule an solve te resulting equation for F. 2. A tangent line is rawn to te perbola c at a point P. (a) Sow tat te mipoint of te line segment cut from tis tangent line b te coorinate aes is P. (b) Sow tat te triangle forme b te tangent line an te coorinate aes alwas as te same area, no matter were P is locate on te perbola. 3. Evaluate lim. l 2 F n m b if if 2 if 2 if 2 if 2 f Draw a iagram sowing two perpenicular lines tat intersect on te -ais an are bot tangent to te parabola 2. Were o tese lines intersect? 5. If c 2, ow man lines troug te point, c are normal lines to te parabola 2? Wat if c 2? 6. Sketc te parabolas 2 an Do ou tink tere is a line tat is tangent to bot curves? If so, fin its equation. If not, w not?

38 4 CHAPTER 2 DERIVATIVES APPLIED PROJECT BUILDING A BETTER ROLLER COASTER L P f Q L Suppose ou are aske to esign te first ascent an rop for a new roller coaster. B stuing potograps of our favorite coasters, ou ecie to make te slope of te ascent.8 an te slope of te rop.6. You ecie to connect tese two straigt stretces L an L 2 wit part of a parabola f a 2 b c, were an f are measure in feet. For te track to be smoot tere can t be abrupt canges in irection, so ou want te linear segments L an L 2 to be tangent to te parabola at te transition points P an Q. (See te figure.) To simplif te equations, ou ecie to place te origin at P.. (a) Suppose te orizontal istance between P an Q is ft. Write equations in a, b, an c tat will ensure tat te track is smoot at te transition points. (b) Solve te equations in part (a) for a, b, an c to fin a formula for f. ; (c) Plot L, f, an L 2 to verif grapicall tat te transitions are smoot. () Fin te ifference in elevation between P an Q. 2. Te solution in Problem migt look smoot, but it migt not feel smoot because te piecewise efine function [consisting of L for, f for, an L 2 for ] oesn t ave a continuous secon erivative. So ou ecie to improve te esign b using a quaratic function q a 2 b c onl on te interval 9 an connecting it to te linear functions b means of two cubic functions: t k 3 l 2 m n p 3 q 2 r s 9 Flason Stuio / Sutterstock CAS (a) Write a sstem of equations in unknowns tat ensure tat te functions an teir first two erivatives agree at te transition points. (b) Solve te equations in part (a) wit a computer algebra sstem to fin formulas for q, t, an. (c) Plot L, t, q,, an L 2, an compare wit te plot in Problem (c). ; Graping calculator or computer require CAS Computer algebra sstem require 2.4 Derivatives of Trigonometric Functions A review of trigonometric functions is given in Appeni D. Before starting tis section, ou migt nee to review te trigonometric functions. In particular, it is important to remember tat wen we talk about te function f efine for all real numbers b f sin it is unerstoo tat sin means te sine of te angle wose raian measure is. A similar convention ols for te oter trigonometric functions cos, tan, csc, sec, an cot. Recall from Section.8 tat all of te trigonometric functions are continuous at ever number in teir omains. If we sketc te grap of te function f sin an use te interpretation of f as te slope of te tangent to te sine curve in orer to sketc te grap of f (see Eercise 6 in Section 2.2), ten it looks as if te grap of f ma be te same as te cosine curve (see Figure ).

39 SECTION 2.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 4 = ƒ= sin π π 2π 2 TEC Visual 2.4 sows an animation of Figure. = fª() π 2 π FIGURE Let s tr to confirm our guess tat if f sin, ten f cos. From te efinition of a erivative, we ave f l f f l sin sin We ave use te aition formula for sine. See Appeni D. l sin cos cos sin sin l l sin sin cos sin cos l sin l cos cos sin cos sin sin cos l l Two of tese four limits are eas to evaluate. Since we regar as a constant wen computing a limit as l, we ave lim sin sin an lim cos cos l l Te limit of sin is not so obvious. In Eample 3 in Section.5 we mae te guess, on te basis of numerical an grapical evience, tat sin 2 lim l We now use a geometric argument to prove Equation 2. Assume first tat lies between an 2. Figure 2(a) sows a sector of a circle wit center O, central angle, an

40 42 CHAPTER 2 DERIVATIVES B D raius. BC is rawn perpenicular to OA. B te efinition of raian measure, we ave arc AB. Also sin sin. From te iagram we see tat BC OB BC AB arc AB E Terefore sin so sin O O (a) C B E A A Let te tangent lines at A an B intersect at E. You can see from Figure 2(b) tat te cir cumference of a circle is smaller tan te lengt of a circumscribe polgon, an so arc. Tus AB AE EB arc AB AE EB AE ED AD OA tan tan FIGURE 2 (b) (In Appeni F te inequalit tan is prove irectl from te efinition of te lengt of an arc witout resorting to geometric intuition as we i ere.) Terefore we ave sin cos so cos sin We know tat lim l an lim l cos, so b te Squeeze Teorem, we ave sin lim l But te function sin is an even function, so its rigt an left limits must be equal. Hence, we ave sin lim l We multipl numerator an enominator b cos in orer to put te function in a form in wic we can use te limits we know. so we ave prove Equation 2. We can euce te value of te remaining limit in cos lim l l l cos sin 2 cos as follows: cos cos cos 2 l cos sin sin lim l cos sin sin lim l l cos (b Equation 2)

41 SECTION 2.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 43 3 cos lim l If we now put te limits 2 an 3 in, we get cos f sin l l sin cos cos sin cos l l So we ave prove te formula for te erivative of te sine function: 4 sin cos Figure 3 sows te graps of te function of Eample an its eriva tive. Notice tat wenever as a orizontal tangent. 5 ª _4 4 v EXAMPLE Differentiate 2 sin. SOLUTION Using te Prouct Rule an Formula 4, we ave 2 sin sin 2 2 cos 2 sin Using te same metos as in te proof of Formula 4, one can prove (see Eercise 2) tat FIGURE 3 _5 5 cos sin Te tangent function can also be ifferentiate b using te efinition of a erivative, but it is easier to use te Quotient Rule togeter wit Formulas 4 an 5: tan sin cos cos sin sin cos cos 2 cos cos sin sin cos 2 cos2 sin 2 cos 2 cos 2 sec2

42 44 CHAPTER 2 DERIVATIVES 6 tan sec2 Te erivatives of te remaining trigonometric functions, csc, sec, an cot, can also be foun easil using te Quotient Rule (see Eercises 7 9). We collect all te ifferentiation formulas for trigonometric functions in te following table. Remember tat te are vali onl wen is measure in raians. Derivatives of Trigonometric Functions Wen ou memorize tis table, it is elpful to notice tat te minus signs go wit te erivatives of te cofunctions, tat is, cosine, cosecant, an cotangent. sin cos cos sin tan sec2 csc csc cot sec sec tan cot csc2 EXAMPLE 2 Differentiate f ave a orizontal tangent? SOLUTION Te Quotient Rule gives sec. For wat values of oes te grap of f tan f tan sec sec tan 2 tan tan sec tan sec sec2 tan 2 3 sec tan tan2 sec 2 tan 2 _3 5 sec tan tan 2 _3 FIGURE 4 Te orizontal tangents in Eample 2 In simplifing te answer we ave use te ientit tan 2 sec 2. Since sec is never, we see tat f wen tan, an tis occurs wen n 4, were n is an integer (see Figure 4). Trigonometric functions are often use in moeling real-worl penomena. In particular, vibrations, waves, elastic motions, an oter quantities tat var in a perioic manner can be escribe using trigonometric functions. In te following eample we iscuss an instance of simple armonic motion.

43 FIGURE 5 4 s SECTION 2.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 45 v EXAMPLE 3 An object at te en of a vertical spring is stretce 4 cm beon its rest position an release at time t. (See Figure 5 an note tat te ownwar irection is positive.) Its position at time t is s f t 4 cos t Fin te velocit an acceleration at time t an use tem to analze te motion of te object. SOLUTION Te velocit an acceleration are 2 s a v s t t 4 cos t 4 cos t 4 sin t t a v t t 4 sin t 4 sin t 4 cos t t _2 FIGURE 6 π 2π t Te object oscillates from te lowest point s 4 cm to te igest point s 4 cm. Te perio of te oscillation is 2, te perio of cos t. Te spee is v 4 sin t, wic is greatest wen sin t, tat is, wen cos t. So te object moves fastest as it passes troug its equilibrium position s. Its spee is wen sin t, tat is, at te ig an low points. Te acceleration a 4 cos t wen s. It as greatest magnitue at te ig an low points. See te graps in Figure 6. EXAMPLE 4 Fin te 27t erivative of cos. SOLUTION Te first few erivatives of f cos are as follows: PS Look for a pattern. f sin f cos f sin f 4 cos f 5 sin We see tat te successive erivatives occur in a ccle of lengt 4 an, in particular, f n cos wenever n is a multiple of 4. Terefore f 24 cos an, ifferentiating tree more times, we ave f 27 sin Our main use for te limit in Equation 2 as been to prove te ifferentiation formula for te sine function. But tis limit is also useful in fining certain oter trigonometric limits, as te following two eamples sow. Note tat sin 7 7 sin. EXAMPLE 5 sin 7 Fin lim. l 4 SOLUTION In orer to appl Equation 2, we first rewrite te function b multipling an iviing b 7: sin 7 7 sin

44 46 CHAPTER 2 DERIVATIVES If we let 7, ten l as l, so b Equation 2 we ave sin 7 lim 7 l 4 4 lim sin 7 l 7 v EXAMPLE 6 Calculate lim cot. l 7 4 lim sin 7 l SOLUTION Here we ivie numerator an enominator b : cos lim cot l l sin l cos cos sin lim cos l lim l sin (b te continuit of cosine an Equation 2) 2.4 Eercises 6 Differentiate.. f cos 2. f s sin 3. f sin 2 cot 4. 2 sec csc 5. sec tan 6. t t 4 sec t tan t 7. c cos t t 2 sin t 8. u a cos u b cot u 9.. sin cos 2 tan sec. f 2. sec 3. t sin t 4. t 7. Prove tat csc csc cot. 8. Prove tat sec sec tan. 9. Prove tat. cot csc2 cos sin sec tan 5. csc cot 6. 2 sin tan 2. Prove, using te efinition of erivative, tat if f cos, ten f sin Fin an equation of te tangent line to te curve at te given point. 2. sec, 3, cos,, 23. cos sin,, 24. tan,, 25. (a) Fin an equation of te tangent line to te curve 2 sin at te point 2,. ; (b) Illustrate part (a) b graping te curve an te tangent line on te same screen. 26. (a) Fin an equation of te tangent line to te curve 3 6 cos at te point 3, 3. ; (b) Illustrate part (a) b graping te curve an te tangent line on te same screen. 27. (a) If f sec, fin f. ; (b) Ceck to see tat our answer to part (a) is reasonable b graping bot an for. f f (a) If f s sin, fin f. ; (b) Ceck to see tat our answer to part (a) is reasonable b graping bot f an f for 2. ; Graping calculator or computer require. Homework Hints available at stewartcalculus.com

45 SECTION 2.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS If H sin, fin H an H. 3. If f t csc t, fin f (a) Use te Quotient Rule to ifferentiate te function f tan sec (b) Simplif te epression for f b writing it in terms of sin an cos, an ten fin f. (c) Sow tat our answers to parts (a) an (b) are equivalent. 32. Suppose f 3 4 an f 3 2, an let t f sin an cos f. Fin (a) t 3 (b) For wat values of oes te grap of f 2sin ave a orizontal tangent? 34. Fin te points on te curve cos 2 sin at wic te tangent is orizontal. 35. A mass on a spring vibrates orizontall on a smoot level surface (see te figure). Its equation of motion is t 8sin t, were t is in secons an in centimeters. (a) Fin te velocit an acceleration at time t. (b) Fin te position, velocit, an acceleration of te mass at time t 2 3. In wat irection is it moving at tat time? equilibrium position ; 36. An elastic ban is ung on a ook an a mass is ung on te lower en of te ban. Wen te mass is pulle ownwar an ten release, it vibrates verticall. Te equation of motion is s 2 cos t 3 sin t, t, were s is measure in centi meters an t in secons. (Take te positive irection to be ownwar.) (a) Fin te velocit an acceleration at time t. (b) Grap te velocit an acceleration functions. (c) Wen oes te mass pass troug te equilibrium position for te first time? () How far from its equilibrium position oes te mass travel? (e) Wen is te spee te greatest? 37. A laer ft long rests against a vertical wall. Let be te angle between te top of te laer an te wall an let be te istance from te bottom of te laer to te wall. If te bottom of te laer slies awa from te wall, ow fast oes cange wit respect to wen 3? 38. An object wit weigt W is ragge along a orizontal plane b a force acting along a rope attace to te object. If te rope makes an angle wit te plane, ten te magnitue of te force is W F sin cos were is a constant calle te coefficient of friction. (a) Fin te rate of cange of F wit respect to. (b) Wen is tis rate of cange equal to? ; (c) If W 5 lb an.6, raw te grap of F as a function of an use it to locate te value of for wic F. Is te value consistent wit our answer to part (b)? Fin te limit. sin lim 4. l tan 6t 4. lim 42. t l sin 2t sin lim 44. l sin 45. lim 46. l tan tan 47. lim 48. l 4 sin cos 49 5 Fin te given erivative b fining te first few erivatives an observing te pattern tat occurs sin 35 sin Fin constants A an B suc tat te function A sin B cos satisfies te ifferential equation 2 sin. 52. (a) Evaluate lim sin. l (b) Evaluate lim sin. l ; (c) Illustrate parts (a) an (b) b graping sin. 53. Differentiate eac trigonometric ientit to obtain a new (or familiar) ientit. (a) tan sin cos (c) sin cos cot csc (b) sin 4 lim l sin 6 cos lim l sin sin 3 sin 5 lim l 2 sin 2 lim l lim l sin 2 2 sec cos

46 48 CHAPTER 2 DERIVATIVES 54. A semicircle wit iameter PQ sits on an isosceles triangle PQR to form a region sape like a two-imensional icecream cone, as sown in te figure. If A is te area of te semicircle an B is te area of te triangle, fin A lim l B 55. Te figure sows a circular arc of lengt s an a cor of lengt, bot subtene b a central angle. Fin s lim l s A( ) P Q B( ) cm cm R ; 56. Let f. s cos 2 (a) Grap f. Wat tpe of iscontinuit oes it appear to ave at? (b) Calculate te left an rigt limits of f at. Do tese values confirm our answer to part (a)? 2.5 Te Cain Rule Suppose ou are aske to ifferentiate te function F s 2 See Section.3 for a review of composite functions. Te ifferentiation formulas ou learne in te previous sections of tis capter o not enable ou to calculate F. Observe tat F is a composite function. In fact, if we let f u su an let u t 2, ten we can write F f t, tat is, F f t. We know ow to ifferentiate bot f an t, so it woul be useful to ave a rule tat tells us ow to fin te erivative of F f t in terms of te erivatives of f an t. It turns out tat te erivative of te composite function f t is te prouct of te erivatives of f an t. Tis fact is one of te most important of te ifferentiation rules an is calle te Cain Rule. It seems plausible if we interpret erivatives as rates of cange. Regar u as te rate of cange of u wit respect to, u as te rate of cange of wit respect to u, an as te rate of cange of wit respect to. If u canges twice as fast as an canges tree times as fast as u, ten it seems reasonable tat canges si times as fast as, an so we epect tat u u Te Cain Rule If t is ifferentiable at an f is ifferentiable at t, ten te composite function F f t efine b F f t is ifferentiable at an F is given b te prouct F f t t In Leibniz notation, if f u an u t are bot ifferentiable functions, ten u u

47 SECTION 2.5 THE CHAIN RULE 49 James Gregor Te first person to formulate te Cain Rule was te Scottis matematician James Gregor ( ), wo also esigne te first practical reflecting telescope. Gregor iscovere te basic ieas of calculus at about te same time as Newton. He became te first Professor of Matematics at te Universit of St. Anrews an later el te same position at te Universit of Einburg. But one ear after accepting tat position e ie at te age of 36. COMMENTS ON THE PROOF OF THE CHAIN RULE Let u be te cange in ucorresponing to a cange of in, tat is, Ten te corresponing cange in is It is tempting to write l u t t f u u f u u l u u l u l u u l u l (Note tat u l as l since t is continuous.) u u Te onl flaw in tis reasoning is tat in it migt appen tat u (even wen ) an, of course, we can t ivie b. Noneteless, tis reasoning oes at least suggest tat te Cain Rule is true. A full proof of te Cain Rule is given at te en of tis section. Te Cain Rule can be written eiter in te prime notation 2 f t f t t or, if f u an u t, in Leibniz notation: 3 u u Equation 3 is eas to remember because if u an u were quotients, ten we coul cancel u. Remember, owever, tat u as not been efine an u soul not be tougt of as an actual quotient. EXAMPLE Fin F if F s 2. SOLUTION (using Equation 2): At te beginning of tis section we epresse F as F f t f t were f u su an t 2. Since f u 2u 2 2su an t 2 we ave F f t t 2 2s 2 s 2

48 5 CHAPTER 2 DERIVATIVES SOLUTION 2 (using Equation 3): If we let u 2 an su, ten F u u 2su 2 2 2s 2 Wen using Formula 3 we soul bear in min tat refers to te erivative of wen is consiere as a function of (calle te erivative of wit respect to ), wereas u refers to te erivative of wen consiere as a function of u (te erivative of wit respect to u). For instance, in Eample, can be consiere as a function of an also as a function of. Note tat ( s 2 ) u ( su ) F s 2 wereas NOTE In using te Cain Rule we work from te outsie to te insie. Formula 2 sas tat we ifferentiate te outer function f [at te inner function t ] an ten we multipl b te erivative of te inner function. s 2 f u u 2su f t f t t outer function evaluate at inner function erivative of outer function evaluate at inner function erivative of inner function v EXAMPLE 2 Differentiate (a) sin 2 an (b) sin 2. SOLUTION (a) If sin 2, ten te outer function is te sine function an te inner function is te squaring function, so te Cain Rule gives sin 2 cos 2 2 outer function 2 cos 2 evaluate at inner function erivative of outer function evaluate at inner function erivative of inner function (b) Note tat sin 2 sin 2. Here te outer function is te squaring function an te inner function is te sine function. So sin 2 2 sin cos inner function erivative of outer function evaluate at inner function erivative of inner function See Reference Page 2 or Appeni D. Te answer can be left as 2sin cos or written as sin 2 (b a trigonometric ientit known as te ouble-angle formula). In Eample 2(a) we combine te Cain Rule wit te rule for ifferentiating te sine function. In general, if sin u, were uis a ifferentiable function of, ten, b te Cain Rule, u u cos u u Tus u sin u cos u

49 SECTION 2.5 THE CHAIN RULE 5 In a similar fasion, all of te formulas for ifferentiating trigonometric functions can be combine wit te Cain Rule. Let s make eplicit te special case of te Cain Rule were te outer function f is a power function. If t n, ten we can write f u u n were u t. B using te Cain Rule an ten te Power Rule, we get u u nu n u n t n t 4 Te Power Rule Combine wit te Cain Rule If n is an real number an u t is ifferentiable, ten Alternativel, u u n n nu t n n t n t Notice tat te erivative in Eample coul be calculate b taking n 2 in Rule 4. EXAMPLE 3 Differentiate 3. SOLUTION Taking u t 3 an n in 4, we ave v EXAMPLE 4 SOLUTION First rewrite f : Fin f if f. s 3 2 f 2 3 Tus EXAMPLE 5 f Fin te erivative of te function t t t 2 9 2t 2 SOLUTION Combining te Power Rule, Cain Rule, an Quotient Rule, we get t t 9 t 2 2t 8 t 2 t 2t 9 t 2 2t 8 2t 2 t 2 45 t 2 8 2t 2 2t

50 52 CHAPTER 2 DERIVATIVES Te graps of te functions an in Eample 6 are sown in Figure. Notice tat is large wen increases rapil an wen as a orizontal tangent. So our answer appears to be reasonable. ª _2 _ FIGURE EXAMPLE 6 Differentiate SOLUTION In tis eample we must use te Prouct Rule before using te Cain Rule: Noticing tat eac term as te common factor , we coul factor it out an write te answer as Te reason for te name Cain Rule becomes clear wen we make a longer cain b aing anoter link. Suppose tat f u, u t, an t, were f, t, an are ifferentiable functions. Ten, to compute te erivative of wit respect to t, we use te Cain Rule twice: t t u u t v EXAMPLE 7 If f sin cos tan, ten f cos cos tan cos tan cos cos tan sin tan tan Notice tat we use te Cain Rule twice. cos cos tan sin tan sec 2 EXAMPLE 8 Differentiate ssec 3. SOLUTION Here te outer function is te square root function, te mile function is te secant function, an te inner function is te cubing function. So we ave 2ssec 3 sec 3 2ssec 3 sec 3 tan sec 3 tan 3 2ssec 3 3

51 How to Prove te Cain Rule SECTION 2.5 THE CHAIN RULE 53 Recall tat if f an canges from a to a, we efine te increment of as Accoring to te efinition of a erivative, we ave So if we enote b te ifference between te ifference quotient an te erivative, we obtain But lim l If we efine to be wen, ten becomes a continuous function of. Tus, for a ifferentiable function f, we can write an is a continuous function of. Tis propert of ifferentiable functions is wat enables us to prove te Cain Rule. PROOF OF THE CHAIN RULE Suppose u t is ifferentiable at a an f u is ifferentiable at b t a. If is an increment in an u an are te corresponing increments in u an, ten we can use Equation 5 to write were l as l. Similarl f a f a lim f a l l f a f a f a f a? f a 5 f a were l as l 6 u t a t a 7 f b u 2 u f b 2 u were 2 l as u l. If we now substitute te epression for u from Equation 6 into Equation 7, we get f b 2 t a so f b 2 t a As l, Equation 6 sows tat u l. So bot l an 2 l as l. Terefore Tis proves te Cain Rule. f b 2 t a l l f b t a f t a t a

52 54 CHAPTER 2 DERIVATIVES 2.5 Eercises 6 Write te composite function in te form f t. [Ientif te inner function u t an te outer function f u.] Ten fin te erivative.. s tan 4. sin cot 5. ssin 6. sin s 7 46 Fin te erivative of te function. 7. F F F s 2. f sec 2. f z z 2 2. f t s 3 tan t 3. cos a a 3 cos 3 5. sec k 6. 3 cot n f t t t 2 3 2t F t 3t 4 2t sins 2 3. f s s 2 s sin cos 24. f s F z z 26. G z r sr cos sin cos F v v 3. sin tan sec 2 m 33. sec 2 tan sin 4 cos cos 2 6 v 3 f t t t cot 2 sin 38. (a s 2 b 2 ) sin sin sin 4. s s 42. s s s 43. t 2r sin r n p 44. cos 4 sin cosssin tan 46. sin Fin te first an secon erivatives of te function. 47. cos cos H t tan 3t 5. 4 s 5 54 Fin an equation of te tangent line to te curve at te given point. 5. 2,, 52. s 3, 2, sin sin,, 54. sin sin 2,, 55. (a) Fin an equation of te tangent line to te curve tan 2 4 at te point,. ; (b) Illustrate part (a) b graping te curve an te tangent line on te same screen (a) Te curve s2 is calle a bullet-nose curve. Fin an equation of te tangent line to tis curve at te point,. ; (b) Illustrate part (a) b graping te curve an te tangent line on te same screen. 57. (a) If f s2 2, fin f. ; (b) Ceck to see tat our answer to part (a) is reasonable b comparing te graps of f an f. ; 58. Te function f sin sin 2,, arises in applications to frequenc moulation (FM) sntesis. (a) Use a grap of f prouce b a graping evice to make a roug sketc of te grap of f. (b) Calculate f an use tis epression, wit a graping evice, to grap f. Compare wit our sketc in part (a). 59. Fin all points on te grap of te function f 2sin sin 2 at wic te tangent line is orizontal. 6. Fin te -coorinates of all points on te curve sin 2 2sin at wic te tangent line is orizontal. 6. If F f t, were f 2 8, f 2 4, f 5 3, t 5 2, an t 5 6, fin F If s4 3f, were f 7 an f 4, fin. 63. A table of values for f, t, f, an t is given. f t (a) If f t, fin. (b) If H t f, fin H. f t ; Graping calculator or computer require CAS Computer algebra sstem require. Homework Hints available at stewartcalculus.com

53 SECTION 2.5 THE CHAIN RULE Let f an t be te functions in Eercise 63. (a) If F f f, fin F 2. (b) If G t t, fin G If f an t are te functions wose graps are sown, let u f t, v t f, an w t t. Fin eac erivative, if it eists. If it oes not eist, eplain w. (a) u (b) v (c) w 66. If f is te function wose grap is sown, let f f an t f 2. Use te grap of f to estimate te value of eac erivative. (a) 2 (b) t If t sf, were te grap of f is sown, evaluate t Suppose f is ifferentiable on an is a real number. Let F f an G f. Fin epressions for (a) F an (b) G. 69. Let r f t, were 2, t 2 3, 4, t 2 5, an f 3 6. Fin r. 7. If t is a twice ifferentiable function an f t 2, fin f in terms of t, t, an t. 7. If F f 3f 4 f, were f an f 2, fin F. 72. If F f f f, were f 2, f 2 3, f 4, f 2 5, an f 3 6, fin F Fin te given erivative b fining te first few eriva tives an observing te pattern tat occurs. f g =ƒ 73. D 3 cos D 35 sin f CAS 75. Te isplacement of a particle on a vibrating string is given b te equation s t 4 sin t were s is measure in centimeters an t in secons. Fin te velocit of te particle after t secons. 76. If te equation of motion of a particle is given b s A cos t, te particle is sai to unergo simple armonic motion. (a) Fin te velocit of te particle at time t. (b) Wen is te velocit? 77. A Cepei variable star is a star wose brigtness alternatel increases an ecreases. Te most easil visible suc star is Delta Cepei, for wic te interval between times of mai - mum brigtness is 5.4 as. Te average brigtness of tis star is 4. an its brigtness canges b.35. In view of tese ata, te brigtness of Delta Cepei at time t, were t is measure in as, as been moele b te function B t sin t (a) Fin te rate of cange of te brigtness after t as. (b) Fin, correct to two ecimal places, te rate of increase after one a. 78. In Eample 4 in Section.3 we arrive at a moel for te lengt of aligt ( in ours) in Pilaelpia on te tt a of te ear: L t sin t 8 Use tis moel to compare ow te number of ours of a - ligt is increasing in Pilaelpia on Marc 2 an Ma A particle moves along a straigt line wit isplacement s t, velocit v t, an acceleration a t. Sow tat a t v t v s Eplain te ifference between te meanings of te erivatives v t an v s. 8. Air is being pumpe into a sperical weater balloon. At an time t, te volume of te balloon is V t an its raius is r t. (a) Wat o te erivatives V r an V t represent? (b) Epress V t in terms of r t. 8. Computer algebra sstems ave commans tat ifferentiate functions, but te form of te answer ma not be convenient an so furter commans ma be necessar to simplif te answer. (a) Use a CAS to fin te erivative in Eample 5 an compare wit te answer in tat eample. Ten use te simplif comman an compare again. (b) Use a CAS to fin te erivative in Eample 6. Wat appens if ou use te simplif comman? Wat ap pens if ou use te factor comman? Wic form of te answer woul be best for locating orizontal tangents?

54 56 CHAPTER 2 DERIVATIVES CAS 82. (a) Use a CAS to ifferentiate te function f 4 4 an to simplif te result. (b) Were oes te grap of f ave orizontal tangents? (c) Grap f an f on te same screen. Are te graps consistent wit our answer to part (b)? 83. Use te Cain Rule to prove te following. (a) Te erivative of an even function is an o function. (b) Te erivative of an o function is an even function. 84. Use te Cain Rule an te Prouct Rule to give an alternative proof of te Quotient Rule. [Hint: Write f t f t.] 85. (a) If n is a positive integer, prove tat sinn cos n n sin n cos n (b) Fin a formula for te erivative of tat is similar to te one in part (a). cos n cos n 86. Suppose f is a curve tat alwas lies above te -ais an never as a orizontal tangent, were f is if ferentiable everwere. For wat value of is te rate of cange of 5 wit respect to eigt times te rate of cange of wit respect to? 87. Use te Cain Rule to sow tat if is measure in egrees, ten sin cos 8 (Tis gives one reason for te convention tat raian measure is alwas use wen ealing wit trigonometric functions in calculus: Te ifferentiation formulas woul not be as simple if we use egree measure.) s (a) Write an use te Cain Rule to sow tat (b) If, fin f an sketc te graps of f an f. Were is f not ifferentiable? (c) If t sin, fin t an sketc te graps of t an t. Were is t not ifferentiable? f sin 89. If f u an u t, were f an t are twice ifferentiable functions, sow tat u 2 u 2 2 u u 2 9. If f u an u t, were f an t possess tir erivatives, fin a formula for 3 3 similar to te one given in Eercise 89. APPLIED PROJECT WHERE SHOULD A PILOT START DESCENT? An approac pat for an aircraft laning is sown in te figure an satisfies te following conitions: =P() ( i) Te cruising altitue is wen escent starts at a orizontal istance from toucown at te origin. ( ii) Te pilot must maintain a constant orizontal spee v trougout escent. ( iii) Te absolute value of te vertical acceleration soul not ecee a constant k (wic is muc less tan te acceleration ue to gravit).. Fin a cubic polnomial P a 3 b 2 c tat satisfies conition ( i) b imposing suitable conitions on P an P at te start of escent an at toucown. 2. Use conitions ( ii) an ( iii) to sow tat 6v 2 2 k 3. Suppose tat an airline ecies not to allow vertical acceleration of a plane to ecee 2 k 86 mi. If te cruising altitue of a plane is 35, ft an te spee is 3 mi, ow far awa from te airport soul te pilot start escent? ; 4. Grap te approac pat if te conitions state in Problem 3 are satisfie. ; Graping calculator or computer require

55 SECTION 2.6 IMPLICIT DIFFERENTIATION Implicit Differentiation Te functions tat we ave met so far can be escribe b epressing one variable eplicitl in terms of anoter variable for eample, s 3 or sin or, in general, f. Some functions, owever, are efine implicitl b a relation between an suc as or In some cases it is possible to solve suc an equation for as an eplicit function (or several functions) of. For instance, if we solve Equation for, we get s25 2, so two of te functions etermine b te implicit Equation l are f s25 2 an t s25 2. Te graps of f an t are te upper an lower semicircles of te cir cle (See Figure.) FIGURE (a) + =25 (b) ƒ=œ 25- (c) =_œ 25- It s not eas to solve Equation 2 for eplicitl as a function of b an. (A computer algebra sstem as no trouble, but te epressions it obtains are ver complicate.) Noneteless, 2 is te equation of a curve calle te folium of Descartes sown in Figure 2 an it implicitl efines as several functions of. Te graps of tree suc functions are sown in Figure 3. Wen we sa tat f is a function efine implicitl b Equation 2, we mean tat te equation 3 f 3 6f is true for all values of in te omain of f. +Á=6 FIGURE 2 Te folium of Descartes FIGURE 3 Graps of tree functions efine b te folium of Descartes

56 58 CHAPTER 2 DERIVATIVES Fortunatel, we on t nee to solve an equation for in terms of in orer to fin te erivative of. Instea we can use te meto of implicit ifferentiation. Tis consists of ifferentiating bot sies of te equation wit respect to an ten solving te resulting equation for. In te eamples an eercises of tis section it is alwas assume tat te given equation etermines implicitl as a ifferentiable function of so tat te meto of implicit ifferentiation can be applie. v EXAMPLE (a) If , fin. (b) Fin an equation of te tangent to te circle at te point 3, 4. SOLUTION (a) Differentiate bot sies of te equation : Remembering tat is a function of an using te Cain Rule, we ave Tus 2 2 Now we solve tis equation for : (b) At te point 3, 4 we ave 3 an 4, so 3 4 An equation of te tangent to te circle at 3, 4 is terefore or SOLUTION 2 (b) Solving te equation , we get s25 2. Te point 3, 4 lies on te upper semicircle s25 2 an so we consier te function f s25 2. Differentiating f using te Cain Rule, we ave f s25 2

57 SECTION 2.6 IMPLICIT DIFFERENTIATION 59 Eample illustrates tat even wen it is possible to solve an equation eplicitl for in terms of, it ma be easier to use implicit ifferentiation. 3 So f 3 3 s an, as in Solution, an equation of te tangent is NOTE Te epression in Solution gives te erivative in terms of bot an. It is correct no matter wic function is etermine b te given equation. For instance, for f s25 2 we ave s25 2 wereas for t s25 2 we ave s25 2 s25 2 v EXAMPLE 2 (a) Fin if (b) Fin te tangent to te folium of Descartes at te point 3, 3. (c) At wat point in te first quarant is te tangent line orizontal? SOLUTION (a) Differentiating bot sies of wit respect to, regaring as a function of, an using te Cain Rule on te term 3 an te Prouct Rule on te term 6, we get or FIGURE 4 4 FIGURE 5 (3, 3) 4 We now solve for : (b) Wen 3, an a glance at Figure 4 confirms tat tis is a reasonable value for te slope at 3, 3. So an equation of te tangent to te folium at 3, 3 is 3 3 or 6 (c) Te tangent line is orizontal if. Using te epression for from part (a), we see tat wen 2 2 (provie tat 2 2 ). Substituting 2 2 in te equation of te curve, we get 3 ( 2 2 ) 3 6( 2 2 ) wic simplifies to Since in te first quarant, we ave 3 6. If , ten Tus te tangent is orizontal at 2 4 3, 2 5 3, wic is approimatel (2.598, 3.748). Looking at Figure 5, we see tat our answer is reasonable.

58 6 CHAPTER 2 DERIVATIVES NOTE 2 Tere is a formula for te tree roots of a cubic equation tat is like te quaratic formula but muc more complicate. If we use tis formula (or a computer algebra sstem) to solve te equation for in terms of, we get tree functions etermine b te equation: Abel an Galois an f s s s s [ f s 3(s s s s )] Te Norwegian matematician Niels Abel prove in 824 tat no general formula can be given for te roots of a fift-egree equation in terms of raicals. Later te Frenc matematician Evariste Galois prove tat it is impossible to fin a general formula for te roots of an nt-egree equation (in terms of algebraic operations on te coefficients) if n is an integer larger tan 4. (Tese are te tree functions wose graps are sown in Figure 3.) You can see tat te meto of implicit ifferentiation saves an enormous amount of work in cases suc as tis. Moreover, implicit ifferentiation works just as easil for equations suc as for wic it is impossible to fin a similar epression for in terms of. EXAMPLE 3 Fin if sin 2 cos. SOLUTION Differentiating implicitl wit respect to an remembering tat is a function of, we get cos 2 sin cos 2 2 (Note tat we ave use te Cain Rule on te left sie an te Prouct Rule an Cain Rule on te rigt sie.) If we collect te terms tat involve, we get _2 2 So cos 2 sin 2 cos cos 2 sin cos 2 cos cos FIGURE 6 _2 Figure 6, rawn wit te implicit-plotting comman of a computer algebra sstem, sows part of te curve sin 2 cos. As a ceck on our calculation, notice tat wen an it appears from te grap tat te slope is approimatel at te origin. Figures 7, 8, an 9 sow tree more curves prouce b a computer algebra sstem wit an implicit-plotting comman. In Eercises 4 42 ou will ave an opportunit to create an eamine unusual curves of tis nature _3 3 _6 6 _9 9 _3 FIGURE 7 ( -)( -4)= ( -4) _6 FIGURE 8 ( -) sin()= -4 _9 FIGURE 9 sin 3= cos 3

59 SECTION 2.6 IMPLICIT DIFFERENTIATION 6 Te following eample sows ow to fin te secon erivative of a function tat is efine implicitl. EXAMPLE 4 Fin if SOLUTION Differentiating te equation implicitl wit respect to, we get Figure sows te grap of te curve of Eample 4. Notice tat it s a stretce an flat tene version of te circle For tis reason it s sometimes calle a fat circle. It starts out ver steep on te left but quickl becomes ver flat. Tis can be seen from te epression Solving for gives To fin we ifferentiate tis epression for using te Quotient Rule an remembering tat is a function of : $+$=6 2 If we now substitute Equation 3 into tis epression, we get FIGURE But te values of an must satisf te original equation So te answer simplifies to Eercises 4 (a) Fin b implicit ifferentiation. (b) Solve te equation eplicitl for an ifferentiate to get in terms of. (c) Ceck tat our solutions to parts (a) an (b) are consistent b substituting te epression for into our solution for part (a) cos s Fin b implicit ifferentiation s s cos cos sin 3. 4 cos sin 4. sin 2 sin 2 5. tan 6. s s 2 8. sin sin 9. cos sin 2. tan 2 2. If f 2 f 3 an f 2, fin f. 22. If t sin t 2, fin t. ; Graping calculator or computer require CAS Computer algebra sstem require. Homework Hints available at stewartcalculus.com

60 62 CHAPTER 2 DERIVATIVES Regar as te inepenent variable an as te epenent variable an use implicit ifferentiation to fin sec tan Use implicit ifferentiation to fin an equation of te tangent line to te curve at te given point. 25. sin 2 cos 2, 26. sin 2 2, ,, (ellipse) ,, 2 (perbola) (carioi) 2, 4, (, 2) ( 3s3, ) (astroi) (3, ) (, 2) (lemniscate) (evil s curve) 33. (a) Te curve wit equation is calle a kample of Euous. Fin an equation of te tangent line to tis curve at te point, 2. ; (b) Illustrate part (a) b graping te curve an te tangent line on a common screen. (If our graping evice will grap implicitl efine curves, ten use tat capabilit. If not, ou can still grap tis curve b graping its upper an lower alves separatel.) 34. (a) Te curve wit equation is calle te Tscirnausen cubic. Fin an equation of te tangent line to tis curve at te point, 2. (b) At wat points oes tis curve ave orizontal tangents? ; (c) Illustrate parts (a) an (b) b graping te curve an te tangent lines on a common screen Fin b implicit ifferentiation s s a 4 CAS CAS 39. If 3, fin te value of at te point were. 4. If 2 3, fin te value of at te point were. 4. Fanciful sapes can be create b using te implicit plotting capabilities of computer algebra sstems. (a) Grap te curve wit equation At ow man points oes tis curve ave orizontal tangents? Estimate te -coorinates of tese points. (b) Fin equations of te tangent lines at te points (, ) an (, 2). (c) Fin te eact -coorinates of te points in part (a). () Create even more fanciful curves b moifing te equation in part (a). 42. (a) Te curve wit equation as been likene to a bouncing wagon. Use a computer algebra sstem to grap tis curve an iscover w. (b) At ow man points oes tis curve ave orizontal tangent lines? Fin te -coorinates of tese points. 43. Fin te points on te lemniscate in Eercise 3 were te tangent is orizontal. 44. Sow b implicit ifferentiation tat te tangent to te ellipse at te point, is 45. Fin an equation of te tangent line to te perbola at te point, a 2 2 b 2 a 2 b 2 2 a 2 2 b Sow tat te sum of te - an -intercepts of an tangent line to te curve s s sc is equal to c. 47. Sow, using implicit ifferentiation, tat an tangent line at a point P to a circle wit center O is perpenicular to te raius OP. 48. Te Power Rule can be prove using implicit ifferentiation for te case were n is a rational number, n p q, an f n is assume beforean to be a ifferentiable function. If p q, ten q p. Use implicit ifferentiation to sow tat p q p q

61 LABORATORY PROJECT FAMILIES OF IMPLICIT CURVES Two curves are ortogonal if teir tangent lines are perpenicular at eac point of intersection. Sow tat te given families of curves are ortogonal trajectories of eac oter; tat is, ever curve in one famil is ortogonal to ever curve in te oter famil. Sketc bot families of curves on te same aes r 2, a b 2 2 a, c 2, 52. a 3, 2 2 b k b 53. Sow tat te ellipse 2 a 2 2 b 2 an te perbola 2 A 2 2 B 2 are ortogonal trajectories if A 2 a 2 an a 2 b 2 A 2 B 2 (so te ellipse an perbola ave te same foci). 54. Fin te value of te number a suc tat te families of curves c an a k 3 are ortogonal trajectories. 55. (a) Te van er Waals equation for n moles of a gas is P n 2 a V 2 V nb nrt were P is te pressure, V is te volume, an T is te temperature of te gas. Te constant R is te universal gas constant an a an b are positive constants tat are caracteristic of a particular gas. If T remains constant, use implicit ifferentiation to fin V P. (b) Fin te rate of cange of volume wit respect to pressure of mole of carbon ioie at a volume of V L an a pressure of P 2.5 atm. Use a L 2 -atm mole 2 an b.4267 L mole. 56. (a) Use implicit ifferentiation to fin if 2 2. CAS (b) Plot te curve in part (a). Wat o ou see? Prove tat wat ou see is correct. (c) In view of part (b), wat can ou sa about te epression for tat ou foun in part (a)? 57. Te equation represents a rotate ellipse, tat is, an ellipse wose aes are not parallel to te coorinate aes. Fin te points at wic tis ellipse crosses te -ais an sow tat te tangent lines at tese points are parallel. 58. (a) Were oes te normal line to te ellipse at te point, intersect te ellipse a secon time? ; (b) Illustrate part (a) b graping te ellipse an te normal line. 59. Fin all points on te curve were te slope of te tangent line is. 6. Fin equations of bot te tangent lines to te ellipse tat pass troug te point 2, Te Bessel function of orer, J, satisfies te ifferential equation for all values of an its value at is J. (a) Fin J. (b) Use implicit ifferentiation to fin J. 62. Te figure sows a lamp locate tree units to te rigt of te -ais an a saow create b te elliptical region If te point 5, is on te ege of te saow, ow far above te -ais is te lamp locate? _5 +4 =5 3? LABORATORY PROJECT CAS FAMILIES OF IMPLICIT CURVES In tis project ou will eplore te canging sapes of implicitl efine curves as ou var te constants in a famil, an etermine wic features are common to all members of te famil.. Consier te famil of curves c (a) B graping te curves wit c an c 2, etermine ow man points of intersection tere are. (You migt ave to zoom in to fin all of tem.) (b) Now a te curves wit c 5 an c to our graps in part (a). Wat o ou notice? Wat about oter values of c? CAS Computer algebra sstem require

62 64 CHAPTER 2 DERIVATIVES 2. (a) Grap several members of te famil of curves 2 2 c 2 2 Describe ow te grap canges as ou cange te value of c. (b) Wat appens to te curve wen c? Describe wat appears on te screen. Can ou prove it algebraicall? (c) Fin b implicit ifferentiation. For te case c, is our epression for consistent wit wat ou iscovere in part (b)? 2.7 Rates of Cange in te Natural an Social Sciences We know tat if f, ten te erivative can be interprete as te rate of cange of wit respect to. In tis section we eamine some of te applications of tis iea to psics, cemistr, biolog, economics, an oter sciences. Let s recall from Section 2. te basic iea bein rates of cange. If canges from to, ten te cange in is 2 an te corresponing cange in is Te ifference quotient 2 f 2 f f 2 f 2 FIGURE P{, fl} Q{, } Î Î m PQ average rate of cange m=fª( )=instantaneous rate of cange is te average rate of cange of wit respect to over te interval, 2 an can be interprete as te slope of te secant line PQ in Figure. Its limit as l is te erivative f, wic can terefore be interprete as te instantaneous rate of cange of wit respect to or te slope of te tangent line at P, f. Using Leibniz notation, we write te process in te form l Wenever te function f as a specific interpretation in one of te sciences, its erivative will ave a specific interpretation as a rate of cange. (As we iscusse in Section 2., te units for are te units for ivie b te units for.) We now look at some of tese interpretations in te natural an social sciences. Psics If s f t is te position function of a particle tat is moving in a straigt line, ten s t represents te average velocit over a time perio t, an v s t represents te instantaneous velocit (te rate of cange of isplacement wit respect to time). Te instantaneous rate of cange of velocit wit respect to time is acceleration: a t v t s t. Tis was iscusse in Sections 2. an 2.2, but now tat we know te ifferentiation formulas, we are able to solve problems involving te motion of objects more easil.

63 v EXAMPLE SECTION 2.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES 65 Te position of a particle is given b te equation s f t t 3 6t 2 9t were t is measure in secons an s in meters. (a) Fin te velocit at time t. (b) Wat is te velocit after 2 s? After 4 s? (c) Wen is te particle at rest? () Wen is te particle moving forwar (tat is, in te positive irection)? (e) Draw a iagram to represent te motion of te particle. (f) Fin te total istance travele b te particle uring te first five secons. (g) Fin te acceleration at time t an after 4 s. () Grap te position, velocit, an acceleration functions for t 5. ( i) Wen is te particle speeing up? Wen is it slowing own? SOLUTION (a) Te velocit function is te erivative of te position function. s f t t 3 6t 2 9t v t s t 3t 2 2t 9 (b) Te velocit after 2 s means te instantaneous velocit wen t 2, tat is, v 2 s t m s t 2 Te velocit after 4 s is v m s (c) Te particle is at rest wen v t, tat is, 3t 2 2t 9 3 t 2 4t 3 3 t t 3 an tis is true wen t or t 3. Tus te particle is at rest after s an after 3 s. () Te particle moves in te positive irection wen v t, tat is, 3t 2 2t 9 3 t t 3 t=3 s= t= s= FIGURE 2 t= s=4 s Tis inequalit is true wen bot factors are positive t 3 or wen bot factors are negative t. Tus te particle moves in te positive irection in te time intervals t an t 3. It moves backwar ( in te negative irection) wen t 3. (e) Using te information from part () we make a scematic sketc in Figure 2 of te motion of te particle back an fort along a line (te s-ais). (f) Because of wat we learne in parts () an (e), we nee to calculate te istances travele uring te time intervals [, ], [, 3], an [3, 5] separatel. Te istance travele in te first secon is From t to t 3 te istance travele is f f 4 4 m f 3 f 4 4 m

64 66 CHAPTER 2 DERIVATIVES 25 a s 5-2 FIGURE 3 From t 3 to t 5 te istance travele is f 5 f 3 2 Te total istance is m. 2 m (g) Te acceleration is te erivative of te velocit function: a t 2 s t 2 v 6t 2 t a m s 2 () Figure 3 sows te graps of s, v, an a. (i) Te particle spees up wen te velocit is positive an increasing ( v an a are bot positive) an also wen te velocit is negative an ecreasing ( v an a are bot negative). In oter wors, te particle spees up wen te velocit an acceleration ave te same sign. (Te particle is puse in te same irection it is moving.) From Figure 3 we see tat tis appens wen t 2 an wen t 3. Te particle slows own wen v an a ave opposite signs, tat is, wen t an wen 2 t 3. Figure 4 summarizes te motion of te particle. TEC In Moule 2.7 ou can see an animation of Figure 4 wit an epression for s tat ou can coose ourself. 5 _5 s a t forwar backwar forwar FIGURE 4 slows own spees up slows own spees up EXAMPLE 2 If a ro or piece of wire is omogeneous, ten its linear ensit is uniform an is efine as te mass per unit lengt m l an measure in kilograms per meter. Suppose, owever, tat te ro is not omogeneous but tat its mass measure from its left en to a point is m f, as sown in Figure 5. FIGURE 5 Tis part of te ro as mass ƒ. Te mass of te part of te ro tat lies between an 2 is given b m f 2 f, so te average ensit of tat part of te ro is average ensit m f 2 f 2

65 SECTION 2.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES 67 If we now let l (tat is, 2 l ), we are computing te average ensit over smaller an smaller intervals. Te linear ensit at is te limit of tese average ensities as l ; tat is, te linear ensit is te rate of cange of mass wit respect to lengt. Smbolicall, 苷 lim l m m 苷 Tus te linear ensit of te ro is te erivative of mass wit respect to lengt. For instance, if m 苷 f 共兲 苷 s, were is measure in meters an m in kilograms, ten te average ensit of te part of te ro given b 艋 艋.2 is m f 共.2兲 f 共兲 s.2 苷 苷.48 kg兾m.2.2 wile te ensit rigt at 苷 is 苷 FIGURE 6 m 冟 苷 苷 2s 冟 苷 苷.5 kg兾m v EXAMPLE 3 A current eists wenever electric carges move. Figure 6 sows part of a wire an electrons moving troug a plane surface, sae re. If Q is te net carge tat passes troug tis surface uring a time perio t, ten te average current uring tis time interval is efine as average current 苷 Q Q2 Q 苷 t t2 t If we take te limit of tis average current over smaller an smaller time intervals, we get wat is calle te current I at a given time t : I 苷 lim t l Q Q 苷 t t Tus te current is te rate at wic carge flows troug a surface. It is measure in units of carge per unit time (often coulombs per secon, calle amperes). Velocit, ensit, an current are not te onl rates of cange tat are important in psics. Oters inclue power (te rate at wic work is one), te rate of eat flow, temperature graient (te rate of cange of temperature wit respect to position), an te rate of eca of a raioactive substance in nuclear psics. Cemistr EXAMPLE 4 A cemical reaction results in te formation of one or more substances (calle proucts) from one or more starting materials (calle reactants). For instance, te equation 2H2 O2 l 2H2 O inicates tat two molecules of rogen an one molecule of ogen form two molecules of water. Let s consier te reaction A BlC

66 68 CHAPTER 2 DERIVATIVES were A an B are te reactants an C is te prouct. Te concentration of a reactant A 23 is te number of moles ( mole 6.22 molecules) per liter an is enote b A. Te concentration varies uring a reaction, so A, B, an C are all functions of time t. Te average rate of reaction of te prouct C over a time interval t t t 2 is C t C t 2 C t t 2 t But cemists are more intereste in te instantaneous rate of reaction, wic is obtaine b taking te limit of te average rate of reaction as te time interval t approaces : C rate of reaction C t l t t Since te concentration of te prouct increases as te reaction procees, te erivative C t will be positive, an so te rate of reaction of C is positive. Te concentrations of te reactants, owever, ecrease uring te reaction, so, to make te rates of reaction of A an B positive numbers, we put minus signs in front of te erivatives A t an B t. Since A an B eac ecrease at te same rate tat C increases, we ave rate of reaction C t A t B t More generall, it turns out tat for a reaction of te form aa bb l cc D we ave a A t b B t c C t D t Te rate of reaction can be etermine from ata an grapical metos. In some cases tere are eplicit formulas for te concentrations as functions of time, wic enable us to compute te rate of reaction (see Eercise 24). EXAMPLE 5 One of te quantities of interest in termonamics is compressibilit. If a given substance is kept at a constant temperature, ten its volume V epens on its pressure P. We can consier te rate of cange of volume wit respect to pressure namel, te erivative V P. As P increases, V ecreases, so V P. Te compressibilit is efine b introucing a minus sign an iviing tis erivative b te volume V: isotermal compressibilit V Tus measures ow fast, per unit volume, te volume of a substance ecreases as te pressure on it increases at constant temperature. For instance, te volume V ( in cubic meters) of a sample of air at 25 C was foun to be relate to te pressure P ( in kilopascals) b te equation V 5.3 P V P

67 SECTION 2.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES 69 Te rate of cange of V wit respect to P wen P 5 kpa is Te compressibilit at tat pressure is Biolog V V P 5.3 P 5 P 2 P m3 kpa V P.22.2 m 3 kpa m 3 P EXAMPLE 6 Let n f t be te number of iniviuals in an animal or plant popu lation at time t. Te cange in te population size between te times t t an t t 2 is n f t 2 f t, an so te average rate of growt uring te time perio t t t 2 is average rate of growt n t f t 2 f t t 2 t Te instantaneous rate of growt is obtaine from tis average rate of growt b letting te time perio t approac : n growt rate t l t n t Strictl speaking, tis is not quite accurate because te actual grap of a population function n f t woul be a step function tat is iscontinuous wenever a birt or eat occurs an terefore not ifferentiable. However, for a large animal or plant population, we can replace te grap b a smoot approimating curve as in Figure 7. n FIGURE 7 A smoot curve approimating a growt function t

68 Ee of Science / Poto Researcers, Inc. 7 CHAPTER 2 DERIVATIVES E. coli bacteria are about 2 micrometers ( m) long an.75 m wie. Te image was prouce wit a scanning electron microscope. To be more specific, consier a population of bacteria in a omogeneous nutrient meium. Suppose tat b sampling te population at certain intervals it is etermine tat te population oubles ever our. If te initial population is n an te time t is measure in ours, ten an, in general, f 2f 2n f 2 2f 2 2 n f 3 2f n f t 2 t n Te population function is n n 2 t. Tis is an eample of an eponential function. In Capter 6 we will iscuss eponential functions in general; at tat time we will be able to compute teir erivatives an tereb etermine te rate of growt of te bacteria population. EXAMPLE 7 Wen we consier te flow of bloo troug a bloo vessel, suc as a vein or arter, we can moel te sape of te bloo vessel b a clinrical tube wit raius R an lengt l as illustrate in Figure 8. R r FIGURE 8 Bloo flow in an arter l Because of friction at te walls of te tube, te velocit v of te bloo is greatest along te central ais of te tube an ecreases as te istance r from te ais increases until v becomes at te wall. Te relationsip between v an r is given b te law of laminar flow iscovere b te Frenc psician Jean-Louis-Marie Poiseuille in 84. Tis law states tat For more etaile information, see W. Nicols an M. O Rourke (es.), McDonal s Bloo Flow in Arteries: Teoretical, Eperimental, an Clinical Principles, 5t e. (New York, 25). v P 4 l R2 r 2 were is te viscosit of te bloo an P is te pressure ifference between te ens of te tube. If P an l are constant, ten v is a function of r wit omain, R. Te average rate of cange of te velocit as we move from r r outwar to r r 2 is given b v r v r 2 v r r 2 r an if we let r l, we obtain te velocit graient, tat is, te instantaneous rate of cange of velocit wit respect to r: Using Equation, we obtain v v velocit graient r l r r v r P Pr 2r 4 l 2 l