.1 DERIVATIVES OF POLYNOIALS AND EXPONENTIAL FUNCTIONS c =c slope=0 0 FIGURE 1 Te grap of ƒ=c is te line =c, so fª()=0. In tis section we learn ow to ifferentiate constant functions, power functions, polnomials, an eponential functions. 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 0, so we must ave f 0. (See Figure 1.) A formal proof, from te efinition of a erivative, is also eas: f l 0 f f In Leibniz notation, we write tis rule as follows. DERIVATIVE OF A CONSTANT FUNCTION c 0 c c l 0 l 0 0 0 = POWER FUNCTIONS We net look at te functions f n, were n is a positive integer. If n 1, te grap of f is te line, wic as slope 1. (See Figure.) So 0 6 0010 slope=1 FIGURE Te grap of ƒ= is te line =, so fª()=1. 1 (You can also verif Equation 1 from te efinition of a erivative.) We ave alrea investigate te cases n an n. In fact, in Section.8 (Eercises 17 an 18) we foun tat 1 For n 4 we fin te erivative of f 4 as follows: Tus f l 0 f f 4 4 6 4 4 4 l 0 4 6 4 4 l 0 l 0 4 6 4 4 4 4 4 4 l 0 1 7
1 7 4 CHAPTER DIFFERENTIATION RULES Comparing te equations in (1), (), an (), we see a pattern emerging. It seems to be a reasonable guess tat, wen n is a positive integer, n n n 1. Tis turns out to be true. THE POWER RULE If n is a positive integer, ten n n n 1 FIRST PROOF Te formula n a n a n 1 n a a n a n 1 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.7.5 for f a an te equation above to write f a l a na n 1 f f a a l a n a n a n 1 n a a n a n 1 l a a n 1 a n a aa n a n 1 SECOND PROOF f l 0 f f n n l 0 N Te Binomial Teorem is given on Reference Page 1. 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 1 n n n 1 n n n 1 n n f l 0 n n 1 n n 1 n n n 1 n l 0 n 1 n n 1 l 0 n n 1 n n n n n 1 because ever term ecept te first as as a factor an terefore approaces 0. We illustrate te Power Rule using various notations in Eample 1. EXAPLE 1 (a) If f 6, ten f 6 5. (b) If 1000, ten 1000 999. (c) If t 4, ten 4t. () r r r t
SECTION.1 DERIVATIVES OF POLYNOIALS AND EXPONENTIAL FUNCTIONS 1 7 5 Wat about power functions wit negative integer eponents? In Eercise 61 we ask ou to verif from te efinition of a erivative tat We can rewrite tis equation as an so te Power Rule is true wen n 1. In fact, we will sow in te net section [Eercise 58(c)] tat it ols for all negative integers. Wat if te eponent is a fraction? In Eample in Section.8 we foun tat wic can be written as 1 1 1 1 s 1 s 1 1 1 Tis sows tat te Power Rule is true even wen n 1. In fact, we will sow in Section.6 tat it is true for all real numbers n. THE POWER RULE (GENERAL VERSION) If n is an real number, ten n n n 1 N Figure sows te function in Eample (b) an its erivative. Notice tat is not ifferentiable at 0 ( is not efine tere). Observe tat is positive wen increases an is negative wen ecreases. ª _ EXAPLE Differentiate: (a) f 1 (b) s SOLUTION In eac case we rewrite te function as a power of. (a) Since f, we use te Power Rule wit n : f 1 _ (b) (s ) 1 1 FIGURE =#œ Te Power Rule enables us to fin tangent lines witout aving to resort to te efinition of a erivative. It also enables us to fin normal lines. Te normal line to a curve C at a 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.)
1 7 6 CHAPTER DIFFERENTIATION RULES V EXAPLE Fin equations of te tangent line an normal line to te curve s at te point 1, 1. Illustrate b graping te curve an tese lines. SOLUTION Te erivative of f s 1 is tangent normal _1 _1 FIGURE 4 f 1 1 s So te slope of te tangent line at (1, 1) is f 1. Terefore an equation of te tangent line is 1 or 1 1 Te normal line is perpenicular to te tangent line, so its slope is te negative reciprocal of, tat is,. Tus an equation of te normal line is 1 1 or 5 We grap te curve an its tangent line an normal line in Figure 4. NEW DERIVATIVES FRO OLD 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. N G EOETRIC INTERPRETATION OF THE CONSTANT ULTIPLE RULE 0 =ƒ =ƒ ultipling b c stretces te grap verticall b a factor of. All te rises ave been ouble but te runs sta te same. So te slopes are ouble, too. THE CONSTANT ULTIPLE RULE If c is a constant an f is a ifferentiable function, ten PROOF Let t cf. Ten EXAPLE 4 (a) 4 4 4 1 (b) t t cf cf t l 0 l 0 l 0 c lim l 0 cf c cf c f f f f f (b Law of limits) 1 1 1 1 1 Te net rule tells us tat te erivative of a sum of functions is te sum of te erivatives.
SECTION.1 DERIVATIVES OF POLYNOIALS AND EXPONENTIAL FUNCTIONS 1 7 7 N Using prime notation, we can write te Sum Rule as f t f t THE SU RULE If f an t are bot ifferentiable, ten f t f t PROOF Let F f t. Ten F F F l 0 f t f t l 0 l 0 l 0 f t f f f f t t t t l 0 (b Law 1) 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 1 t an appling te Sum Rule an te Constant ultiple Rule, we get te following formula. THE DIFFERENCE RULE If f an t are bot ifferentiable, ten f t f t Te Constant ultiple Rule, te Sum Rule, an te Difference Rule can be combine wit te Power Rule to ifferentiate an polnomial, as te following eamples emonstrate. EXAPLE 5 8 1 5 4 4 10 6 5 8 1 5 4 4 10 6 5 8 7 1 5 4 4 4 10 6 1 0 8 7 60 4 16 0 6
1 7 8 CHAPTER DIFFERENTIATION RULES (0, 4) V EXAPLE 6 Fin te points on te curve 4 6 4 were te tangent line is orizontal. SOLUTION Horizontal tangents occur were te erivative is zero. We ave {_œ, _5} 0 {œ, _5} 4 6 4 4 1 0 4 FIGURE 5 Te curve =$-6@+4 an its orizontal tangents Tus 0 if 0 or 0, tat is, s. So te given curve as orizontal tangents wen 0, s, an s. Te corresponing points are 0, 4, (s, 5), an ( s, 5). (See Figure 5.) EXAPLE 7 Te equation of motion of a particle is s t 5t t 4, were s is measure in centimeters an t in secons. Fin te acceleration as a function of time. Wat is te acceleration after secons? SOLUTION Te velocit an acceleration are v t s t 6t 10t a t v t 1t 10 Te acceleration after s is a 14 cm s. EXPONENTIAL FUNCTIONS Let s tr to compute te erivative of te eponential function f a using te efinition of a erivative: Te factor a f l 0 f f a a a a a 1 l 0 l 0 oesn t epen on, so we can take it in front of te limit: f a lim l 0 a 1 a a l 0 Notice tat te limit is te value of te erivative of f at 0, tat is, a 1 lim f 0 l 0 Terefore we ave sown tat if te eponential function f a is ifferentiable at 0, ten it is ifferentiable everwere an 4 f f 0 a Tis equation sas tat te rate of cange of an eponential function is proportional to te function itself. (Te slope is proportional to te eigt.)
SECTION.1 DERIVATIVES OF POLYNOIALS AND EXPONENTIAL FUNCTIONS 1 7 9 1 1 0.1 0.7177 1.161 0.01 0.6956 1.1047 0.001 0.694 1.099 0.0001 0.69 1.0987 Numerical evience for te eistence of f 0 is given in te table at te left for te cases a an a. (Values are state correct to four ecimal places.) It appears tat te limits eist an for a, for a, 1 f 0 0.69 l 0 1 f 0 1.10 l 0 In fact, it can be prove tat tese limits eist an, correct to si ecimal places, te values are 0.69147 0 1.09861 0 Tus, from Equation 4, we ave 5 0.69 1.10 Of all possible coices for te base a in Equation 4, te simplest ifferentiation formula occurs wen f 0 1. In view of te estimates of f 0 for a an a, it seems reasonable tat tere is a number a between an for wic f 0 1. It is traitional to enote tis value b te letter e. (In fact, tat is ow we introuce e in Section 1.5.) Tus we ave te following efinition. N In Eercise 1 we will see tat e lies between.7 an.8. Later we will be able to sow tat, correct to five ecimal places, e.7188 DEFINITION OF THE NUBER e e is te number suc tat e 1 lim 1 l 0 Geometricall, tis means tat of all te possible eponential functions a, te function f e is te one wose tangent line at ( 0, 1 as a slope f 0 tat is eactl 1. (See Figures 6 an 7.) = {, e } slope=e = =e 1 =e 1 slope=1 0 0 FIGURE 6 FIGURE 7 If we put a e an, terefore, f 0 1 in Equation 4, it becomes te following important ifferentiation formula.
1 8 0 CHAPTER DIFFERENTIATION RULES TEC Visual.1 uses te slope-a-scope to illustrate tis formula. DERIVATIVE OF THE NATURAL EXPONENTIAL FUNCTION e e Tus te eponential function f e as te propert tat it is its own erivative. Te geometrical significance of tis fact is tat te slope of a tangent line to te curve e is equal to te -coorinate of te point (see Figure 7). V EXAPLE 8 If f e, fin f an f. Compare te graps of f an f. SOLUTION Using te Difference Rule, we ave f f e e e 1 _1.5 fª 1.5 In Section.8 we efine te secon erivative as te erivative of f, so FIGURE 8 = FIGURE 9 _1 (ln, ) = 1 0 1 f e 1 e 1 e Te function f an its erivative f are grape in Figure 8. Notice tat f as a orizontal tangent wen 0; tis correspons to te fact tat f 0 0. Notice also tat, for 0, f is positive an f is increasing. Wen 0, f is negative an f is ecreasing. EXAPLE 9 At wat point on te curve e is te tangent line parallel to te line? SOLUTION Since e, we ave e. Let te -coorinate of te point in question be a. Ten te slope of te tangent line at tat point is e a. Tis tangent line will be parallel to te line if it as te same slope, tat is,. Equating slopes, we get e a a ln Terefore te require point is a, e a ln,. (See Figure 9.).1 EXERCISES 1. (a) How is te number e efine? (b) Use a calculator to estimate te values of te limits.7 1 lim l 0 an.8 1 lim l 0 correct to two ecimal places. Wat can ou conclue about te value of e?. (a) Sketc, b an, te grap of te function f e, paing particular attention to ow te grap crosses te -ais. Wat fact allows ou to o tis? (b) Wat tpes of functions are f e an t e? Compare te ifferentiation formulas for f an t. (c) Wic of te two functions in part (b) grows more rapil wen is large? Differentiate te function.. f 186.5 4. f s0 5. f t 6. F 4 8 t 7. f 4 6 8. f t 1 t 6 t 4 t
SECTION.1 DERIVATIVES OF POLYNOIALS AND EXPONENTIAL FUNCTIONS 1 8 1 9. f t 1 4 t 4 8 10. 11. 5 1. 1. V r 4 r 14. 15. A s 1 16. 17. G s e 18. 19. 0. 1. a b c.. s 5 F ( 1 ) 5 4 s 4. 5. 4 6. 5e R t 5t 5 B c 6 s f t st s 1 st s 1 t u s u su (b) Using te grap in part (a) to estimate slopes, make a roug sketc, b an, of te grap of f. (See Eample 1 in Section.8.) (c) Calculate f an use tis epression, wit a graping evice, to grap f. Compare wit our sketc in part (b). ; 44. (a) Use a graping calculator or computer to grap te function t e in te viewing rectangle 1, 4 b 8, 8. (b) Using te grap in part (a) to estimate slopes, make a roug sketc, b an, of te grap of t. (See Eample 1 in Section.8.) (c) Calculate t an use tis epression, wit a graping evice, to grap t. Compare wit our sketc in part (b). 45 46 Fin te first an secon erivatives of te function. 45. f 4 16 46. G r sr s r 7. H 8. ae v b v c 1 v 9. u s 5 t 4st 5 0. 1. z A Be 10 4 Fin an equation of te tangent line to te curve at te given point.. v s 1 s e 1 1. s 4, 1, 1 4. 4, 5 6 Fin equations of te tangent line an normal line to te curve at te given point. 5. 4 e, 0, 6. 1, ; 7 8 Fin an equation of te tangent line to te curve at te given point. Illustrate b graping te curve an te tangent line on te same screen. 7., 1, 8. s, ; 9 4 Fin f. Compare te graps of f an f an use tem to eplain w our answer is reasonable. 9. f e 5 40. f 5 0 50 41. f 4. f 1 15 5 1, 9 1, 0 1, ; 4. (a) Use a graping calculator or computer to grap te function f 4 6 7 0 in te viewing rectangle, 5 b 10, 50. ; 47 48 Fin te first an secon erivatives of te function. Ceck to see tat our answers are reasonable b comparing te graps of f, f, an f. 47. f 5 4 48. f e 49. Te equation of motion of a particle is s t t, were s is in meters an t is in secons. Fin (a) te velocit an acceleration as functions of t, (b) te acceleration after s, an (c) te acceleration wen te velocit is 0. 50. Te equation of motion of a particle is s t 7t 4t 1, 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 1 s. ; (c) Grap te position, velocit, an acceleration functions on te same screen. 51. Fin te points on te curve 1 1 were te tangent is orizontal. 5. For wat values of oes te grap of f ave a orizontal tangent? 5. Sow tat te curve 6 5 as no tangent line wit slope 4. 54. Fin an equation of te tangent line to te curve s tat is parallel to te line 1. 55. Fin equations of bot lines tat are tangent to te curve 1 an are parallel to te line 1 1. ; 56. At wat point on te curve 1 e is te tangent line parallel to te line 5? Illustrate b graping te curve an bot lines. 57. Fin an equation of te normal line to te parabola 5 4 tat is parallel to te line 5.
1 8 CHAPTER DIFFERENTIATION RULES 58. Were oes te normal line to te parabola at te point (1, 0) intersect te parabola a secon time? Illustrate wit a sketc. 59. Draw a iagram to sow tat tere are two tangent lines to te parabola tat pass troug te point 0, 4. Fin te coorinates of te points were tese tangent lines intersect te parabola. 60. (a) Fin equations of bot lines troug te point, tat are tangent to te parabola. (b) Sow tat tere is no line troug te point, 7 tat is tangent to te parabola. Ten raw a iagram to see w. 61. Use te efinition of a erivative to sow tat if f 1, ten f 1. (Tis proves te Power Rule for te case n 1.) 6. Fin te nt erivative of eac function b calculating te first few erivatives an observing te pattern tat occurs. (a) f n (b) f 1 6. Fin a secon-egree polnomial P suc tat P 5, P, an P. 64. Te equation 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 B C satisfies tis equation. (Differential equations will be stuie in etail in Capter 9.) 65. Fin a cubic function a b c wose grap as orizontal tangents at te points, 6 an, 0. 66. Fin a parabola wit equation a b c tat as slope 4 at 1, slope 8 at 1, an passes troug te point, 15. 67. Let Is f ifferentiable at 1? Sketc te graps of f an f. 68. At wat numbers is te following function t ifferentiable? t 1 f if 1 if 1 1 if 1 if 1 if 1 Give a formula for t an sketc te graps of t an t. 69. (a) For wat values of is te function ifferentiable? Fin a formula for f. (b) Sketc te graps of f an f. 70. Were is te function ifferentiable? Give a formula for an sketc te graps of an. 71. Fin te parabola wit equation a b wose tangent line at (1, 1) as equation. 7. Suppose te curve 4 a b c as a tangent line wen 0 wit equation 1 an a tangent line wen 1 wit equation. Fin te values of a, b, c, an. 7. For wat values of a an b is te line b tangent to te parabola a wen? 74. Fin te value of c suc tat te line 6 is tangent to te curve cs. 75. Let f Fin te values of m an b tat make f ifferentiable everwere. 76. 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. 1000 1 77. Evaluate lim. l 1 1 1 m b if if f 9 78. Draw a iagram sowing two perpenicular lines tat intersect on te -ais an are bot tangent to te parabola. Were o tese lines intersect? 79. If c 1, ow man lines troug te point 0, c are normal lines to te parabola? Wat if c 1? 80. Sketc te parabolas an. Do ou tink tere is a line tat is tangent to bot curves? If so, fin its equation. If not, w not? A P P L I E D P R O J E C T BUILDING A BETTER ROLLER COASTER 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 0.8 an te slope of te rop 1.6. You ecie to connect tese two straigt stretces L 1 an L wit part of a parabola f a 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
SECTION. THE PRODUCT AND QUOTIENT RULES 1 8 L P f Q L segments L 1 an L 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. 1. (a) Suppose te orizontal istance between P an Q is 100 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 1, f, an L to verif grapicall tat te transitions are smoot. () Fin te ifference in elevation between P an Q.. Te solution in Problem 1 migt look smoot, but it migt not feel smoot because te piecewise efine function [consisting of L 1 for 0, f for 0 100, an L for 100] oesn t ave a continuous secon erivative. So ou ecie to improve te esign b using a quaratic function q a b c onl on te interval 10 90 an connecting it to te linear functions b means of two cubic functions: t k l m n 0 10 p q r s 90 100 CAS (a) Write a sstem of equations in 11 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, t, q,, an, an compare wit te plot in Problem 1(c). L 1 L. THE PRODUCT AND QUOTIENT RULES Te formulas of tis section enable us to ifferentiate new functions forme from ol functions b multiplication or ivision. THE PRODUCT RULE Î u Î u Îu Î Îu 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. Ten te Power Rule gives f 1 an t. But ft, so ft. Tus ft f t. Te correct formula was iscovere b Leibniz (soon after is false start) an is calle te Prouct Rule. Before stating te Prouct Rule, let s see ow we migt iscover it. We start b assuming tat u f an v t are bot positive ifferentiable functions. Ten we can interpret te prouct uv as an area of a rectangle (see Figure 1). If canges b an amount, ten te corresponing canges in u an v are u Îu FIGURE 1 Te geometr of te Prouct Rule u f f v t t an te new value of te prouct, u u v v, can be interprete as te area of te large rectangle in Figure 1 (provie tat u an v appen to be positive). Te cange in te area of te rectangle is 1 uv u u v v uv u v v u u v te sum of te tree sae areas