3 DIFFERENTIATION. Differential calculus is the study of the derivative, and differentiation is the process of PREVIEW VERSION NOT FINAL

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1 3 DIFFERENTIATION Differential calculus is te stu of te erivative, an ifferentiation is te process of computing erivatives. Wat is a erivative? Tere are tree equall important answers: A erivative is a rate of cange, it is te slope of a tangent line, an (more formall), it is te limit of a ifference quotient, as we will eplain sortl. In tis capter, we eplore all tree facets of te erivative an evelop te basic rules of ifferentiation. Wen ou master tese tecniques, ou will possess one of te most useful an fleible tools tat matematics as to offer. Calculus is te founation for all of our unerstaning of motion, incluing te aeronamic principles tat mae supersonic fligt possible. REMINDER A secant line is an line troug two points on a curve or grap. 3. Definition of te Derivative We begin wit two questions: Wat is te precise efinition of a tangent line? An ow can we compute its slope? To answer tese questions, let s return to te relationsip between tangent an secant lines first mentione in Section.. Te secant line troug istinct points P = (a, f (a)) an Q = (, f ()) on te grap of a function f () as slope [Figure (A)] f = f () f (a) a were f = f () f (a) an = a f () f (a) Te epression is calle te ifference quotient. a Q = (, f()) P = (a, f(a)) Δf = f() f(a) P = (a, f(a)) Tangent = a FIGURE Te secant line as slope f/. Our goal is to compute te slope of te tangent line at (a, f (a)). a (A) a (B) Now observe wat appens as Q approaces P or, equivalentl, as approaces a. Figure suggests tat te secant lines get progressivel closer to te tangent line. If we imagine Q moving towar P, ten te secant line appears to rotate into te tangent line as in (D). Terefore, we ma epect te slopes of te secant lines to approac te slope of te tangent line. Base on tis intuition, we efine te erivative f (a) (wic is rea f prime of a ) as te limit f f () f (a) (a) = lim a }{{ a } Limit of slopes of secant lines 0

2 SECTION 3. Definition of te Derivative P Q P Q P Q P Q a (A) FIGURE Te secant lines approac te tangent line as Q approaces P. a (B) a (C) a (D) Tere is anoter wa of writing te ifference quotient using a new variable : = a P Q f(a + ) f(a) a = a + FIGURE 3 Te ifference quotient can be written in terms of. We ave = a + an, for = a (Figure 3), f () f (a) a = f (a + ) f (a) Te variable approaces 0 as a, so we can rewrite te erivative as f (a) = lim 0 f (a + ) f (a) Eac wa of writing te erivative is useful. Te version using is often more convenient in computations. Te erivative of f () at = a is te limit of te if- DEFINITION Te Derivative ference quotients (if it eists): f (a) = lim 0 f (a + ) f (a) Wen te limit eists, we sa tat f is ifferentiable at = a. An equivalent efinition of te erivative is f (a) = lim a f () f (a) a We can now efine te tangent line in a precise wa, as te line of slope f (a) troug P = (a, f (a)). REMINDER Te equation of te line troug P = (a, b) of slope m in point-slope form: b = m( a) DEFINITION Tangent Line Assume tat f () is ifferentiable at = a. Te tangent line to te grap of = f () at P = (a, f (a)) is te line troug P of slope f (a). Te equation of te tangent line in point-slope form is f (a) = f (a)( a) 3

3 CHAPTER 3 DIFFERENTIATION (5, 5) = = FIGURE 4 Tangent line to = at = 5. Isaac Newton referre to calculus as te meto of fluions (from te Latin wor for flow ), but te term ifferential calculus, introuce in its Latin form calculus ifferentialis b Gottfrie Wilelm Leibniz, eventuall won out an was aopte universall. EXAMPLE Equation of a Tangent Line Fin an equation of te tangent line to te grap of f () = at = 5. Solution First, we must compute f (5). We are free to use eiter Eq. () or Eq. (). Using Eq. (), we ave f f () f(5) 5 (5) = lim = lim = lim ( 5)( + 5) 5 5 = lim ( + 5) = 0 5 Net, we appl Eq. (3) wit a = 5. Because f(5) = 5, an equation of te tangent line is 5 = 0( 5), or, in slope-intercept form: = 0 5 (Figure 4). Te net two eamples illustrate ifferentiation (te process of computing te erivative) using Eq. (). For clarit, we break up te computations into tree steps. EXAMPLE Compute f (3), were f () = 8. Solution Using Eq. (), we write te ifference quotient at a = 3 as f (a + ) f (a) = f(3 + ) f(3) ( = 0) Step. Write out te numerator of te ifference quotient. f(3 + ) f(3) = ( (3 + ) 8(3 + ) ) ( 3 8(3) ) Step. Divie b an simplif. Step 3. Compute te limit. f(3 + ) f(3) = ( ( ) (4 + 8) ) (9 4) = = ( ) = = } {{} Cancel f f(3 + ) f(3) (3) = lim = lim ( ) = FIGURE 5 Grap of f () =. Te tangent line at = as equation = 4 +. EXAMPLE 3 Sketc te grap of f () = an te tangent line at =. (a) Base on te sketc, o ou epect f () to be positive or negative? (b) Fin an equation of te tangent line at =. Solution Te grap an tangent line at = are sown in Figure 5. (a) We see tat te tangent line as negative slope, so f () must be negative. (b) We compute f () in tree steps as before. Step. Write out te numerator of te ifference quotient. f( + ) f() = + = ( + ) + ( + ) = ( + ) Step. Divie b an simplif. f( + ) f() = ( ) = ( + ) ( + )

4 SECTION 3. Definition of te Derivative 3 Step 3. Compute te limit. f f( + ) f() () = lim = lim 0 0 ( + ) = 4 Te function value is f() =, so te tangent line passes troug (, ) an as equation 4 f() = m + b In slope-intercept form, = 4 +. = ( ) 4 4 FIGURE 6 Te erivative of f () = m + b is f (a) = m for all a. 4 f() = b 4 FIGURE 7 Te erivative of a constant function f () = b is f (a) = 0 for all a. Te grap of a linear function f () = m + b (were m an b are constants) is a line of slope m. Te tangent line at an point coincies wit te line itself (Figure 6), so we soul epect tat f (a) = m for all a. Let s ceck tis b computing te erivative: f f (a + ) f (a) (m(a + ) + b) (ma + b) (a) = lim = lim 0 0 m = lim 0 = lim m = m 0 If m = 0, ten f () = b is constant an f (a) = 0 (Figure 7). In summar, THEOREM Derivative of Linear an Constant Functions If f () = m + b is a linear function, ten f (a) = m for all a. If f () = b is a constant function, ten f (a) = 0 for all a. EXAMPLE 4 Fin te erivative of f () = 9 5 at = an = 5. Solution We ave f (a) = 9 for all a. Hence, f () = f (5) = 9. a a + FIGURE 8 Wen is small, te secant line as nearl te same slope as te tangent line. P Q Secant Tangent Estimating te Derivative Approimations to te erivative are useful in situations were we cannot evaluate f (a) eactl. Since te erivative is te limit of ifference quotients, te ifference quotient soul give a goo numerical approimation wen is sufficientl small: f (a) f (a + ) f (a) if is small Grapicall, tis sas tat for small, te slope of te secant line is nearl equal to te slope of te tangent line (Figure 8). EXAMPLE 5 Estimate te erivative of f () = sin at = π 6. Solution We calculate te ifference quotient for several small values of : sin ( π 6 + ) sin π 6 = sin( π 6 + ) 0.5 Table on te net page suggests tat te limit as a ecimal epansion beginning In oter wors, f ( π 6 )

5 4 CHAPTER 3 DIFFERENTIATION TABLE Values of te Difference Quotient for Small >0 sin ( π 6 + ) 0.5 <0 sin ( π 6 + ) Secant ( < 0) Tangent Secant ( > 0) In te net eample, we use grapical reasoning to etermine te accurac of te estimates obtaine in Eample 5. EXAMPLE 6 Determining Accurac Grapicall Let f () = sin. Sow tat te approimation f ( ) π is accurate to four ecimal places. π 6 = sin FIGURE 9 Te tangent line is squeeze in between te secant lines wit >0 an <0. Tis tecnique of estimating an unknown quantit b sowing tat it lies between two known values ( squeezing it ) is use frequentl in calculus. Solution Observe in Figure 9 tat te position of te secant line relative to te tangent line epens on weter is positive or negative. Wen >0, te slope of te secant line is smaller tan te slope of te tangent line, but it is larger wen <0. Tis tells us tat te ifference quotients in te secon column of Table are smaller tan f ( ) π 6 an tose in te fourt column are greater tan f ( ) π 6. From te last line in Table we ma conclue tat ( f π ) It follows tat te estimate f ( π 6 ) is accurate to four ecimal places. In Section 3.6, we will see tat te eact value is f ( π 6 ) = cos ( π6 ) = 3/ , just about miwa between an CONCEPTUAL INSIGHT Are Limits Reall Necessar? It is natural to ask weter limits are reall necessar. Te tangent line is eas to visualize. Is tere peraps a better or simpler wa to fin its equation? Histor gives one answer: Te metos of calculus base on limits ave stoo te test of time an are use more wiel toa tan ever before. Histor asie, we can see irectl w limits pla suc a crucial role. Te slope of a line can be compute if te coorinates of two points P = (, ) an Q = (, ) on te line are known: Slope of line = Tis formula cannot be applie to te tangent line because we know onl tat it passes troug te single point P = (a, f (a)). Limits provie an ingenious wa aroun tis obstacle. We coose a point Q = (a +, f (a + )) on te grap near P an form te secant line. Te slope of tis secant line is just an approimation to te slope of te tangent line: f (a + ) f (a) Slope of secant line = slope of tangent line But tis approimation improves as 0, an b taking te limit, we convert our approimations into te eact slope.

6 SECTION 3. Definition of te Derivative 5 3. SUMMARY Te ifference quotient: f (a + ) f (a) Te ifference quotient is te slope of te secant line troug te points P = (a, f (a)) an Q = (a +, f (a + )) on te grap of f (). Te erivative f (a) is efine b te following equivalent limits: f f (a + ) f (a) f () f (a) (a) = lim = lim 0 a a If te limit eists, we sa tat f is ifferentiable at = a. B efinition, te tangent line at P = (a, f (a)) is te line troug P wit slope f (a) [assuming tat f (a) eists]. Equation of te tangent line in point-slope form: f (a) = f (a)( a) To calculate f (a) using te limit efinition: Step. Write out te numerator of te ifference quotient. Step. Divie b an simplif. Step 3. Compute te erivative b taking te limit. For small values of, we ave te estimate f (a) f (a + ) f (a). 3. EXERCISES Preliminar Questions. Wic of te lines in Figure 0 are tangent to te curve? A B FIGURE 0. Wat are te two was of writing te ifference quotient? Eercises. Let f () = 5. Sow tat f(3 + ) = Ten sow tat f(3 + ) f(3) C = an compute f (3) b taking te limit as 0.. Let f () = 3 5. Sow tat te secant line troug (,f()) an ( +, f ( + )) as slope + 5. Ten use tis formula to compute te slope of: D f (a + ) f (a) 3. Fin a an suc tat is equal to te slope of te secant line between (3,f(3)) an (5,f(5)). 4. Wic erivative is approimate b tan( π )? Wat o te following quantities represent in terms of te grap of f () = sin? sin.3 sin 0.9 (a) sin.3 sin 0.9 (b) (c) f (0.9) 0.4 (a) Te secant line troug (,f()) an (3,f(3)) (b) Te tangent line at = (b taking a limit) In Eercises 3 6, compute f (a) in two was, using Eq. () an Eq. (). 3. f () = + 9, a = 0 4. f () = + 9, a = 5. f () = , a = 6. f () = 3, a =

7 6 CHAPTER 3 DIFFERENTIATION In Eercises 7 0, refer to Figure. 7. Fin te slope of te secant line troug (,f()) an (.5,f(.5)). Is it larger or smaller tan f ()? Eplain. f( + ) f() 8. Estimate for = 0.5. Wat oes tis quantit represent? Is it larger or smaller tan f ()? Eplain. 9. Estimate f () an f (). 0. Fin a value of for wic f( + ) f() f() = FIGURE In Eercises 4, refer to Figure.. Determine f (a) for a =,, 4, 7.. For wic values of is f () < 0? 3. Wic is larger, f (5.5) or f (6.5)? 4. Sow tat f (3) oes not eist FIGURE Grap of f (). In Eercises 5 8, use te limit efinition to calculate te erivative of te linear function. 5. f () = f () = 7. g(t) = 8 3t 8. k(z) = 4z + 9. Fin an equation of te tangent line at = 3, assuming tat f(3) = 5 an f (3) =? 0. Fin f(3) an f (3), assuming tat te tangent line to = f () at a = 3 as equation = Describe te tangent line at an arbitrar point on te curve = Suppose tat f( + ) f() = Calculate: (a) Te slope of te secant line troug (,f()) an (6,f(6)) (b) f () 3. Let f () =. Does f( + ) equal + or +? Compute te ifference quotient at a = wit = Let f () =. Does f(5 + ) equal 5 + or 5 +? Compute te ifference quotient at a = 5 wit =. 5. Let f () = /. Compute f (5) b sowing tat f(5 + ) f(5) = ( ) 6. Fin an equation of te tangent line to te grap of f () = / at = 9. In Eercises 7 44, use te limit efinition to compute f (a) an fin an equation of te tangent line. 7. f () = + 0, a = 3 8. f () = 4, a = 9. f (t) = t t, a = f () = 8 3, a = 3. f () = 3 +, a = 0 3. f (t) = t 3 + 4t, a = f () =, a = f () = +, a = f () =, a = 36. f (t) = + 3 t, a = 37. f () = + 4, a = 38. f (t) = 3t + 5, a = 39. f () =, a = f () =, a = f (t) = t +, a = 3 4. f () =, a = 43. f () = +, a = f (t) = t 3, a = 45. Figure 3 isplas ata collecte b te biologist Julian Hule ( ) on te average antler weigt W of male re eer as a function of age t. Estimate te erivative at t = 4. For wic values of t is te slope of te tangent line equal to zero? For wic values is it negative? Antler weigt W (kg) Age (ears) FIGURE Figure 4(A) sows te grap of f () =. Te close-up in Figure 4(B) sows tat te grap is nearl a straigt line near = 6. Estimate te slope of tis line an take it as an estimate for f (6). Ten compute f (6) an compare wit our estimate. t

8 SECTION 3. Definition of te Derivative (A) Grap of = FIGURE (B) Zoom view near (6, 4) 47. Let f () = 4 +. (a) Plot f () over [, ]. Ten zoom in near = 0 until te grap appears straigt, an estimate te slope f (0). (b) Use (a) to fin an approimate equation to te tangent line at = 0. Plot tis line an f () on te same set of aes. 48. Let f () = cot. Estimate f ( π ) grapicall b zooming in on a plot of f () near = π. 49. Determine te intervals along te -ais on wic te erivative in Figure 5 is positive FIGURE Sketc te grap of f () = sin on [0,π] an guess te value of f ( π ). Ten calculate te ifference quotient at = π for two small positive an negative values of. Are tese calculations consistent wit our guess? In Eercises 5 56, eac limit represents a erivative f (a). Fin f () an a. (5 + ) lim 0 sin ( π 53. lim 6 + ) lim lim lim lim Appl te meto of Eample 6 to f () = sin to etermine f ( π 4 ) accuratel to four ecimal places. 58. Appl te meto of Eample 6 to f () = cos to etermine f ( π 5 ) accuratel to four ecimal places. Use a grap of f () to eplain ow te meto works in tis case. 59. For eac grap in Figure 6, etermine weter f () is larger or smaller tan te slope of te secant line between = an = + for >0. Eplain. = f() (A) (B) FIGURE 6 = f () 60. Refer to te grap of f () = in Figure 7. (a) Eplain grapicall w, for >0, f( ) f(0) f (0) f () f(0) (b) Use (a) to sow tat f (0) (c) Similarl, compute f () to four ecimal places for =,, 3, 4. () Now compute te ratios f ()/f (0) for =,, 3, 4. Can ou guess an approimate formula for f ()? 3 FIGURE 7 Grap of f () =. 6. Sketc te grap of f () = 5/ on [0, 6]. (a) Use te sketc to justif te inequalities for >0: f(4) f(4 ) f (4) f(4 + ) f(4) (b) Use (a) to compute f (4) to four ecimal places. (c) Use a graping utilit to plot f () an te tangent line at = 4, using our estimate for f (4). 6. Verif tat P = (, ) lies on te graps of bot f () = /( + ) an L() = + m( ) for ever slope m. Plot f () an L() on te same aes for several values of m until ou fin a value of m for wic = L() appears tangent to te grap of f (). Wat is our estimate for f ()? 63. Use a plot of f () = to estimate te value c suc tat f (c) = 0. Fin c to sufficient accurac so tat f (c + ) f (c) for =±0.00

9 8 CHAPTER 3 DIFFERENTIATION 64. Plot f () = an = + a on te same set of aes for several values of a until te line becomes tangent to te grap. Ten estimate te value c suc tat f (c) =. In Eercises 65 7, estimate erivatives using te smmetric ifference quotient (SDQ), efine as te average of te ifference quotients at an : ( f (a + ) f (a) + ) f (a ) f (a) = f (a + ) f (a ) Te SDQ usuall gives a better approimation to te erivative tan te ifference quotient. 65. Te vapor pressure of water at temperature T (in kelvins) is te atmosperic pressure P at wic no net evaporation takes place. Use te following table to estimate P (T ) for T = 303, 33, 33, 333, 343 b computing te SDQ given b Eq. (4) wit = 0. T (K) P (atm) Use te SDQ wit = ear to estimate te rate of cange in U.S. etanol prouction P in te ears 000, 00, 004, 006 (Figure 8). Epress our answer in te correct units. P (billions of gallons) FIGURE 8 U.S. Etanol Prouction In Eercises 67 68, traffic spee S along a certain roa (in km/) varies as a function of traffic ensit q (number of cars per km of roa). Use te following ata to answer te questions: 67. Estimate S (80). q (ensit) S (spee) Eplain w V = qs, calle traffic volume, is equal to te number of cars passing a point per our. Use te ata to estimate V (80). Eercises 69 7: Te current (in amperes) at time t (in secons) flowing in te circuit in Figure 9 is given b Kircoff s Law: i(t) = Cv (t) + R v(t) were v(t) is te voltage (in volts), C tecapacitance (in faras), an R te resistance (in oms, ). v + i R FIGURE Calculate te current at t = 3 if v(t) = 0.5t + 4V were C = 0.0 F an R = Use te following ata to estimate v (0) (b an SDQ). Ten estimate i(0), assuming C = 0.03 an R =,000. t v(t) Assume tat R = 00 but C is unknown. Use te following ata to estimate v (4) (b an SDQ) an euce an approimate value for te capacitance C. t v(t) i(t) C Furter Insigts an Callenges 7. Te SDQ usuall approimates te erivative muc more closel tan oes te orinar ifference quotient. Let f () = an a = 0. Compute te SDQ wit = 0.00 an te orinar ifference quotients wit =±0.00. Compare wit te actual value, wic is f (0) = ln. 73. Eplain ow te smmetric ifference quotient efine b Eq. (4) can be interprete as te slope of a secant line. 74. Wic of te two functions in Figure 0 satisfies te inequalit f (a + ) f (a ) for >0? Eplain in terms of secant lines. f (a + ) f (a)

10 SECTION 3. Te Derivative as a Function Sow tat if f () is a quaratic polnomial, ten te SDQ at = a (for an = 0) is equal to f (a). Eplain te grapical meaning of tis result. a a 76. Let f () =. Compute f () b taking te limit of te SDQs (wit a = ) as 0. (A) FIGURE 0 (B) 3. Te Derivative as a Function Often, te omain of f () is clear from te contet. If so, we usuall o not mention te omain eplicitl. In te previous section, we compute te erivative f (a) for specific values of a. It is also useful to view te erivative as a function f () wose value at = a is f (a). Te function f () is still efine as a limit, but te fie number a is replace b te variable : f f ( + ) f () () = lim 0 If = f (), we also write or () for f (). Te omain of f () consists of all values of in te omain of f () for wic te limit in Eq. () eists. We sa tat f () is ifferentiable on (a, b) if f () eists for all in (a, b). Wen f () eists for all in te interval or intervals on wic f () is efine, we sa simpl tat f () is ifferentiable. f() = 3 = ( 3, 9) 3 0 FIGURE Grap of f () = 3. EXAMPLE Prove tat f () = 3 is ifferentiable. Compute f () an fin an equation of te tangent line at = 3. Solution We compute f () in tree steps as in te previous section. Step. Write out te numerator of te ifference quotient. f ( + ) f () = ( ( + ) 3 ( + ) ) ( 3 ) Step. Divie b an simplif. f ( + ) f () Step 3. Compute te limit. = ( ) ( 3 ) = = ( ) (factor out ) = ( ) = ( = 0) f f ( + ) f () () = lim = lim ( ) = In tis limit, is treate as a constant because it oes not cange as 0. We see tat te limit eists for all, so f () is ifferentiable an f () = 3. Now evaluate: f( 3) = ( 3) 3 ( 3) = 9 f ( 3) = 3( 3) = 5 An equation of te tangent line at = 3 is 9 = 5( + 3) (Figure ).

11 30 CHAPTER 3 DIFFERENTIATION EXAMPLE Prove tat = is ifferentiable an calculate. Solution Te omain of f () = is { : = 0}, so assume tat = 0. We compute f () irectl, witout te separate steps of te previous eample: f ( + ) f () = lim = lim 0 0 = lim 0 ( + ) ( + ) ( ) ( + ) = lim 0 ( + ) = lim ( + ) ( ( + ) ) = lim 0 ( + ) 0 = + 0 ( + 0) = 4 = 3 + ( + ) (cancel ) Te limit eists for all = 0, so is ifferentiable an = 3. Leibniz Notation Te prime notation an f () was introuce b te Frenc matematician Josep Louis Lagrange (736 83). Tere is anoter stanar notation for te erivative tat we owe to Leibniz (Figure ): f or In Eample, we sowe tat te erivative of = is = 3. In Leibniz notation, we woul write = 3 or = 3 FIGURE Gottfrie Wilelm von Leibniz (646 76), German pilosoper an scientist. Newton an Leibniz (pronounce Libe-nitz ) are often regare as te inventors of calculus (working inepenentl). It is more accurate to creit tem wit eveloping calculus into a general an funamental iscipline, because man particular results of calculus a been iscovere previousl b oter matematicians. To specif te value of te erivative for a fie value of, sa, = 4, we write f or =4 You soul not tink of / as te fraction ivie b. Te epressions an are calle ifferentials. Te pla a role in some situations (in linear approimation an in more avance calculus). At tis stage, we treat tem merel as smbols wit no inepenent meaning. CONCEPTUAL INSIGHT Leibniz notation is wiel use for several reasons. First, it remins us tat te erivative f/, altoug not itself a ratio, is in fact a limit of ratios f /. Secon, te notation specifies te inepenent variable. Tis is useful wen variables oter tan are use. For eample, if te inepenent variable is t, we write f/t. Tir, we often tink of / as an operator tat performs ifferentiation on functions. In oter wors, we appl te operator / to f to obtain te erivative f/. We will see oter avantages of Leibniz notation wen we iscuss te Cain Rule in Section 3.7. =4 A main goal of tis capter is to evelop te basic rules of ifferentiation. Tese rules enable us to fin erivatives witout computing limits.

12 SECTION 3. Te Derivative as a Function 3 THEOREM Te Power Rule For all eponents n, Te Power Rule is vali for all eponents. We prove it ere for a wole number n (see Eercise 95 for a negative integer n an p. 83 for arbitrar n). n = n n Proof Assume tat n is a wole number an let f () = n. Ten f n a n (a) = lim a a To simplif te ifference quotient, we nee to generalize te following ientities: Te generalization is a = ( a)( + a) 3 a 3 = ( a )( + a + a ) 4 a 4 = ( a )( 3 + a + a + a 3) n a n = ( a) ( n + n a + n 3 a + +a n + a n ) To verif Eq. (), observe tat te rigt-an sie is equal to ( n + n a + n 3 a + +a n + a n ) a ( n + n a + n 3 a + +a n + a n ) Wen we carr out te multiplications, all terms cancel ecept te first an te last, so onl n a n remains, as require. Equation () gives us Terefore, n a n a = n + n a + n 3 a + +a n + a n }{{} n terms f (a) = lim a ( n + n a + n 3 a + +a n + a n ) = a n + a n a + a n 3 a + +aa n + a n (n terms) = na n Tis proves tat f (a) = na n, wic we ma also write as f () = n n. We make a few remarks before proceeing: ( = a) 3 CAUTION Te Power Rule applies onl to te power functions = n. It oes not appl to eponential functions suc as =. Te erivative of = is not. We will stu te erivatives of eponential functions later in tis section. It ma be elpful to remember te Power Rule in wors: To ifferentiate n, bring own te eponent an subtract one (from te eponent). eponent = (eponent) eponent Te Power Rule is vali for all eponents, weter negative, fractional, or irrational: 3/5 = 3 5 8/5, =

13 3 CHAPTER 3 DIFFERENTIATION Te Power Rule can be applie wit an variable, not just. For eample, z z = z, t t0 = 0t 9, r r/ = r / Net, we state te Linearit Rules for erivatives, wic are analogous to te linearit laws for limits. THEOREM Linearit Rules Sum an Difference Rules: Assume tat f an g are ifferentiable. Ten f + g an f g are ifferentiable, an (f + g) = f + g, (f g) = f g Constant Multiple Rule: For an constant c, cf is ifferentiable an (cf ) = cf Proof To prove te Sum Rule, we use te efinition (f + g) () = lim 0 (f ( + ) + g( + )) (f () + g()) Tis ifference quotient is equal to a sum ( = 0): (f ( + ) + g( + )) (f () + g()) Terefore, b te Sum Law for limits, = f ( + ) f () (f + g) f ( + ) f () g( + ) g() () = lim + lim 0 0 = f () + g () + g( + ) g() as claime. Te Difference an Constant Multiple Rules are prove similarl f(t) FIGURE 3 Grap of f (t) = t 3 t + 4. Tangent lines at t =± are orizontal. 4 t EXAMPLE 3 Fin te points on te grap of f (t) = t 3 t + 4 were te tangent line is orizontal (Figure 3). Solution We calculate te erivative: f t = t ( t 3 t + 4 ) = t t3 t (t) + 4 (Sum an Difference Rules) t = t t3 t + 0 (Constant Multiple Rule) t = 3t (Power Rule) Note in te secon line tat te erivative of te constant 4 is zero. Te tangent line is orizontal at points were te slope f (t) is zero, so we solve f (t) = 3t = 0 t =± Now f() = an f( ) = 0. Hence, te tangent lines are orizontal at (, ) an (, 0).

14 EXAMPLE 4 Calculate g t SECTION 3. Te Derivative as a Function 33, were g(t) = t 3 + t t 4/5. t= Solution We ifferentiate term-b-term using te Power Rule witout justifing te intermeiate steps. Writing t as t /, we ave g t = t ( t 3 + t / t 4/5) ( ) ( = 3t 4 + t / 4 ) t 9/5 5 6 Here tangent lines ave negative slope g t = 3t 4 + t / t 9/5 = t= 5 = 6 5 Te erivative f () gives us important information about te grap of f (). For eample, te sign of f () tells us weter te tangent line as positive or negative slope, an te magnitue of f () reveals ow steep te slope is. 6 (A) Grap of f () = f () > 0 f () > 0 f () < 0 EXAMPLE 5 Grapical Insigt How is te grap of f () = relate to te erivative f () = ? Solution Te erivative f () = = 3( 6)( ) is negative for <<6 an positive elsewere [Figure 4(B)]. Te following table summarizes tis sign information [Figure 4(A)]: 6 Propert of f () Propert of te Grap of f () FIGURE (B) Grap of te erivative f () = f () < 0 for <<6 Tangent line as negative slope for <<6. f () = f (6) = 0 Tangent line is orizontal at = an = 6. f () > 0 for < an >6 Tangent line as positive slope for < an >6. Note also tat f () as becomes large. Tis correspons to te fact tat te tangent lines to te grap of f () get steeper as grows large. EXAMPLE 6 Ientifing te Derivative Te grap of f () is sown in Figure 5(A). Wic grap (B) or (C), is te grap of f ()? FIGURE 5 (A) Grap of f() (B) (C) Slope of Tangent Line Were Negative (0, ) an (4, 7) Zero =, 4, 7 Positive (, 4) an (7, ) Solution In Figure 5(A) we see tat te tangent lines to te grap ave negative slope on te intervals (0, ) an (4, 7). Terefore f () is negative on tese intervals. Similarl (see te table in te margin), te tangent lines ave positive slope (an f () is positive) on te intervals (, 4) an (7, ). Onl (C) as tese properties, so (C) is te grap of f ().

15 34 CHAPTER 3 DIFFERENTIATION Te Derivative of e Te number e was introuce informall in Section.6. Now tat we ave te erivative in our arsenal, we can efine e as follows: e is te unique number for wic te eponential function f () = e is its own erivative. To justif tis efinition, we must prove tat a number wit tis propert eists. In some was, te number e is complicate : It is irrational an it cannot be efine witout using limits. However, te elegant formula e = e sows tat e is simple from te point of view of calculus an tat e is simpler tan te seemingl more natural eponential functions an 0. THEOREM 3 Te Number e Tere is a unique positive real number e wit te propert e = e 4 Te number e is irrational, wit approimate value e.78. Proof We sall take for grante a few plausible facts wose proofs are somewat tecnical. Te first fact is tat f () = b is ifferentiable for all b>0. Assuming tis, let us compute its erivative: f ( + ) f () = b+ b f f ( + ) f () b (b ) () = lim = lim 0 0 = b lim 0 = b b b ( b ) = b (b ) Notice tat we took te factor b outsie te limit. Tis is legitimate because b oes not epen on. Denote te value of te limit on te rigt b m(b): ( b ) m(b) = lim 5 0 Wat we ave sown, ten, is tat te erivative of b is proportional to b : b = m(b) b 6 Before continuing, let s investigate m(b) numericall using Eq. (5). EXAMPLE 7 Estimate m(b) numericall for b =,.5, 3, an 0. Solution We create a table of values of ifference quotients to estimate m(b). (.5) m() 0.69 m(.5) 0.9 m(3).0 m(0).30

16 SECTION 3. Te Derivative as a Function 35 In man books, e is enote ep(). Wenever we refer to te eponential function witout specifing te base, te reference is to f () = e. Te number e as been compute to an accurac of more tan 00 billion igits. To 0 places, e = Since m(.5) 0.9 an m(3).0, tere must eist a number b between.5 an 3 suc tat m(b) =. Tis follows from te Intermeiate Value Teorem (if we assume te fact tat m(b) is a continuous function of b). If we also use te fact tat m(b) is an increasing function of b, we ma conclue tat tere is precisel one number b suc tat m(b) =. Tis is te number e. Using infinite series (see Eercise 87 in Section 0.7), we can sow tat e is irrational an we can compute its value to an esire egree of accurac. For most purposes, te approimation e.78 is aequate. e 3.5 GRAPHICAL INSIGHT Te grap of f () = b passes troug (0, ) because b 0 = (Figure 6). Te number m(b) is simpl te slope of te tangent line at (0, ): = m(b) b 0 = m(b) =0 b Tese tangent lines become steeper as b increases, an b = e is te unique value for wic te tangent line as slope. In Section 3.9, we will sow more generall tat m(b) = ln b, te natural logaritm of b. FIGURE 6 Te tangent lines to = b at = 0 grow steeper as b increases. EXAMPLE 8 Fin te tangent line to te grap of f () = 3e 5 at =. Solution We compute bot f () an f(): 4 3 f() = 3e 5 =.7.7 f () = (3e 5 ) = 3 e 5 = 3e 0 f () = 3e 0().7 f() = 3e 5( ).7 An equation of te tangent line is = f() + f ()( ). Using tese approimate values, we write te equation as (Figure 7) =.7 +.7( ) or = FIGURE 7 CONCEPTUAL INSIGHT Wat precisel o we mean b b? We ave taken for grante tat b is meaningful for all real numbers, but we never specifie ow b is efine wen is irrational. If n is a wole number, b n is simpl te prouct b b b(n times), an for an rational number = m/n, b = b m/n = ( b /n) m ( n ) m = b Wen is irrational, tis efinition oes not appl an b cannot be efine irectl in terms of roots an powers of b. However, it makes sense to view b m/n as an approimation to b wen m/n is a rational number close to. For eample, 3 soul be approimatel equal to because.44 is a goo rational approimation to. Formall, ten, we ma efine b as a limit over rational numbers m/n approacing : b = lim m/n bm/n We can sow tat tis limit eists an tat te function f () = b tus efine is not onl continuous but also ifferentiable (see Eercise 80, in Section 5.7).

17 36 CHAPTER 3 DIFFERENTIATION L Differentiabilit, Continuit, an Local Linearit In te rest of tis section, we eamine te concept of ifferentiabilit more closel. We begin b proving tat a ifferentiable function is necessaril continuous. In particular, a ifferentiable function cannot ave an jumps. Figure 8 sows w: Altoug te secant lines from te rigt approac te line L (wic is tangent to te rigt alf of te grap), te secant lines from te left approac te vertical (an teir slopes ten to ). = f() c FIGURE 8 Secant lines at a jump iscontinuit. THEOREM 4 Differentiabilit Implies Continuit f is continuous at = c. If f is ifferentiable at = c, ten Proof B efinition, if f is ifferentiable at = c, ten te following limit eists: f (c) = lim c f () f (c) c We must prove tat lim c f () = f (c), because tis is te efinition of continuit at = c. To relate te two limits, consier te equation (vali for = c) f () f (c) = ( c) Bot factors on te rigt approac a limit as c, so ( ( c) ( ) lim f () f (c) = lim c c = ( lim f () f (c) c f () f (c) c ) ( ( c) lim c c = 0 f (c) = 0 ) f () f (c) c b te Prouct Law for limits. Te Sum Law now iels te esire conclusion: lim c f () = lim (f () f (c)) + lim f (c) = 0 + f (c) = f (c) c c Most of te functions encountere in tis tet are ifferentiable, but eceptions eist, as te net eample sows. ) All ifferentiable functions are continuous b Teorem 4, but Eample 9 sows tat te converse is false. A continuous function is not necessaril ifferentiable. EXAMPLE 9 Continuous But Not Differentiable Sow tat f () = is continuous but not ifferentiable at = 0. Solution Te function f () is continuous at = 0 because lim 0 =0 = f(0). On te oter an, f f(0 + ) f(0) (0) = lim = lim = lim Tis limit oes not eist [an ence f () is not ifferentiable at = 0] because { = if >0 if <0 an tus te one-sie limits are not equal: lim = an lim 0+ 0 =

18 SECTION 3. Te Derivative as a Function 37 GRAPHICAL INSIGHT Differentiabilit as an important grapical interpretation in terms of local linearit. We sa tat f is locall linear at = a if te grap looks more an more like a straigt line as we zoom in on te point (a, f (a)). In tis contet, te ajective linear means resembling a line, an local inicates tat we are concerne onl wit te beavior of te grap near (a, f (a)). Te grap of a locall linear function ma be ver wav or nonlinear, as in Figure 9. But as soon as we zoom in on a sufficientl small piece of te grap, it begins to appear straigt. Not onl oes te grap look like a line as we zoom in on a point, but as Figure 9 suggests, te zoom line is te tangent line. Tus, te relation between ifferentiabilit an local linearit can be epresse as follows: If f (a) eists, ten f is locall linear at = a: As we zoom in on te point (a, f (a)), te grap becomes nearl inistinguisable from its tangent line. Tangent FIGURE 9 Local linearit: Te grap looks more an more like te tangent line as we zoom in on a point. Local linearit gives us a grapical wa to unerstan w f () = is not ifferentiable at = 0 (as sown in Eample 9). Figure 0 sows tat te grap of f () = as a corner at = 0, an tis corner oes not isappear, no matter ow closel we zoom in on te origin. Since te grap oes not straigten out uner zooming, f () is not locall linear at = 0, an we cannot epect f (0) to eist FIGURE 0 Te grap of f () = is not locall linear at = 0. Te corner oes not isappear wen we zoom in on te origin. Anoter wa tat a continuous function can fail to be ifferentiable is if te tangent line eists but is vertical (in wic case te slope of te tangent line is unefine). EXAMPLE 0 Vertical Tangents Sow tat f () = /3 is not ifferentiable at = 0. Solution Te limit efining f (0) is infinite: f () f(0) /3 0 /3 lim = lim = lim = lim = 0 /3 Terefore, f (0) oes not eist (Figure ).

19 38 CHAPTER 3 DIFFERENTIATION 0.5 As a final remark, we mention tat tere are more complicate was in wic a continuous function can fail to be ifferentiable. Figure sows te grap of f () = sin. If we efine f(0) = 0, ten f is continuous but not ifferentiable at = 0. Te secant lines keep oscillating an never settle own to a limiting position (see Eercise 97) FIGURE Te tangent line to te grap of f () = /3 at te origin is te (vertical) -ais. Te erivative f (0) oes not eist FIGURE (A) Grap of f() = sin (B) Secant lines o not settle own to a limiting position. 3. SUMMARY Te erivative f () is te function wose value at = a is te erivative f (a). We ave several ifferent notations for te erivative of = f ():, (), f (), Te value of te erivative at = a is written (a), f (a),, =a Te Power Rule ols for all eponents n: n = n n, f f Te Linearit Rules allow us to ifferentiate term b term: =a Sum Rule: (f + g) = f + g, Constant Multiple Rule: (cf ) = cf Te erivative of b is proportional to b : b = m(b)b b, were m(b) = lim 0 Te number e.78 is efine b te propert m(e) =, so tat e = e Differentiabilit implies continuit: If f () is ifferentiable at = a, ten f () is continuous at = a. However, tere eist continuous functions tat are not ifferentiable. If f (a) eists, ten f is locall linear in te following sense: As we zoom in on te point (a, f (a)), te grap becomes nearl inistinguisable from its tangent line.

20 SECTION 3. Te Derivative as a Function EXERCISES Preliminar Questions. Wat is te slope of te tangent line troug te point (,f()) if f () = 3?. Evaluate (f g) () an (3f + g) () assuming tat f () = 3 an g () = To wic of te following oes te Power Rule appl? (a) f () = (b) f () = e (c) f () = e () f () = e (e) f () = (f) f () = 4/5 4. Coose (a) or (b). Te erivative oes not eist if te tangent line is: (a) orizontal (b) vertical. 5. Wic propert istinguises f () = e from all oter eponential functions g() = b? Eercises In Eercises 6, compute f () using te limit efinition.. f () = 3 7. f () = f () = 3 4. f () = 5. f () = 6. f () = / In Eercises 7 4, use te Power Rule to compute te erivative = t t 3 t=4 9. t t/3 0. t=8 t t /5 t= t t /3 t t π In Eercises 5 8, compute f () an fin an equation of tetangent line to te grap at = a. 5. f () = 4, a = 6. f () =, a = 5 7. f () = 5 3, a = 4 8. f () = 3, a = 8 9. Calculate: (a) e (b) t (5t 8et ) Hint for (c): Write e t 3 as e 3 e t. (c) t et 3 0. Fin an equation of te tangent line to = 4e at =. 7. f (s) = 4 s + 3 s 8. W() = /3 9. g() = e 30. f () = 3e 3 3. (t) = 5e t 3 3. f () = 9 /3 + 8e In Eercises 33 36, calculate te erivative b epaning or simplifing te function. 33. P (s) = (4s 3) 34. Q(r) = ( r)(3r + 5) 35. g() = + 4 / 36. s(t) = t t / In Eercises 37 4, calculate te erivative inicate. T 37., T = 3C /3 P 38., P = 7 C C=8 V V = V 39. s, z z= s = 4z 6z r t, t=4 r = t e t 4. R W, W = R = W π p, =4 p = 7e 43. Matc te functions in graps (A) (D) wit teir erivatives (I) (III) in Figure 3. Note tat two of te functions ave te same erivative. Eplain w. In Eercises 3, calculate te erivative. (A) (B) (C) (D). f () = f () = f () = 4 5/ f () = 5/ / + 5. g(z) = 7z 5/4 + z (t) = 6 t + t (I) (II) FIGURE 3 (III)

21 40 CHAPTER 3 DIFFERENTIATION 44. Of te two functions f an g in Figure 4, wic is te erivative of te oter? Justif our answer. FIGURE 4 f() g() 45. Assign te labels f (), g(), an () to te graps in Figure 5 in suc a wa tat f () = g() an g () = (). (A) (B) (C) FIGURE Accoring to te peak oil teor, first propose in 956 b geopsicist M. Hubbert, te total amount of crue oil Q(t) prouce worlwie up to time t as a grap like tat in Figure 6. (a) Sketc te erivative Q (t) for 900 t 50. Wat oes Q (t) represent? (b) In wic ear (approimatel) oes Q (t) take on its maimum value? (c) Wat is L = lim Q(t)? An wat is its interpretation? t () Wat is te value of lim t Q (t)? Q (trillions of barrels) t (ear) FIGURE 6 Total oil prouction up to time t 47. Use te table of values of f () to etermine wic of (A) or (B) in Figure 7 is te grap of f (). Eplain f () (A) (B) FIGURE 7 Wic is te grap of f ()? 48. Let R be a variable an r a constant. Compute te erivatives: (a) R R (b) R r (c) R r R Compute te erivatives, were c is a constant. (a) t ct3 (b) z (5z + 4cz ) (c) (9c 3 4c) 50. Fin te points on te grap of f () = 3 were te tangent line is orizontal. 5. Fin te points on te grap of = at wic te slope of te tangent line is equal to Fin te values of were = 3 an = + 5 ave parallel tangent lines. 53. Determine a an b suc tat p() = + a + b satisfies p() = 0 an p () = Fin all values of suc tat te tangent line to = is steeper tan te tangent line to = Let f () = Sow tat f () 3 for all an tat, for ever m> 3, tere are precisel two points were f () = m. Inicate te position of tese points an te corresponing tangent lines for one value of m in a sketc of te grap of f (). 56. Sow tat te tangent lines to = 3 3 at = a an at = b are parallel if a = b or a + b =. 57. Compute te erivative of f () = 3/ using te limit efinition. Hint: Sow tat f ( + ) f () = ( + )3 3 ( ) ( + ) Use te limit efinition of m(b) to approimate m(4). Ten estimate te slope of te tangent line to = 4 at = 0 an =. 59. Let f () = e. Use te limit efinition to compute f (0), an fin te equation of te tangent line at = Te average spee (in meters per secon) of a gas molecule is 8RT v avg = πm

22 SECTION 3. Te Derivative as a Function 4 were T is te temperature (in kelvins), M is te molar mass (in kilograms per mole), an R = 8.3. Calculate v avg /T at T = 300 K for ogen, wic as a molar mass of 0.03 kg/mol. 6. Biologists ave observe tat te pulse rate P (in beats per minute) in animals is relate to bo mass (in kilograms) b te approimate formula P = 00m /4. Tis is one of man allometric scaling laws prevalent in biolog. Is P/m an increasing or ecreasing function of m? Fin an equation of te tangent line at te points on te grap in Figure 8 tat represent goat (m = 33) an man (m = 68). (A) (I) Pulse (beats/min) 00 Guinea pig (B) (II) 00 Goat Man FIGURE Cattle Mass (kg) 6. Some stuies suggest tat kine mass K in mammals (in kilograms) is relate to bo mass m (in kilograms) b te approimate formula K = 0.007m Calculate K/m at m = 68. Ten calculate te erivative wit respect to m of te relative kine-to-mass ratio K/m at m = Te Clausius Claperon Law relates te vapor pressure of water P (in atmosperes) to te temperature T (in kelvins): P T = k P T were k is a constant. Estimate P/T for T = 303, 33, 33, 333, 343 using te ata an te approimation P T P (T + 0) P (T 0) 0 T (K) P (atm) Do our estimates seem to confirm te Clausius Claperon Law? Wat is te approimate value of k? (C) (III) FIGURE Make a roug sketc of te grap of te erivative of te function in Figure 0(A). 68. Grap te erivative of te function in Figure 0(B), omitting points were te erivative is not efine. 3 = (A) (B) FIGURE Sketc te grap of f () =. Ten sow tat f (0) eists. 70. Determine te values of at wic te function in Figure is: (a) iscontinuous, an (b) nonifferentiable Let L be te tangent line to te perbola = at = a, were a>0. Sow tat te area of te triangle boune b L an te coorinate aes oes not epen on a. 65. In te setting of Eercise 64, sow tat te point of tangenc is te mipoint of te segment of L ling in te first quarant. 66. Matc functions (A) (C) wit teir erivatives (I) (III) in Figure 9. FIGURE 3 4

23 4 CHAPTER 3 DIFFERENTIATION In Eercises 7 76, fin te points c (if an) suc tat f (c) oes not eist. 7. f () = 7. f () =[] 73. f () = /3 74. f () = 3/ 75. f () = 76. f () = In Eercises 77 8, zoom in on a plot of f () at te point (a, f (a)) an state weter or not f () appears to be ifferentiable at = a. If it is nonifferentiable, state weter te tangent line appears to be vertical or oes not eist. 77. f () = ( ), a = f () = ( 3) 5/3, a = f () = ( 3) /3, a = f () = sin( /3 ), a = 0 8. f () = sin, a = 0 8. f () = sin, a = Plot te erivative f () of f () = 3 0 for > 0 (set te bouns of te viewing bo appropriatel) an observe tat f () > 0. Wat oes te positivit of f () tell us about te grap of f () itself? Plot f () an confirm tis conclusion. 84. Fin te coorinates of te point P in Figure at wic te tangent line passes troug (5, 0). f () = 9 9 P FIGURE Grap of f () = 9. Eercises refer to Figure 3. Lengt QR is calle te subtangent at P, an lengt RT is calle te subnormal. 85. Calculate te subtangent of f () = + 3 at =. 86. Sow tat te subtangent of f () = e is everwere equal to. 87. Prove in general tat te subnormal at P is f ()f (). 88. Sow tat PQas lengt f () + f (). Q P = (, f()) R T FIGURE 3 = f() Tangent line 89. Prove te following teorem of Apollonius of Perga (te Greek matematician born in 6 bce wo gave te parabola, ellipse, an perbola teir names): Te subtangent of te parabola = at = a is equal to a/. 90. Sow tat te subtangent to = 3 at = a is equal to 3 a. 9. Formulate an prove a generalization of Eercise 90 for = n. Furter Insigts an Callenges 9. Two small arces ave te sape of parabolas. Te first is given b f () = for an te secon b g() = 4 ( 4) for 6. A boar is place on top of tese arces so it rests on bot (Figure 4). Wat is te slope of te boar? Hint: Fin te tangent line to = f () tat intersects = g() in eactl one point. FIGURE A vase is forme b rotating = aroun te -ais. If we rop in a marble, it will eiter touc te bottom point of te vase or be suspene above te bottom b toucing te sies (Figure 5). How small must te marble be to touc te bottom? FIGURE Let f () be a ifferentiable function, an set g() = f ( + c), were c is a constant. Use te limit efinition to sow tat g () = f ( + c). Eplain tis result grapicall, recalling tat te grap of g() is obtaine b sifting te grap of f () c units to te left (if c>0) or rigt (if c<0). 95. Negative Eponents Let n be a wole number. Use te Power Rule for n to calculate te erivative of f () = n b sowing tat f ( + ) f () = ( + ) n n n ( + ) n

24 SECTION 3.3 Prouct an Quotient Rules Verif te Power Rule for te eponent /n, were n is a positive integer, using te following trick: Rewrite te ifference quotient for = /n at = b in terms of u = (b + ) /n an a = b /n. 97. Infinitel Rapi Oscillations Define sin = 0 f () = 0 = 0 Sow tat f () is continuous at = 0 but f (0) oes not eist (see Figure ). 98. For wic value of λ oes te equation e = λ ave a unique solution? For wic values of λ oes it ave at least one solution? For intuition, plot = e an te line = λ. REMINDER Te prouct function fg is efine b (fg)() = f () g(). 3.3 Prouct an Quotient Rules Tis section covers te Prouct Rule an Quotient Rule for computing erivatives. Tese two rules, togeter wit te Cain Rule an implicit ifferentiation (covere in later sections), make up an etremel effective ifferentiation toolkit. If f an g are ifferentiable functions, ten fg is iffer- THEOREM Prouct Rule entiable an (fg) () = f () g () + g() f () It ma be elpful to remember te Prouct Rule in wors: Te erivative of a prouct is equal to te first function times te erivative of te secon function plus te secon function times te erivative of te first function: First (Secon) + Secon (First) We prove te Prouct Rule after presenting tree eamples. EXAMPLE Fin te erivative of () = (9 + ). Solution Tis function is a prouct: First {}}{ Secon {}}{ () = (9 + ) B te Prouct Rule (in Leibniz notation), () = First {}}{ Secon {}}{ (9 + ) + First Secon {}} {{}}{ (9 + ) ( ) = ( )(9) + (9 + )() = EXAMPLE Fin te erivative of = ( + )( 3/ + ). Solution Use te Prouct Rule: Note ow te prime notation is use in te solution to Eample. We write ( 3/ + ) to enote te erivative of 3/ +, etc. First (Secon) + Secon (First) ( {}}{ = + )( 3/ + ) ( + 3/ + )( + ) = ( + )( 3 /) + ( 3/ + )( ) (compute te erivatives) = 3 / + 3 / / = 3 / + / (simplif)

25 44 CHAPTER 3 DIFFERENTIATION In te previous two eamples, we coul ave avoie te Prouct Rule b epaning te function. Tus, te result of Eample can be obtaine as follows: = ( + )( 3/ + ) = 3/ + + / + = ( 3/ + + / + ) = 3 / + / In te net eample, te function cannot be epane, so we must use te Prouct Rule (or go back to te limit efinition of te erivative). EXAMPLE 3 Calculate t t e t. Solution Use te Prouct Rule an te formula t et = e t : t t e t = First (Secon) + Secon (First) {}}{ t t et + e t t t = t e t + e t (t) = (t + t)e t Proof of te Prouct Rule Accoring to te limit efinition of te erivative, f( + )(g( + ) g()) (fg) () = lim 0 f ( + )g( + ) f ()g() g( + ) g() f() f( + ) g()( f( + ) f()) We can interpret te numerator as te area of te sae region in Figure : Te area of te larger rectangle f ( + )g( + ) minus te area of te smaller rectangle f ()g(). Tis sae region is te union of two rectangular strips, so we obtain te following ientit (wic ou can ceck irectl): f ( + )g( + ) f ()g() = f ( + ) ( g( + ) g() ) + g() ( f ( + ) f () ) FIGURE Now use tis ientit to write (fg) () as a sum of two limits: (fg) () = lim f ( + ) 0 g( + ) g() } {{ } Sow tat tis equals f ()g (). + lim g() 0 f ( + ) f () } {{ } Sow tat tis equals g()f (). Te use of te Sum Law is vali, provie tat eac limit on te rigt eists. To ceck tat te first limit eists an to evaluate it, we note tat f () is continuous (because it is ifferentiable) an tat g() is ifferentiable. + ) g() g( + ) g() lim f ( + )g( = lim f ( + ) lim = f () g () Te secon limit is similar: f ( + ) f () f ( + ) f () lim g() = g() lim = g() f () Using Eq. () an Eq. (3) in Eq. (), we conclue tat fg is ifferentiable an tat (fg) () = f ()g () + g()f () as claime.

26 SECTION 3.3 Prouct an Quotient Rules 45 CONCEPTUAL INSIGHT Te Prouct Rule was first state b te 9-ear-ol Leibniz in 675, te ear e evelope some of is major ieas on calculus. To ocument is process of iscover for posterit, e recore is tougts an struggles, te moments of inspiration as well as te mistakes. In a manuscript ate November, 675, Leibniz suggests incorrectl tat (fg) equals f g. He ten catces is error b taking f () = g() = an noticing tat (fg) () = ( ) = is not equal to f ()g () = = Ten as later, on November, Leibniz writes own te correct Prouct Rule an comments Now tis is a reall notewort teorem. Wit te benefit of insigt, we can point out tat Leibniz migt ave avoie is error if e a pai attention to units. Suppose f (t) an g(t) represent istances in meters, were t is time in secons. Ten (fg) as units of m /s. Tis cannot equal f g, wic as units of (m/s)(m/s) = m /s. Te net teorem states te rule for ifferentiating quotients. Note, in particular, tat (f/g) is not equal to te quotient f /g. REMINDER Te quotient function f/g is efine b ( ) f () = f () g g() THEOREM Quotient Rule If f an g are ifferentiable functions, ten f/g is ifferentiable for all suc tat g() = 0, an ( ) f () = g()f () f ()g () g g() Te numerator in te Quotient Rule is equal to te bottom times te erivative of te top minus te top times te erivative of te bottom: Bottom (Top) Top (Bottom) Bottom Te proof is similar to tat of te Prouct Rule (see Eercises 58 60). EXAMPLE 4 Compute te erivative of f () = Solution Appl te Quotient Rule: Bottom Top {}}{{}}{ f ( + ) () () = Top {}}{ () Bottom {}}{ ( + ) +. ( + ) = ( + )() ()() ( + ) = + ( + ) = ( + ) EXAMPLE 5 Calculate e t t e t + t. Solution Use te Quotient Rule an te formula (e t ) = e t : e t t e t + t = (et + t)(e t ) e t (e t + t) (e t + t) = (et + t)e t e t (e t + ) (t )et (e t + t) = (e t + t)

27 46 CHAPTER 3 DIFFERENTIATION EXAMPLE 6 Fin te tangent line to te grap of f () = Solution f () = At =, ( 3 ) = + Bottom {}}{ (4 3 + ) An equation of te tangent line at (, 5) is Top {}} { (3 + ) Top {}}{ (3 + ) (4 3 + ) = (43 + )(6 + ) (3 + )( ) (4 3 + ) at =. Bottom {}}{ (4 3 + ) = ( ) ( ) (4 3 + ) = (4 3 + ) f() = 3 + = f () = 5 = 5 5 = ( ) or = R V FIGURE Apparatus of resistance R attace to a batter of voltage V. P r FIGURE 3 Grap of power versus resistance: P = V R (R + r) R R EXAMPLE 7 Power Delivere b a Batter Te power tat a batter supplies to an apparatus suc as a laptop epens on te internal resistance of te batter. For a batter of voltage V an internal resistance r, te total power elivere to an apparatus of resistance R (Figure ) is P = V R (R + r) (a) Calculate P/R, assuming tat V an r are constants. (b) Were, in te grap of P versus R, is te tangent line orizontal? Solution (a) Because V is a constant, we obtain (using te Quotient Rule) P R = V ( ) R (R + r) R R (R + r) = V R R R (R + r) (R + r) 4 4 We ave R R =, an Rr = 0 because r is a constant. Tus, R (R + r) = R (R + rr + r ) = R R + r R R + R r Using Eq. (5) in Eq. (4), we obtain = R + r + 0 = (R + r) 5 P R = V (R + r) R(R + r) (R + r) R (R + r) 4 = V (R + r) 3 = V r R (R + r) 3 6 (b) Te tangent line is orizontal wen te erivative is zero. We see from Eq. (6) tat te erivative is zero wen r R = 0 tat is, wen R = r.

28 SECTION 3.3 Prouct an Quotient Rules 47 GRAPHICAL INSIGHT Figure 3 sows tat te point were te tangent line is orizontal is te maimum point on te grap. Tis proves an important result in circuit esign: Maimum power is elivere wen te resistance of te loa (apparatus) is equal to te internal resistance of te batter. 3.3 SUMMARY Two basic rules of ifferentiation: Prouct Rule: (fg) = fg + gf Quotient Rule: ( f g ) = gf fg g Remember: Te erivative of fg is not equal to f g. Similarl, te erivative of f/g is not equal to f /g. 3.3 EXERCISES Preliminar Questions. Are te following statements true or false? If false, state te correct version. (a) fg enotes te function wose value at is f (g()). (b) f/g enotes te function wose value at is f ()/g(). (c) Te erivative of te prouct is te prouct of te erivatives. () (e) (fg) = f(4)g (4) g(4)f (4) =4 (fg) = f(0)g (0) + g(0)f (0) =0. Fin (f/g) () if f() = f () = g() = an g () = Fin g() if f() = 0, f () =, an (fg) () = 0. Eercises In Eercises 6, use te Prouct Rule to calculate te erivative.. f () = 3 ( + ). f () = (3 5)( 3) 3. f () = e 4. f () = ( 9)(4e + ) 5. s, s=4 (s) = (s / + s)(7 s ) 6. t, t= = (t 8t )(e t + t ) In Eercises 7, use te Quotient Rule to calculate te erivative. 7. f () = g 9. t, g(t) = t + t= t 8. f () = w 0. z, w = z z=9 z + z. g() = + e. f () = e + In Eercises 3 6, calculate te erivative in two was. First use te Prouct or Quotient Rule; ten rewrite te function algebraicall an appl te Power Rule irectl. 3. f (t) = (t + )(t ) 4. f () = (3 + ) 5. (t) = t t 6. g() = In Eercises 7 38, calculate te erivative. 7. f () = ( 3 + 5)( ) 8. f () = (4e )( 3 + ) 9., = z 0. =3 + 0, z = = 3 +. f () = ( + )( ). f () = 95/

29 48 CHAPTER 3 DIFFERENTIATION 3. 5., = 4 4 = 5 z, z = = 3 + t 7. (t) = (t + )(t + ) 8. f () = 3/( /) 4. f () = 4 + e + 6. f () = f (t) = 3 / 5 / 30. () = π ( ) 46. Plot te erivative of f () = /( + ) over [ 4, 4]. Use te grap to etermine te intervals on wic f () > 0 an f () < 0. Ten plot f () an escribe ow te sign of f () is reflecte in te grap of f (). 47. Plot f () = /( ) (in a suitabl boune viewing bo). Use te plot to etermine weter f () is positive or negative on its omain { : = ±}. Ten compute f () an confirm our conclusion algebraicall. 48. Let P = V R/(R + r) as in Eample 7. Calculate P/r, assuming tat r is variable an R is constant. 3. f () = ( + 3)( )( 5) 3. f () = e ( + )( + 4) 33. f () = e + ( )( ) 35. g(z) = z 4 z z z g() = e+ + e e + Hint: Simplif first. ( ) 36. (a + b)(ab + ) (a,b constants) ( ) t t t ( constant) ( ) a + b 38. (a, b, c, constants) c + In Eercises 39 4, calculate te erivative using te values: 39. (fg) (4) an (f/g) (4). 40. F (4), were F () = f (). 4. G (4), were G() = g(). 4. H (4), were H () = 43. Calculate F (0), were f(4) f (4) g(4) g (4) 0 5 g()f (). F () = Hint: Do not calculate F (). Instea, write F () = f ()/g() an epress F (0) irectl in terms of f(0), f (0), g(0), g (0). 44. Procee as in Eercise 43 to calculate F (0), were F () = ( + + 4/3 + 5/3) Use te Prouct Rule to calculate e. 49. Fin a>0 suc tat te tangent line to te grap of f () = e passes troug te origin (Figure 4). a FIGURE 4 at = a f() = e 50. Current I (amperes), voltage V (volts), an resistance R (oms) in a circuit are relate b Om s Law, I = V /R. (a) Calculate R I R=6 if V is constant wit value V = 4. (b) Calculate V R R=6 if I is constant wit value I = Te revenue per mont earne b te Couture cloting cain at time t is R(t) = N(t)S(t), were N(t) is te number of stores an S(t) is average revenue per store per mont. Couture embarks on a two-part campaign: (A) to buil new stores at a rate of 5 stores per mont, an (B) to use avertising to increase average revenue per store at a rate of $0,000 per mont. Assume tat N(0) = 50 an S(0) = $50,000. (a) Sow tat total revenue will increase at te rate R t = 5S(t) + 0,000N(t) Note tat te two terms in te Prouct Rule correspon to te separate effects of increasing te number of stores on te one an, an te average revenue per store on te oter. (b) Calculate R t = 0. t=0 (c) If Couture can implement onl one leg (A or B) of its epansion at t = 0, wic coice will grow revenue most rapil? 5. Te tip spee ratio of a turbine (Figure 5) is te ratio R = T/W, were T is te spee of te tip of a blae an W is te spee of te win. (Engineers ave foun empiricall tat a turbine wit n blaes

30 SECTION 3.3 Prouct an Quotient Rules 49 etracts maimum power from te win wen R = π/n.) Calculate R/t (t in minutes) if W = 35 km/ an W ecreases at a rate of 4 km/ per minute, an te tip spee as constant value T = 50 km/. 53. Te curve = /( + ) is calle te witc of Agnesi (Figure 6) after te Italian matematician Maria Agnesi (78 799), wo wrote one of te first books on calculus. Tis strange name is te result of a mistranslation of te Italian wor la versiera, meaning tat wic turns. Fin equations of te tangent lines at =±. 3 3 FIGURE 6 Te witc of Agnesi. 54. Let f () = g() =. Sow tat (f/g) = f /g. 55. Use te Prouct Rule to sow tat (f ) = ff. FIGURE 5 Turbines on a win farm 56. Sow tat (f 3 ) = 3f f. Furter Insigts an Callenges 57. Let f, g, be ifferentiable functions. Sow tat (fg) () is equal to f ()g() () + f ()g ()() + f ()g()() Hint: Write fg as f (g). 58. Prove te Quotient Rule using te limit efinition of te erivative. 59. Derivative of te Reciprocal Use te limit efinition to prove ( ) = f () f () f 7 () Hint: Sow tat te ifference quotient for /f () is equal to f () f ( + ) f ()f ( + ) 60. Prove te Quotient Rule using Eq. (7) an te Prouct Rule. 6. Use te limit efinition of te erivative to prove te following special case of te Prouct Rule: (f ()) = f () + f () 6. Carr out Maria Agnesi s proof of te Quotient Rule from er book on calculus, publise in 748: Assume tat f, g, an = f/g are ifferentiable. Compute te erivative of g = f using te Prouct Rule, an solve for. 63. Te Power Rule Revisite If ou are familiar wit proof b inuction, use inuction to prove te Power Rule for all wole numbers n. Sow tat te Power Rule ols for n = ; ten write n as n an use te Prouct Rule. Eercises 64 65: A basic fact of algebra states tat c is a root of a polnomial f () if an onl if f () = ( c)g() for some polnomial g(). Wesa tat c is a multiple root if f () = ( c) (), were () is a polnomial. 64. Sow tat c is a multiple root of f () if an onl if c is a root of bot f () an f (). 65. Use Eercise 64 to etermine weter c = is a multiple root: (a) (b) Figure 7 is te grap of a polnomial wit roots at A, B, an C. Wic of tese is a multiple root? Eplain our reasoning using Eercise 64. A B FIGURE Accoring to Eq. (6) in Section 3., b = m(b) b. Use te Prouct Rule to sow tat m(ab) = m(a) + m(b). C

31 50 CHAPTER 3 DIFFERENTIATION 3.4 Rates of Cange Recall te notation for te average rate of cange of a function = f () over an interval [ 0, ]: = cange in = f ( ) f ( 0 ) = cange in = 0 Average Rate of Cange = = f ( ) f ( 0 ) 0 We usuall omit te wor instantaneous an refer to te erivative simpl as te rate of cange. Tis is sorter an also more accurate wen applie to general rates, because te term instantaneous woul seem to refer onl to rates wit respect to time. In our prior iscussion in Section., limits an erivatives a not et been introuce. Now tat we ave tem at our isposal, we can efine te instantaneous rate of cange of wit respect to at = 0 : Instantaneous Rate of Cange = f ( 0 ) = lim 0 = lim f ( ) f ( 0 ) 0 0 Keep in min te geometric interpretations: Te average rate of cange is te slope of te secant line (Figure ), an te instantaneous rate of cange is te slope of te tangent line (Figure ). (, f( )) ( 0, f( 0 )) ( 0, f ( 0 )) 0 FIGURE Te average rate of cange over [ 0, ] is te slope of te secant line. FIGURE Te instantaneous rate of cange at 0 is te slope of te tangent line. 0 TABLE Data from Mars Patfiner Mission, Jul 997 Time Temperature ( C) 5: : 7.6 6: : : : : : :34 4 Leibniz notation / is particularl convenient because it specifies tat we are consiering te rate of cange of wit respect to te inepenent variable. Te rate / is measure in units of per unit of. For eample, te rate of cange of temperature wit respect to time as units suc as egrees per minute, wereas te rate of cange of temperature wit respect to altitue as units suc as egrees per kilometer. EXAMPLE Table contains ata on te temperature T on te surface of Mars at Martian time t, collecte b te NASA Patfiner space probe. (a) Calculate te average rate of cange of temperature T from 6: am to 9:05 am. (b) Use Figure 3 to estimate te rate of cange at t = :8 pm. Solution (a) Te time interval [6:, 9:05] as lengt, 54 min, or t =.9. Accoring to Table, te cange in temperature over tis time interval is T = 44.3 ( 7.6) = 7.3 C

32 SECTION 3.4 Rates of Cange T ( C) 0 (:8,.3) A 80 0:00 4:48 9:36 4:4 9: 0:00 FIGURE 3 Temperature variation on te surface of Mars on Jul 6, 997. B Eq. (), A/r is equal to te circumference πr. We can eplain tis intuitivel as follows: Up to a small error, te area A of te ban of wit r in Figure 4 is equal to te circumference πr times te wit r. Terefore, A πr r an A r = lim A r 0 r = πr t Te average rate of cange is te ratio T t = C/ (b) Te rate of cange is te erivative T/t, wic is equal to te slope of te tangent line troug te point (:8,.3) in Figure 3. To estimate te slope, we must coose a secon point on te tangent line. Let s use te point labele A, wose coorinates are approimatel (4:48, 5). Te time interval from 4:48 am to :8 pm as lengt 7, 40 min, or t = 7.67, an T t = slope of tangent line.3 ( 5) C/ EXAMPLE Let A = πr be te area of a circle of raius r. (a) Compute A/r at r = an r = 5. (b) W is A/r larger at r = 5? Solution Te rate of cange of area wit respect to raius is te erivative (a) We ave A r = π().57 r= A r = r (πr ) = πr an A r = π(5) 3.4 r=5 (b) Te erivative A/r measures ow te area of te circle canges wen r increases. Figure 4 sows tat wen te raius increases b r, te area increases b a ban of tickness r. Te area of te ban is greater at r = 5 tan at r =. Terefore, te erivative is larger (an te tangent line is steeper) at r = 5. In general, for a fie r, te cange in area A is greater wen r is larger. Area Tangent at r = 5 Tangent at r = r = r = 5 FIGURE 4 Te pink bans represent te cange in area wen r is increase b r. 5 r r r Te Effect of a One-Unit Cange For small values of, te ifference quotient is close to te erivative itself: f ( 0 ) f ( 0 + ) f ( 0 ) Tis approimation generall improves as gets smaller, but in some applications, te approimation is alrea useful wit =. Setting = in Eq. () gives f ( 0 ) f ( 0 + ) f ( 0 ) 3 In oter wors, f ( 0 ) is approimatel equal tote cange in f cause b aone-unit cange in wen = 0.

33 5 CHAPTER 3 DIFFERENTIATION EXAMPLE 3 Stopping Distance For spees s between 30 an 75 mp, te stopping istance of an automobile after te brakes are applie is approimatel F (s) =.s s ft. For s = 60 mp: (a) Estimate te cange in stopping istance if te spee is increase b mp. (b) Compare our estimate wit te actual increase in stopping istance. Solution (a) We ave Using Eq. (3), we estimate F (s) = s (.s s ) =. + 0.s ft/mp F (60) =. + 6 = 7. ft/mp F(6) F(60) F }{{} (60) = 7. ft Cange in stopping istance Tus, wen ou increase our spee from 60 to 6 mp, our stopping istance increases b rougl 7 ft. (b) Te actual cange in stopping istance is F(6) F(60) = = 7.5, so te estimate in (a) is fairl accurate. Altoug C() is meaningful onl wen is a wole number, economists often treat C() as a ifferentiable function of so tat te tecniques of calculus can be applie. Marginal Cost in Economics Let C() enote te ollar cost (incluing labor an parts) of proucing units of a particular prouct. Te number of units manufacture is calle te prouction level. To stu te relation between costs an prouction, economists efine te marginal cost at prouction level 0 as te cost of proucing one aitional unit: Marginal cost = C( 0 + ) C( 0 ) In tis setting, Eq. (3) usuall gives a goo approimation, so we take C ( 0 ) as an estimate of te marginal cost. 5,000 0,000 5,000 Total cost ($) Number of passengers FIGURE 5 Cost of an air fligt. Te slopes of te tangent lines are ecreasing, so marginal cost is ecreasing. EXAMPLE 4 Cost of an Air Fligt Compan ata suggest tat te total ollar cost of a certain fligt is approimatel C() = , were is te number of passengers (Figure 5). (a) Estimate te marginal cost of an aitional passenger if te fligt alrea as 50 passengers. (b) Compare our estimate wit te actual cost of an aitional passenger. (c) Is it more epensive to a a passenger wen = 50 or wen = 00? Solution Te erivative is C () = (a) We estimate te marginal cost at = 50 b te erivative C (50) = 0.005(50) 0.76(50) + 0 = Tus, it costs approimatel $39.75 to a one aitional passenger. (b) Te actual cost of aing one aitional passenger is C(5) C(50),77.0,37.50 = Our estimate of $39.75 is close enoug for practical purposes.

34 SECTION 3.4 Rates of Cange 53 (c) Te marginal cost at = 00 is approimatel C (00) = 0.005(00) 0.76(00) + 0 = 8 Since > 8, it is more epensive to a a passenger wen = 50. In is famous tetbook Lectures on Psics, Nobel laureate Ricar Fenman (98 988) uses a ialogue to make a point about instantaneous velocit: Policeman: M frien, ou were going 75 miles an our. Driver: Tat s impossible, sir, I was traveling for onl seven minutes s (km) t () FIGURE 6 Grap of istance versus time. v (m/s) 30 0 Linear Motion Recall tat linear motion is motion along a straigt line. Tis inclues orizontal motion along a straigt igwa an vertical motion of a falling object. Let s(t) enote te position or istance from te origin at time t. Velocit is te rate of cange of position wit respect to time: v(t) = velocit = s t Te sign of v(t) inicates te irection of motion. For eample, if s(t) is te eigt above groun, ten v(t) > 0 inicates tat te object is rising. Spee is efine as te absolute value of velocit v(t). Figure 6 sows te position of a car as a function of time. Remember tat te eigt of te grap represents te car s istance from te point of origin. Te slope of te tangent line is te velocit. Here are some facts we can glean from te grap: Speeing up or slowing own? Te tangent lines get steeper in te interval [0, ], so tecarwas speeing up uring te first our. Te get flatter in te interval [, ], so te car slowe own in te secon our. Staning still Te grap is orizontal over [, 3] (peraps te river stoppe at a restaurant for an our). Returning to te same spot Te grap rises an falls in te interval [3, 4], inicating tat te river returne to te restaurant (peraps se left er cell pone tere). Average velocit Te grap rises more over [0, ] tan over [3, 5], so te average velocit was greater over te first two ours tan over te last two ours. EXAMPLE 5 A truck enters te off-ramp of a igwa at t = 0. Its position after t secons is s(t) = 5t 0.3t 3 m for 0 t 5. (a) How fast is te truck going at te moment it enters te off-ramp? (b) Is te truck speeing up or slowing own? FIGURE 7 Grap of velocit v(t) = 5 0.9t. t (s) Solution Te truck s velocit at time t is v(t) = t (5t 0.3t3 ) = 5 0.9t. (a) Te truck enters te off-ramp wit velocit v(0) = 5 m/s. (b) Since v(t) = 5 0.9t is ecreasing (Figure 7), te truck is slowing own. Motion Uner te Influence of Gravit Galileo iscovere tat te eigt s(t) an velocit v(t) at time t (secons) of an object tosse verticall in te air near te eart s surface are given b te formulas Galileo s formulas are vali onl wen air resistance is negligible. We assume tis to be te case in all eamples. s(t) = s 0 + v 0 t gt, Te constants s 0 an v 0 are te initial values: s 0 = s(0), te position at time t = 0. v 0 = v(0), te velocit at t = 0. v(t) = s t = v 0 gt 4

35 54 CHAPTER 3 DIFFERENTIATION s (m) Maimum eigt t (s) FIGURE 8 Maimum eigt occurs wen s (t) = v(t) = 0, were te tangent line is orizontal. Galileo s formulas: s(t) = s 0 + v 0 t gt, v(t) = s t = v 0 gt g is te acceleration ue to gravit on te surface of te eart (negative because te up irection is positive), were g 9.8 m/s or g 3 ft/s A simple observation enables us to fin te object s maimum eigt. Since velocit is positive as te object rises an negative as it falls back to eart, te object reaces its maimum eigt at te moment of transition, wen it is no longer rising an as not et begun to fall. At tat moment, its velocit is zero. In oter wors, te maimumeigt is attaine wen v(t) = 0. At tis moment, te tangent line to te grap of s(t) is orizontal (Figure 8). EXAMPLE 6 Fining te Maimum Heigt A stone is sot wit a slingsot verticall upwar wit an initial velocit of 50 m/s from an initial eigt of 0 m. (a) Fin te velocit at t = an at t = 7. Eplain te cange in sign. (b) Wat is te stone s maimum eigt an wen oes it reac tat eigt? Solution Appl Eq. (4) wit s 0 = 0, v 0 = 50, an g = 9.8: (a) Terefore, s(t) = t 4.9t, v(t) = t v() = () = 30.4 m/s, v(7) = (7) = 8.6 m/s At t =, te stone is rising an its velocit v() is positive (Figure 8). At t = 7, te stone is alrea on te wa own an its velocit v(7) is negative. (b) Maimum eigt is attaine wen te velocit is zero, so we solve v(t) = t = 0 t = Te stone reaces maimum eigt at t = 5. s. Its maimum eigt is s(5.) = (5.) 4.9(5.) 37.6 m How important are units? In September 999, te $5 million Mars Climate Orbiter spacecraft burne up in te Martian atmospere before completing its scientific mission. Accoring to Artur Stepenson, NASA cairman of te Mars Climate Orbiter Mission Failure Investigation Boar, 999, Te root cause of te loss of te spacecraft was te faile translation of Englis units into metric units in a segment of groun-base, navigation-relate mission software. In te previous eample, we specifie te initial values of position an velocit. In te net eample, te goal is to etermine initial velocit. EXAMPLE 7 Fining Initial Conitions Wat initial velocit v 0 is require for a bullet, fire verticall from groun level, to reac a maimum eigt of km? Solution We nee a formula for maimum eigt as a function of initial velocit v 0. Te initial eigt is s 0 = 0, so te bullet s eigt is s(t) = v 0 t gt b Galileo s formula. Maimum eigt is attaine wen te velocit is zero: v(t) = v 0 gt = 0 t = v 0 g Te maimum eigt is te value of s(t) at t = v 0 /g: ( ) ( ) v0 v0 s = v 0 ( ) g g g v0 = v 0 g g v0 g = v 0 g Now we can solve for v 0 using te value g = 9.8 m/s (note tat km = 000 m). In Eq. (5), istance must be in meters because our value of g as units of m/s. Maimum eigt = v 0 g = v 0 (9.8) = 000 m 5 Tis iels v 0 = ()(9.8) m/s. In realit, te initial velocit woul ave to be consierabl greater to overcome air resistance.

36 SECTION 3.4 Rates of Cange 55 FIGURE 9 Apparatus of te tpe use b Galileo to stu te motion of falling objects. HISTORICAL PERSPECTIVE Galileo Galilei (564 64) iscovere te laws of motion for falling objects on te eart s surface aroun 600. Tis pave te wa for Newton s general laws of motion. How i Galileo arrive at is formulas? Te motion of a falling object is too rapi to measure irectl, witout moern potograpic or electronic apparatus. To get aroun tis ifficult, Galileo eperimente wit balls rolling own an incline (Figure 9). For a sufficientl flat incline, e was able to measure te motion wit a water clock an foun tat te velocit of te rolling ball is proportional to time. He ten reasone tat motion in free-fall is just a faster version of motion own an incline an euce te formula v(t) = gt for falling objects (assuming zero initial velocit). Prior to Galileo, it a been assume incorrectl tat eav objects fall more rapil tan ligter ones. Galileo realize tat tis was not true (as long as air resistance is negligible), an inee, te formula v(t) = gt sows tat te velocit epens on time but not on te weigt of te object. Interestingl, 300 ears later, anoter great psicist, Albert Einstein, was eepl puzzle b Galileo s iscover tat all objects fall at te same rate regarless of teir weigt. He calle tis te Principle of Equivalence an sougt to unerstan w it was true. In 96, after a ecae of intensive work, Einstein evelope te General Teor of Relativit, wic finall gave a full eplanation of te Principle of Equivalence in terms of te geometr of space an time. 3.4 SUMMARY Te (instantaneous) rate of cange of = f () wit respect to at = 0 is efine as te erivative f ( 0 ) = lim 0 = lim f ( ) f ( 0 ) 0 0 Te rate / is measure in units of per unit of. For linear motion, velocit v(t) is te rate of cange of position s(t) wit respect to time tat is, v(t) = s (t). In some applications, f ( 0 ) provies a goo estimate of te cange in f ue to a oneunit increase in wen = 0 : f ( 0 ) f ( 0 + ) f ( 0 ) Marginal cost is te cost of proucing one aitional unit. If C()is te cost of proucing units, ten te marginal cost at prouction level 0 is C( 0 + ) C( 0 ). Te erivative C ( 0 ) is often a goo estimate for marginal cost. Galileo s formulas for an object rising or falling uner te influence of gravit near te eart s surface (s 0 = initial position, v 0 = initial velocit): s(t) = s 0 + v 0 t gt, v(t) = v 0 gt were g 9.8 m/s, or g 3 ft/s. Maimum eigt is attaine wen v(t) = EXERCISES Preliminar Questions. Wic units migt be use for eac rate of cange? (a) Pressure (in atmosperes) in a water tank wit respect to ept (b) Te rate of a cemical reaction (cange in concentration wit respect to time wit concentration in moles per liter)

37 56 CHAPTER 3 DIFFERENTIATION. Two trains travel from New Orleans to Mempis in 4 ours. Te first train travels at a constant velocit of 90 mp, but te velocit of te secon train varies. Wat was te secon train s average velocit uring te trip? 3. Estimate f(6), assuming tat f(5) = 43, f (5) = Eercises In Eercises 8, fin te rate of cange.. Area of a square wit respect to its sie s wen s = 5.. Volume of a cube wit respect to its sie s wen s = Cube root 3 wit respect to wen =, 8, Te reciprocal / wit respect to wen =,, Te iameter of a circle wit respect to raius. 6. Surface area A of a spere wit respect to raius r (A = 4πr ). 4. Te population P (t) of Freeonia in 009 was P(009) = 5 million. (a) Wat is te meaning of P (009)? (b) Estimate P(00) if P (009) = Te velocit (in cm/s) of bloo molecules flowing troug a capillar of raius cm is v = r, were r is te istance from te molecule to te center of te capillar. Fin te rate of cange of velocit wit respect to r wen r = cm. 6. Figure isplas te voltage V across a capacitor as a function of time wile te capacitor is being carge. Estimate te rate of cange of voltage at t = 0 s. Inicate te values in our calculation an inclue proper units. Does voltage cange more quickl or more slowl as time goes on? Eplain in terms of tangent lines. 7. Volume V of a cliner wit respect to raius if te eigt is equal to te raius. V (volts) 5 8. Spee of soun v (in m/s) wit respect to air temperature T (in kelvins), were v = 0 T. In Eercises 9, refer to Figure 0, te grap of istance versus time for a car trip. 9. Fin te average velocit over eac interval. (a) [0, 0.5] (b) [0.5, ] (c) [,.5] () [, ] 0. At wat time is velocit at a maimum? FIGURE 40 t (s). Matc te escriptions (i) (iii) wit te intervals (a) (c). (i) Velocit increasing (ii) Velocit ecreasing (iii) Velocit negative (a) [0, 0.5] (b) [.5, 3] (c) [.5, ] Distance (km) t () FIGURE 0 Grap of istance versus time for a car trip.. Use te ata from Table in Eample to calculate te average rate of cange of Martian temperature T wit respect to time t over te interval from 8:36 am to 9:34 am. 3. Use Figure 3 from Eample to estimate te instantaneous rate of cange of Martian temperature wit respect to time (in egrees Celsius per our) at t = 4 am. 4. Te temperature (in C) of an object at time t (in minutes) is T (t) = 3 8 t 5t + 80 for 0 t 0. At wat rate is te object cooling at t = 0? (Give correct units.) 7. Use Figure to estimate T/ at = 30 an 70, were T is atmosperic temperature (in egrees Celsius) an is altitue (in kilometers). Were is T/ equal to zero? T ( C) Tropospere Stratospere Mesospere Termospere FIGURE Atmosperic temperature versus altitue. (km) 8. Te eart eerts a gravitational force of F (r) = ( )/r newtons on an object wit a mass of 75 kg locate r meters from te center of te eart. Fin te rate of cange of force wit respect to istance r at te surface of te eart. 9. Calculate te rate of cange of escape velocit v esc = ( )r / m/s wit respect to istance r from te center of te eart.

38 SECTION 3.4 Rates of Cange Te power elivere b a batter to an apparatus of resistance R (in oms) is P =.5R/(R + 0.5) watts. Fin te rate of cange of power wit respect to resistance for R = 3 an R = 5.. Te position of a particle moving in a straigt line uring a 5-s trip is s(t) = t t + 0 cm. Fin a time t at wic te instantaneous velocit is equal to te average velocit for te entire trip.. Te eigt (in meters) of a elicopter at time t (in minutes) is s(t) = 600t 3t 3 for 0 t. (a) Plot s(t) an velocit v(t). (b) Fin te velocit at t = 8 an t = 0. (c) Fin te maimum eigt of te elicopter. 3. A particle moving along a line as position s(t) = t 4 8t m at time t secons. At wic times oes te particle pass troug te origin? At wic times is te particle instantaneousl motionless (tat is, it as zero velocit)? 4. Plot te position of te particle in Eercise 3. Wat is te fartest istance to te left of te origin attaine b te particle? 5. Abullet is fire in te air verticall from groun level wit an initial velocit 00 m/s. Fin te bullet s maimum velocit an maimum eigt. 6. Fin te velocit of an object roppe from a eigt of 300 m at te moment it its te groun. 7. A ball tosse in te air verticall from groun level returns to eart 4 s later. Fin te initial velocit an maimum eigt of te ball. 8. Olivia is gazing out a winow from te tent floor of a builing wen a bucket (roppe b a winow waser) passes b. Se notes tat it its te groun.5 s later. Determine te floor from wic te bucket was roppe if eac floor is 5 m ig an te winow is in te mile of te tent floor. Neglect air friction. 9. Sow tat for an object falling accoring to Galileo s formula, te average velocit over an time interval [t,t ] is equal to te average of te instantaneous velocities at t an t. (a) Te average rate of cange of I wit respect to R for te interval from R = 8 to R = 8. (b) Te rate of cange of I wit respect to R wen R = 8 (c) Te rate of cange of R wit respect to I wen I = Etan fins tat wit ours of tutoring, e is able to answer correctl S() percent of te problems on a mat eam. Wic woul ou epect to be larger: S (3) or S (30)? Eplain. 34. Suppose θ(t) measures te angle between a clock s minute an our ans. Wat is θ (t) at 3 o clock? 35. To etermine rug osages, octors estimate a person s bo surface area (BSA) (in meters square) using te formula BSA = m/60, were is te eigt in centimeters an m te mass in kilograms. Calculate te rate of cange of BSA wit respect to mass for a person of constant eigt = 80. Wat is tis rate at m = 70 an m = 80? Epress our result in te correct units. Does BSA increase more rapil wit respect to mass at lower or iger bo mass? 36. Te atmosperic CO level A(t) at Mauna Loa, Hawaii at time t (in parts per million b volume) is recore b te Scripps Institution of Oceanograp. Te values for te monts Januar December 007 were 38.45, , 384.3, 386.6, , , , 38.78, , 380.8, 38.33, (a) Assuming tat te measurements were mae on te first of eac mont, estimate A (t) on te 5t of te monts Januar November. (b) In wic monts i A (t) take on its largest an smallest values? (c) In wic mont was te CO level most nearl constant? 37. Te tangent lines to te grap of f () = grow steeper as increases. At wat rate o te slopes of te tangent lines increase? 38. Figure 3 sows te eigt of a mass oscillating at te en of a spring. troug one ccle of te oscillation. Sketc te grap of velocit as a function of time. 30. An object falls uner te influence of gravit near te eart s surface. Wic of te following statements is true? Eplain. (a) Distance travele increases b equal amounts in equal time intervals. (b) Velocit increases b equal amounts in equal time intervals. (c) Te erivative of velocit increases wit time. FIGURE 3 t 3. B Faraa s Law, if a conucting wire of lengt l meters moves at velocit v m/s perpenicular to a magnetic fiel of strengt B (in teslas), a voltage of size V = Blv is inuce in te wire. Assume tat B = an l = 0.5. (a) Calculate V/v. (b) Fin te rate of cange of V wit respect to time t if v = 4t Te voltage V, current I, an resistance R in a circuit are relate b Om s Law: V = IR, were te units are volts, amperes, an oms. Assume tat voltage is constant wit V = volts. Calculate (specifing units): In Eercises 39 46, use Eq. (3) to estimate te unit cange. 39. Estimate an Compare our estimates wit te actual values. 40. Estimate f(4) f(3) if f () =. Ten estimate f(4), assuming tat f(3) =. 4. Let F (s) =.s s be te stopping istance as in Eample 3. Calculate F(65) an estimate te increase in stopping istance if spee is increase from 65 to 66 mp. Compare our estimate wit te actual increase.

39 58 CHAPTER 3 DIFFERENTIATION 4. Accoring to Kleiber s Law, te metabolic rate P (in kilocalories per a) an bo mass m (in kilograms) of an animal are relate b a tree-quarter-power law P = 73.3m 3/4. Estimate te increase in metabolic rate wen bo mass increases from 60 to 6 kg. 43. Te ollar cost of proucing bagels is C() = (/000) 3. Determine te cost of proucing 000 bagels an estimate te cost of te 00st bagel. Compare our estimate wit te actual cost of te 00st bagel. 44. Suppose te ollar cost of proucing vieo cameras is C() = (a) Estimate te marginal cost at prouction level = 5000 an compare it wit te actual cost C(500) C(5000). (b) Compare te marginal cost at = 5000 wit te average cost per camera, efine as C()/. 45. Deman for a commoit generall ecreases as te price is raise. Suppose tat te eman for oil (per capita per ear) is D(p) = 900/p barrels, were p is te ollar price per barrel. Fin te eman wen p = $40. Estimate te ecrease in eman if p rises to $4 an te increase if p eclines to $ Te reprouction rate f of te fruit fl Drosopila melanogaster, grown in bottles in a laborator, ecreases wit te number p of flies in te bottle. A researcer as foun te number of offspring per female per a to be approimatel f (p) = (34 0.6p)p (a) Calculate f(5) an f (5). (b) Estimate te ecrease in ail offspring per female wen p is increase from 5 to 6. Is tis estimate larger or smaller tan te actual value f(6) f(5)? (c) Plot f (p) for 5 p 5 an verif tat f (p) is a ecreasing function of p. Do ou epect f (p) to be positive or negative? Plot f (p) an confirm our epectation. 47. Accoring to Stevens Law in pscolog, te perceive magnitue of a stimulus is proportional (approimatel) to a power of te actual intensit I of te stimulus. Eperiments sow tat te perceive brigtness B of a ligt satisfies B = ki /3, were I is te ligt intensit, wereas te perceive eaviness H of a weigt W satisfies H = kw 3/ (k is a constant tat is ifferent in te two cases). Compute B/I an H/W an state weter te are increasing or ecreasing functions. Ten eplain te following statements: (a) A one-unit increase in ligt intensit is felt more strongl wen I is small tan wen I is large. (b) Aing anoter poun to a loa W is felt more strongl wen W is large tan wen W is small. 48. Let M(t) be te mass (in kilograms) of a plant as a function of time (in ears). Recent stuies b Niklas an Enquist ave suggeste tat a remarkabl wie range of plants (from algae an grass to palm trees) obe a tree-quarter-power growt law tat is, M/t = CM 3/4 for some constant C. (a) If a tree as a growt rate of 6 kg/r wen M = 00 kg, wat is its growt rate wen M = 5 kg? (b) If M = 0.5 kg, ow muc more mass must te plant acquire to ouble its growt rate? Furter Insigts an Callenges Eercises 49 5: Te Lorenz curve = F (r) is use b economists to stu income istribution in a given countr (see Figure 4). B efinition, F (r) is te fraction of te total income tat goes to te bottom rt part of te population, were 0 r. For eample, if F(0.4) = 0.45, ten te bottom 40% of ouseols receive 4.5% of te total income. Note tat F(0) = 0 an F() =. 49. Our goal is to fin an interpretation for F (r). Te average income for a group of ouseols is te total income going to te group ivie b te number of ouseols in te group. Te national average income is A = T /N, were N is te total number of ouseols an T is te total income earne b te entire population. (a) Sow tat te average income among ouseols in te bottom rt part is equal to (F (r)/r)a. (b) Sow more generall tat te average income of ouseols belonging to an interval [r, r + r] is equal to ( ) F (r + r) F (r) A r (c) Let 0 r. A ouseol belongs to te 00rt percentile if its income is greater tan or equal to te income of 00r % of all ouseols. Pass to te limit as r 0 in (b) to erive te following interpretation: A ouseol in te 00rt percentile as income F (r)a. In particular, a ouseol in te 00rt percentile receives more tan te national average if F (r) > an less if F (r) <. () For te Lorenz curves L an L in Figure 4(B), wat percentage of ouseols ave above-average income? 50. Te following table provies values of F (r) for Sween in 004. Assume tat te national average income was A = 30, 000 euros. r F (r) (a) Wat was te average income in te lowest 40% of ouseols? (b) Sow tat te average income of te ouseols belonging to te interval [0.4, 0.6] was 6,700 euros. (c) Estimate F (0.5). Estimate te income of ouseols in te 50t percentile? Was it greater or less tan te national average? 5. Use Eercise 49 (c) to prove: (a) F (r) is an increasing function of r. (b) Income is istribute equall (all ouseols ave te same income) if an onl if F (r) = r for 0 r.

40 SECTION 3.5 Higer Derivatives 59 F(r) r (A) Lorenz curve for Sween in 004 F(r).0 (c) Using Eq. (3), sow tat for n 00, te nt website receive at most 00 more visitors tan te (n + )st website. In Eercises 53 54, te average cost per unit at prouction level is efine as C avg () = C()/,were C()is te cost function. Average cost is ameasure of te efficienc of te prouction process. 53. Sow tat C avg () is equal to te slope of te line troug te origin an te point (, C()) on te grap of C(). Using tis interpretation, etermine weter average cost or marginal cost is greater at points A, B, C, D in Figure 5. C D P L L (B) Two Lorenz curves: Te tangent lines at P an Q ave slope. FIGURE 4 Q r A B C Prouction level FIGURE 5 Grap of C(). 54. Te cost in ollars of proucing alarm clocks is C() = were is in units of 000. (a) Calculate te average cost at = 4, 6, 8, an 0. (b) Use te grapical interpretation of average cost to fin te prouction level 0 at wic average cost is lowest. Wat is te relation between average cost an marginal cost at 0 (see Figure 6)? 5. Stuies of Internet usage sow tat website popularit is escribe quite well b Zipf s Law, accoring to wic te nt most popular website receives rougl te fraction /n of all visits. Suppose tat on a particular a, te nt most popular site a approimatel V (n) = 0 6 /n visitors (for n 5,000). (a) Verif tat te top 50 websites receive nearl 45% of te visits. Hint: Let T (N) enote te sum of V (n) for n N. Use a computer algebra sstem to compute T(45) an T(5,000). (b) Verif, b numerical eperimentation, tat wen Eq. (3) is use to estimate V (n + ) V (n), te error in te estimate ecreases as n grows larger. Fin (again, b eperimentation) an N suc tat te error is at most 0 for n N. C (ollars) 5,000 0,000 5, FIGURE 6 Cost function C() = Higer Derivatives Higer erivatives are obtaine b repeatel ifferentiating a function = f (). If f is ifferentiable, ten te secon erivative, enote f or, is te erivative f () = ( f () ) Te secon erivative is te rate of cange of f (). Te net eample igligts te ifference between te first an secon erivatives.

41 60 CHAPTER 3 DIFFERENTIATION E (0 6 kw) t FIGURE Houseol energ consumption E(t) in German in million kilowatt-ours. EXAMPLE Figure an Table escribe te total ouseol energ consumption E(t) in German in ear t. Discuss E (t) an E (t). TABLE Houseol Energ Consumption in German Year Consumption (0 6 kw) Yearl increase Solution We will sow tat E (t) is positive but E (t) is negative. Accoring to Table, te consumption eac ear was greater tan te previous ear, so te rate of cange E (t) is certainl positive. However, te amount of increase ecline from.6 million in 003 to 0. in 006. So altoug E (t) is positive, E (t) ecreases from one ear to te net, an terefore its rate of cange E (t) is negative. Figure supports tis conclusion: Te slopes of te segments in te grap are ecreasing. / as units of per unit of. / as units of / per unit of or units of per unit of square. Te process of ifferentiation can be continue, provie tat te erivatives eist. Te tir erivative, enote f () or f (3) (), is te erivative of f (). More generall, te nt erivative f (n) () is te erivative of te (n )st erivative. We call f () te zeroet erivative an f () te first erivative. In Leibniz notation, we write f, f, 3 f 3, 4 f 4,... EXAMPLE Calculate f ( ) for f () = Solution We must calculate te first tree erivatives: f () = ( ) = f () = ( ) = f () = ( ) = At =, f ( ) = = 348. Polnomials ave a special propert: Teir iger erivatives are eventuall te zero function. More precisel, if f () is a polnomial of egree k, ten f (n) () is zero for n>k. Table illustrates tis propert for f () = 5. B contrast, te iger erivatives of a nonpolnomial function are never te zero function (see Eercise 85, Section 4.9). TABLE Derivatives of 5 f () f () f () f () f (4) () f (5) () f (6) ()

42 SECTION 3.5 Higer Derivatives 6 EXAMPLE 3 Calculate te first four erivatives of =. Ten fin te pattern an etermine a general formula for (n). Solution B te Power Rule, () =, = 3, = (3) 4, (4) = (3)(4) 5 Tus REMINDER n-factorial is te number n! =n(n )(n ) ()().! =,! =()() =, 3! =(3)()() = 6 B convention, we set 0! =. It is not alwas possible to fin a simple formula for te iger erivatives of a function. In most cases, te become increasingl complicate. We see tat (n) () is equal to ±n! n. Now observe tat te sign alternates. Since te o-orer erivatives occur wit a minus sign, te sign of (n) () is ( ) n. In general, terefore, (n) () = ( ) n n! n. EXAMPLE 4 Calculate te first tree erivatives of f () = e. Ten etermine a general formula for f (n) (). Solution Use te Prouct Rule: f () = (e ) = e + e = ( + )e f () = ( ( + )e ) = ( + )e + e = ( + )e f () = ( ( + )e ) = ( + )e + e = ( + 3)e We see tat f n () = f n () + e, wic leas to te general formula f (n) () = ( + n)e One familiar secon erivative is acceleration. An object in linear motion wit position s(t) at time t as velocit v(t) = s (t) an acceleration a(t) = v (t) = s (t). Tus, acceleration is te rate at wic velocit canges an is measure in units of velocit per unit of time or istance per time square suc as m/s. Heigt (m) 7 EXAMPLE 5 Acceleration Due to Gravit Fin te acceleration a(t) of a ball tosse verticall in te air from groun level wit an initial velocit of m/s. How oes a(t) escribe te cange in te ball s velocit as it rises an falls? (A).45 t (s) Solution Te ball s eigt at time t is s(t) = s 0 + v 0 t 4.9t m b Galileo s formula [Figure (A)]. In our case, s 0 = 0 an v 0 =, so s(t) = t 4.9t m. Terefore, v(t) = s (t) = 9.8t m/s an te ball s acceleration is a(t) = s (t) = ( 9.8t) = 9.8 m/s t Velocit (m/s) As epecte, te acceleration is constant wit value g = 9.8 m/s. As te ball rises an falls, its velocit ecreases from to m/s at te constant rate g [Figure (B)]..45 (B) FIGURE Heigt an velocit of a ball tosse verticall wit initial velocit m/s. t (s) GRAPHICAL INSIGHT Can we visualize te rate represente b f ()? Te secon erivative is te rate at wic f () is canging, so f () is large if te slopes of te tangent lines cange rapil, as in Figure 3(A) on te net page. Similarl, f () is small if te slopes of te tangent lines cange slowl in tis case, te curve is relativel flat, as in Figure 3(B). If f is a linear function [Figure 3(C)], ten te tangent line oes not cange at all an f () = 0. Tus, f () measures te bening or concavit of te grap.

43 6 CHAPTER 3 DIFFERENTIATION (A) Large secon erivative: Tangent lines turn rapil. FIGURE 3 (B) Smaller secon erivative: Tangent lines turn slowl. (C) Secon erivative is zero: Tangent line oes not cange. EXAMPLE 6 Ientif curves I an II in Figure 4(B) as te graps of f () or f () for te function f () in Figure 4(A). Solution Te slopes of te tangent lines to te grap of f () are increasing on te interval [a,b]. Terefore f () is an increasing function an its grap must be II. Since f () is te rate of cange of f (), f () is positive an its grap must be I. Slopes of tangent lines increasing I II a b a b (A) Grap of f() (B) Grap of first two erivatives FIGURE SUMMARY Te iger erivatives f,f,f,...are efine b successive ifferentiation: f () = f (), f () = f (),... Te nt erivative is enote f (n) (). Te secon erivative plas an important role: It is te rate at wic f canges. Grapicall, f measures ow fast te tangent lines cange irection an tus measures te bening of te grap. If s(t) is te position of an object at time t, ten s (t) is velocit an s (t) is acceleration. 3.5 EXERCISES Preliminar Questions. On September 4, 003, te Wall Street Journal printe te ealine Stocks Go Higer, Toug te Pace of Teir Gains Slows. Reprase tis ealine as a statement about te first an secon time erivatives of stock prices an sketc a possible grap.. True or false? Te tir erivative of position wit respect to time is zero for an object falling to eart uner te influence of gravit. Eplain. 3. Wic tpe of polnomial satisfies f () = 0 for all? 4. Wat is te milliont erivative of f () = e?

44 SECTION 3.5 Higer Derivatives 63 Eercises In Eercises 6, calculate an.. = 4. = 7 3. = = 4t 3 9t = 4 3 πr3 6. = 7. = 0t 4/5 6t /3 8. = 9/5 9. = z 4 z 0. = 5t 3 + 7t 8/3. = θ (θ + 7). = ( + )( 3 + ) 3. = 4 4. = In Eercises 3 36, fin a general formula for f (n) (). 3. f () = 3. f () = ( + ) 33. f () = / 34. f () = 3/ 35. f () = e 36. f () = e 37. (a) Fin te acceleration at time t = 5 min of a elicopter wose eigt is s(t) = 300t 4t 3 m. (b) Plot te acceleration (t) for 0 t 6. How oes tis grap sow tat te elicopter is slowing own uring tis time interval? 38. Fin an equation of te tangent to te grap of = f () at = 3, were f () = Figure 5 sows f, f, an f. Determine wic is wic. 5. = 5 e 6. = e In Eercises 7 6, calculate te erivative inicate. 7. f (4) (), f () = 4 8. g ( ), g(t) = 4t 5 9. t, = 4t 3 + 3t t= f t 4 4 t 4, f (t) = 6t 9 t 5 t=, = t 3/4. f (4), f (t) = t t t=6 3. f ( 3), f () = 4e 3 4. f (), f (t) = t t + 5. (), (w) = we w 6. g (0), g(s) = es s + 7. Calculate (k) (0) for 0 k 5, were = 4 + a 3 + b + c + (wit a, b, c, te constants). 8. Wic of te following satisf f (k) () = 0 for all k 6? (a) f () = (b) f () = 3 (c) f () = () f () = 6 (e) f () = 9/5 (f) f () = Use te result in Eample 3 to fin Calculate te first five erivatives of f () =. (a) Sow tat f (n) () is a multiple of n+/. (b) Sow tat f (n) () alternates in sign as ( ) n for n. (c) Fin a formula for f (n) () for n. Hint: Verif tat te coefficient is ± 3 5 n 3 n. (A) 3 3 (B) FIGURE Te secon erivative f is sown in Figure 6. Wic of (A) or (B) is te grap of f an wic is f? f () (A) FIGURE 6 4. Figure 7 sows te grap of te position s of an object as a function of time t. Determine te intervals on wic te acceleration is positive. Position Time FIGURE 7 4. Fin a polnomial f () tat satisfies te equation f () + f () =. 43. Fin a value of n suc tat = n e satisfies te equation = ( 3). 40 (C) (B) 3

45 64 CHAPTER 3 DIFFERENTIATION 44. Wic of te following escriptions coul not appl to Figure 8? Eplain. (a) Grap of acceleration wen velocit is constant (b) Grap of velocit wen acceleration is constant (c) Grap of position wen acceleration is zero Position to its initial position after te ole is rille. Sketc possible graps of te rill bit s vertical velocit an acceleration. Label te point were te bit enters te seet metal. In Eercises 48 49, refer to te following. In a 997 stu, Boarman an Lave relate te traffic spee S on a two-lane roa to traffic ensit Q (number of cars per mile of roa) b te formula S =,88Q 0.05Q for 60 Q 400 (Figure 9). Time FIGURE Accoring to one moel tat takes into account air resistance, te acceleration a(t) (in m/s ) of a skiver of mass m in free fall satisfies a(t) = k m v(t), were v(t) is velocit (negative since te object is falling) an k is a constant. Suppose tat m = 75 kg an k = 4 kg/m. (a) Wat is te object s velocit wen a(t) = 4.9? (b) Wat is te object s velocit wen a(t) = 0? Tis velocit is te object s terminal velocit. 46. Accoring to one moel tat attempts to account for air resistance, te istance s(t) (in meters) travele b a falling rainrop satisfies s t = g D ( ) s t were D is te rainrop iameter an g = 9.8 m/s. Terminal velocit v term is efine as te velocit at wic te rop as zero acceleration (one can sow tat velocit approaces v term as time procees). (a) Sow tat v term = 000gD. (b) Fin v term for rops of iameter 0 3 m an 0 4 m. (c) In tis moel, o rainrops accelerate more rapil at iger or lower velocities? 47. A servomotor controls te vertical movement of a rill bit tat will rill a pattern of oles in seet metal. Te maimum vertical spee of te rill bit is 4 in./s, an wile rilling te ole, it must move no more tan.6 in./s to avoi warping te metal. During a ccle, te bit begins an ens at rest, quickl approaces te seet metal, an quickl returns 48. Calculate S/Q an S/Q. 49. (a) Eplain intuitivel w we soul epect tat S/Q < 0. S (mp) Q FIGURE 9 Spee as a function of traffic ensit. (b) Sow tat S/Q > 0. Ten use te fact tat S/Q < 0 an S/Q > 0 to justif te following statement: A one-unit increase in traffic ensit slows own traffic more wen Q is small tan wen Q is large. (c) S/Q > 0? Plot S/Q. Wic propert of tis grap sows tat 50. Use a computer algebra sstem to compute f (k) () for k =,, 3 for te following functions. (a) f () = ( + 3 ) 5/3 4 (b) f () = Let f () = +. Use a computer algebra sstem to compute te f (k) () for k 4. Can ou fin a general formula for f (k) ()? Furter Insigts an Callenges 5. Fin te 00t erivative of p() = ( ) 0 ( + ) ( ) 53. Wat is p (99) () for p() as in Eercise 5? 54. Use te Prouct Rule twice to fin a formula for (fg) in terms of f an g an teir first an secon erivatives. 55. Use te Prouct Rule to fin a formula for (fg) an compare our result wit te epansion of (a + b) 3. Ten tr to guess te general formula for (fg) (n). 56. Compute f ( + ) + f ( ) f () f () = lim 0 for te following functions: (a) f () = (b) f () = (c) f () = 3 Base on tese eamples, wat o ou tink te limit f represents?

46 SECTION 3.6 Trigonometric Functions 65 CAUTION In Teorem we are ifferentiating wit respect to measure in raians. Te erivatives of sine an cosine wit respect to egrees involves an etra, unwiel factor of π/80 (see Eample 7 in Section 3.7). 3.6 Trigonometric Functions We can use te rules evelope so far to ifferentiate functions involving powers of, but we cannot et anle te trigonometric functions. Wat is missing are te formulas for te erivatives of sin an cos. Fortunatel, teir erivatives are simple eac is te erivative of te oter up to a sign. Recall our convention: Angles are measure in raians, unless oterwise specifie. THEOREM Derivative of Sine an Cosine are ifferentiable an Te functions = sin an = cos sin = cos an cos = sin Proof We must go back to te efinition of te erivative: REMINDER Aition formula for sin : sin( + ) = sin cos + cos sin sin( + ) sin sin = lim 0 We cannot cancel te b rewriting te ifference quotient, but we can use te aition formula (see marginal note) to write te numerator as a sum of two terms: sin( + ) sin = sin cos + cos sin sin = (sin cos sin ) + cos sin = sin (cos ) + cos sin (aition formula) Tis gives us sin sin( + ) sin sin( + ) sin = lim 0 = sin (cos ) + cos sin sin (cos ) cos sin = lim + lim 0 0 cos sin = (sin ) lim + (cos ) lim 0 0 }{{}}{{} Tis equals 0. Tis equals. Here, we can take sin an cos outsie te limits in Eq. () because te o not epen on. Te two limits are given b Teorem in Section.6, cos sin lim = 0 an lim 0 0 = Terefore, Eq. () reuces to te formula sin = cos, as esire. Te formula cos = sin is prove similarl (see Eercise 53). EXAMPLE Calculate f (), were f () = cos. Solution B te Prouct Rule, f () = cos + cos = ( sin ) + cos = cos sin f () = (cos sin ) = sin ( (sin ) + sin ) = sin cos

47 66 CHAPTER 3 DIFFERENTIATION GRAPHICAL INSIGHT Te formula (sin ) = cos is mae plausible wen we compare te graps in Figure. Te tangent lines to te grap of = sin ave positive slope on te interval ( π, π ), an on tis interval, te erivative = cos is positive. Similarl, te tangent lines ave negative slope on te interval ( π, 3π ), were = cos is negative. Te tangent lines are orizontal at = π, π, 3π, were cos = 0. = sin π π 3π FIGURE Compare te graps of = sin an its erivative = cos. = cos REMINDER Te stanar trigonometric functions are efine in Section.4. Te erivatives of te oter stanar trigonometric functions can be compute using te Quotient Rule. We erive te formula for (tan ) in Eample an leave te remaining formulas for te eercises (Eercises 35 37). THEOREM Derivatives of Stanar Trigonometric Functions tan = sec, cot = csc, sec = sec tan csc = csc cot π π π 3π EXAMPLE Verif te formula tan = sec (Figure ). Solution Use te Quotient Rule an te ientit cos + sin = : = tan tan = = ( ) sin = cos (sin ) sin (cos ) cos cos cos cos sin ( sin ) cos π π π = sec 3π = cos + sin cos = cos = sec EXAMPLE 3 Fin te tangent line to te grap of = tan θ sec θ at θ = π 4. Solution B te Prouct Rule, FIGURE Graps of = tan an its erivative = sec. = tan θ(sec θ) + sec θ(tan θ) = tan θ(sec θ tan θ)+ sec θ sec θ = tan θ sec θ + sec 3 θ

48 SECTION 3.6 Trigonometric Functions 67 π π 4 5 = tan θ sec θ π 4 π θ Now use te values sec π 4 = an tan π 4 = to compute ( π ) ( π ) ( π ) = tan sec = ( π ) 4 ( = tan π ) ( π ) sec 4 4 ( + sec 3 π ) = + = 3 4 An equation of te tangent line (Figure 3) is = 3 ( θ π 4 ). FIGURE 3 Tangent line to = tan θ sec θ at θ = π SUMMARY Basic trigonometric erivatives: sin = cos, cos = sin Aitional formulas: tan = sec, cot = csc, sec = sec tan csc = csc cot 3.6 EXERCISES Preliminar Questions. Determine te sign (+ or ) tat iels te correct formula for te following: (a) (sin + cos ) =±sin ± cos (b) sec =±sec tan (c) cot =±csc Eercises In Eercises 4, fin an equation of tetangent line atte point inicate.. = sin, = π 4. = cos, = π 3. Wic of te following functions can be ifferentiate using te rules we ave covere so far? (a) = 3 cos cot (b) = cos( ) (c) = e sin 3. Compute (sin + cos ) witout using te erivative formulas for sin an cos. 4. How is te aition formula use in eriving te formula (sin ) = cos? 3. f () = ( 4 4 ) sec 4. f (z) = z tan z 5. = sec θ θ 6. G(z) = tan z cot z 3. = tan, = π 4 4. = sec, = π 6 In Eercises 5 4, compute te erivative. 7. R() = 3 cos 4 sin 8. f () = sin + 5. f () = sin cos 6. f () = cos 7. f () = sin 8. f () = 9 sec + cot 9. H (t) = sin t sec t 0. (t) = 9 csc t + t cot t. f (θ) = tan θ sec θ. k(θ) = θ sin θ 9. f () = + tan tan 0. f (θ) = θ tan θ sec θ. f () = e sin. (t) = e t csc t 3. f (θ) = e θ (5 sin θ 4 tan θ) 4. f () = e cos

49 68 CHAPTER 3 DIFFERENTIATION In Eercises 5 34, fin an equation of te tangent line at te point specifie. 5. = 3 + cos, = 0 6. = tan θ, θ = π 6 7. = sin + 3 cos, = 0 8. = sin t + cos t, t = π 3 9. = (sin θ + cos θ), θ = π = csc cot, = π 4 3. = e cos, = 0 3. = e cos, = π = e t ( cos t), t = π 34. = e θ sec θ, θ = π 4 In Eercises 35 37, use Teorem to verif te formula. 35. cot = csc 36. sec = sec tan 37. csc = csc cot 38. Sow tat bot = sin an = cos satisf =. In Eercises 39 4, calculate te iger erivative. 39. f (θ), f (θ) = θ sin θ 40. t cos t 4.,, = tan 4.,, = e t sin t 43. Calculate te first five erivatives of f () = cos. Ten etermine f (8) an f (37). 44. Fin (57), were = sin. 45. Fin te values of between 0 an π were te tangent line to te grap of = sin cos is orizontal. 46. Plot te grap f (θ) = sec θ + csc θ over [0, π] an etermine te number of solutions to f (θ) = 0 in tis interval grapicall. Ten compute f (θ) an fin te solutions. 47. Let g(t) = t sin t. (a) Plot te grap of g wit a graping utilit for 0 t 4π. (b) Sow tat te slope of te tangent line is nonnegative. Verif tis on our grap. (c) For wic values of t in te given range is te tangent line orizontal? 48. Let f () = (sin )/ for = 0 an f(0) =. (a) Plot f () on [ 3π, 3π]. (b) Sow tat f (c) = 0 if c = tan c. Use te numerical root finer on a computer algebra sstem to fin a goo approimation to te smallest positive value c 0 suc tat f (c 0 ) = 0. (c) Verif tat te orizontal line = f (c 0 ) is tangent to te grap of = f () at = c 0 b plotting tem on te same set of aes. 49. Sow tat no tangent line to te grap of f () = tan as zero slope. Wat is te least slope of a tangent line? Justif b sketcing te grap of (tan ). 50. Te eigt at time t (in secons) of a mass, oscillating at te en of a spring, is s(t) = sin t cm. Fin te velocit an acceleration at t = π 3 s. 5. Te orizontal range R of a projectile launce from groun level at an angle θ an initial velocit v 0 m/s is R = (v0 /9.8) sin θ cos θ. Calculate R/θ. If θ = 7π/4, will te range increase or ecrease if te angle is increase sligtl? Base our answer on te sign of te erivative. 5. Sow tat if π <θ<π, ten te istance along te -ais between θ an te point were te tangent line intersects te -ais is equal to tan θ (Figure 4). = sin π θ FIGURE 4 π tan θ Furter Insigts an Callenges 53. Use te limit efinition of te erivative an te aition law for te cosine function to prove tat (cos ) = sin. 54. Use te aition formula for te tangent tan( + ) = tan + tan + tan tan to compute (tan ) irectl as a limit of te ifference quotients. You will also nee to sow tat lim tan 0 =. 55. Verif te following ientit an use it to give anoter proof of te formula (sin ) = cos. ) ( ) sin( + ) sin = cos ( + sin Hint: Use te aition formula to prove tat sin(a + b) sin(a b) = cos a sin b. 56. Sow tat a nonzero polnomial function = f () cannot satisf te equation =. Use tis to prove tat neiter sin nor cos is a polnomial. Can ou tink of anoter wa to reac tis conclusion b consiering limits as? 57. Let f () = sin an g() = cos. (a) Sow tat f () = g() + sin an g () = f () + cos. (b) Verif tat f () = f () + cos an g () = g() sin.

50 SECTION 3.7 Te Cain Rule 69 (c) B furter eperimentation, tr to fin formulas for all iger erivatives of f an g. Hint: Te kt erivative epens on weter k = 4n, 4n +, 4n +, or 4n Figure 5 sows te geometr bein te erivative formula (sin θ) = cos θ. Segments BA an BD are parallel to te - an -aes. Let sin θ = sin(θ + ) sin θ. Verif te following statements. (a) sin θ = BC (b) BDA = θ Hint: OA AD. (c) BD = (cos θ)ad Now eplain te following intuitive argument: If is small, ten BC BD an AD, so sin θ (cos θ) an (sin θ) = cos θ. O θ D C B FIGURE 5 A 3.7 Te Cain Rule Te Cain Rule is use to ifferentiate composite functions suc as = cos( 3 ) an = 4 +. Recall tat a composite function is obtaine b plugging one function into anoter. Te composite of f an g, enote f g, is efine b (f g)() = f ( g() ) For convenience, we call f te outsie function an g te insie function. Often, we write te composite function as f (u), were u = g(). For eample, = cos( 3 ) is te function = cos u, were u = 3. In verbal form, te Cain Rule sas ( f (g()) ) = outsie (insie) insie A proof of te Cain Rule is given at te en of tis section. THEOREM Cain Rule If f an g are ifferentiable, ten te composite function (f g)() = f (g()) is ifferentiable an ( ) f (g()) = f ( g() ) g () EXAMPLE Calculate te erivative of = cos( 3 ). Solution As note above, = cos( 3 ) is a composite f (g()) were f (u) = cos u, u = g() = 3 f (u) = sin u, g () = 3 Note tat f (g()) = sin( 3 ), so b te Cain Rule, cos(3 ) = sin( 3 ) }{{} f (g()) (3 ) = 3 }{{} sin( 3 ) g ()

51 70 CHAPTER 3 DIFFERENTIATION EXAMPLE Calculate te erivative of = 4 +. Solution Te function = 4 + is a composite f (g()) were f (u) = u /, u = g() = 4 + f (u) = u /, g () = 4 3 Note tat f (g()) = (4 + ) /, so b te Cain Rule, 4 + = (4 + ) / }{{} f (g()) (4 3 ) = }{{} g () EXAMPLE 3 Calculate ( ) for = tan Solution Te outsie function is f (u) = tan u. Because f (u) = sec u, te Cain Rule gives us Now, b te Quotient Rule, ( ) ( ) ( ) tan = sec }{{} Derivative of insie function ( ) ( + ) = ( + ) + ( + ) = ( + ) We obtain ( ) ( ) ( ) sec tan = sec + + ( + ) = + ( + ) It is instructive to write te Cain Rule in Leibniz notation. Let Ten, b te Cain Rule, = f (u) = f (g()) or = f (u) g () = f u u = u u

52 SECTION 3.7 Te Cain Rule 7 Cristiaan Hugens (69 695), one of te greatest scientists of is age, was Leibniz s teacer in matematics an psics. He amire Isaac Newton greatl but i not accept Newton s teor of gravitation. He referre to it as te improbable principle of attraction, because it i not eplain ow two masses separate b a istance coul influence eac oter. CONCEPTUAL INSIGHT In Leibniz notation, it appears as if we are multipling fractions an te Cain Rule is simpl a matter of canceling te u. Since te smbolic epressions /u an u/ are not fractions, tis oes not make sense literall, but it oes suggest tat erivatives beave as ifte were fractions (tis is reasonable because a erivative is a limit of fractions, namel of te ifference quotients). Leibniz s form also empasizes a ke aspect of te Cain Rule: Rates of cange multipl. To illustrate, suppose tat (tanks to our knowlege of calculus) our salar increases twice as fast as our frien s. If our frien s salar increases $4000 per ear, our salar will increase at te rate of 4000 or $8000 per ear. In terms of erivatives, (our salar) t = (our salar) salar) (frien s (frien s salar) t $8000/r = $4000/r EXAMPLE 4 Imagine a spere wose raius r increases at a rate of 3 cm/s. At wat rate is te volume V of te spere increasing wen r = 0 cm? Solution Because we are aske to etermine te rate at wic V is increasing, we must fin V/t. Wat we are given is te rate r/t, namel r/t = 3 cm/s. Te Cain Rule allows us to epress V/t in terms of V/r an r/t: V = }{{} t Rate of cange of volume wit respect to time V }{{} r Rate of cange of volume wit respect to raius r t }{{} Rate of cange of raius wit respect to time To compute V/r, we use te formula for te volume of a spere, V = 4 3 πr3 : V r = ( ) 4 r 3 πr3 = 4πr Because r/t = 3, we obtain V t = V r r t = 4πr (3) = πr For r = 0, V t = (π)0 = 00π 3770 cm 3 /s r=0 We now iscuss some important special cases of te Cain Rule. THEOREM General Power an Eponential Rules g()n = n(g()) n g () (for an number n) eg() = g ()e g() If g() is ifferentiable, ten Proof Let f (u) = u n. Ten g() n = f (g()), an te Cain Rule iels g()n = f (g())g () = n(g()) n g ()

53 7 CHAPTER 3 DIFFERENTIATION On te oter an, e g() = (g()), were (u) = e u. We obtain eg() = (g())g () = e g() g () = g ()e g() EXAMPLE 5 General Power an Eponential Rules Fin te erivatives of (a) = ( ) /3 an (b) = e cos t. Solution Appl g()n = ng() n g () in (A) an eg() = g ()e g() in (B). (a) ( ) /3 = 3 ( ) 4/3 ( ) = 3 ( ) 4/3 ( + 7) (b) t ecos t = e cos t t cos t = (sin t)ecos t Te Cain Rule applie to f (k + b) iels anoter important special case: f (k + b) = f (k + b) (k + b) = kf (k + b) If f () is ifferentiable, ten for an con- THEOREM 3 Sifting an Scaling Rule stants k an b, f (k + b) = kf (k + b) Slope f (c) c π = f() = sin π For eample, ( sin + π ) ( = cos + π ) 4 4 (9 )5 = (9)(5)(9 ) 4 = 45(9 ) 4 sin( 4t) = 4cos( 4t) t Slope f (c) = f() = sin t e7 5t = 5e 7 5t c/ π π FIGURE Te erivative of f() at = c/ is twice te erivative of f () at = c. GRAPHICAL INSIGHT To unerstan Teorem 3 grapicall, recall tat te graps of f (k + b) an f () are relate b sifting an scaling (Section.). For eample, if k>, ten te grap of f (k + b) is a compresse version of te grap of f () tat is steeper b a factor of k. Figure illustrates a case wit k =. Wen te insie function is itself a composite function, we appl te Cain Rule more tan once, as in te net eample.

54 SECTION 3.7 Te Cain Rule 73 EXAMPLE 6 Using te Cain Rule Twice Calculate + +. Solution First appl te Cain Rule wit insie function u = + + : ( + ( + ) /) / ( = + ( + ) /) / ( + ( + ) /) Ten appl te Cain Rule again to te remaining erivative: ( + ( + ) /) / = ( + ( + ) /) ( ) / ( + ) / () = ( + ) / ( + ( + ) /) / Accoring to our convention, sin enotes te sine of raians, an wit tis convention, te formula (sin ) = cos ols. In te net eample, we erive a formula for te erivative of te sine function wen is measure in egrees. EXAMPLE 7 Trigonometric Derivatives in Degrees Calculate te erivative of te sine function as a function of egrees rater tan raians. Solution To solve tis problem, it is convenient to use an unerline to inicate a function of egrees rater tan raians. For eample, sin = sine of egrees A similar calculation sows tat te factor π appears in te formulas for te 80 erivatives of te oter stanar trigonometric functions wit respect to egrees. For eample, ( π ) tan = sec 80 Te functions sin an sin are ifferent, but te are relate b ( π ) sin = sin 80 because egrees correspons to 80 π raians. B Teorem 3, sin = ( π ) ( π ) ( π ) ( π ) sin 80 = cos = cos 80 Proof of te Cain Rule Te ifference quotient for te composite f g is f (g( + )) f (g()) ( = 0) Our goal is to sow tat (f g) is te prouct of f (g()) an g (), so it makes sense to write te ifference quotient as a prouct: f (g( + )) f (g()) = f (g( + )) f (g()) g( + ) g() g( + ) g()

55 74 CHAPTER 3 DIFFERENTIATION Tis is legitimate onl if te enominator g( + ) g() is nonzero. Terefore, to continue our proof, we make te etra assumption tat g( + ) g() = 0 for all near but not equal to 0. Tis assumption is not necessar, but witout it, te argument is more tecnical (see Eercise 05). Uner our assumption, we ma use Eq. () to write (f g) () as a prouct of two limits: (f g) f (g( + )) f (g()) g( + ) g() () = lim lim 0 g( + ) g() 0 }{{}}{{} Sow tat tis equals f (g()). Tis is g (). Te secon limit on te rigt is g (). Te Cain Rule will follow if we sow tat te first limit equals f (g()). To verif tis, set k = g( + ) g() Ten g( + ) = g() + k an f (g( + )) f (g()) g( + ) g() = f (g() + k) f (g()) k Te function g() is continuous because it is ifferentiable. Terefore, g( + ) tens to g() an k = g( + ) g() tens to zero as 0. Tus, we ma rewrite te limit in terms of k to obtain te esire result: f (g( + )) f (g()) f (g() + k) f (g()) lim = lim = f (g()) 0 g( + ) g() k 0 k 3.7 SUMMARY Te Cain Rule epresses (f g) in terms of f an g : (f (g())) = f (g()) g () In Leibniz notation: = u General Power Rule: General Eponential Rule: Sifting an Scaling Rule: u, were = f (u) an u = g() g()n = n(g()) n g () eg() = g ()e g() f (k + b) = kf (k + b)

56 SECTION 3.7 Te Cain Rule EXERCISES Preliminar Questions. Ientif te outsie an insie functions for eac of tese composite functions. (a) = (b) = tan( + ) (c) = sec 5 () = ( + e ) 4. Wic of te following can be ifferentiate easil witout using te Cain Rule? (a) = tan(7 + ) (b) = + Eercises In Eercises 4, fill in a table of te following tpe: f (g()) f (u) f (g()) g () (f g). f (u) = u 3/, g() = 4 +. f (u) = u 3, g() = f (u) = tan u, g() = 4 4. f (u) = u 4 + u, g() = cos In Eercises 5 6, write te function as a composite f (g()) an compute te erivative using te Cain Rule. 5. = ( + sin ) 4 6. = cos( 3 ) 7. Calculate cos u for te following coices of u(): (a) u = 9 (b) u = (c) u = tan 8. Calculate f ( + ) for te following coices of f (u): (a) f (u) = sin u (b) f (u) = 3u 3/ (c) f (u) = u u 9. Compute f f u if = an u = Compute f if f (u) = u, u() = 5, an u () = 5. = In Eercises, use te General Power Rule or te Sifting an Scaling Rule to compute te erivative.. = ( 4 + 5) 3. = ( ) 3 3. = = (4 3 ) 5 5. = ( + 9) 6. = ( ) 4/3 7. = cos 4 θ 8. = cos(9θ + 4) 9. = ( cos θ + 5 sin θ) 9 0. = sin. = e. = e 8+9 (c) = sec (e) = e () = cos (f) = e sin 3. Wic is te erivative of f(5)? (a) 5f () (b) 5f (5) (c) f (5) 4. Suppose tat f (4) = g(4) = g (4) =. Do we ave enoug information to compute F (4), were F () = f (g())? If not, wat is missing? In Eercises 3 6, compute te erivative of f g. 3. f (u) = sin u, g() = + 4. f (u) = u +, g() = sin 5. f (u) = e u, g() = + 6. f (u) = u u, g() = csc In Eercises 7 8, fin te erivatives of f (g()) an g(f ()). 7. f (u) = cos u, u = g() = + 8. f (u) = u 3, u = g() = + In Eercises 9 4, use te Cain Rule to fin te erivative. 9. = sin( ) 30. = sin 3. = t = (t + 3t + ) 5/ 33. = ( 4 3 ) /3 34. = ( + ) 3/ 35. = 37. = sec ( ) = cos 3 (θ) 38. = tan(θ 4θ) 39. = tan(θ + cos θ) 40. = e 4. = e 9t 4. = cos 3 (e 4θ ) In Eercises 43 7, fin te erivative using te appropriate rule or combination of rules. 43. = tan( + 4) 44. = sin( + 4) 45. = cos( 3) 46. = sin( ) cos( ) 47. = (4t + 9) / 48. = (z + ) 4 (z ) = ( 3 + cos ) = sin(cos(sin )) 5. = sin cos 5. = (9 (5 4 ) 7 ) 3

57 76 CHAPTER 3 DIFFERENTIATION 53. = (cos 6 + sin ) / 54. = ( + )/ = tan 3 + tan( 3 ) 56. = 4 3 cos 57. = z + z 58. = (cos cos + 7) = cos( + ) + cos 60. = sec( t 9) 6. = cot 7 ( 5 ) 6. = cos(/) + ( 63. = + cot 5 ( )) 64. = 4e + 7e 65. = (e 3 + 3e ) = cos(te t ) 67. = e ( ++3) 68. = e e 69. = = = (k + b) /3 ; k an b an constants 7. = kt 4 + b ; k, b constants, not bot zero In Eercises 73 76, compute te iger erivative sin( ) (9 )8 76. ( + 9) sin() 77. Te average molecular velocit v of a gas in a certain container is given b v = 9 T m/s, were T is te temperature in kelvins. Te temperature is relate to te pressure (in atmosperes) b T = 00P. Fin v P. P = Te power P in a circuit is P = Ri, were R is te resistance an i is te current. Fin P/t at t = 3 if R = 000 an i varies accoring to i = sin(4πt) (time in secons). 79. An epaning spere as raius r = 0.4t cm at time t (in secons). Let V be te spere s volume. Fin V/t wen (a) r = 3 an (b) t = A 005 stu b te Fiseries Researc Services in Abereen, Scotlan, suggests tat te average lengt of te species Clupea arengus (Atlantic erring) as a function of age t (in ears) can be moele b L(t) = 3( e 0.37t ) cm for 0 t 3. See Figure. (a) How fast is te lengt canging at age t = 6 ears? (b) At wat age is te lengt canging at a rate of 5 cm/r? L (cm) t (ear) FIGURE Average lengt of te species Clupea arengus 8. A 999 stu b Starke an Scarneccia evelope te following moel for te average weigt (in kilograms) at age t (in ears) of cannel catfis in te Lower Yellowstone River (Figure 3): W(t) = ( e t ) Fin te rate at wic weigt is canging at age t = 0. Lower Yellowstone River W (kg) t (ear) FIGURE 3 Average weigt of cannel catfis at age t 8. Te functions in Eercises 80 an 8 are eamples of te von Bertalanff growt function M(t) = ( a + (b a)e kmt ) /m (m = 0) introuce in te 930s b Austrian-born biologist Karl Luwig von Bertalanff. Calculate M (0) in terms of te constants a, b, k an m. 83. Wit notation as in Eample 7, calculate (a) θ sin θ ( ) (b) θ + tan θ θ=60 θ θ= Assume tat f(0) =, f (0) = 3, (0) =, (0) = 7 Calculate te erivatives of te following functions at = 0: (a) (f ()) 3 (b) f(7) (c) f(4)(5) 85. Compute te erivative of (sin ) at = π 6, assuming tat (0.5) = Let F () = f (g()), were te graps of f an g are sown in Figure 4. Estimate g () an f (g()) an compute F ().