Issuu

Applied Calculus For Business Economics And The Social And Life Sciences 11th Edition Hoffmann Solutions Manual Visit to Download in Full: https://testbankdeal.com/download/applied-calculus-for-busi ness-economics-and-the-social-and-life-sciences-11th-edition-hoffmann-solutions-ma nual/

Chapter 2 Differentiation: Basic Concepts

2.1 The Derivative

1. If f(x)=4,then f(x + h)=4.

Thedifferencequotient(DQ)is ()()440. fxhfx hh +−− == 0 ()lim()()0 h fxfxhfx h →

Theslopeofthelinetangenttothegraph of f at x =1is(1)7. ′ −=− f

+− ′ = =

Theslopeis(0)0. mf ′ ==

2. f(x)=3

Thedifferencequotientis ()()3(3)00 +−−−− === fxhfx hhh

Then 0 ()lim00. → ′ == h fx

Theslopeofthelinetangenttothegraph of f at x =1is(1)0. ′ = f

3. If f(x)=5x 3,then f(x + h)=5(x + h)3.

Thedifferencequotient(DQ)is

5. If2()235, fxxx=−+ then 2 ()2()3()5. fxhxhxh +=+−++

+−++−+ +− = =+− 0 ()lim()()43 h fxfxhfxx h →

[2()3()5][235] 42()3 xhxhxx hh xhhh h xh

+− ′ = =− Theslopeis(0)3. mf ′ ==−

2 ()1

fxhfxxhx hh h h

()()[5()3][53] 5 5

+− ′ = =

+−+−−− = = = 0 ()lim()()5 h fxfxhfx h →

Theslopeis(2)5. mf ′ ==

4. f(x)=27x

Thedifferencequotientis

=− Thedifferencequotientis 22 222 2 ()()(()1)(1) 211 2 2 fxhfxxhx hh xhxhx h hxhxh h +−+−−− = ++−−+ = + ==+

Then

+−−+−− = −−−+ = ==−

fxhfxxhx hh xhx h h h

()()(27())(27)

27727 7 7

Then 0 ()lim(7)7. → ′ =−=− h fx

Thedifferencequotient(DQ)is ()() fxhfx h +− 22 2 423

0 ()lim(2)2

© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Applied Calculus For Business Economics And The Social And Life Sciences 11th Edition Hoffmann Visit TestBankDeal.com to get complete for all chapters

fxx

h fxxhx → ′ =+= . Theslopeofthelinetangenttothegraph

f at1 x =− is(1)2 f ′ −=−

7. If3()1,fxx=− then

9. If2(),gt t = then2(). gthth += +

10.

11. If 1 (),Hu u = then 1 ().Huh uh +=

88 Chapter2.Differentiation:BasicConcepts

3 22 3223 ()()1 (2)()1 331 fxhxh xxhhxh xxhxhh +=+− =+++− =+++− 32233 223 22 22 ()() 331(1) 33 (33) 33 fxhfx h xxhxhhx h xhxhh h hxxhh h xxhh +− +++−−− = ++ = ++ = =++ 0 22 0 2 ()lim()() lim33 3 h h fxfxhfx h xxhh x → → +− ′ = =++ = Theslopeis2(2)3(2)12. mf ′ ===

Thedifferencequotientis 33 32233 223 22 ()()(())() 33 33 33 +−−+−− = −−−−+ = = =−−− fxhfxxhx hh xxhxhhx h xhxhh h xxhh Then 222 0

→

=−−−=− h fxxxhhx

8. ()3 =− fxx

()lim(33)3.

′

=− f

Theslopeofthelinetangenttothegraph of f at x =1is(1)3. ′

22 22 ()() () () 22() ()() 2 () tht tht gthgt hh tth htth tth htth tth + + +− = + =⋅ + −+ = + = + 02 ()lim()()2 h gtgthgtht → +− ′ = =− Theslopeis18. 2 mg  ′ ==−  

Thedifferencequotient(DQ)is

2 1 ()fx x = Thedifferencequotientis 22 222 222 222 11 () 11 2 (2) (2) 2 222 222 ()() 2 (2) 2 (2) xhx xhxhx xxhxh xhxhx fxhfx hh h h hxh hxhxhx xh xhxhx + ++ −++ ++ +− = = = = ++ = ++ Then 0222 4 3 2 ()lim (2) 2 2 h fxxh xhxhx x x x → ′ = ++ = =− Theslopeofthelinetangenttothegraph of f at2 x = is 1

f ′ =−

(2)4

+ Thedifferencequotientis

Then 0 11 ()lim 2 → ′ = = ++ h fx xhxx

Theslopeofthelinetangenttothegraph of f at x =9is 1 (9). 6 ′ = f

13. If f(x)=2,then f(x + h)=2. Thedifferencequotient(DQ)is ()()220. fxhfx hh +−− ==

Theslopeofthetangentiszeroforall valuesof x.Since f(13)=2. y 2=0(x 13),or y =2.

14. For()3 fx = ,

forall x.Soatthepoint4 c =− ,theslope ofthetangentlineis(4)0 mf ′ =−= .The point(4,3)isonthetangentlinesoby thepoint-slopeformulatheequationofthe tangentlineis30[(4)] yx−=−− or 3 y =

15. If f(x)=72x,then f(x + h)=72(x + h).

Thedifferencequotient(DQ)is ()()[72()][72] 2

+− ′ = =−

fxhfxxhx hh +−−+−− = =− 0 ()lim()()2 h fxfxhfx h →

Theslopeofthelineis(5)2. mf ′ ==−

Since f(5)=3,(5,3)isapointonthe curveandtheequationofthetangentline is y (3)=2(x 5)or y =2x +7.

16. For f(x)=3x, 0 0

forall x.Soatthepoint c =1,theslopeof

Chapter2.Differentiation:BasicConcepts 89 ( ) ( ) ( ) ( ) ( ) 11 ()() () 1 uhu fxhfx h uuh huuh uuhuuh huuhuuh uuh huuhuuh h huuhuuh uuhuuh + +− + =⋅ + ++ −+ = + ++ −+ = +++ = +++ = +++ ( ) () () 0 0 ()lim()() 1 lim 1 1 2 1 2 h h Hufxhfx h uuhuuh uuuu uu uu → → +− ′ = = +++ = ⋅+ = =− Theslopeis11 (4). 2(4)416 mH ′ ==−=−

Thedifferencequotientis ( ) ( ) ()() 1 +− +− = +−++ = ⋅ ++ +− = ++ = ++ = ++ fxhfx h xhx h xhxxhx hxhx xhx hxhx h hxhx xhx

12. () = fxx

()lim()()lim00

hh fxfxhfx h →→ +− ′ = ==

()lim()()33

lim0 hh fxfxhfxhh→→ +−− ′ = ==

()lim()()

+−

= +− = =

fxfxhfx

li333 m 3 → →

′

h h

h xhx h

thetangentlineis(1)3. ′ ==mf Thepoint (1,3)isonthetangentlinesobythe point-slopeformulatheequationofthe tangentlineis y 3=3(x 1)or y =3x

17. If2 (),fxx = then2 ()(). fxhxh +=+ Thedifferencequotient(DQ)is

Since f(1)=2,(1,2)isapointonthe curveandtheequationofthetangentline is22((1)) 24 yx

Theslopeofthelineis(1)2. mf ′ == Since f(1)=1,(1,1)isapointonthe curveandtheequationofthetangentline is y

20. For 2()3fx x = , 0 22 0 022 3

forall x.Atthepoint1 c = ,theslopeof thetangentlineis(1)6 mf ′ ==− .The point(1,1)isonthetangentlinesobythe point-slopeformulatheequationofthe tangentlineis(1)6(1) yx−−=−− or 65yx=−+

19. If2(),fx x =− then2(). fxhxh += +

Thedifferencequotient(DQ)is

21. Firstweobtainthederivativeof Thedifferencequotientis

90 Chapter2.Differentiation:BasicConcepts

22 2 ()()() 2 2 fxhfxxhx hh xhh h xh +−+− = + = =+ 0 ()lim()()2 h fxfxhfxx h → +− ′ = =

1=2(x 1)or y =2x 1.

=−

0 22 0 0 ()lim()() lim(23())(23) lim(63) 6 h h h fxfxhfx h xhx h xh x → → → +− ′ = −+−− = =−− =−

18. For2 ()23 fxx

22 22 ()() () () 22() ()() 2 () xhx xhx fxhfx hh xxh hxxh xxh hxxh xxh + + +− = + + =⋅ + −++ = + = + 02 ()lim()()2 h fxfxhfxhx → +− ′ = =

=−=

Theslopeofthelineis(1)2. mf ′

yx −=−− =+

()lim()() 33 lim() lim63 ()

h h h fxfxhfx h xhx h xh xhx x → → → +− ′ = + = = + = Atthepoint 1 2 c = ,theslopeofthe tangentlineis 1 48 2 mf  ′ ==−   .The point 1 ,12 2    isonthetangentlinesoby thepoint-slopeformulatheequationofthe tangentlineis 1 1248 2 yx  −=−−  or4836 yx=−+

Soatthepoint c =1,theslopeofthe

tangentlineis 1 (1). 2 ′ =− f Thepoint(1, 1)isonthetangentlinesobythepointslopeformula,theequationofthetangent lineis 1 1(1) 2 −=−−yx or 13 22 =−+yx

23. If 3 1 (),fx x = then 3 1 (). () fxh xh += +

Thedifferencequotient(DQ)is

Theslopeis4

Further, f(1)=1sotheequationoftheline is y 1=3(x 1),or y =3x +4.

24. FromExercise7ofthissection

()32fxx ′ = .Atthepoint1 c = ,theslope

Chapter2.Differentiation:BasicConcepts 91 Then Nowsince 11 ()2. 2 fx xx  ′ ==  

equationofthetangentlineis 1 4(4), 2 yx−=− or 1 2. 2 yx=+ 22. For 1 (), = fx x ( ) ( ) () 0 11 0 02 02 02 2 3/2 ()lim()() lim lim lim() 1 lim 1 2 1 2 → + → → → → +− ′ = = −+ = + −+ = +++ = +++ = = h xhx h h h h fxfxhfx h h xxh hxxh xxh hxxhxxh xxhxxh xx x

Theslopeis1(4), 2

′ == f(4)=4,the

0 22 033 2 33 4 ()lim()() lim33 () 3 () 3 h h fxfxhfx h xxhh xxh x xx x → → +− ′ = = + = =−

mf ′ ==−=−

3 (1)3. (1)

ofthetangentlineis(1)3 mf ′ == .The point(1,0)isonthetangentlinesobythe point-slopeformulatheequationofthe tangentlineis03(1) yx−=− or 33yx=− .

25. If y = f(x)=3,then f(x + h)=3. Thedifferencequotient(DQ)is

Thedifferencequotient(DQ)is 22 2

26. For f(x)=17, dy dx at014 = x is

+− +−+−− = = =−−

lim()()12 h dyfxhfx x dxh → +− = =−

fxhfx h xhxhxx h hxhh h xh 0

3 dy dx = when x =1.

(1)lim(1)(1) lim((1)2(1))(12(1))

0 2 0

+−++−+ = = =

fxhfxxhx hh h h lim()()lim33

+− = ==

3 dy dx = when x =1.

(3)lim(3)(3) lim(62(3))(62(3))

0 0 0

Thedifferencequotient(DQ)is

92 Chapter2.Differentiation:BasicConcepts

()()330.

0 lim()()0

fxhfx hh +−− ==

h dyfxhfx dxh → +−

dy dx = when x =2.

0 0 0 (14)lim(14)(14) lim17(17) lim0 0 → → → +− ′ = = = = h h h

h h

ffhf h

27. If y = f(x)=3x +5,then f(x + h)=3(x + h)+5=3x +3h +5. Thedifferencequotient(DQ)is ()()335(35) 3 3

hh dyfxhfx dxh→→

2 lim 2 h h h ffhf h h h h h → → → +− ′ = −+−− = = =−

28. For()62 fxx =− , dy dx at03 x = is

29. If y = f(x)= x(1 x),or2 (),fxxx =− then2 ()()(). fxhxhxh +=+−+

()() [()()][] 2 12

lim 0 → → → +− ′ = +−+−− = = = h h h

30. For2()2, =− fxxxdy dx at01 = x is

ffhf h hh h h h

31. If 1 (),yfxx x ==− then 1 (). fxhxhxh +=+− +

32. For 1 ()2fx x = , dy dx at03 x =− is

33. (a) If2 (),fxx = then2(2)(2)4 f −=−= and2(1.9)(1.9)3.61. f −=−= The slopeofthesecantlinejoiningthe points(2,4)and(1.9,3.61)onthe graphof f

(b) If2 (),fxx = then 222 ()()2. fxhxhxxhh +=+=++ Thedifferencequotient(DQ)is

Theanswerispart(a)isarelatively goodapproximationtotheslopeofthe tangentline.

35. (a) If3 (),fxx = then f(1)=1, 3 (1.1)(1.1)1.331. f ==

Theslopeofthesecantlinejoiningthe points(1,1)and(1.1,1.331)onthe graphof f is

(b) If3 (),fxx = then 3 ()(). fxhxh +=+

Thedifferencequotient(DQ)is

Theslopeofthetangentlineatthe point(2,4)onthegraphof f is tan(2)2(2)4.mf ′ =−=−=−

Theslopeistan(1)3. mf ′ == Noticethatthisslopewas approximatedbytheslopeofthe secantinpart(a).

Chapter2.Differentiation:BasicConcepts 93 0 2 0 2 lim()() lim1 1 h h dyfxhfx dxh xhx x → → +− = =++ =+ When x =1,2

dy dx =+=

(1)12.

0 11 2(3)2(3) 0 0 (3)lim(3)(3) lim 1 lim 5(5) 1 25 h h h h ffhf h h h → −−+−− → → −+−− ′ −= = = =

21 sec 21 3.614 3.9. 1.9(2) yy m xx == =−

222 2 ()()2 2 (2) 2

hh xhh h hxh h xh +−++− = + = + = =+ 0 0 ()lim()() lim2 2 h h fxfxhfx h xh x → → +− ′ = =+ =

fxhfxxxhhx

34. (a) () () () 1 2 1 2 1122 22 1 2 3 4 1 2 (0) 0 2(2(0)0) 0 3 2 =    = = = ff m

0 2 0 0 (0)lim(0)(0) lim20 lim(2) 2 → → → +− ′ = = =− = h h h ffhf h hh h h

(b)

21 sec 21 1.3311 3.31. 1.11 yy m

===

33 223 22 ()()() 33 33 fxhfxxhx hh xhxhh h xxhh +−+− = ++ = =++ 2 0 ()lim()()3 h fxfxhfxx h → +− ′ = =

Theanswerinpart(a)isarelatively goodapproximationtotheslopeofthe tangentline.

xx Since f(0)=0and fxfx xx = =− =−

()() . fxfx ()()0 0 13 16 0.8125.

161616256

(b) If2()3, fxxx =− then 2 ()3()(). fxhxhxh +=+−+

39. (a) If1 (), 1 t st t = + theaveragerateof changeof s is21

94 Chapter2.Differentiation:BasicConcepts 36. (a) () 1 2 1 2 1 21 111 1 2 1 2 11 32 1 2 (1) (1) 1 3 ff m = = = =− (b) 0 11 1111 0 0 (1)lim(1)(1) lim 1 lim 2(2) 1 4 h h h h h ffhf h h h → −+ −+−−− → → −+−− ′ −= = = =−

2 11113 3,

37. (a) If2()3, fxxx =− theaveragerateof changeof f is21 21 f  =−=−  13 21256 1 2116

22 222 2 ()() 3()()(3) 3633 63 631. fxhfx h xhxhxx h xxhhxhxx h xhhh h xh +− +−+−− = + + −−−+ = +− = =+− 0 ()lim(631)61 h fxxhx → ′ =+−=− Theinstantaneousrateofchangeat x =0is(0)1. f ′ =− Noticethatthis rateisestimatedbytheaverageratein part (a). 38. (a) () () () 1 2 ave1 2 11 22 1 2 1 2 (0) 0 120(12(0)) 00 0 ff f = = = = (b) 0 0 0 (0)lim(0)(0) lim(12)0 lim(12) 1 h h h ffhf h hh h h → → → +− ′ = = =− = Theanswerinpartaisnotavery goodapproximationtotheaverage rateofchange.

Thedifferencequotient(DQ)is

tt Since 1 2 1 2 11 3 21 s  −==− −+  and 1 2 0131 (0)1,4. 010 s −−+ ==−= + © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

()()stst

+−− +++ +− = Multiplyingnumeratorand denominatorby(t + h +1)(t +1). (1) 1 1 1 2 3

(b) ()

(a) ss s

11 (1)(1) = = = =

shs s h h h hh hh h hh h

h h h h h

→ → → → →

0 0 0 0 0

(1)lim(1)(1) 11 lim 1111 lim 11 11 lim 11 1 lim 11 1 2

+− ′ = +− = +−++ = ⋅ ++ +− = ++ = ++ = Theanswerinpartaisarelatively goodapproximationtothe instantaneousrateofchange.

41. (a) Theaveragerateoftemperature changebetween0 t and0th + hours aftermidnight.Theinstantaneousrate oftemperaturechange0 t hoursafter midnight.

(b) Theaveragerateofchangeinblood alcohollevelbetween0 t and 0 th + hoursafterconsumption.The instantaneousrateofchangeinblood alcohollevel0 t hoursafter consumption.

for

distribution in any

Chapter2.Differentiation:BasicConcepts 95

22 (1)(1)(1)(1) (1)(1)

2 (1)(1) 2 (1)(1) thttth htht tthtthtthtth htht h htht tht +−+−−++ = +++ +−++−−−−+++ = +++ = +++ = +++ 02 22 ()lim(1)(1)(1) h st thtt → ′ = = +++ + Theinstantaneousrateofchange when 1 2 t =− is ()12 2 12 8. 21 s  ′ −==  −+ Noticethattheestimategivenbythe averagerateinpart

(b) If1 (), 1 t st t = + then ()1 (). ()1 sthth th +− += ++ Thedifferencequotient(DQ)is 11 ()()11 tht sthsttht hh differs significantly. 40. (a) () 1 4 ave1 4 1 4 3 4 1 2 3 4

(c) Theaveragerateofchangeofthe 30-yearfixedmortgageratebetween 0t and0th + yearsafter2005.The instantaneousrateofchangeof 30-yearfixedmortgagerate0 t years after2005.

42. (a) ...theaveragerateofchangeof revenuewhentheproductionlevel changesfrom0 x to0xh + units. ...theinstantaneousrateofchangeof revenuewhentheproductionlevel is0 x units.

website,

part.

(b) ...theaveragerateofchangeinthe fuellevel,inlb/ft,astherockettravels between0 x and0xh + feetabovethe ground.

2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized

sale or

manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted

in whole or

...theinstantaneousrateinfuellevel whentherocketis0 x feetabovethe ground.

(c) ...theaveragerateofchangein volumeofthegrowthasthedrug dosagechangesfrom0 x to0xh + mg. ...theinstantaneousrateinthe growth’svolumewhen0 x mgofthe drughavebeeninjected.

43. P(x)=4,000(15 x)(x 2)

(a) Thedifferencequotient(DQ)is

(b) Theaveragerateas q increasesfrom q =0to q =20is 2 (20)(0)[70(20)(20)]0

(b) ()0Px ′ = when4,000(172x)=0. 17 8.5, 2 x == or850units.

When()0, Px ′ = thelinetangentto thegraphof P ishorizontal.Sincethe graphof P isaparabolawhichopens down,thishorizontaltangentindicates amaximumprofit.

44. (a) Profit=(numbersold)(profitoneach) Profitoneach

=sellingpricecosttoobtain ()(120)(50) =−−Pppp

Since q =120 p, p =120 q. P(q)= q[(120 q)50] or 2 ()(70)70. =−=− Pqqqqq

Since(20) ′ P ispositive,profitis increasing. 45.

(a) As x increasesfrom10to11,the averagerateofchangeis or$2,940perunit.

96 Chapter2.Differentiation:BasicConcepts

2 ()() [4,000(15())(()2)] [4,000(15)(2)]

4,000(172) 4,000(172) PxhPx h xhxh h xx h xhxhxx h hxhh h xh +− −++− = −−+−−−− = = =−− 0 ()lim()() 4,000(172) h PxhPx Px h x → +− ′ = =−

4,000[(15)(2)(15)(2)]

2020 $50perrecorder

= PP

P Thedifferencequotientis 22 222 2 ()() [70()()][70] 7070270 702 702 +− +−+−− = +−−−−+ = = =−− PqhPq h qhqhqq h qhqqhhqq h hqhh h qh 0 ()lim()()702 → +− ′ = =− h PqPqhPqq h (20)702(20)$30′ =−= P per

recorder.

Theaveragerateofchangeiscloseto thisvalueandisanestimateofthis instantaneousrateofchange. Sinceispositive,thecostwill increase.

Chapter2.Differentiation:BasicConcepts 97 (b) So,thedifferencequotient(DQ)is or$2,900perunit

46. (a) () ave (3,100)(3,025) 3,1003,025 3,1003,1003,1003,025 75 3,100103155 75 28.01 QQQ = = = ≈ Theaveragerateofchangeinoutputisabout28unitsperworker-hour. © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Theinstantaneousrateofchangeis28.2unitsperworker-hour.

WritingExerciseAnswerswillvary.

Theinstantaneousrateofchangeis$80perunitwhen x =4.Theexpenditureisdecreasing

49. When t =30, 65503 50304 dV dt ≈= Inthe“longrun,”therateatwhich V ischangingwithrespecttotimeisgettingsmallerandsmaller, decreasingtozero.

98 Chapter2.Differentiation:BasicConcepts (b) ( ) ( ) 0 0 0 0 0

lim3,1003,0253,1003,025 lim3,1003,025553,02555 3,02555 lim3,100(3,0253,025) 3,02555 lim3,100 3,02555 3,100 110 28.2 h h h h h QQhQ h h h hh hh h hh h → → → → → +− ′ = +− = +− ++ = ⋅ ++ +− = ++ = ++ = ≈

(3,025)lim(3,025)(3,025)

2 ()() (35200) 35200 ExxDx xx xx =⋅ =−+ =−+ (b) ave 22 (5)(4) 54 35(5)200(5)(35(4)200(4)) 125240 115 EE E = =−+−−+ =− =− Theaveragechangeinconsumerexpendituresis$115perunit. (c) 0 22 0 2 0 0 (4)lim(4)(4) lim35(4)200(4)(35(4)200(4)) lim3580 lim(3580) 80 h h h h EhE E h hh h hh h h → → → → +− ′ = −+++−−+ = = =−− =−

47.

48. (a)

when x =4.

50. Answerswillvary.Drawingatangentlineateachoftheindicatedpointsonthecurveshowsthe populationisgrowingatapproximately10/dayafter20daysand8/dayafter36days.Thetangent lineslopeissteepestbetween24and30daysatapproximately27days.

51. When h =1,000meters,

When h =2,000meters,0C/meter.

Sincethelinetangenttothegraphat h =2,000ishorizontal,itsslopeiszero.

52.

(a) Theaveragerateofchangeis. Since and,

Thepopulation’saveragerateofchangefor2010–2012iszero.

(b) Tofindtheinstantaneousrate,calculate P’(2). sothedifferencequotient(DQ)is

For2012, t =2,sotheinstantaneousrateofchangeis P’(2)=–12(2)+12 =–12,oradecreaseof12,000people/year.

Chapter2.Differentiation:BasicConcepts 99

60 2,0001,000 6 1,000 0.006C/meter dT dh ≈ = =−°

dT dh =°

Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any

This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in

or part.

by McGraw-Hill

manner.

whole

53. ()4.44.92 Httt

=−

(a) Thedifferencequotient(DQ)is

After1second, H ischangingatarateof(1)4.49.8(1)5.4m/sec, H ′ =−=− wherethenegative representsthat H isdecreasing.

(b) ()0Ht ′ = when4.49.8t =0,or t ≈ 0.449seconds.

Thisrepresentsthetimewhentheheightisnotchanging(neitherincreasingnordecreasing). Thatis,thisrepresentsthehighestpointinthejump.

Again,thenegativerepresentsthat H isdecreasing.

54. (a) If P(t)representsthebloodpressurefunctionthen(0.7)80 P ≈ ,(0.75)77 P ≈ ,and(0.8)85 P ≈ Theaveragerateofchangeon[0.7,0.75]isapproximately77806 0.5 =− mm/secwhileon [0.75,0.8]theaveragerateofchangeisabout 8577 16 0.5 = mm/sec.Therateofchangeis greaterinmagnitudeintheperiodfollowingtheburstofblood.

(b) Writingexerciseanswerswillvary.

100 Chapter2.Differentiation:BasicConcepts

2 4.44.90 (4.44.9)0 4.44.90 44 0.898seconds 49 tt tt t t −= −= −= =≈ Atthistime,therateofchangeis 4444 4.49.8 4949 4.4m/sec. H  ′ =−   =−

55. 2 ()0.00090.1317.81 Dppp =−++

()() . DpDp pp

Since 2 (60) 0.0009(60)0.13(60)17.81

()() DphDp h +− 2 2 22 2 2 [0.0009()0.13() 17.81 (0.00090.1317.81)] [0.00090.00180.0009 0.130.1317.81 0.00090.1317.81] 0.00180.00090.13 0.00180.00090.13 phph pp h pphh ph pp h phhh h ph −+++ + −−++ +++ +−− = −−+ = =−−+ 0 () lim(0.00180.00090.13) 0.00180.13 h Dx ph p → ′ =−−+ =−+ Theinstantaneousrateofchange when p =60is (60)0.0018(60)0.13 0.022mm D ′ =−+ = permmofmercury.Since(60) D ′ is positive,thepressureisincreasing when p =60.

p p −+=

≈

56. (a) Therocketis feetabove ground.

(b) Theaveragevelocitybetween0and 40secondsisgivenby feet/second.

57. (a)

So,thedifferencequotient(DQ)is

2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Chapter2.Differentiation:BasicConcepts 101

22.37 D

= and 2

D =−++ =

(a) Theaveragerateofchangeis 21 21 = =

=−++

(61) 0.0009(61)0.13(61)17.81 22.3911,

22.391122.37 6160 0.0211mmpermmofmercury

(b) 2 ()0.0009()0.13 Dphph+=−++ (p + h)+17.81

So,thedifferencequotient(DQ)is

Atthispressure,thediameteris neitherincreasingnordecreasing.

Multiplyingthenumeratorand denominatorby gives

(b) At1 x =− ,3(1)25 y =−−=− and (1,5)isapointonthetangentline. Usingthepoint-slopeformulawith

yx=− isthegraph of2 yx = shifteddown3units.Sothe graphsareparallelandtheirtangent lineshavethesameslopesforany valueof x.Thisaccounts geometricallyforthefactthattheir derivativesareidentical.

(b) Since25 yx=+ istheparabola 2 yx = shiftedup5unitsandthe constantappearstohavenoeffecton thederivative,thederivativeofthe function25 yx=+ isalso2x.

102 Chapter2.Differentiation:BasicConcepts (b) (c) 58.

0 0 ()lim(3()2)(32) lim3 3 h h fxxhx h h h → → +−−− ′ = = =

(a)

32yx

m = gives(5)3((1)) yx−−=−− or

=−

2 ()().

22 2 ()()() 2 2 fxhfxxhx hh xhh h xh +−+− = + = =+ 0 () lim()() 2 h dyfx dx fxhfx h x → ′ = +− = = For2()3,yfxx==− 2 ()()3. fxhxh+=+− Thedifferencequotient(DQ)is 222[()3](3)2 2 xhxxhh hh xh +−−−+ = =+ 0 () lim()() 2 h dyfx dx fxhfx h x → ′ = +− = =

fxhxh +=+ Thedifferencequotient(DQ)is

Thegraphof23

for

on a website,

2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized

sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted

in whole or part.

60. (a) For2()3 fxxx =+ ,thederivativeis

(d) Thederivativeof() fx isthesumof thederivativeof() gx and() hx

(a) For2 (),yfxx ==

For3 (),yfxx == 3

()(). fxhxh +=+

Thedifferencequotient(DQ)is

(b) Thepatternseemstobethatthe derivativeof x raisedtoapower() n x isthatpowertimes x raisedtothe powerdecreasedbyone1(). n nx So, thederivativeofthefunction4

63. When x <0,thedifferencequotient(DQ) is()()()()

When x >0,thedifferencequotient(DQ) is()()()1. fxhfxxhx hh +−+− ==

So, 0 ()lim11. h fx → ′ ==

Sincethereisasharpcornerat x =0 (graphchangesfrom y = x to y = x),the graphmakesanabruptchangeindirection at x =0.So, f isnotdifferentiableat x =0.

Chapter2.Differentiation:BasicConcepts 103

22 0 222 0 0 () lim[()3()](3) lim2333 lim(23)23 → → → ′ +++−+ = ++++−− = =++=+ h h h fx xhxhxx h xhxhxhxx h xhx

22 0 222 0 0 ()lim() 2 lim lim(2) 2 h h h xhx gxh xhxhx h xh x → → → +− ′ = ++− = =+ = Whilefor()3hxx = ,thederivativeis 00 ()lim3()33lim3 hh hxxhxh hh→→ +− ′ = ==

(b) For2 () gxx = ,thederivativeis

61.

2 ()().

Thedifferencequotient(DQ)is 22 2 ()()() 2 2 fxhfxxhx hh xhh h xh +−+− = + = =+ 0 () lim()() 2 h dyfx dx fxhfx h x → ′ = +− = =

fxhxh +=+

33223

hh xxhh +−++ = =++ 0 2 () lim()() 3 h dyfx dx fxhfx h x → ′ = +− = =

22 ()33 33 xhxxhxhh

is3 4 x andthederivativeofthe function27 yx = is2627. x

0 0 0 0 lim[()]() lim lim lim ,aconstant. h h h h dymxhbmxb dxh mxmhbmxb h mh h m m → → → → ++−+ = ++−− = = = =

yx =

62. If ymxb =+ then

fxhfxxhx hh h h +−−+−− = = =− So, 0 ()lim11. h fx → ′ =−=−

64. (a) Writeanynumber x as xch =+ .If thevalueof x isapproaching c,then h isapproaching0andviceversa.Thus theindicatedlimitisthesameasthe limitinthedefinitionofthe derivative.Lessformally,notethatif xc ≠ then()() fxfc xc istheslopeof asecantline.s x approaches c the slopesofthesecantlinesapproachthe slopeofthetangentat c

(b)

usingpart(a)forthefirstlimitonthe right.

=−

2 11 1(1)(1) limlim2 11xx xxx xx++ →→ −+ = = 2 11 1(1)(1) limlim2 11xx xxx xx→→ −+ = =−

66. UsingtheTRACEfeatureofacalculator withthegraphof3220.84yxx=−+ showsapeakat0 x = andavalleyat

0.2667 x = .Notethepeaksandvalleys arehardtoseeonthegraphunlessasmall rectanglesuchas[0.3,0.5] × [3.8,4.1]is used.

=−

fxfc fxfc fxfc

→ →→ → =−

xc xcxc xc

65. Toshowthat 21 ()1 x fx x = isnot differentiableat x =1, press y =andinput2(abs(1))/(1) xx for1 y =

TheabsisundertheNUMmenuinthe mathapplication.

Usewindowdimensions[4,4]1by [4,4]1

67. Tofindtheslopeoflinetangenttothe graphof2()23 fxxxx =+− at x =3.85,fillinthetablebelow.

The x + h rowcanbefilledinmanually. For f(x),press y =andinput

( ) ^22(3) xxx +− for1 y = Usewindowdimensions[1,10]1by [1,10]1.

Usethevaluefunctionunderthecalc menuandenter x =3.85tofind

f(x)=4.37310.

For f(x + h),usethevaluefunctionunder thecalcmenuandenter x =3.83tofind

f(x + h)=4.35192.Repeatthisprocessfor x =3.84,3.849,3.85,3.851,3.86,and 3.87.

The()()

fxhfx h +− canbefilledin

104 Chapter2.Differentiation:BasicConcepts

 =  

lim()()lim()

 =   ′ =⋅ =

lim[()()] lim()()() xc xc fxfc fxfc xc xc → →

lim[()()]

()0 0 xc xcxc fxfc fxfc xc xc fc → →→

solim()() xc fxfc → = meaning f(x)is continuousat xc =

PressGraph Weseethat f isnotdefinedat x =1.There canbenopointoftangency.

manuallygiventhattherestofthetableis nowcomplete.So, slope(3.85)1.059. f ′ =≈

2.2 Techniques of Differentiation

1. Sincethederivativeofanyconstantis

6. 7/3 yx = 7/314/377 33 dy xx dx ==

7. 3.7 yx = 3.712.73.73.7 dy xx dx ==

(Note: y =2isahorizontallineandall horizontallineshaveaslopeofzero,so

Chapter2.Differentiation:BasicConcepts 105 h 0.020.010.001 x + h 3.833.843.849 f(x)4.373104.373104.37310 f(x + h)4.351924.362514.37204 ()() fxhfx h +− 1.0591.0591.059 00.0010.010.02 3.853.8513.863.87 4.373104.373104.373104.37310 4.373104.374154.383684.39426 undefined1.0581.0581.058

zero, 2 0 y dy dx =− =

y =3 0 = dy dx

505 yx dydd x dxdxdx dy dx =− =− =−= 4. y =2x +7 2(1)02 =−+=− dy dx 5. 4 yx = 415 5 4 44 dy xx dxx =−=−=−

dx mustbezero.) 2.

(5)(3)

=−−=

=π

=π=π

1.2 1.212.2 4 0(1.2)1.2 =−

yx dy xx dx 9. 2 yr

21 (2)2 dy rr dx

dr ==

221/2yxx ==⋅ 12. 433/422 yxx == 3/411/4 4 333 2 422 dy xx dxx  = ==   13. 91/2 9 yt t == 1/21 3/2 3/23 1 9 2 1 9 2 99 or 22 dyt dx t tt  =−   =−  =−− © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

10. 43 π 3 yr = 422 π(3)4π 3 dy rr

11.

14. 2 2 33 22 yt t ==

3 3 33 2(2)dyt dtt =−=

=++

yxx dyddd xx dxdxdxdx dy x dx =+

=−+−

=−+

dy xx dx xx

=−++

=−++ =−+

yxxx dydddd xxx dxdxdxdxdx dy xx dx

512 ()(5)()(12) 9401

fxxx xx

11 ()8610 42 231

′ =⋅−⋅−+ =−−

()0.02(3)0.30.060.3 fxxx ′ =−+=−+

uu ′ =−+− =−+

32 ()4(0.07)3(1.21)30 0.283.633

106 Chapter2.Differentiation:BasicConcepts

15. 2 2 23 ()(2)(3) 22

16. 53 3496yxxx =−+− 42 42 3(5)4(3)9(1)0 15129

17. 98 98 87

18. 8611 ()2 42 fxxxx=−−+ 75 75

19. 3 ()0.020.3 fxxx =−+ 3 ()(0.02)(0.3) dd fxxx dxdx ′ =−+

21. 121/2 11211/21 233/2 233/2 ()()() 1 12 2 1 12 2 121 , 2 dydddttt dtdtdtdt ttt ttt ttt =+−  =−+−−−  =−−+ =−−+ or233 121 2 ttt −−+ 22. 123 23 3222 32 33 yxxx xxx =−+=−+ 234 234 234 2 (1)(3)(2)(2)(3)3 342 342 dy xxx dx xxx xxx  =−−−+−   =−+− =−+− 23. 33/23/2 3 1 (), fxxxx x =+=+ 3/23/2 3/213/21 1/25/2 1/2 5/25 ()()() 33 22 33 22 3333 ,or 2222 dd fxxx dxdx xx xx xx xx ′ =+ =+ =− =−− 24. 3 3/21/2 4 ()22 242 ftt t tt =+− =+− 3/211/21 1/23/2 3 31 ()(2)(4)0 22 32 2 3 fttt tt t t ′ =+− =− =− © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

20. 43 ()0.071.2135.2 fuuuu =−+−

fuuu

28.

=−−+− =−−+

29. 32 2 531 3103 yxxx dy xx dx

At x =1,10. dy dx = Theequationofthe tangentlineat(1,8)is y +8=10(x +1),or y =10x +2.

30. Given53352yxxx=−−+ andthepoint (1,5),then42595 dy xx dx =−− andthe slopeofthetangentlineat1 x = is 42 5(1)9(1)59 m =−−=− .Theequation ofthetangentlineisthen (5)9(1)yx−−=−− or94 yx=−+ .

31. 1211/2112 yxx xx =−+=−+ 23/2 23/2 11 dy xx dxxx =−=−

At 7 4,, 4    1 16 dy dx =− Theequationof thetangentlineis71(4), 416 yx−=−− or 1 2. 16 yx=−+

32. Given32 2 16 yxx x =−+ andthepoint (4,7),then3 332 2 2 dy xx dxx =−− andthe slopeofthetangentlineat4 x = is 3 33211 42(4) 22 4 m = −=− .Theequation ofthetangentlineisthen 11

(7)(4) 2 yx −−=−− or 11 15 2 yx=−+

33. 232 2 ()(32)23 623 yxxxxxx dy xx dx

=−+=+− =+−

At x =1,1. dy dx = Theequationofthe

Chapter2.Differentiation:BasicConcepts 107

111 2, 1633 xx yx xx xxxxx =−+−++ =−+−++

1.22.1 1.22.1 75 75 yxx xx =+=−+ 1.212.11 2.21.1 1.1 2.2 1.2(7)2.1(5) 8.410.5 8.4 10.5 dy xx dx xx x x =−−+ =+ =+ 27. 52 3 52 33 2 21 4 4 4 4 xx y x xx xx x x xx = =− =− =− 21 11 2 2 ()(4) 24(1) 24 4 2 dydd xx dxdxdx xx xx x x =− =−− =+ =+

25. 2 3/2

213/22 21 163 3

26.

23532 (67)67 yxxxxxx =−+=−+ 42 51814 dy xxx dx =−+

tangentlineat(1,2)is y 2=1(x +1), or y = x +3.

34. Given43 2 yxx x =−+ andthepoint

(1,4),then32 13 8 2 dy x dxxx =−− andthe slopeofthetangentlineat1 x = is

3 2 8(1)139

2(1)2 1 m =−−= .Theequation ofthetangentlineisthen94(1) 2 yx−=− or 91 22yx=−

'(2)481193 44 mf==+= .Theequationof thetangentlineis19366(4)

1 ()22 2 ()6

fxxxx x fxx x

35. 332 2 2 3

=−+=−+ ′ =−−

At x =1,(1)4. f ′ −=− Further, y = f (1)=3.Theequationofthetangent lineat(1,3)is y 3=4(x +1),or y =4x 1.

36. 432 ()326;2 fxxxxx =−+−=

32 ()494 fxxxx ′ =−+

(2)1624866 f =−+−=− so(2,6)isa pointonthetangentline.Theslopeis

(2)323684 mf ′ ==−+= .Theequation ofthetangentlineis(6)4(2) yx−−=− or 414yx=−

Further,84 (2)4. 33 yf==−+= The equationofthetangentlineat4

40. 3/2 ()(1);4 fxxxxxx =−=−= 3 ()1 2 fxx ′ =−

1 () 2 ()1

fxxxx x fx x

37. 2 2 3

=−=− ′ =+

At x =1,(1)3. f ′ = Further, y = f (1)=0. Theequationofthetangentlineat(1,0)is y 0=3(x 1),or y =3x 3.

38. 3 ();4 fxxxx=+=

()321 2 fxx x ′ =+

(4)64266 f =+= so(4,66)isapointon thetangentline.Theslopeis

(4)844 f =−= so(4,4)isapointonthe tangentline.Theslopeis (4)312. mf ′ ==−= Theequationofthe tangentlineis42(4) yx−=− or 24yx=−

41. 4 3 ()231 ()83 fxxx fxx =++ ′ =+

Therateofchangeof f at x =1is

(1)5. f ′ −=−

42. 3 ()35;2 fxxxx=−+= 2 ()33 (2)3(4)39 fxx f ′ =− ′ =−=

108 Chapter2.Differentiation:BasicConcepts

193 127 4 yx=− . 39. 331/2 2 1/2 11 ()88 33 ()8 2

fxx x =−+=−+⋅ ′ =−+ At x =2, (2)48 22 18 4 22 1 42 2 3. f ′ =−+ =−+ =−+⋅ =−

4 yx−=− or

fxxxxx

2, 3    is 4

33(2),yx−=−− or 22

3 yx=−+

43. 1/22 2 1/23

fxxxxxx x fx xx

=−+=−+ ′ =−−

1 () 12 ()1 2

Therateofchangeof f at x =1is

3 (1). 2 f ′ =−

48. ()11; =+=+ fxxxx x f(1)=1+1=2

2 2 1 ()11;(1)110 ′′ =−=−=−=fxxf x

At c =1,therelativerateofchangeis (1)00 (1)2

′ == f f

fx x f

′ =+ ′ =+=

44. ()5;4 fxxxx=+= 1 ()5 2 121 (4)52(2)4

45. 1/2

+ = =+ =+ =+ 1/2 1 () 2 fx x ′ =

() 1 1

fxxx x xx xx x x

Therateofchangeof f at x =1is 1 (1). 2 f ′ =

49. 2 1/22 3/22

=+ =⋅+ =+ 1/2 33 ()22 22 fxxxxx ′ =+=+

() fxxxx xxx xx

Therelativerateofchangeis () 3 2 22 ()2234 . ()22 fxxxxx fxxxxxxx + ′ + =⋅= + +

′ + = = +

When x =4, () 2 ()344(4)11 (4)24 2444 fx f

50. 11 ()(4)41; =−=−fxxxx 41 31 ()33 f =−= 24 ()4;(3)9 ′′ =−=−fxxf

46. 2 ();1 fxxxx x =−= 2 ()23 2 (1)237 22

fxx x f

′ =− ′ =−−=−

47. 32 2 ()254 ()610 fxxx fxxx =−+ ′ =−

Therelativerateofchangeis 2 32 ()610 . ()254 fxxx fxxx

′ = −+

At c =3,therelativerateofchangeis 4 9 1 3

′ ==− f f

(3)4 . (3)3

51. (a) 2 ()0.11020 ()0.210 Attt Att =++ ′ =+

Intheyear2008,therateofchangeis

(4)0.810 A′ =+ or$10,800peryear.

(b) A(4)=(0.1)(16)+40+20=61.6,so thepercentagerateofchangeis

(100)(10.8)17.53.

61.6 % =

′ ==− −+

When x =1,(1)6104. (1)254 f f

52. (a) Since32()615 fxxxx =−++ isthe numberofradiosassembled x hours after8:00A.M.,therateofassembly after x hoursis

Chapter2.Differentiation:BasicConcepts 109

2 ()31215fxxx ′ =−++ radiosper hour.

(b) Therateofassemblyat9:00A.M. (1 x = )is (1)3121524 f ′ =−++= radiosper hour.

53. (a) 2 ()2040600 Txxx=++ dollars Therateofchangeofpropertytaxis ()4040dollars/year.Txx ′ =+ Intheyear2008, x =0, (0)40dollars/year. T ′ =

(b) Intheyear2012, x =4and T(4)=$1,080.Intheyear2008, x =0 and T(0)=$600. Thechangeinpropertytaxis T(4) T(0)=$480.

54. 2 ()2,300125517Mx xx =+− 23 ()1251034Mx xx ′ =−+ 23 (9)12510340.125 99 M ′ =−+≈− .Salesare decreasingatarateofapproximately 1/8motorcycleper$1,000ofadvertising.

4.0(# 4.0(250)11,200 250 4,8004.0

x x x x    = +      =+

So,thecostfunctionis 9,800 ()4. Cxx x =+

(b) Therateofchangeofthecostis ().Cx ′ 1 2

()9,8004 ()9,8004dollars/miperhr Cxxx Cx x

=+ ′ =−+

When x =40, (40)2.125dollars/miperhr. C ′ =−

Since(40) C ′ isnegative,thecostis decreasing.

56. (a) Since2()1004005,000 Cttt=++ is thecirculation t yearsfromnow,the rateofchangeofthecirculationin t yearsis

()200400Ctt ′ =+ newspapersper year.

(b) Therateofchangeofthecirculation 5yearsfromnowis

(5)200(5)4001,400 C ′ =+= newspap ersperyear.Thecirculationis increasing.

(6)(5)11,0009,500 1,500newspapers. CC−=− =

costdriverhrs) mi x x  =   =

57. (a) SinceGary’sstartingsalaryis$45,000 andhegetsaraiseof$2,000peryear, hissalary t yearsfromnowwillbe dollars.

Thepercentagerateofchangeofthis salary t yearsfromnowis

110 Chapter2.Differentiation:BasicConcepts

55. (a) Cost=costdriver+costgasoline 20(# 250 20 5,000

costgasoline gals) dollars

(b) Thepercentagerateofchangeafter1 yearis

2 ()18500Nxx ′ =+ commutersper week.After8weeksthisrateis 2 (8)18(8)5001652 N ′ =+= usersper week.

(b) Theactualchangeinusageduringthe 8thweekis (8)(7)15,07213,558 1,514riders. NN−=− =

(c) Inthelongrun,approaches0. Thatis,thepercentagerateofGary’s salarywillapproach0(eventhough Gary’ssalarywillcontinuetoincrease ataconstantrate.)

58. Let G(t)betheGDPinbillionsofdollars where t isyearsand0 t = represents1997. SincetheGDPisgrowingataconstant rate, G(t)isalinearfunctionpassing throughthepoints(0,125)and(8,155)

Then

()15512515125125 804 Gttt =+=+ .

In2012,15 t = andthemodelpredictsa GDPof(15)181.25 G = billiondollars.

59. (a) f(x)=6x +582 TherateofchangeofSATscoresis

()6.fx ′ =−

(b) Therateofchangeisconstant,sothe dropwillnotvaryfromyeartoyear. Therateofchangeisnegative,sothe scoresaredeclining.

60. (a) Since3()65008,000Nxxx=++ is thenumberofpeopleusingrapid transitafter x weeks,therateatwhich systemuseischangingafter x weeks is

′ = +

x Px x

100(9)100(20) (9)2(9)4(9)5,000

3/2

2 100()200(100)200 ()100 (100) Ptt Ptt t ′ + = = + +

Chapter2.Differentiation:BasicConcepts 111



  

61. (a) 3/2 ()245,000Pxxx=++ isthe population x monthsfromnow.The rateofpopulationgrowthis 1/2 1/2 ()2432 26peoplepermonth. +

Ninemonthsfromnow,the

populationwillbechangingattherate of1/2(9)26(9)20peopleper P ′ =+= month.

2,000 5,126 0.39

P P

= ++ = =

(b) Thepercentagerateatwhichthe populationwillbechanging9months fromnowis

′

Education.

is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any

This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

62. (a) 22 ()20010,000(100) Ptttt =++=+ ()22002(100)Pttt ′ =+=+ Thepercentagerateofchangeis

2013 by McGraw-Hill

This

manner.

(b) Thepercentagerateofchanges approaches0since 200 lim0. 100 tt →∞ = +

63. 331/2()105105 Nttttttt =++=++

Therateofchangeoftheinfected populationis

2 1/2 1 ()305 2 people/day. Ntt t ′ =++

Onthe9thday,(9)2,435people/day. N ′ =

64. (a) 3 43 ()5,175(8) 5,1758 Nttt tt =−− =−+

32 (3)4(3)83(3)108 N ′ =−+⋅= people perweek.

(b) Thepercentagerateofchangeof N is givenby 32 43 100()100(424)

is T(0.442) ≈ 42.8°C.

Thebird’stemperaturestartsat

T(0)=37.1 °C,increasesto

T(0.442)=42.8°C,andthenbeginsto decrease.

66. (a) Usingthegraph,the x-value(taxrate) thatappearstocorrespondtoa y-value (percentagereduction)of50is150,or ataxrateof150dollarsperton carbon.

()5,1758 Nttt Nttt

′ −+ = −+ Agraphofthisfunctionshowsthatit neverexceeds25%.

65. (a) 32 ()68.0730.9812.52 37.1 Ttttt =−++ + 2

()204.2161.9612.52 Tttt ′ =−++

()Tt ′ representstherateatwhichthe bird’stemperatureischangingafter t days,measuredin °Cperday.

(b) (0)12.52C/day T ′ =° since(0) T ′ is positive,thebird’stemperatureis increasing.

(0.713)47.12C/day T ′ ≈−°

Since(0.713) T ′ isnegative,thebird’s temperatureisdecreasing.

(b) Usingthepoints(200,60)and(300, 80),fromthegraph,therateofchange isapproximately orincreasingatapproximately0.2% perdollar.(Answerswillvary dependingonthechoiceof h.)

67. (a) 2 ()0.050.13.4 ()0.10.1 PPM PPM/year Qttt Qtt =++ ′ =+

Therateofchangeof Q at t =1is (1)0.2PPM/year. Q ′ =

(b) Q(1)=3.55PPM, Q(0)=3.40,and Q(1) Q(0)=0.15PPM.

68.

22 441 π π 339 N PNT kTk  = =    µµ 22 2 2 4π 4π 99 dPNN T dtkkT  = −=    µµ

tt t t = ≈

=−++ −±−−

0204.2161.9612.52

61.96(61.96)4(204.21)(12.52)

2(204.21)

Thebird’stemperaturewhen t =0.442

112 Chapter2.Differentiation:BasicConcepts

0.442days

69. Since g representstheaccelerationdueto gravityfortheplanetourspyison,the formulafortherock’sheightis Sincehethrowstherockfromground

level,.Also,sinceitreturnstothe groundafter5seconds,

73. (a) 43 ()48 stttt =−+ for0 ≤ t ≤ 4 32 ()4128 vttt=−+ and 2 ()1224 attt =−

(b) Tofindalltimeingivenintervalwhen stationary,

Therockreachesitsmaximumheight halfwaythroughitstrip,orwhen. So, Substituting

So,ourspyisonMars.

70. (a) 2 ()26 sttt=−+ for02 t ≤≤ ()22 ()2 vtt at =− =

(b) Theparticleisstationarywhen ()220vtt=−= whichisattime1 t = .

71. (a) 2 ()325 sttt=+− for0 ≤ t ≤ 1 v(t)=6t +2and a(t)=6

(b) 6t +2=0at t =3.Theparticleisnot stationarybetween t =0and t =1.

72. (a) 32 ()91525 stttt=−++ for06 t ≤≤ 2 ()318153(1)(5) ()6186(3) vttttt attt =−+=−− =−=−

(b) Theparticleisstationarywhen ()3(1)(5)0 vttt=−−= whichisat times 1 t = and5 t =

74. (a) Sincetheinitialvelocityis00 V = feet persecond,theinitialheightis 0144 H = feetand32 g = feetper secondpersecond,theheightofthe stoneattime t is 2 00 2

1 ()2 16144.

=−++ =−+ Thestonehitsthegroundwhen 2 ()161440Htt=−+= ,thatiswhen 29 t = orafter3 t = seconds.

HtgtVtH t

(b) Thevelocityattime t isgivenby ()32 Htt =− .Whenthestonehitsthe ground,itsvelocityis(3)96 H ′ =− feetpersecond.

75. (a) Ifafter2secondstheballpassesyou onthewaydown,then0 (2),HH = where200 ()16 HttVtH =−++ So,2000 16(2)()(2), VHH−++= 0 6420, V −+= or032. sec ft V =

(b) Theheightofthebuildingis0 H feet. Frompart (a) youknowthat 2 0 ()1632. HtttH =−++ Moreover, H(4)=0sincetheballhitstheground after4seconds.So,

Chapter2.Differentiation:BasicConcepts 113

2 0 16(4)32(4)0, H −++= or 0128feet. H =

()3232. sec ft Htt ′ =−+

After2seconds,thespeedwillbe (2)32feet H ′ =− persecond,where theminussignindicatesthatthe directionofmotionisdown.

(d) Thespeedatwhichtheballhitsthe groundis(4)96. ft sec H ′ =−

76. Let(x, y)beapointonthecurvewhere thetangentlinegoesthrough(0,0).Then theslopeofthetangentlineisequalto 0 0 yy xx = .Theslopeisalsogivenby

()24fxx ′ =− .Thus24 y x x =− or 2 24 yxx =−

Since(x, y)isapointonthecurve,we musthave2425 yxx=−+ .Settingthe twoexpressionsfor y equaltoeachother

78. (a) If4 () fxx = then 4 432234

()() 464 fxhxh xxhxhxhh +=+ =++++

32234()()464 fxhfxxhxhxhh +−=+++ and

()()3223 464 fxhfxxxhxhh h +− =+++

gives 22 2 42524 25 5

−+=− = =±

xxxx x x

If5 x =− ,then70 y = ,theslopeis

14andthetangentlineis14yx =−

If5 x = ,then30 y = ,theslopeis6and thetangentlineis6yx = .

77. ()2 fxaxbxc =++

Since f (0)=0, c =0and2(). fxaxbx =+

Since f (5)=0,0=25a +5b

Further,sincetheslopeofthetangentis1 when x =2,(2)1. f ′ =

()2

12(2)4 fxaxb abab ′ =+ =+=+

Now,solvethesystem:0=25a +5b and 1=4a + b.Since14a = b,using substitution

0255(14) 025520 055

aa aa a

= +− =+− =+ or a =1and b =14(1)=5.

So,2()5. fxxx =−+

(b) If() n fxx = then 1221()()(1)... 2 nnnnnn nn fxhxhxnxhxhnxhh +=+=++ +++

1221()()(1)... 2 nnnn nn fxhfxnxhxhnxhh +−=+ +++

114 Chapter2.Differentiation:BasicConcepts

material

for

for sale or distribution in any

distributed,

2013 by McGraw-Hill Education. This is proprietary

solely

authorized instructor use. Not authorized

manner. This document may not

copied, scanned, duplicated, forwarded,

or posted on a website, in whole or part.

2.3 Product and Quotient Rules; Higher-Order Derivatives

Chapter2.Differentiation:BasicConcepts 115 and 1221()()(1) ... 2 nnnnfxhfxnnnxxhnxhh h +− =++++ (c) Frompart(b) ()()122 (1) 2 fxhfxnnn nn nxhxh h +−  =+ ++   Thefirsttermontherightdoesnotinvolve h whilethesecondtermapproaches0as0 h → . Thus1 0 []lim()(). nn h dfxhfx xnx dxh → +− = = 79. ()() fgx ′ + 0 0 0 00 lim()()()() lim()()[()()] lim()()()() lim()()()() lim ()() h h h hh fgxhfgx h fxhgxhfxgx h fxhfxgxhgx h fxhfxgxhgx hh fxgx → → → →→ ++−+ = +++−+ = +−++− = +− +− = + =+′′

1. f(x)=(2x +1)(3x 2), ()(21)(32) (32)(21) (31)(3)(32)(2) 121 d fxxx dx d xx dx xx x ′ =+− +−+ =++− =− 2. ()(5)(12) fxxx =−− ()(5)(12)(12)(5) 2(5)1(12) 114 dd fxxxxx dxdx xx x =−−+−− =−−+− =− © 2013 by

for

not be

distributed, or

on a website, in

or part.

McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized

sale or distribution in any manner. This document may

copied, scanned, duplicated, forwarded,

posted

whole

116 Chapter2.Differentiation:BasicConcepts 3. y =10(3u +1)(15u), ()() 10(31)(15) 10(31)1515(31) 10[(31)(5)(15)(3)] 30020 dy du d uu du dd uuuu dudu uu u =+−  =+−+−+   =+−+− =−− 4. 2 400(15)(32)yxx =−− 2 22 2 2 400(15)(32) 400(15)(32)(32)(15) 400(15)(3)(32)(2) 400(9445) dyd xx dxdx dd xxxx dxdx xxx xx =−−   =−−+−−  =−+−−  =−++ 5. 53 53 53 2 42 53 2 11 ()(21) 3 1 (21) 11 3(21)1 1 (56) 411 24 33 3 d fxxxxdxx d xxx xdx xx x xxx x xxx x   ′ =−+−      +−−+       =−++       +−−     =−+++ 6. 3 ()3(525)(2) fxxxxx =−−++ () 32 5/231/2 1/2 1 ()3(525)22(152) 2 10515 12092430 22 fxxxxxx x xxxx x   ′ =−−++++−     =−−++−− © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Chapter2.Differentiation:BasicConcepts 117 7. 1 2, x y x + = 2 2 2 (2)(1)(1)(2) (2) (2)(1)(1)(1) (2) 3 (2) dd dxdx dyxxxx dxx xx x x −+−+− = −−+ = =− 8. 23 54 x y x = + 2 2 2 (54)(23)(23)(54) (54) 2(54)5(23) (54) 23 (54) dd dyxxxx dxdx dxx xx x x +−−−+ = + +−− = + = + 9. 2 (), 2 ftt t = 22 22 2 22 2 22 (2)()(2) () (2) (2)(1)()(2) (2) 2 (2) dd dtdt tttt ft t ttt t t t ′ = = = 10. 1 ()2fx x = 22()(2)(0)1(1)1 (2)(2) fxx xx ′ = = 11. 3 5, y x = + 2 2 2 (5)(3)3(5) (5) (5)(0)3(1) (5) 3 (5) dd dxdx dyxx dxx x x x +−+ = + +− = + =− + © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

118 Chapter2.Differentiation:BasicConcepts 12. 2 2 1 1 t y t + = 22 2222 (1)(2)(1)(2)4 (1)(1) dyttttt dttt −−+− = = 13. 2 2 ()32, 251 fxxx xx −+ = +− 22 22 22 22 2 22 2 22 2 22 (251)(32) () (251) (32)(251) (251) (251)(23) (251) (32)(45) (251) 11107 (251) d dx d dx xxxx fx xx xxxx xx xxx xx xxx xx xx xx +−−+ ′ = +− −++− +− +−− = +− −++ +− = +− 14. 2 ()(1)(4) 21 xxx gx x ++− = 22 2 32 2 (21)[1(1)(4)(21)](1)(4)(2) () (21) 49611 (21) xxxxxxxx gx x xxx x  −−+++−+−++−  ′ = −+−− = 15. 2 ()(25)(25)(25) fxxxx =+=++ ()(25)(25) (25)(25) 2(25)(25) 2(25)(5) 2050 10(25) d fxxx dx d xx dx d xx dx x x x ′ =++ +++ =++ =+ =+ =+ 16. 2 2 2 11 ()2 fxxx xx  =+=++   3 2 ()2 fxx x ′ =− © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19.

When x =0, y =4and17. dy dx = Theequationofthetangentlineat(0,4)is y +4=17(x 0),or y =17x 4.

20. 2 (31)(2) yxxx =+−−

2 (31)(1)(23)(2) yxxxx ′ =+−−++−

At01 x = ,(3)(1)3 y == and(3)(1)(5)(1)2 y ′ =−+= .Theequationofthetangentlineisthen 32(1)yx−=− or21 yx=+

x y x dy dxx

= + = +

21. 2

23 3 (23)

When x =1, y =1and3. dy dx = Theequationofthetangentlineat(1,1)is

y +1=3(x +1),or y =3x +2.

Chapter2.Differentiation:BasicConcepts 119

tt ++ == ++ ( ) 1/2 1/2 21/2 21/2 2 121/2 2 2 21/251/2 2 21/2 5/23/2 1/22 53 2 (25)() ()(25) () (25) (25)2()(2) (25) 2102 (25)2 42025 2(25) 42025 2(25) d dt d dt t t ttt ttt gt t tttt t tttt tt ttt tt ttt tt ++ −++ ′ = + ++−+ = + +−+ = + +−+ = + +−+ = +

hxxx xx =+ −+ 22 2222 22 2222 ()(1)(1)(2)(1)(1)(4)(2) (1)(1) 181 (1)(1) hxxxxxxx xx xxx xx −−+−−− ′ = + −+ =+ −+

17. 221/2 ()2525 tttt gt

18. 22

() 11

(51)(43) 3017 yxx dy x dx =−+ =+

22. 7 52 x y x + =

2 (52)(1)(7)(2) (52) xx y x −−+− ′ =

At00 x = , 7 5 y = and 2 51419 525 y + ′ ==

Theequationofthetangentlineisthen 719(0) 525 yx−=− or 197 255yx=+

23. () 2 1/22 3(2) (3)(2) yxxx xxx

=+− =+−

23/2 1/2 153 32 2 dy xx dxx =−−++

When x =1, y =4and11 2 dy dx =−

Theequationofthetangentlineat(1,4)is114(1), 2 yx−=−− or 1119 . 22yx=−+

24. 2 ()(1)(87) fxxxx=−−+

fxxxxx xx xx

′ =⋅−++−− =−+ =−−

2 2 ()1(87)(1)(28) 31815 3(1)(5)

()0fx ′ = when x =1and x =5.

2 (1)(11)(1817)0 f =−−⋅+=

2 (5)(51)(5857)32 f =−−⋅+=−

Thetangentlinesat(1,0)and(5,32)arehorizontal.

25. 2 2 2

fxxxx fxxxxx x

()(1)(2) ()(1)(21)(2)(1) 33

=+−− ′ =+−+−− =−

Since() fx ′ representstheslopeofthe tangentlineandtheslopeofahorizontal lineiszero,needtosolve

2 0333(1)(1) xxx =−=+− or x =1,1.

When x =1, f (1)=0andwhen x =1, f (1)=4.So,thetangentlineis horizontalatthepoints(1,0)and (1,4).

120 Chapter2.Differentiation:BasicConcepts

26. 2 2 1 () 1 fxxx xx +− = −+ 22 22 2 22 22 () (21)(1)(1)(21) (1) 24 (1) 2(2) (1) fx xxxxxx xx xx xx xx xx ′ +−+−+−− = −+ −+ = −+ = −+ ()0fx ′ = when x =0and x =2 2 2 (0)0011 001 f +− ==− −+ 2 2 (2)2215 2213 f +− == −+ © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Thetangentlinesat(0,1)and 5 2, 3 

are horizontal.

=+ =+

31. 2

yx x dyx dxx

3 24 1(24)(0)3(4) (24)

+ = ++ ′ = ++

27. 2 2 22

1 () 1 2 () (1)

fxx xx fxxx xx

Since() fx ′ representstheslopeofthe tangentlineandtheslopeofahorizontal lineiszero,needtosolve 2 22 2

When x =0,2 12 14. (2) dy dx =+=

32. 235yxx=+− 23yx ′ =+

At0,3 xy ′ == sotheslopeofthe

2 0 (1) 02(2)0,2. or

= ++ =−−=−+=−

xx xx xxxxx

When x =0, f (0)=1andwhen x =2, 1 (2). 3 f −=− So,thetangentlineis horizontalatthepoints(0,1)and 1 2,. 3

28. 2 (2)() yxxx =++

21 (2)12 2 dy xxxx dxx

perpendicularlineis 1 3 m =− .The perpendicularlinepassesthroughthe point(0,5)andsohasequation 1 5 3 yx=−−

=+− =+−+−

29. 23 223 (3)(52) (3)(6)(52)(2) yxx dy xxxx dx

When x =1, (13)(6)(52)(2)18. dy dx =+−+−=−

30. 21 35 x y x = + 22 2(35)3(21)13 (35)(35) dyxx dxxx +−− = = ++

At01 x = ,2 1313 864 dy dx ==

33. 11/2 21/2

yxxx x dy dxxx

=−=− =−

2 2 21 2

When x =1, 15 2. 22 dy dx =−−=−

Theslopeofalineperpendiculartothe

tangentlineat x =1is2 5

Theequationofthenormallineat(1,1)is 2 1(1), 5 yx−=− or 23 . 55yx=+

34. ()(3)1 yxx =+− () 1 (3)1 2 yxx x  ′ =+−+−

At1,2 xy ′ ==− sotheslopeofthe

perpendicularlineis 1 2 m = .The perpendicularlinepassesthroughthe point(1,0)andsohasequation

11 22yx=−

Chapter2.Differentiation:BasicConcepts 121

 

  



 

 =++=  

()

=++++

At04 x = , 1 (18)18(6)70.5 4 dy dx

 

is proprietary material

for authorized instructor use. Not authorized for sale or distribution in any

This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

2013 by McGraw-Hill Education. This

solely

manner.

38.

x dyxx

x y = =

When x =1, 2 (23)(5)(57)(3)31. (23) dy dx −−+−

Theslopeofalineperpendiculartothe tangentlineat x =1is 1 . 31

Theequationofthenormallineat (1,12)is112(1), 31 yx+=−− or 1371 3131yx

36. 37. Usingtheproductrule,

Sinceand,

122 Chapter2.Differentiation:BasicConcepts

35. 2 dxx

57 23 (23)(5)(57)(3) (23) + = −−+− =

=−−

Substituting x =3,

Usingthequotientrule,andnotingthatthe derivativeofthenumeratorrequiresthe productrulefortheterm,

Chapter2.Differentiation:BasicConcepts 123

39.

Substituting, () ()()()()() () ()()() [] () () [] ()() ( () () [] ) () 2 2 3 2 2 2 31531111 1 315 1113 315 8311 1 8 113 8 gg h g gg h g  ′ +−−+−     ′ −=    −+−−     ′ −−−+−  ′ −=   Sinceand, 40. (a) 2 253yxx=−− 45yx ′ =− (b)

=+− (21)(1)(2)(3)45yxxx ′ =++−=− 41. (a) 3 32 6 32 6 4 23 ()(2)(23)(3) 49 49 x y x dyxxx dxx xx x x x = = −+ = −+ = (b) 3 43 4 4 (23)() (23)(3)()(2) 3(23)2 49 yxx dy xxx dx xx x x x =− =−−+ −−+ = −+ = (c) 23 34 34 4 23 49 49 49 yxx dy xx dx xx x x =− =−+ =+ −+ =

105 ()56274 fxxxx =−−+ 94 83 ()503027 ()450120 fxxx fxxx ′ =−− ′′ =− 43.

()4962

()212186 ()82418 fxxxxx fxxxx fxxx =−+−− ′ =−+− ′′ =−+ 44. 2 311 5 32 yx xx =+++ 1/233/251 6 26 dy xxx dx =−− 2 3/245/2 2 3/245/2 51 18 44 5181 44 dy xxx dx xxx =−++ =−++ © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

(21)(3)yxx

42.

532 42 3 2

attherateof (3)15.72 p ′ =− dollarsperthousand calculators.

(b) ()() Rxxpx = 222

(3)46.29 R ′ = sorevenueisincreasing attherateof$46.29perthousand

50. (a) Sinceprofitequalsrevenueminuscost andrevenueequalspricetimesthe quantitysold,theprofitfunction P(p)is givenby

Whenthepriceis$12perbottle, (12)1.133 P ′ =− .Theprofitis decreasing.

124 Chapter2.Differentiation:BasicConcepts 45. 11/21/221 22 36 yxxxx =−+− 21/23/2221 2 3212 dy dxy xxx ′ = =−−++ 2 33/25/2 2 33/25/2 421 348 421 348 dy yxxx dx xxx ′′ ==+− =+− 46. 21()2 yxxx x  =−−  2 2 2 11 ()2(21)2 641 dy xxxx dxx x xx   =−++−−    =−− 2 2 124 dy x dx =−

3 (21)(35)yxxx =+−+ 32 32 (21)(3)(35)(32) 1215127 dy dxy xxxx xxx ′ = =+−+++ =+++ 2 2 2363012dy yxx dx ′′ ==++ 48. (a) 2 ()1,000 0.38 px x = + 2 22 22 ()(0.38)(0)1,000(0.6) (0.38) 600 (0.38) xx px x x x +− ′ = + = + whenthelevelofproductionis3,000 (3 x = )calculators,demandischanging

47.

′′=+  = +  ++  −+ =

2 22 ()()()(1) 6001,000 (0.38)0.38 3008,000 (0.38) Rxxpxpx x x xx x x

t = +

2 ()(40.3)(2000)(2000)(0.3) (40.3) tt St t +− ′ = + Therateofchangeintheyear2010is

Rewritethefunctionas 4 2,000 (). 0.3 t St = + Since40 t → as t → +∞,salesapproach 2,0006,666.67

calculators. 49. ()200040.3 t St

(a)

(b)

0.3 thousand, ≈ or approximately$6,666,667inthelong run.

2 ()()() 500(0.23200) 3 PppBpCp p pp p =− =−++ + (b) 2 2

(3) 1500 0.43 (3) Ppppp p p p +− ′ = + =−− +

()(3)500500(1)0.43

51. 2 2 ()10055 1030 tt Pt tt

 ++ =  ++   (a)

2 2 22

ttt ttt Pt tt

(1030)(25) ()100(55)(210) (1030)

+++ −+++ ′ = ++

Therateofchangeafter5weeksis

(255030)(105) (5)100(25255)(1010) (255030) P

+++ −+++ ′ = ++ (5)4.31%perweek. P ′ =

Since(5) P ′ ispositive,thepercentage isincreasing.

(b) Rewritethefunctionas

Substitutethisintothesecondequation toget Then, Currently,when x =0, or$17.50perunit

53. (a)

1 ()100. 1 tt tt pt ++ = ++

55 1030

Sinceispositive,therevenueis currentlyincreasing.

Since25510 ,, tt t and230 t allgotozero as t → +∞,thepercentageapproaches 100%inthelongrun,sotherateof changeapproaches0.

52. (a) (b) Therateofchangeoftheworker’srateis thesecondderivative

(b)

At9:00a.m., t =1and

Revenue=(#units)(sellingprice)

x =#units=10004t,where t is measuredinweeks p =sellingprice=5+0.05t

So,

Tofind R(x),solvethefirstequationfor t

54. (a)

Sinceisnegative,theaverage revenueiscurrentlydecreasing.

Since x isthelevelofproductionattime t, x =400–2t or t =200–0.5x and

Chapter2.Differentiation:BasicConcepts 125

for

2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized

sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

126 Chapter2.Differentiation:BasicConcepts

C(x)=10,000.Therefore, and When t =0, x =400 and Sotheprofitisdecreasingattherateof $120/unit.

Theaverageprofitwillbedecreasingat therateofabout0.31dollarsperunit perunit. 55. 1/2 22()100100 0.0390.039 xx Px xx == ++ (a) ()21/2 1 2 1/2 22 (0.039) (()(0.06) )100 (0.039) xx xx Px x + ′ = + Therateofchangeofpercentagepollutionwhen16milliondollarsarespentis 21/2 1 2 22 1/2 22 (16) [0.03(16)9](16) 100 [0.03(16)9] (16)[0.06(16)] [0.03(16)9] 0.63 P ′   +   =  +      +  =− percent Since(16) P ′ isnegative,thepercentageisdecreasing. (b) ()0Px ′ = when 21/2 1/2 1 0(0.039)2 ()(0.06) xx xx  =+   or x =10milliondollars. Testingonevaluelessthan10andonevaluegreaterthan10shows() Px ′ isincreasingwhen0 < x <10,anddecreasingwhen x >10. © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Weknow

(b)

Chapter2.Differentiation:BasicConcepts 127 56. (a) 2 ()2410 1 t Pt t + = + 2 22 2 22 22 ()(1)(24)(2410)(2) (1) 242024 (1) 4(23)(32) (1) ttt Pt t tt t tt t +−+ ′ = + −−+ = + −+− = + (1)5 P ′ =− sothepopulationisdecreasingat1 t = . (b) Thepopulationrateofchangeis0at 2 3 t = ,positivefor 2 3 t < andnegativefor 2 3 t > .The populationbeginstodeclineafter 2 3 t = or40minutesaftertheintroductionofthetoxin. 57. 123 3()FKMM =− (a) 2 2 1 3(23) 2 3 dF S dM KMM KMM = =− =− (b) 12 (26)2 33 dS KMKM dM =−=− istheratethesensitivityischanging. 58. (a) 2 2 () 316 t Ct t = + 22 2222()()(316)22(6)326 (316)(316) tttt RtCt tt +−− ′ == = ++ R(t)ischangingattherate 22222 2423 ()(316)(12)(326)(2)(316)(6)36(16) (316)(316) ttttttt Rt tt +−−−+ ′ = = ++ (b) (1)26361 C ′ = ,theconcentrationisincreasingatthistime. (c) R(t)ispositiveandtheconcentrationisincreasinguntil()0 Rt = orwhen23260 t −= .This occurswhen42.3 3 t =≈ hours(ignoringthenegativesolution.)Theconcentrationbeginsto declineafterroughly2.3hours. © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

(d) Theconcentrationischangingatadecliningratewhen

orwhen236(16)0 tt −< (assuming0 t > ).Thisoccurswhen04 t <<

59. ()2061Pt t =− +

(a)

(b) orincreasingatarateof1,500peopleperyear

So,thepopulationwillactuallyincreaseby1thousandpeople.

(d) orincreasingatarateof60peopleperyear

(e) Inthelongrun, 2 6 lim()lim0 (1) tt Pt t →∞→∞ ′ = = +

So,thepopulationgrowthapproacheszero.

60. (a) () m Ax px Bx = +

128 Chapter2.Differentiation:BasicConcepts

2 23 ()36(16)0 (316) tt

′ = <

1 2 2 ()()()()() () ((1)) () mm m m m BxAAxmx px Bx ABmx Bx +− ′ = + +− = + © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

AmxmxBm px Bx

−−+

′′ = + ()0px ′′ = when(1) 1 m Bm x m + =

61. (a) 53 ()357 sttt=−− 4242 ()151515() vttttt =−=−

32 ()15(42)30(21) attttt=−=−

(b) a(t)=0when230(21)0, tt −= or t =0and2.

2 t =

62. (a) 43 ()253 stttt=−+−

)=0when6

(b) ()0at = when4 t = 65. 5 ()105 1 Dtt t =+− +

5 4

Chapter2.Differentiation:BasicConcepts 129

(b) 1 3 (1)(1) () () 

mm m 

32 ()()8151 vtsttt ′ ==−+ 2 ()()()2430 6(45) atvtsttt tt ′′′ ===− =−

a(t

t +14=0,or 7 3 t =

(b) ()0at = at0 t = and

63. 32 ()72 stttt=−+++ (a) 2 ()3141 ()614 vttt att =−++ =−+

(b)

64. (a) 5/22 ()4153 stttt=−+− 3/2 1/2 ()()10301 ()()()1530

vtsttt atvtstt

′ ==−+ ′′′ ===−

Applied calculus for business economics and the social and life sciences 11th edition hoffmann solut

Chapter 2

Differentiation: Basic Concepts

53. ()4.44.92 Httt

fxhfx h +− canbefilledin

2.2 Techniques of Differentiation

2.3 Product and Quotient Rules; Higher-Order Derivatives