devexpress asp.net pdf viewer : Cut pages from pdf SDK software project winforms .net html UWP Calculus8-part768

4.6 ExponentialandLogarithmicfunctions
81
fromtheexample: log
a
(a
5
)=5,log
a
(a
3
)=3,log
a
(a
8
)=8. Solog
a
(a
5
a
3
)=log
a
(a
8
)=
8=5+3=log
a
(a
5
)+log
a
(a
3
). Nowlet’smakethisabitmoregeneral. SupposeAand
B aretwonumbers,A=a
x
,andB=a
y
. Thenlog
a
(AB)=log
a
(a
x
a
y
)=log
a
(a
x+y
)=
x+y=log
a
(A)+log
a
(B).
Nowconsider(a
5
)
3
=a
5
a
5
a
5
=a
5+5+5
=a
53
=a
15
. Againit’sclearthat t more
generally(a
m
)
n
=a
mn
,andagainthisgivesusafactaboutlogarithms. IfA=a
x
then
A
y
=(a
x
)
y
=a
xy
,solog
a
(A
y
) =xy =ylog
a
(A)|the exponent canbe \pulledout in
front."
Wehavecheatedabitintheprevioustwoparagraphs.Itisobviousthata
5
=aaaaa
anda
3
=aaaandthattherestoftheexamplefollows;likewiseforthesecondexample.
But when n we e consider an n exponential function n a
x
we can’t be e limited d to substituting
integersforx. Whatdoesa
2:5
ora
1:3
ora
mean?Andisitreallytruethata
2:5
a
1:3
=
a
2:5 1:3
? The e answertotherstquestionisactuallyquitedicult,sowewillevadeit;
theanswertothesecondquestionis\yes."
We’llevadethefullanswertothehardquestion,butwehavetoknowsomethingabout
exponentialfunctions. Youneedrsttounderstandthatsinceit’snot\obvious"what2
x
shouldmean,wearereallyfreetomakeitmeanwhateverwewant,solongaswekeepthe
behaviorthatisobvious,namely,whenxisapositiveinteger. Whatelsedowewantto
betrueabout2
x
? Wewantthepropertiesoftheprevioustwoparagraphstobetruefor
allexponents:2
x
2
y
=2
x+y
and(2
x
)
y
=2
xy
.
Afterthepositiveintegers,thenexteasiestnumbertounderstandis0: 2
0
=1. You
havepresumablylearnedthisfactinthepast;whyisittrue? Itistruepreciselybecause
we want t 2
a
2
b
= 2
a+b
to be e true about the function n 2
x
. We e need it to be true that
2
0
2
x
=2
0+x
=2
x
,andthisonlyworksif2
0
=1. Thesameargumentimpliesthata
0
=1
foranya.
Thenexteasiestsetofnumberstounderstandisthenegativeintegers: forexample,
2
3
=1=2
3
.Weknowthatwhatever2
3
meansitmustbethat2
3
2
3
=2
3+3
=2
0
=1,
whichmeansthat2
3
mustbe1=2
3
. Infact,bythesameargument,onceweknowwhat
2
x
meansforsomevalueofx,2
x
mustbe1=2
x
andmoregenerallya =1=ax.
Next,consideranexponent1=q,whereq isapositiveinteger. . Wewantittobetrue
that(2
x
)
y
=2
xy
,so(2
1=q
)
q
=2.Thismeansthat2
1=q
isaq-throotof2,2
1=q
=
q
p
2.This
isallweneedtounderstandthat2
p=q
=(2
1=q
)
p
=(
q
p
2)
p
anda
p=q
=(a
1=q
)
p
=(
q
p
a)
p
.
What’s left is the hard d part: what t does2
x
mean when n x x cannot be e written as s a
fraction,likex=
p
2 orx=? What t weknowsofarishowto assignmeaning to2
x
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82
Chapter 4 4 Transcendental l Functions
wheneverx=p=q;ifweweretographthiswe’dseesomethinglikethis:
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Butthisisapoorpicture,becauseyoucan’tseethatthe\curve"isreallyawholelot
ofindividualpoints,above therationalnumbersonthex-axis. . Thereare e reallyalotof
\holes"inthecurve,abovex=,forexample. But(thisisthehardpart)itispossible
toprovethattheholescanbe\lledin",andthattheresultingfunction,called2
x
,really
doeshavethepropertieswewant,namelythat2
x
2
y
=2
x+y
and(2
x
)
y
=2
xy
.
Exercises 4.6.
1. Expandlog
10
((x+45)
7
(x 2)).
2. Expandlog
2
x
3
3x 5+(7=x)
.
3. Writelog
2
3x+17log
2
(x 2) 2log
2
(x
2
+4x+1)asasinglelogarithm.
4. Solvelog
2
(1+
p
x)=6forx.
5. Solve2
x
2
=8forx.
6. Solvelog
2
(log
3
(x))=1forx.
Aswith the sine, we don’t knowanything about derivativesthat allowsusto compute
thederivativesoftheexponentialandlogarithmicfunctionswithoutgoingbacktobasics.
Let’sdoalittleworkwiththedenitionagain:
d
dx
a
x
= lim
x!0
a
x+x
a
x
x
= lim
x!0
a
x
a
x
a
x
x
= lim
x!0
a
x
ax  1
x
=a
x
lim
x!0
a
x
1
x
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4.7 Derivatives s ofthe exponentialandlogarithmicfunctions
83
Therearetwointerestingthingstonotehere:Asinthecaseofthesinefunctionweareleft
withalimitthatinvolvesxbutnotx,whichmeansthatwhatever lim
x!0
(a
x
1)=x
is,weknowthatitisanumber,thatis,aconstant.Thismeansthata
x
hasaremarkable
property: itsderivativeisaconstanttimesitself.
Weearlierremarkedthatthehardestlimitwewouldcomputeis lim
x!0
sinx=x=1;we
nowhavealimitthat isjusta bittoohard toincludehere. In n fact thehard partisto
seethat lim
x!0
(a
x
1)=xevenexists|doesthisfractionreallygetcloserandcloserto
somexedvalue?Yesitdoes,butwewillnotprovethisfact.
Wecan lookat someexamples. Consider(2
x
1)=x forsomesmallvaluesofx: 1,
0:828427124, 0:756828460, , 0:724061864, 0:70838051, 0:70070877 7 when x is1, , 1=2, , 1=4,
1=8,1=16,1=32,respectively. Itlookslikethisissettlinginaround0:7,whichturnsout
to betrue(butthe limit isnotexactly0:7). Considernext(3
x
1)=x: 2,1:464101616,
1:264296052,1:177621520,1:13720773,1:11768854,atthesamevaluesofx. Itturnsout
tobetruethatinthelimitthisisabout1:1. Twoexamplesdon’testablishapattern,but
ifyoudomoreexamplesyouwillndthat thelimitvariesdirectlywith thevalueofa:
biggera,biggerlimit;smallera,smallerlimit. Aswecanalreadysee,someoftheselimits
willbelessthan1andsomelargerthan1.Somewherebetweena=2anda=3thelimit
willbeexactly1;thevalueatwhichthishappensiscallede,sothat
lim
x!0
e
x
1
x
=1:
Asyoumightguessfromourtwoexamples,eiscloserto3thanto2,andinfacte2:718.
Nowweseethatthefunctione
x
hasatrulyremarkableproperty:
d
dx
e
x
= lim
x!0
e
x+x
e
x
x
= lim
x!0
exex  ex
x
= lim
x!0
e
x
e
x
1
x
=e
x
lim
x!0
e
x
1
x
=e
x
Thatis,e
x
isitsownderivative,orinotherwordstheslopeofe
x
isthesameasitsheight,
or the e same e asitssecond d coordinate: The e function f(x) ) = e
x
goesthrough the point
(z;e
z
)andhasslopee
z
there,nomatterwhatzis. Itissometimesconvenienttoexpress
thefunctione
x
withoutanexponent,sincecomplicatedexponentscanbehardtoread. In
suchcasesweuseexp(x),e.g.,exp(1+x
2
)insteadofe
1+x
2
.
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84
Chapter 4 4 Transcendental l Functions
Whataboutthelogarithmfunction? Thistooishard,butasthecosinefunctionwas
easierto dooncethesinewasdone,sothe logarithmiseasierto donowthat we know
thederivativeoftheexponentialfunction. Let’sstartwithlog
e
x,whichasyouprobably
knowisoftenabbreviatedlnxandcalledthe\naturallogarithm"function.
Considertherelationshipbetweenthetwofunctions,namely,that theyareinverses,
thatone\undoes"theother.Graphicallythismeansthattheyhavethesamegraphexcept
thatoneis\ ipped"or\re ected"throughtheliney=x,asshowningure4.7.1.
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Figure4.7.1
Theexponentialandlogarithmfunctions.
Thismeansthattheslopesofthesetwofunctionsarecloselyrelatedaswell:Forexample,
theslopeofe
x
iseatx=1;atthecorrespondingpointontheln(x)curve,theslopemust
be1=e,becausethe\rise"andthe\run"havebeeninterchanged. Sincetheslopeofe
x
is
eatthepoint(1;e),theslopeofln(x)is1=eatthepoint(e;1).
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Figure 4.7.2
Slopeoftheexponentialandlogarithmfunctions.
Moregenerally,weknowthattheslopeofe
x
ise
z
atthepoint(z;e
z
),sotheslopeof
ln(x)is1=e
z
at(e
z
;z),asindicatedingure4.7.2.Inotherwords,theslopeoflnxisthe
reciprocaloftherstcoordinateatanypoint;thismeansthattheslopeoflnxat(x;lnx)
is1=x.Theupshotis:
d
dx
lnx=
1
x
:
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4.7 Derivatives s ofthe exponentialandlogarithmicfunctions
85
Wehavediscussedthisfromthepointofviewofthegraphs,whichiseasytounderstand
butisnotnormallyconsideredarigorousproof|itistooeasytobeledastraybypictures
thatseemreasonablebutthatmisssomehardpoint. Itispossibletodothisderivation
withoutresortingtopictures,andindeedwewillseeanalternateapproachsoon.
Notethatlnxisdenedonlyforx>0.Itissometimesusefultoconsiderthefunction
lnjxj,afunctiondenedforx6=0. Whenx<0,lnjxj=ln( x)and
d
dx
lnjxj=
d
dx
ln( x)=
1
x
( 1)=
1
x
:
Thuswhetherxispositiveornegative,thederivativeisthesame.
Whataboutthefunctionsa
x
andlog
a
x? Weknowthatthederivativeofa
x
issome
constanttimesa
x
itself,butwhatconstant? Rememberthat\thelogarithmistheexpo-
nent"andyouwillseethata=e
lna
. Then
a
x
=(e
lna
)
x
=e
xlna
;
andwecancomputethederivativeusingthechainrule:
d
dx
a
x
=
d
dx
(e
lna
)
x
=
d
dx
e
xlna
=(lna)e
xlna
=(lna)a
x
:
The constant issimply y lna. Likewise e we e can compute the derivative of the logarithm
functionlog
a
x.Since
x=e
lnx
wecantakethelogarithmbaseaofbothsidestoget
log
a
(x)=log
a
(e
lnx
)=lnxlog
a
e:
Then
d
dx
log
a
x=
1
x
log
a
e:
Thisisaperfectlygoodanswer,butwecanimproveitslightly. Since
a=e
lna
log
a
(a)=log
a
(e
lna
)=lnalog
a
e
1=lnalog
a
e
1
lna
=log
a
e;
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86
Chapter 4 4 Transcendental l Functions
wecanreplacelog
a
etoget
d
dx
log
a
x=
1
xlna
:
Youmayifyouwishmemorizetheformulas
d
dx
a
x
=(lna)a
x
and
d
dx
log
a
x=
1
xlna
:
Becausethe\trick"a=e
lna
isoftenuseful,andsometimesessential,itmaybebetterto
rememberthetrick,nottheformula.
EXAMPLE4.7.1
Computethederivativeoff(x)=2
x
.
d
dx
2
x
=
d
dx
(e
ln2
)
x
=
d
dx
e
xln2
=
d
dx
xln2
e
xln2
=(ln2)e
xln2
=2
x
ln2
EXAMPLE4.7.2
Computethederivativeoff(x)=2
x
2
=2
(x
2
)
.
d
dx
2
x
2
=
d
dx
e
x
2
ln2
=
d
dx
x
2
ln2
e
x
2
ln2
=(2ln2)xe
x
2
ln2
=(2ln2)x2
x
2
EXAMPLE 4.7.3
Compute thederivative off(x)=x
x
. At t rst thisappearstobe
anewkindof function: it t isnot aconstant powerof x,anditdoesnot seemtobe an
exponentialfunction,sincethebaseisnotconstant. Butinfactitisnoharderthanthe
previousexample.
d
dx
x
x
=
d
dx
e
xlnx
=
d
dx
xlnx
e
xlnx
=(x
1
x
+lnx)x
x
=(1+lnx)x
x
4.8 ImplicitDierentiation
87
EXAMPLE 4.7.4
Recallthat we have notjustied thepowerrule except whenthe
exponent isapositive ornegativeinteger. We e canuse theexponentialfunctiontotake
careofotherexponents.
d
dx
x
r
=
d
dx
e
rlnx
=
d
dx
rlnx
e
rlnx
=(r
1
x
)x
r
=rx
r 1
Exercises4.7.
In1{19,ndthederivativesofthefunctions.
1. 3
x
2
)
2.
sinx
ex
)
3. (e
x
)
2
)
4. sin(e
x
))
5. e
sinx
)
6. x
sinx
)
7. x
3
e
x
)
8. x+2
x
)
9. (1=3)
x
2
)
10. e
4x
=x)
11. ln(x
3
+3x))
12. ln(cos(x)))
13.
p
ln(x2)=x)
14. ln(sec(x)+tan(x)))
15. x
cos(x)
)
16. xlnx
17. ln(ln(3x))
18.
1+ln(3x
2
)
1+ln(4x)
19.
x
8
(x 23)
1=2
27x6(4x 6)8
20. Findthevalueofasothatthetangentlinetoy=ln(x)atx=aisalinethroughtheorigin.
Sketchtheresultingsituation. )
21. Iff(x)=ln(x
3
+2)computef
0
(e
1=3
).
Aswehaveseen,thereisacloserelationshipbetweenthederivativesofe
x
andlnxbecause
thesefunctionsare inverses. Ratherthan n relying onpicturesforourunderstanding,we
wouldliketobeable toexploitthisrelationshipcomputationally. Infactthistechnique
canhelpusndderivativesinmanysituations,notjustwhenweseekthederivativeofan
inversefunction.
88
Chapter 4 4 Transcendental l Functions
Wewillbeginbyillustratingthetechniquetondwhatwealreadyknow,thederivative
of lnx. Let’s s write e y y = = lnx x and then x x = e
lnx
= e
y
, that t is, , x = e
y
. We e say y that
thisequationdenesthe functiony =lnx implicitlybecause whileit isnot an explicit
expressiony=:::,itistruethatifx=e
y
thenyisinfactthenaturallogarithmfunction.
Now,forthetimebeing,pretendthatallweknowofy isthat x=e
y
;what canwesay
aboutderivatives?Wecantakethederivativeofbothsidesoftheequation:
d
dx
x=
d
dx
e
y
:
Thenusingthechainruleontherighthandside:
1=
d
dx
y
e
y
=y
0
e
y
:
Thenwecansolvefory0:
y
0
=
1
ey
=
1
x
:
There isone little dicultyhere. Tousethe e chain ruletocomputed=dx(e
y
)=y
0
e
y
we
needtoknowthatthefunctionyhasaderivative. Allwehaveshownisthatif f ithasa
derivativethenthatderivativemustbe1=x. Whenusingthismethodwewillalwayshave
toassumethatthedesiredderivativeexists,butfortunatelythisisasafeassumptionfor
mostsuchproblems.
Theexampley=lnxinvolvedaninversefunctiondenedimplicitly,butotherfunc-
tionscanbedenedimplicitly,andsometimesasingleequationcanbeusedtoimplicitly
dene more thanone function. Here’sa a familiarexample. The e equation r
2
= x
2
+y
2
describesacircleofradiusr.Thecircleisnotafunctiony=f(x)becauseforsomevalues
ofxtherearetwocorrespondingvaluesofy. Ifwewanttoworkwithafunction,wecan
breakthecircleintotwopieces,theupperandlowersemicircles,eachofwhichisafunction.
Let’scallthesey=U(x)andy =L(x);infact thisisafairlysimpleexample,and it’s
possibletogiveexplicitexpressionsforthese:U(x)=
p
r2 x2 andL(x)= 
p
r2 x.
Butit’ssomewhateasier,andquiteuseful,toviewbothfunctionsasgivenimplicitlyby
r
2
=x
2
+y
2
: bothr
2
=x
2
+U(x)
2
andr
2
=x
2
+L(x)
2
aretrue,andwecanthinkof
r
2
=x
2
+y
2
asdeningbothU(x)andL(x).
Nowwe can take the derivativeof both sidesasbefore, , remembering g that y isnot
simplyavariablebutafunction|inthiscase,yiseitherU(x)orL(x)butwe’renotyet
specifyingwhichone. Whenwetakethederivativewejusthavetoremembertoapplythe
4.8 ImplicitDierentiation
89
chainrulewhereyappears.
d
dx
r
2
=
d
dx
(x
2
+y
2
)
0=2x+2yy
0
y
0
=
2x
2y
x
y
Nowwehaveanexpressionfory
0
,butit containsyaswellasx. Thismeansthat t ifwe
wanttocomputeyforsomeparticularvalueofxwe’llhavetoknoworcomputeyatthat
valueofxaswell. ItisatthispointthatwewillneedtoknowwhetheryisU(x)orL(x).
OccasionallyitwillturnoutthatwecanavoidexplicituseofU(x)orL(x)bythenature
oftheproblem
EXAMPLE4.8.1 Findtheslopeofthecircle4=x
2
+y
2
atthepoint(1; 
p
3).Since
weknowboththexandycoordinatesofthepointofinterest,wedonotneedtoexplicitly
recognizethatthispointisonL(x),andwedonotneedtouseL(x)tocomputey|but
wecould.Usingthecalculationofy
0
fromabove,
y
0
x
y
1
p
3
=
1
p
3
:
Itisinstructivetocomparethisapproachtoothers.
Wemight haverecognizedatthestartthat(1; 
p
3)isonthefunctiony=L(x)=
p
4 x2. WecouldthentakethederivativeofL(x),usingthepowerruleandthechain
rule,toget
L
0
(x)= 
1
2
(4 x
2
)
1=2
( 2x)=
x
p
4 x2
:
ThenwecouldcomputeL
0
(1)=1=
p
3bysubstitutingx=1.
Alternately,wecouldrealizethatthepointisonL(x),butusethefactthaty
0
= x=y.
SincethepointisonL(x)wecanreplaceybyL(x)toget
y
0
x
L(x)
=
x
p
4 x2
;
withoutcomputingthederivativeofL(x)explicitly.Thenwesubstitutex=1andgetthe
sameanswerasbefore.
InthecaseofthecircleitispossibletondthefunctionsU(x)andL(x)explicitly,but
therearepotentialadvantagestousingimplicitdierentiationanyway. Insomecasesitis
moredicultorimpossibletondanexplicitformulaforyandimplicitdierentiationis
theonlywaytondthederivative.
90
Chapter 4 4 Transcendental l Functions
EXAMPLE4.8.2 Findthederivativeofanyfunctiondenedimplicitlybyyx
2
+e
y
=x.
Wetreatyasanunspeciedfunctionandusethechainrule:
d
dx
(yx
2
+e
y
)=
d
dx
x
(y2x+y
0
x
2
)+y
0
e
y
=1
y
0
x
2
+y
0
e
y
=1 2xy
y
0
(x
2
+e
y
)=1 2xy
y
0
=
1 2xy
x2+ey
Youmightthinkthatthestepinwhichwesolvefory
0
couldsometimesbedicult|
after all, , we’re e using implicit dierentiation here because we can’t solve the equation
yx
2
+e
y
=xfory,somaybeaftertakingthederivativewegetsomethingthatishardto
solvefory
0
. Infact,thisneverhappens. Alloccurrencesy
0
comefromapplyingthechain
rule,andwheneverthechainruleisuseditdepositsasingley
0
multipliedbysomeother
expression. So o it willalwaysbepossible togroup thetermscontaining ytogetherand
factoroutthey
0
,justasinthepreviousexample. Ifyouevergetanythingmoredicult
youhavemadeamistakeandshouldxitbeforetryingtocontinue.
Itissometimesthecasethatasituationleadsnaturallytoanequationthatdenesa
functionimplicitly.
EXAMPLE4.8.3 Considerallthepoints(x;y)thathavethepropertythatthedistance
from(x;y)to(x
1
;y
1
)plusthedistancefrom(x;y)to(x
2
;y
2
)is2a(aissomeconstant).
Thesepointsformanellipse,whichlikeacircleisnot afunctionbutcanviewedastwo
functionspastedtogether. Becauseweknowhowtowritedownthedistancebetweentwo
points,wecanwritedownanimplicitequationfortheellipse:
p
(x x
1
)2+(y y
1
)2+
p
(x x
2
)2+(y y
2
)2=2a:
Thenwecanuseimplicitdierentiationtondtheslopeoftheellipseatanypoint,though
thecomputationisrathermessy.
EXAMPLE 4.8.4
Wehavealreadyjustiedthe powerrulebyusingthe exponential
function,butwecouldalsodoitforrationalexponentsbyusingimplicitdierentiation.
Supposethaty=x
m=n
,wheremandnarepositiveintegers. Wecanwritethisimplicitly
asy
n
=x
m
,thenbecausewejustiedthepowerruleforintegers,wecantakethederivative
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