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4.8 ImplicitDierentiation
91
ofeachside:
ny
n 1
y
0
=mx
m 1
y
0
=
m
n
x
m 1
yn 1
y
0
=
m
n
x
m 1
(xm=n)n 1
y
0
=
m
n
x
m 1 (m=n)(n 1)
y
0
=
m
n
x
m 1 m+(m=n)
y
0
=
m
n
x
(m=n) 1
Exercises4.8.
Inexercises1{8,ndaformulaforthederivativey
0
atthepoint(x;y):
1. y
2
=1+x
2
)
2. x
2
+xy+y
2
=7)
3. x
3
+xy
2
=y
3
+yx
2
)
4. 4cosxsiny=1)
5.
p
x+
p
y=9)
6. tan(x=y)=x+y)
7. sin(x+y)=xy)
8.
1
x
+
1
y
=7)
9. Ahyperbolapassingthrough(8;6)consistsofallpointswhosedistancefromtheoriginisa
constantmorethanitsdistancefromthepoint (5,2). . Findtheslopeofthetangentlineto
thehyperbolaat(8;6). )
10. Computey
0
fortheellipseofexample4.8.3.
11. Ify=log
a
xthena
y
=x. Useimplicitdierentiationtondy
0
.
12. Thegraphoftheequationx
2
xy+y
2
=9isanellipse. Findthelinestangenttothiscurve
atthetwopointswhereitintersectsthex-axis. Showthattheselinesareparallel. )
13. Repeatthepreviousproblemforthepointsatwhichtheellipseintersectsthey-axis.)
14. Findthepointsontheellipsefromtheprevioustwoproblemswheretheslopeishorizontal
andwhereitisvertical. )
15. Findanequationforthetangentlinetox
4
=y
2
+x
2
at(2;
p
12). (Thiscurveisthekampyle
ofEudoxus.) )
16. Findanequationforthetangentlinetox
2=3
+y
2=3
=a
2=3
atapoint(x
1
;y
1
)onthecurve,
withx
1
6=0andy
1
6=0. (Thiscurveisanastroid.) )
17. Findanequationforthetangentlineto(x
2
+y
2
)
2
=x
2
y
2
atapoint(x
1
;y
1
)onthecurve,
withx
1
6=0; 1;1. . (Thiscurveisalemniscate.) )
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92
Chapter 4 4 Transcendental l Functions
Denition. Two o curves are orthogonal if at each point of intersection, the e angle e between
theirtangentlinesis=2. Twofamiliesofcurves,AandB,areorthogonaltrajectoriesofeach
otherifgivenanycurveC inAandanycurveDinBthecurvesC andDareorthogonal. . For
example,thefamilyofhorizontallinesintheplaneisorthogonaltothefamilyofverticallinesin
theplane.
18. Show w that x
2
y
2
= 5 is orthogonal l to o 4x
2
+9y
2
= 72. (Hint: You u need d to nd d the
intersectionpoints of the two curves andthenshowthat theproductofthederivatives at
eachintersectionpointis 1.)
19. Showthatx
2
+y
2
=r
2
isorthogonaltoy=mx.Concludethatthefamilyofcirclescentered
attheoriginisanorthogonaltrajectoryofthefamilyoflinesthatpassthroughtheorigin.
Notethatthere is atechnicalissuewhenm=0. . The e circles failtobedierentiable
whentheycrossthex-axis. However,thecirclesareorthogonaltothex-axis. Explainwhy.
Likewise,theverticallinethroughtheoriginrequiresaseparateargument.
20. Fork6=0andc6=0showthaty
2
x
2
=kisorthogonaltoyx=c.Inthecasewherekand
carebothzero,thecurvesintersectat theorigin. . Arethecurves s y
2
x
2
=0andyx=0
orthogonaltoeachother?
21. Supposethatm6=0. . Showthatthefamilyofcurvesfy=mx+bjb2Rgisorthogonalto
thefamilyofcurvesfy= (x=m)+cjc2Rg.
The trigonometric functions frequently arise in problems, , and often n it t is s necessary to
invertthefunctions,forexample,tondananglewithaspeciedsine. Ofcourse,there
aremanyangleswiththesamesine,sothesinefunctiondoesn’tactuallyhaveaninverse
thatreliably\undoes"thesinefunction. Ifyouknowthatsinx=0:5,youcan’treverse
thistodiscoverx,thatis,youcan’tsolveforx,asthereareinnitelymanyangleswith
sine0:5. Nevertheless,it t isusefultohavesomethinglikeaninversetothesine,however
imperfect. Theusualapproach h istopickoutsomecollectionof anglesthatproduceall
possible values s ofthe sine e exactly y once. If f we \discard" all otherangles, the resulting
functiondoeshaveaproperinverse.
Thesinetakesonallvaluesbetween 1and1exactlyonceontheinterval[ =2;=2].
Ifwetruncatethesine,keepingonlytheinterval[ =2;=2],asshowningure4.9.1,then
thistruncated sine hasaninversefunction. Wecallthisthe e inverse sine orthearcsine,
andwritey=arcsin(x).
Recallthat afunction anditsinverse undo each otherin eitherorder, , forexample,
(
3
p
x)
3
=xand
3
p
x3=x. Thisdoesnotworkwiththesineandthe\inversesine"because
theinversesine istheinverseof thetruncatedsinefunction,notthe realsinefunction.
Itistruethatsin(arcsin(x))=x,thatis,thesineundoesthearcsine. Itisnottruethat
thearcsineundoesthesine,forexample,sin(5=6)=1=2andarcsin(1=2)==6,sodoing
rstthesinethenthearcsinedoesnotgetusbackwherewestarted.Thisisbecause5=6
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4.9 InverseTrigonometric c Functions
93
1
1
=2
3=2
2
=2
3=2
2
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1
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=2
=2
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=2
=2
1
1
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Figure4.9.1
Thesine,thetruncatedsine,theinversesine.
isnotinthedomainofthetruncatedsine. Ifwestartwithananglebetween =2and
=2thenthearcsinedoesreversethesine:sin(=6)=1=2andarcsin(1=2)==6.
Whatisthederivativeofthearcsine? Sincethisisaninversefunction,wecandiscover
thederivativebyusingimplicitdierentiation.Supposey=arcsin(x).Then
sin(y)=sin(arcsin(x))=x:
Nowtakingthederivativeofbothsides,weget
y
0
cosy=1
y
0
=
1
cosy
Asweexpectwhenusingimplicitdierentiation,y appearsontherighthandside here.
Wewouldcertainlyprefertohavey
0
writtenintermsofx,andasinthecaseoflnxwe
can actuallydo that here. Since e sin
2
y+cos
2
y =1, cos
2
y =1 sin
2
y = 1 x
2
. So
cosy=
p
1 x,butwhichisit|plusorminus?Itcouldingeneralbeeither,butthis
isn’t\ingeneral": sincey=arcsin(x)weknowthat =2y=2,andthecosineof
anangleinthisintervalisalwayspositive. Thuscosy=
p
1 x2 and
d
dx
arcsin(x)=
1
p
1 x2
:
Notethatthisagreeswithgure4.9.1: thegraphofthearcsinehaspositiveslopeevery-
where.
Wecandosomethingsimilarforthecosine. Aswiththesine,wemustrsttruncate
the cosine e so o that it can be inverted, , as s shown n in n gure 4.9.2. Then n we use e implicit
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94
Chapter 4 4 TranscendentalFunctions
1
1
=2
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1
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Figure4.9.2
Thetruncatedcosine,theinversecosine.
dierentiationtondthat
d
dx
arccos(x)=
1
p
1 x2
:
Notethatthetruncatedcosineusesadierentintervalthanthetruncatedsine,sothatif
y=arccos(x)weknowthat0y. Thecomputationofthederivativeofthearccosine
isleftasanexercise.
Finallywelookat thetangent;the othertrigonometricfunctionsalsohave \partial
inverses"but the sine, , cosine e andtangent are enough formost purposes. The e tangent,
truncated tangent and d inverse tangent t are e shown in n gure 4.9.3; the e derivative e of f the
arctangentisleftasanexercise.
=2
=2
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=2
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.
.
.
=2
=2
..............................................
......................................
......................
.............
..........
........
.....
.....
....
....
...
..
...
...
..
..
..
..
..
..
..
..
..
..
.
..
..
..
..
..
..
.
..
..
..
..
..
..
..
..
..
..
...
..
...
...
...
....
.....
.....
.......
.........
............
...................
...............................
............................................................
Figure 4.9.3
Thetangent,thetruncatedtangent,theinversetangent.
Exercises 4.9.
1. Showthatthederivativeofarccosxis 
1
p
1 x2
.
2. Showthatthederivativeofarctanxis
1
1+x2
.
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4.10 Limits s revisited
95
3. Theinverseofcot t isusuallydenedsothattherangeofarccotis(0;). . Sketchthegraph
ofy=arccotx. Intheprocessyouwillmakeitclearwhatthedomainofarccotis. Findthe
derivativeofthearccotangent. )
4. Showthatarccotx+arctanx==2.
5. Findthederivativeofarcsin(x
2
). )
6. Findthederivativeofarctan(e
x
).)
7. Findthederivativeofarccos(sinx
3
))
8. Findthederivativeofln((arcsinx)
2
))
9. Findthederivativeofarccose
x
)
10. Findthederivativeofarcsinx+arccosx)
11. Findthederivativeoflog
5
(arctan(x
x
)))
Wehavedenedandusedtheconceptoflimit,primarilyinourdevelopmentofthederiva-
tive. Recallthat lim
x!a
f(x)=Listrueif,inaprecisesense,f(x)getscloserandcloserto
Lasx getscloserandclosertoa. Whilesomelimitsare e easytosee,otherstake some
ingenuity;inparticular,thelimitsthatdenederivativesarealwaysdicultontheirface,
sincein
lim
x!0
f(x+x) f(x)
x
boththenumeratoranddenominatorapproachzero. Typicallythisdicultycanbe e re-
solvedwhenfisa\nice"functionandwearetryingtocomputeaderivative.Occasionally
suchlimitsareinterestingforotherreasons,andthelimitofafractioninwhichbothnu-
meratoranddenominatorapproachzerocan be dicult toanalyze. Nowthat t we have
thederivativeavailable,thereisanothertechniquethatcansometimesbehelpfulinsuch
circumstances.
Before weintroducethe technique,wewillalsoexpand ourconcept oflimit,intwo
ways.Whenthelimitoff(x)asxapproachesadoesnotexist,itmaybeusefultonotein
whatwayitdoesnotexist. Wehavealreadytalkedaboutonesuchcase:one-sidedlimits.
Anothercaseiswhen\f goestoinnity". . Wealsowilloccasionallywanttoknowwhat
happenstof whenx\goestoinnity".
EXAMPLE 4.10.1
Whathappensto 1=x asx goesto0? Fromtheright,1=x x gets
biggerandbigger,orgoestoinnity.Fromtheleftitgoestonegativeinnity.
EXAMPLE 4.10.2
What happenstothefunctioncos(1=x) asx goestoinnity? ? It
seemsclearthatasxgetslargerandlarger,1=xgetscloserandclosertozero,socos(1=x)
shouldbegettingcloserandclosertocos(0)=1.
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96
Chapter 4 4 Transcendental l Functions
As with h ordinary y limits, these e concepts s can n be made e precise. Roughly, , we e want
lim
x!a
f(x)=1tomeanthatwecanmakef(x)arbitrarilylargebymakingxcloseenough
to a, , and lim
x!1
f(x) =L L should mean we can make f(x) as s close e aswe want to L by
makingxlargeenough. Comparethisdenitiontothedenitionoflimit t insection2.3,
denition2.3.2.
DEFINITION4.10.3
Iffisafunction,wesaythat lim
x!a
f(x)=1ifforeveryN>0
there isa >0 such thatwheneverjx aj<,f(x) ) >N. We e canextendthisin the
obviouswaystodene lim
x!a
f(x)= 1, lim
x!a
f(x)=1,and lim
x!a+
f(x)=1.
DEFINITION4.10.4 Limitatinnity
Iffisafunction,wesaythat lim
x!1
f(x)=L
ifforevery>0there isan N >0 sothat wheneverx>N,jf(x) Lj<. We e may
similarly dene lim
x! 1
f(x) = = L, and using g the idea of the previous s denition, we e may
dene lim
x!1
f(x)=1.
Weincludethesedenitionsforcompleteness,butwewillnotexplorethemindetail.
Suceittosaythatsuchlimitsbehaveinmuchthesamewaythatordinarylimitsdo;in
particulartherearesomeanalogsoftheorem2.3.6.
Nowconsiderthislimit:
lim
x!
x
2
2
sinx
:
As x approaches , , both the e numerator and denominator approach h zero, , so o it is not
obviouswhat,ifanything,thequotientapproaches.Wecanoftencomputesuchlimitsby
applicationofthefollowingtheorem.
THEOREM 4.10.5
L’H^opital’s Rule
For\sucientlynice"functionsf(x) and
g(x),if lim
x!a
f(x)=0= lim
x!a
g(x)orboth lim
x!a
f(x)=1andlim
x!a
g(x)=1,andif
lim
x!a
f
0
(x)
g0(x)
exists,then lim
x!a
f(x)
g(x)
=lim
x!a
f
0
(x)
g0(x)
.Thisremainstrueif\x!a"isreplacedby
\x!1"or\x! 1".
Thistheoremissomewhatdiculttoprove,inpartbecauseitincorporatessomany
dierentpossibilities,sowewillnotproveithere. Wealsowillnotneedtoworryabout
theprecisedenitionof\sucientlynice",asthefunctionsweencounterwillbesuitable.
EXAMPLE4.10.6
Compute lim
x!
x
2
2
sinx
intwoways.
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4.10 Limits s revisited
97
FirstweuseL’H^opital’sRule: Sincethenumeratoranddenominatorbothapproach
zero,
lim
x!
x
2
2
sinx
=lim
x!
2x
cosx
;
providedthe latterexists. But t infact thisisan easylimit,sincethe denominatornow
approaches 1,so
lim
x!
x
2
2
sinx
=
2
1
= 2:
Wedon’treallyneedL’H^opital’sRuletodothislimit. Rewriteitas
lim
x!
(x+)
x 
sinx
andnotethat
lim
x!
x 
sinx
=lim
x!
x 
sin(x )
=lim
x!0
x
sinx
sincex approacheszeroasxapproaches. . Now
lim
x!
(x+)
x 
sinx
= lim
x!
(x+)lim
x!0
x
sinx
=2( 1)= 2
asbefore.
EXAMPLE4.10.7
Compute lim
x!1
2x
2
3x+7
x2+47x+1
intwoways.
Asxgoestoinnityboththenumeratoranddenominatorgotoinnity,sowemay
applyL’H^opital’sRule:
lim
x!1
2x
2
3x+7
x2+47x+1
= lim
x!1
4x 3
2x+47
:
Inthesecondquotient,itisstillthecasethatthenumeratoranddenominatorbothgoto
innity,soweareallowedtouseL’H^opital’sRuleagain:
lim
x!1
4x 3
2x+47
= lim
x!1
4
2
=2:
Sotheoriginallimitis2aswell.
Again,wedon’treallyneedL’H^opital’sRule,andinfactamoreelementaryapproach
iseasier|wedividethenumeratoranddenominatorbyx
2
:
lim
x!1
2x
2
3x+7
x2+47x+1
= lim
x!1
2x
2
3x+7
x2+47x+1
1
x2
1
x2
= lim
x!1
3
x
+
7
x2
1+
47
x
+
1
x2
:
Nowasxapproachesinnity,allthequotientswithsomepowerofxinthedenominator
approachzero,leaving2inthenumeratorand1inthedenominator,sothelimitagainis
2.
98
Chapter 4 4 TranscendentalFunctions
EXAMPLE4.10.8
Compute lim
x!0
secx 1
sinx
.
Boththenumeratoranddenominatorapproachzero,soapplyingL’H^opital’sRule:
lim
x!0
secx 1
sinx
=lim
x!0
secxtanx
cosx
=
10
1
=0:
EXAMPLE4.10.9
Compute lim
x!0+
xlnx.
Thisdoesn’tappeartobesuitableforL’H^opital’sRule,butitalsoisnot\obvious".
Asxapproacheszero,lnxgoesto 1,sotheproductlookslike(somethingverysmall)
(somethingverylargeandnegative).Butthiscouldbeanything:itdependsonhowsmall
andhowlarge. Forexample,consider(x
2
)(1=x),(x)(1=x),and(x)(1=x
2
).Asxapproaches
zero, eachof these e is(somethingverysmall)(somethingverylarge), , yet t the limits s are
respectivelyzero,1,and1.
WecaninfactturnthisintoaL’H^opital’sRuleproblem:
xlnx=
lnx
1=x
=
lnx
1
:
Nowasxapproacheszero,both thenumeratoranddenominatorapproachinnity(one
1andone+1,butonlythesizeisimportant).UsingL’H^opital’sRule:
lim
x!0+
lnx
1
= lim
x!0+
1=x
2
= lim
x!0+
1
x
( x
2
)= lim
x!0+
x=0:
Onewaytointerpretthisisthatsince lim
x!0+
xlnx=0,thexapproacheszeromuchfaster
thanthelnxapproaches 1.
Exercises 4.10.
Computethelimits.
1. lim
x!0
cosx 1
sinx
)
2. lim
x!1
e
x
x3
)
3. lim
x!1
p
x+x 
p
x x)
4. lim
x!1
lnx
x
)
5. lim
x!1
lnx
p
x
)
6. lim
x!1
e
x
+e
x
ex e x
)
7. lim
x!0
p
9+x 3
x
)
8. lim
t!1
+
(1=t) 1
t 2t+1
)
9. lim
x!2
p
x+2
4 x2
)
10. lim
t!1
t+5 2=t 1=t
3
3t+12 1=t2
)
11. lim
y!1
p
y+1+
p
y 1
y
)
12. lim
x!1
p
x 1
3
p
x 1
)
4.10 Limits s revisited
99
13. lim
x!0
(1 x)
1=4
1
x
)
14. lim
t!0
t+
1
t
((4 t)
3=2
8))
15. lim
t!0
+
1
t
+
1
p
t
(
p
t+1 1))
16. lim
x!0
x
2
p
2x+1 1
)
17. lim
u!1
(u 1)
3
(1=u) u+3u 3
)
18. lim
x!0
2+(1=x)
3 (2=x)
)
19. lim
x!0+
1+5=
p
x
2+1=
p
x
)
20. lim
x!0+
3+x
1=2
+x
1
2+4x 1=2
)
21. lim
x!1
x+x
1=2
+x
1=3
x2=3 +x1=4
)
22. lim
t!1
q
t
t+1
q
4t+1
t+2
)
23. lim
t!1
t
t 1
q
t
t 1
)
24.
lim
x! 1
x+x
1
1+
p
1 x
)
25. lim
x!=2
cosx
(=2) x
)
26. lim
x!0
e
x
1
x
)
27. lim
x!0
x
2
e x 1
)
28. lim
x!1
lnx
x 1
)
29. lim
x!0
ln(x
2
+1)
x
)
30. lim
x!1
xlnx
x 1
)
31. lim
x!0
sin(2x)
ln(x+1)
)
32. lim
x!1
x
1=4
1
x
)
33. lim
x!1+
p
x
x 1
)
34. lim
x!1
p
x 1
x 1
)
35. lim
x!1
x
1
+x
1=2
x+x 1=2
)
36. lim
x!1
x+x
2
2x+x 2
)
37. lim
x!1
5+x
1
1+2x 1
)
38. lim
x!1
4x
p
2x+1
)
39. lim
x!0
3x
2
+x+2
x 4
)
40. lim
x!0
p
x+1 1
p
x+4 2
)
41. lim
x!0
p
x+1 1
p
x+2 2
)
42. lim
x!0+
p
x+1+1
p
x+1 1
)
43. lim
x!0
p
x+1 1
p
x+1 1
)
44. lim
x!1
(x+5)
1
2x
+
1
x+2
)
45. lim
x!0+
(x+5)
1
2x
+
1
x+2
)
46. lim
x!1
(x+5)
1
2x
+
1
x+2
)
47. lim
x!2
x
3
6x 2
x+4
)
48. lim
x!2
x
3
6x 2
x 4x
)
49. lim
x!1+
x
3
+4x+8
2x 2
)
100
Chapter 4 4 Transcendental l Functions
50. Thefunctionf(x)=
x
p
x+1
has twohorizontalasymptotes. . Findthemandgivearough
sketchoff withitshorizontalasymptotes. )
Thehyperbolicfunctionsappearwithsomefrequencyinapplications,andarequitesimilar
inmanyrespectstothetrigonometricfunctions. Thisisabitsurprisinggivenourinitial
denitions.
DEFINITION4.11.1
Thehyperbolic cosineisthefunction
coshx=
e
x
+e
x
2
;
andthehyperbolic sineisthefunction
sinhx=
e
x
e
x
2
:
Noticethatcoshiseven(thatis,cosh( x)=cosh(x))whilesinhisodd(sinh( x)=
sinh(x)),andcoshx+sinhx=e
x
. Also,forallx,coshx>0,whilesinhx=0ifand
onlyife
x
e
x
=0,whichistruepreciselywhenx=0.
LEMMA4.11.2
Therangeofcoshxis[1;1).
Proof. Lety=coshx.Wesolveforx:
y=
e+e x
2
2y=e
x
+e
x
2ye
x
=e
2x
+1
0=e
2x
2ye
x
+1
e
x
=
2y
p
4y2 4
2
e
x
=y
p
y2 1
Fromthelastequation,weseey
2
1,andsincey0,itfollowsthaty1.
Nowsupposey1,soy
p
y2 1>0. . Thenx=ln(y
p
y2 1)isarealnumber,
andy=coshx,soyisintherangeofcosh(x).
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