﻿
4.9 ImplicitDiﬀerentiation
87
factoroutthey
,justasinthepreviousexample. Ifyouevergetanythingmorediﬃcult
youhavemadeamistakeandshouldﬁxitbeforetryingtocontinue.
Itissometimesthecasethatasituationleadsnaturallytoanequationthatdeﬁnesa
functionimplicitly.
EXAMPLE4.18
Considerallthepoints(x,y)thathavethepropertythatthedistance
from(x,y)to(x
1
,y
1
)plusthedistancefrom(x,y)to(x
2
,y
2
)is2a(aissomeconstant).
Thesepointsformanellipse,whichlikeacircleisnot afunctionbut canviewedastwo
functionspastedtogether.Becauseweknowhowtowritedownthedistancebetweentwo
points,wecanwritedownanimplicitequationfortheellipse:
p
(x−x
1
)2+(y−y
1
)2+
p
(x−x
2
)2+(y−y
2
)=2a.
Thenwecanuseimplicitdiﬀerentiationtoﬁndtheslopeoftheellipseatanypoint.
EXAMPLE 4.19
We have alreadyjustiﬁed thepower rulebyusing the exponential
function,butwecouldalsodoitforrationalexponentsbyusingimplicitdiﬀerentiation.
Supposethaty=x
m/n
,wheremandnarepositiveintegers.Wecanwritethisimplicitly
asy
n
=x
m
,thenbecausewejustiﬁedthepowerruleforintegers,wecantakethederivative
ofeachside:
ny
n−1
y
=mx
m−1
y
=
m
n
x
m−1
yn−1
y
=
m
n
xm−1
(xm/n)n−1
y
=
m
n
x
m−1−(m/n)(n−1)
y
=
m
n
x
m−1−m+(m/n)
y
=
m
n
x
(m/n)−1
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88
Chapter4 TranscendentalFunctions
Exercises
Inexercises1–8,ﬁndaformulaforthederivativey
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.
x+
y=9
6. tan(x/y)=x+y
7. sin(x+y)=xy
8.
1
x
+
1
y
=7
9. Ahyperbolapassingthrough(8,6)consistsofallpointswhosedistancefromtheoriginisa
constant morethanitsdistancefromthepoint(5,2). . Findtheslopeofthetangentlineto
thehyperbolaat(8,6).
10. Computey
fortheellipseofexample4.18.
11. Thegraphoftheequationx
2
−xy+y
2
=9isanellipse.Findthelinestangenttothiscurve
atthetwopointswhereitintersectsthex-axis.Showthattheselinesareparallel.
12. Repeatthepreviousproblemforthepointsatwhichtheellipseintersectsthey-axis.
13. Findthepointsontheellipsefromtheprevioustwoproblemswheretheslopeishorizontal
andwhereitisvertical.
14. Findanequationforthetangentlinetox
4
=y
2
+x
2
at(2,
12).(Thiscurveisthekampyle
ofEudoxus.)
15. Findanequationforthetangentlinetox
2/3
+y
2/3
=a
2/3
atapoint(x
1
,y
1
)onthecurve,
withx
1
=0andy
1
=0.(Thiscurveisanastroid.)
16. Findanequationforthetangentlineto(x
2
+y
2
)
2
=x
2
−y
2
atapoint(x
1
,y
1
)onthecurve,
withx
1
=0,−1,1.(Thiscurveisalemniscate.)
Deﬁnition. Two o curves s are e orthogonal if at each point t of f intersection, the angle between
their tangent linesis π/2. . Two o familiesofcurves, A andB, areorthogonal trajectories of
eachotherifgivenanycurveCinAandanycurveDinBthecurvesC andDareorthogonal.
Forexample,thefamilyofhorizontallinesintheplaneisorthogonaltothefamilyofverticallines
intheplane.
17. Show w that x
2
−y
2
= 5 is orthogonal l to 4x
2
+9y
2
= 72. . (Hint: : You u need to o ﬁnd d the
intersectionpoints ofthe twocurvesandthenshow that the product of the derivatives at
eachintersectionpointis−1.)
18. Showthatx
2
+y
2
=r
2
isorthogonaltoy=mx.Concludethatthefamilyofcirclescentered
attheoriginisanorthogonaltrajectoryofthefamilyoflinesthatpassthroughtheorigin.
Notethatthereisatechnicalissuewhenm=0. Thecircles s failtobediﬀerentiablewhen
theycrossthex-axis.However,thecirclesareorthogonaltothex-axis.Explainwhy.
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4.10 Inverse e TrigonometricFunctions
89
19. For r k,c= 0 show that y
2
−x
2
= k is orthogonalto yx = c. . Inthe e case where k k and d c
are both zero,the curves intersectatthe origin. . Are e thecurves y
2
−x
2
=0 andyx= 0
orthogonaltoeachother?
20. Supposethatm=0. Showthatthefamilyofcurves{y=mx+b|b∈R}isorthogonalto
thefamilyofcurves{y=−(x/m)+c|c∈R}.
The trigonometric c functions frequently arise in problems, , and d often it t is necessary to
invertthefunctions,forexample,toﬁndananglewithaspeciﬁedsine. Ofcourse,there
aremanyangleswiththesamesine,sothesinefunctiondoesn’tactuallyhaveaninverse
thatreliably“undoes”thesinefunction. Ifyouknowthatsinx=0.5,youcan’treverse
thistodiscoverx,thatis,youcan’tsolveforx,asthereareinﬁnitelymanyangleswith
sine0.5. Nevertheless,itisusefultohavesomethinglikeaninversetothesine,however
imperfect. Theusualapproachistopickoutsomecollectionof f anglesthatproduce all
possible valuesof the sine exactlyonce. . If f we “discard” all other angles, the resulting
functiondoeshaveaproperinverse.
Thesinetakesonallvaluesbetween−1and1exactlyonceontheinterval[−π/2,π/2].
Ifwetruncatethesine,keepingonlytheinterval[−π/2,π/2],asshowninﬁgure4.5,then
thistruncatedsine hasaninversefunction. . Wecallthisthe e inversesineor thearcsine,
andwritey=arcsin(x).
−1
1
π/2
π
3π/2
−π/2
−π
−3π/2
−2π
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Figure4.5
Thesine,thetruncatedsine,theinversesine.
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90
Chapter4 TranscendentalFunctions
Recallthata function and itsinverseundoeachother ineitherorder,for example,
(
3
x)
3
=xand
3
x3=x. Thisdoesnotworkwiththesineandthe“inversesine”because
theinversesineistheinverseof thetruncatedsine function,not therealsine function.
Itistruethatsin(arcsin(x))=x,thatis,thesineundoesthearcsine. Itisnottruethat
thearcsineundoesthesine,forexample,sin(5π/6)=1/2andarcsin(1/2)=π/6,sodoing
ﬁrstthesinethenthearcsinedoesnotgetusbackwherewestarted.Thisisbecause5π/6
isnotinthedomainofthetruncatedsine. Ifwestart t withananglebetween−π/2and
π/2thenthearcsinedoesreversethesine:sin(π/6)=1/2andarcsin(1/2)=π/6.
Whatisthederivativeofthearcsine?Sincethisisaninversefunction,wecandiscover
thederivativebyusingimplicitdiﬀerentiation.Supposey=arcsin(x). Then
sin(y)=sin(arcsin(x))=x.
Nowtakingthederivativeofbothsides,weget
y
cosy=1
y
=
1
cosy
Asweexpect whenusingimplicit diﬀerentiation,yappearsontheright handsidehere.
Wewouldcertainlyprefer tohavey writtenintermsofx,andasinthecaseoflnxwe
can actually do that here. . Since e sin
2
y+cos
2
y =1, , cos
2
y = 1−sin
2
y =1−x
2
. So
cosy=±
1−x,butwhichisit—plusorminus? Itcouldingeneralbeeither,butthis
isn’t“ingeneral”: sincey=arcsin(x)weknowthat t −π/2≤y≤π/2,andthecosineof
anangleinthisintervalisalwayspositive. Thuscosy=
1−xand
d
dx
arcsin(x)=
1
1−x2
.
Notethatthisagreeswithﬁgure4.5:thegraphofthearcsinehaspositiveslopeeverywhere.
Wecandosomethingsimilarforthecosine.Aswiththesine,wemustﬁrsttruncatethe
cosinesothatitcanbeinverted,asshowninﬁgure4.6.Thenweuseimplicitdiﬀerentiation
toﬁndthat
d
dx
arccos(x)=
−1
1−x2
.
Notethatthetruncatedcosineusesadiﬀerentintervalthanthetruncatedsine,sothatif
y=arccos(x)weknowthat0≤y≤π. Thecomputationofthederivativeofthearccosine
isleftasanexercise.
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4.10 InverseTrigonometricFunctions
91
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π
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Figure 4.6
Thetruncatedcosine,theinversecosine.
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−π/2
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Figure 4.7
Thetangent,thetruncatedtangent,theinversetangent.
Finallywe lookat thetangent;the other trigonometric functionsalso have “partial
inverses” but the sine, , cosine e and tangent are enoughfor most purposes. . The e tangent,
truncatedtangentandinversetangentareshowninﬁgure4.7;thederivativeofthearct-
angentisleftasanexercise.
Exercises
1. Showthatthederivativeofarccosxis−
1
1−x2
.
2. Showthatthederivativeofarctanxis
1
1+x2
.
3. Findthederivativeofarccotx,theinversecotangent.
4. Findthederivativeofarcsin(x
2
).
5. Findthederivativeofarctan(e
x
).
6. Theinverseofcot t isusuallydeﬁnedsothattherangeofarccotis(0,π). . Sketchthegraph
ofy=arccotx. Intheprocessyouwillmakeitclearwhatthedomainofarccotis.
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92
Chapter4 TranscendentalFunctions
7. Usetheprecedingexercisetoshowthatarccotx+arctanx=π/2.
8. Findthederivativeofarccos(sinx
3
)
9. Findthederivativeofln((arcsinx)
2
)
10. Findthederivativeofarccose
x
11. Findthederivativeofarcsinx+arccosx
12. Findthederivativeoflog
5
(arctan(x
x
))
5
CurveSketching
Whetherweareinterestedinafunctionasapurelymathematicalobjectorinconnection
with some application n to o the e realworld, , it t isoften useful l to o know what t the graph h of
thefunctionlookslike. Wecanobtaina a goodpictureofthegraphusingcertaincrucial
informationprovidedbyderivativesofthefunctionandcertainlimits.
A local maximum m point t ona function isa point (x,y) onthe graph ofthe function
whose y coordinateislarger thanallother y coordinateson thegraph at points“close
to”(x,y). Moreprecisely,(x,f(x))isalocalmaximumifthereisaninterval(a,b)with
a<x <b and f(x) ≥f(z) foreveryz z in (a,b). . Similarly,(x,y) ) isalocal minimum
pointifithaslocallythesmallestycoordinate. Againbeingmoreprecise: (x,f(x))isa
localminimumifthereisaninterval(a,b)witha<x<bandf(x)≤f(z)foreveryzin
(a,b).Alocalextremumiseitheralocalminimumoralocalmaximum.
Localmaximumandminimumpointsarequitedistinctiveonthegraphofafunction,
andarethereforeusefulinunderstandingtheshapeofthegraph.Inmanyappliedproblems
wewanttoﬁndthelargestorsmallestvaluethatafunctionachieves(forexample,wemight
wanttoﬁndtheminimumcostatwhichsometaskcanbeperformed)andsoidentifying
maximumandminimumpointswillbeusefulforappliedproblemsaswell. Someexamples
oflocalmaximumandminimumpointsareshowninﬁgure5.1.
93
94
Chapter5 CurveSketching
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A
A
B
Figure 5.1
Somelocalmaximumpoints(A)andminimumpoints(B).
If (x,f(x)) isapoint wheref(x) reachesa localmaximumorminimum, and ifthe
derivativeoff existsatx,thenthegraphhasatangentlineandthetangentlinemustbe
horizontal. Thisisimportantenoughtostateasatheorem:
THEOREM5.1
Fermat’sTheorem
Iff(x)hasalocalextremumatx=aandf
isdiﬀerentiableata,thenf
(a)=0.
Thus,the only y pointsat t whicha function can have a localmaximumor minimum
arepointsatwhichthederivativeiszero,asinthelefthandgraphinﬁgure 5.1,or the
derivativeisundeﬁned,asintherighthandgraph.Anyvalueofxforwhichf
(x)iszeroor
undeﬁnediscalledacriticalvalueforf.Whenlookingforlocalmaximumandminimum
points,youarelikelytomaketwosortsofmistakes: Youmayforgetthatamaximumor
minimumcanoccur wherethederivativedoesnotexist,andsoforgetto checkwhether
thederivativeexistseverywhere.Youmightalsoassumethatanyplacethatthederivative
iszeroisalocalmaximumorminimumpoint,butthisisnottrue.Aportionofthegraph
off(x)=x
3
isshowninﬁgure5.2. Thederivativeoff isf
(x)=3x
2
,andf
(0)=0,but
thereisneitheramaximumnorminimumat(0,0).
Sincethederivativeiszeroorundeﬁnedatbothlocalmaximumandlocalminimum
points,weneedawaytodeterminewhich,ifeither,actuallyoccurs.Themostelementary
approach,butonethatisoftentediousordiﬃcult,istotestdirectlywhethertheycoor-
dinates“near”thepotentialmaximumorminimumareaboveorbelowthey coordinate
atthepointofinterest. Ofcourse,therearetoomanypoints“near”thepointtotest,but
alittlethoughtshowsweneedonlytesttwoprovidedweknowthatf iscontinuous(recall
thatthismeansthatthegraphoffhasnojumpsorgaps).
Suppose, for example, , that t we have identiﬁed d three points s at which f is zero or
nonexistent: (x
1
,y
1
),(x
2
,y
2
),(x
3
,y
3
),andx
1
<x
2
<x
3
(seeﬁgure 5.3). . Supposethat
5.1 MaximaandMinima
95
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Figure 5.2
Nomaximumorminimumeventhoughthederivativeiszero.
wecomputethevalueoff(a)forx
1
<a<x
2
,andthatf(a)<f(x
2
). Whatcanwesay
about the graphbetweenaandx
2
? Couldtherebea a point (b,f(b)),a<b<x
2
with
f(b)>f(x
2
)? No: ifthere e were,thegraphwouldgo upfrom(a,f(a)) to(b,f(b))then
downto(x
2
,f(x
2
))andsomewhereinbetweenwouldhavealocalmaximumpoint.Butat
thatlocalmaximumpointthederivativeoffwouldbezeroornonexistent,yetwealready
knowthatthederivativeiszeroornonexistentonlyatx
1
,x
2
,andx
3
.Theupshotisthat
onecomputationtellsusthat(x
2
,f(x
2
))hasthelargestycoordinateofanypointonthe
graphnearx
2
andtotheleftofx
2
. Wecanperformthesametestontheright. Ifweﬁnd
thatonbothsidesofx
2
thevaluesaresmaller,thenthere must bealocalmaximumat
(x
2
,f(x
2
));ifweﬁndthatonbothsidesofx
2
thevaluesarelarger,thentheremustbea
localminimumat(x
2
,f(x
2
));ifweﬁndoneofeach,thenthereisneitheralocalmaximum
orminimumatx
2
.
x
1
a
b
x
2
x
3
Figure5.3
Testingforamaximumorminimum.
Itisnot alwayseasytocomputethe value ofafunctionata particularpoint. . The
taskismadeeasierbytheavailabilityofcalculatorsandcomputers,buttheyhavetheir
owndrawbacks—theydonotalwaysallowustodistinguishbetweenvaluesthatarevery
96
Chapter5 CurveSketching
closetogether. Nevertheless,becausethismethodisconceptuallysimpleandsometimes
easytoperform,youshouldalwaysconsiderit.
EXAMPLE5.2
Findalllocalmaximumandminimumpointsforthefunctionf(x)=
x
3
−x. The e derivative e is f
(x) = 3x
2
−1. Thisis s deﬁned d everywhere and is s zero o at
x=±
3/3. Lookingﬁrstatx=
3/3,weseethatf(
3/3)=−2
3/9. Nowwetest
two pointson either side of x x =
3/3, making g sure that t neither isfarther r away y than
the nearest criticalvalue; ; since
3 < < 3,
3/3 <1 and we canuse x =0 and x =1.
Sincef(0)=0>−2
3/9andf(1) =0>−2
3/9,theremust bea localminimumat
x=
3/3. Forx=−
3/3,weseethatf(
3/3)=2
3/9. Thistimewecanusex=0
and x = −1, and we ﬁnd that f(−1) ) = f(0) = 0 0 < < 2
3/9, so o there e must t be a local
maximumatx=−
3/3.
Ofcoursethisexample ismade verysimple byour choiceof pointstotest,namely
x=−1,0,1. Wecouldhaveusedothervalues,say−5/4,1/3,and3/4,butthiswould
havemadethecalculationsconsiderablymoretedious.
EXAMPLE5.3
Findalllocalmaximumandminimumpointsforf(x)=sinx+cosx.
The derivative is s f
(x) = cosx−sinx. . This s is always deﬁned d and d is s zero o whenever
cosx=sinx. Recallingthatthecosxandsinxarethexandycoordinatesofpointsona
unitcircle,weseethatcosx=sinxwhenxisπ/4,π/4±π,π/4±2π,π/4±3π,etc.Since
bothsineandcosinehaveaperiodof2π,weneedonlydeterminethestatusofx=π/4
andx=5π/4. We e canuse 0andπ/2 totestthe criticalvaluex =π/4. . We e ﬁnd that
f(π/4)=
2,f(0)=1<
2andf(π/2)=1,sothereisalocalmaximumwhenx=π/4
andalsowhenx=π/4±2π,π/4±4π,etc.
Weuseπand2πtotestthecriticalvaluex=5π/4.Therelevantvaluesaref(5π/4)=
2,f(π)=−1>−
2,f(2π)=1>−
2,sothere isalocalminimumat x=5π/4,
5π/4±2π,5π/4±4π,etc.