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3.5 The e ChainRule
57
EXAMPLE3.8
Formthetwopossiblecompositionsoff(x)=
xandg(x)=625−x
2
.
First,f(g(x))=
625−x2;secondg(f(x))=625−(
x)
2
=625−x.
Supposewewantthederivativeoff(g(x)).Again,let’ssetupthederivativeandplay
somealgebraictricks:
d
dx
f(g(x))= lim
∆x→0
f(g(x+∆x))−f(g(x))
∆x
= lim
∆x→0
f(g(x+∆x))−f(g(x))
g(x+∆x))−g(x)
g(x+∆x))−g(x)
∆x
Nowweseeimmediatelythatthesecondfractionturnsintog(x)whenwetakethelimit.
denominator,g(x+∆x))−g(x),isa change in thevalue of g,so let’sabbreviate it as
∆g=g(x+∆x))−g(x),whichalsomeansg(x+∆x)=g(x)+∆g. Thisgivesus
lim
∆x→0
f(g(x)+∆g)−f(g(x))
∆g
.
As∆xgoesto0,itisalsotruethat∆ggoesto0,becauseg(x+∆x)goestog(x).Sowe
canrewritethislimitas
lim
∆g→0
f(g(x)+∆g)−f(g(x))
∆g
.
(g(x)),thatis,thefunctionf
(x)with
xreplacedbyg(x). Ifthisallwithstandsscrutiny,wethenget
d
dx
f(g(x))=f
(g(x))g
(x).
Unfortunately,thereisasmallﬂawintheargument.Recallthatwhatwemeanbylim
∆x→0
involveswhathappenswhen∆xiscloseto0butnotequalto0. Thequaliﬁcationisvery
important,sincewemustbeabletodivideby∆x.Butwhen∆xiscloseto0butnotequal
to0,∆g=g(x+∆x))−g(x)iscloseto0butpossiblyequalto0. Thismeansitdoesn’t
reallymakesenseto divide by∆g. . Fortunately,itispossible e torecasttheargument to
avoidthisdiﬃculty, but t it isa bit tricky;wewillnot include the details,which canbe
foundinmanycalculusbooks.
ThechainrulehasaparticularlysimpleexpressionifweusetheLeibniznotationfor
thederivative. Thequantityf
(g(x))isthederivativeoff withxreplacedbyg;thiscan
bewrittendf/dg. Asusual,g
(x)=dg/dx.Thenthechainrulebecomes
df
dx
=
df
dg
dg
dx
.
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58
Chapter3 RulesFor r FindingDerivatives
Thislooksliketrivialarithmetic,butitisnot: dg/dxisnotafraction,thatis,notliteral
division,butasinglesymbolthatmeansg
(x).Nevertheless,itturnsoutthatwhatlooks
liketrivialarithmetic,andisthereforeeasytoremember,isreallytrue.
It willtakeabit ofpractice tomakethe useofthechainrulecome naturally—it is
morecomplicatedthantheearlierdiﬀerentiationruleswehaveseen.
EXAMPLE 3.9
Compute the e derivative e of
625−x2. We e already know that the
625−x2,computed directlyfromthe limit. . Inthecontextofthe e chain
rule, we have e f(x) ) =
x, g(x) ) = 625−x
2
. We e know that f
(x) = (1/2)x
−1/2
, so
f
(g(x))=(1/2)(625−x
2
)
−1/2
. Notethatthisisatwostepcomputation: : ﬁrstcompute
f
(x),thenreplacexbyg(x).Sinceg
(x)=−2xwehave
f
(g(x))g
(x)=
1
2
625−x2
(−2x)=
−x
625−x2
.
EXAMPLE 3.10
Computethe derivative of1/
625−x2. Thisisa a quotientwitha
constantnumerator,sowecouldusethequotientrule,butitissimplertousethechain
rule. Thefunctionis(625−x
2
)
−1/2
,thecompositionoff(x)=x
−1/2
andg(x)=625−x
2
.
Wecomputef
(x)=(−1/2)x
−3/2
usingthepowerrule,andthen
f
(g(x))g
(x)=
−1
2(625−x2)3/2
(−2x)=
x
(625−x2)3/2
.
Inpractice,ofcourse,youwillneedtousemorethanoneoftheruleswehavedeveloped
tocomputethederivativeofacomplicatedfunction.
EXAMPLE 3.11
Computethederivativeof
f(x)=
x
2
−1
x
x2+1
.
The“last” operationhereisdivision,soto getstartedwe needtouse thequotient rule
ﬁrst. Thisgives
f
(x)=
(x
2
−1)
x
x2+1−(x
2
−1)(x
x2+1)
x2(x2+1)
=
2x
2
x2+1−(x
2
−1)(x
x2+1)
x2(x2+1)
.
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3.5 The e ChainRule
59
Nowwe needto compute thederivative ofx
x2+1. Thisisa a product,sowe use the
productrule:
d
dx
x
p
x2+1=x
d
dx
p
x2+1+
p
x2+1.
Finally,weusethechainrule:
d
dx
p
x2+1=
d
dx
(x
2
+1)
1/2
=
1
2
(x
2
+1)
−1/2
(2x)=
x
x2+1
.
Andputtingitalltogether:
f
(x)=
2x2
x2+1−(x2−1)(x
x2+1)
x2(x2+1)
.
=
2x
2
x2+1−(x
2
−1)
µ
x
x
x2+1
+
x2+1
x2(x2+1)
.
Thiscanbesimpliﬁedofcourse,butwehavedoneallthecalculus,sothatonlyalgebrais
left.
EXAMPLE 3.12
Compute the derivative of
q
1+
p
1+
x. Here e we have a more
complicated chain of f compositions, , so o we e use the chain n rule twice. . At t the outermost
“layer”wehavethefunctiong(x)=1+
p
1+
xpluggedintof(x)=
x,soapplying
thechainruleoncegives
d
dx
r
1+
q
1+
x=
1
2
µ
1+
q
1+
x
−1/2
d
dx
µ
1+
q
1+
x
.
Nowweneedthederivativeof
p
1+
x.Usingthechainruleagain:
d
dx
q
1+
x=
1
2
¡
1+
x
¢
−1/2
1
2
x
−1/2
.
Sotheoriginalderivativeis
d
dx
r
1+
q
1+
x=
1
2
µ
1+
q
1+
x
−1/2
1
2
¡
1+
x
¢
−1/2
1
2
x
−1/2
.
=
1
8
x
p
1+
x
q
1+
p
1+
x
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60
Chapter3 RulesFor r FindingDerivatives
Usingthechainrule,thepowerrule,andtheproductrule,itispossibletoavoidusing
thequotientruleentirely.
EXAMPLE3.13
Computethederivativeoff(x)=
x
3
x2+1
.Writef(x)=x
3
(x
2
+1)
−1
,
then
f
(x)=x
3
d
dx
(x
2
+1)
−1
+3x
2
(x
2
+1)
−1
=x
3
(−1)(x
2
+1)
−2
(2x)+3x
2
(x
2
+1)
−1
=−2x
4
(x
2
+1)
−2
+3x
2
(x
2
+1)
−1
=
−2x
4
(x2+1)2
+
3x
2
x2+1
=
−2x
4
(x2+1)2
+
3x
2
(x
2
+1)
(x2+1)2
=
−2x4+3x4+3x2
(x2+1)2
=
x4+3x2
(x2+1)2
forfewermemorizedformulas.
Exercises
theseinmorethanonewayifpossible.
1. x
4
−3x
3
+(1/2)x
2
+7x−π
2. x
3
−2x
2
+4
x
3. (x
2
+1)
3
4. x
169−x
5. (x
2
−4x+5)
25−x
6.
r−x2,risaconstant
7.
1+x
8.
1
p
5−
x
9. (1+3x)
2
10.
(x
2
+x+1)
(1−x)
11.
25−x2
x
12.
r
169
x
−x
13.
p
x−x−(1/x)
14. 100/(100−x
2
)
3/2
15.
3
x+x
16.
q
(x+1)+
p
1+(x+1)
17. (x+8)
5
18. (4−x)
3
19. (x
2
+5)
3
20. (6−2x
2
)
3
21. (1−4x
3
)
−2
22. 5(x+1−1/x)
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3.5 The e ChainRule
61
23. 4(2x
2
−x+3)
−2
24.
1
1+1/x
25.
−3
4x−2x+1
26. (x
2
+1)(5−2x)/2
27. (3x
2
+1)(2x−4)
3
28.
x+1
x−1
29.
x
2
−1
x+1
30.
(x−1)(x−2)
x−3
31.
2x
−1
−x
−2
3x−1 −4x−2
32. 3(x
2
+1)(2x
2
−1)(2x+3)
33.
1
(2x+1)(x−3)
34. ((2x+1)
−1
+3)
−1
35. (2x+1)
3
(x
2
+1)
2
36. Findanequationforthetangentlinetof(x)=(x−2)
1/3
/(x
3
+4x−1)
2
atx=1.
37. Findanequationforthetangentlinetoy=9x
−2
at(3,1).
38. Findanequationforthetangentlineto(x
2
−4x+5)
25−xat(3,8).
39. Findanequationforthetangentlineto
(x
2
+x+1)
(1−x)
at(2,−7).
40. Findanequationforthetangentlineto
q
(x+1)+
p
1+(x+1)at(1,
p
4+
5).
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4
TranscendentalFunctions
Sofarwehaveusedonlyalgebraicfunctionsasexampleswhenﬁndingderivatives,thatis,
multiplication,division,andraisingtoconstantpowers. Bothintheoryandpracticethere
are other functions,called transcendental,that are veryuseful. . Most t important among
these are e the trigonometric functions, the inverse trigonometric functions, exponential
functions,andlogarithms.
Whenyouﬁrstencounteredthetrigonometricfunctionsitwasprobablyinthecontextof
“triangletrigonometry,”deﬁning,forexample,thesineofanangleasthe“sideopposite
overthehypotenuse.” Whilethiswillstillbeusefulinaninformalway,weneedtousea
moreexpansivedeﬁnitionofthetrigonometricfunctions. Firstanimportantnote: : while
degreemeasureofanglesissometimesconvenientbecauseitissofamiliar,itturnsoutto
beill-suitedtomathematicalcalculation,so(almost)everythingwedowillbeintermsof
63
64
Chapter4 TranscendentalFunctions
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.
x
(cosx,sinx)
y
A
B
Anangle,x,at thecenterofthecircleisassociatedwithanarcofthecircle whichit is
said tosubtend. . Inthe e ﬁgure,thisarc istheportion ofthecirclefrompoint(1,0) to
point A. . The e length of thisarc isthe radian measure ofthe angle x; ; thefact t that the
radian measure isan actualgeometriclengthislargelyresponsible forthe usefulnessof
measureofthefullcircularangle(thatis,ofthe360degreeangle)is2π.
Whileananglewithaparticularmeasurecanappearanywherearoundthecircle,we
need a ﬁxed, conventionallocation so that we canuse the coordinate systemto deﬁne
propertiesofthe angle. . Thestandard d convention istoplacethe startingradiusfor the
angleonthepositivex-axis,andtomeasurepositiveanglescounterclockwisearoundthe
circle. Intheﬁgure,xisthestandardlocationoftheangleπ/6,thatis,thelengthofthe
arcfrom(1,0)toAisπ/6.Theangleyinthepictureis−π/6,becausethedistancefrom
(1,0)toBalongthecircleisalsoπ/6,butinaclockwisedirection.
Nowthefundamentaltrigonometricdeﬁnitionsare: thecosineofxandthesineofx
aretheﬁrstandsecondcoordinatesofthepointA,asindicatedintheﬁgure.Theanglex
showncanbeviewedasanangleofarighttriangle,meaningtheusualtriangledeﬁnitions
ofthesineandcosinealsomakesense. Sincethehypotenuseofthetriangleis1,the“side
oppositeoverhypotenuse”deﬁnitionofthesineisthesecondcoordinateofpointAover
1,whichisjustthesecondcoordinate;inotherwords,bothmethodsgivethesamevalue
forthesine.
Thesimpletriangledeﬁnitionsworkonlyforanglesthatcan“ﬁt”inarighttriangle,
namely,anglesbetween0andπ/2. Thecoordinatedeﬁnitions,ontheotherhand,apply
4.1 TrigonometricFunctions
65
toanyangles,asindicatedinthisﬁgure:
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x
A
(cosx,sinx)
Theanglexsubtendstheheavyarcintheﬁgure,thatis,x=7π/6. Bothcoordinatesof
pointAinthisﬁgurearenegative,sothesineandcosineof7π/6arebothnegative.
Theremainingtrigonometricfunctionscanbemosteasilydeﬁnedintermsofthesine
andcosine,asusual:
tanx=
sinx
cosx
cotx=
cosx
sinx
secx=
1
cosx
cscx=
1
sinx
andtheycanalsobedeﬁnedasthecorrespondingratiosofcoordinates.
Althoughthetrigonometricfunctionsaredeﬁnedintermsoftheunitcircle,theunit
circle diagramisnot what we normallyconsider the graphof a trigonometric function.
(Theunitcircleisthegraphof,well,thecircle.) Wecaneasilygetaqualitativelycorrect
ideaofthegraphsofthetrigonometricfunctionsfromtheunitcirclediagram. Consider
thesinefunction,y=sinx. Asxincreasesfrom0intheunitcirclediagram,thesecond
coordinate of f the e point t A A goes s from0 to o a maximumof 1, , then n back to 0, then n to o a
minimumof −1, then backto 0, and then it t obviously repeatsitself. . So o the graph of
y=sinxmustlooksomethinglikethis:
−1
1
π/2
π
3π/2
−π/2
−π
−3π/2
−2π
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