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3.3 TheProductRule
61
g
0
(x)must actuallyexistforthistomakesense. . Wealsoreplaced lim
x!0
f(x+x)with
f(x)|whyisthisjustied?
Whatwereallyneedtoknowhereisthat lim
x!0
f(x+x)=f(x),orinthelanguage
ofsection2.5,thatf iscontinuousatx. Wealreadyknowthatf
0
(x)exists(orthewhole
approach,writing thederivativeof fg intermsoff
0
and g
0
,doesn’t make sense). This
turnsouttoimplythatf iscontinuousaswell. Here’swhy:
lim
x!0
f(x+x)= lim
x!0
(f(x+x) f(x)+f(x))
= lim
x!0
f(x+x) f(x)
x
x+ lim
x!0
f(x)
=f
0
(x)0+f(x)=f(x)
Tosummarize:theproductrulesaysthat
d
dx
(f(x)g(x))=f(x)g
0
(x)+f
0
(x)g(x):
Returningtotheexamplewestartedwith,letf(x)=(x
2
+1)(x
3
3x).Thenf
0
(x)=
(x
2
+1)(3x
2
3)+(2x)(x
3
3x)=3x
4
3x
2
+3x
2
3+2x
4
6x
2
=5x
4
6x
2
3,
asbefore. Inthiscaseitisprobablysimplertomultiplyf(x)outrst,thencomputethe
derivative;here’sanexampleforwhichwereallyneedtheproductrule.
EXAMPLE 3.3.1
Compute thederivative off(x)=x
2
p
625 x2. Wehavealready
computed
d
dx
p
625 x=
x
p
625 x2
.Now
f
0
(x)=x
2
x
p
625 x2
+2x
p
625 x2=
x
3
+2x(625 x
2
)
p
625 x2
=
3x
3
+1250x
p
625 x2
:
Exercises3.3.
In1{4,ndthederivativesofthefunctionsusingtheproductrule.
1. x
3
(x
3
5x+10))
2. (x
2
+5x 3)(x
5
6x
3
+3x
2
7x+1))
3.
p
x
p
625 x)
4.
p
625 x2
x20
)
5. Usethe e product rule to computethe derivativeoff(x)= (2x 3)
2
. Sketchthefunction.
Findanequationofthetangentlinetothecurveatx=2.Sketchthetangentlineatx=2.
)
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62
Chapter 3 3 Rulesfor r FindingDerivatives
6. Supposethatf,g,andharedierentiablefunctions.Showthat(fgh)
0
(x)=f
0
(x)g(x)h(x)+
f(x)g
0
(x)h(x)+f(x)g(x)h
0
(x).
7. Stateandprovearuletocompute(fghi)
0
(x),similartotheruleinthepreviousproblem.
Productnotation. Supposef
1
;f
2
;:::f
n
arefunctions. Theproductofallthesefunctions
canbewritten
Yn
k=1
f
k
:
Thisissimilartotheuseof
X
todenoteasum.Forexample,
Y5
k=1
f
k
=f
1
f
2
f
3
f
4
f
5
and
Yn
k=1
k=12:::n=n!:
Wesometimesusesomewhatmorecomplicatedconditions;forexample
Yn
k=1;k6=j
f
k
denotestheproductoff
1
throughf
n
exceptforf
j
. Forexample,
Y5
k=1;k6=4
x
k
=xx
2
x
3
x
5
=x
11
:
8. Thegeneralized d product rulesaysthatiff
1
;f
2
;:::;f
n
aredierentiablefunctions atx
then
d
dx
Yn
k=1
f
k
(x)=
Xn
j=1
0
@
f
0
j
(x)
Yn
k=1;k6=j
f
k
(x)
1
A
:
Verifythatthisisthesameasyouranswertothepreviousproblemwhenn=4,andwrite
outwhatthissayswhenn=5.
Whatisthederivativeof(x
2
+1)=(x
3
3x)? Moregenerally,we’dliketohaveaformula
to compute the derivative off(x)=g(x) if we alreadyknowf
0
(x) and g
0
(x). Instead d of
attackingthisproblemhead-on,let’snoticethatwe’vealreadydonepartoftheproblem:
f(x)=g(x)=f(x)(1=g(x)),thatis,thisis\really"aproduct,andwecancomputethe
derivativeifweknowf
0
(x)and(1=g(x))
0
. Soreallythe e onlynewbitofinformationwe
needis(1=g(x))intermsofg0(x). Aswiththeproductrule,let’ssetthisupandseehow
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3.4 TheQuotient t Rule
63
farwecanget:
d
dx
1
g(x)
= lim
x!0
1
g(x+x)
1
g(x)
x
= lim
x!0
g(x) g(x+x)
g(x+x)g(x)
x
= lim
x!0
g(x) g(x+x)
g(x+x)g(x)x
= lim
x!0
g(x+x) g(x)
x
1
g(x+x)g(x)
g0(x)
g(x)2
Nowwecanputthistogetherwiththeproductrule:
d
dx
f(x)
g(x)
=f(x)
g
0
(x)
g(x)2
+f
0
(x)
1
g(x)
=
f(x)g
0
(x)+f
0
(x)g(x)
g(x)2
=
f
0
(x)g(x) f(x)g
0
(x)
g(x)2
:
EXAMPLE3.4.1
Computethederivativeof(x
2
+1)=(x
3
3x).
d
dx
x
2
+1
x3 3x
=
2x(x
3
3x) (x
2
+1)(3x
2
3)
(x3 3x)2
=
x
4
6x
2
+3
(x3 3x)2
:
Itisoftenpossibletocalculatederivativesinmorethanoneway,aswehavealready
seen. Sinceeveryquotientcan n be written asa product,itisalwayspossible tousethe
productruletocomputethederivative,thoughitisnotalwayssimpler.
EXAMPLE3.4.2 Findthederivativeof
p
625 x2=
p
xintwoways:usingthequotient
rule,andusingtheproductrule.
Quotientrule:
d
dx
p
625 x2
p
x
=
p
x( x=
p
625 x2) 
p
625 x21=(2
p
x)
x
:
Notethatwehaveused
p
x=x
1=2
tocomputethederivativeof
p
xbythepowerrule.
Productrule:
d
dx
p
625 x2x
1=2
=
p
625 x2
1
2
x
3=2
+
x
p
625 x2
x
1=2
:
Withabitofalgebra,bothofthesesimplifyto
x2+625
2
p
625 x2x3=2
:
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64
Chapter 3 3 Rulesfor r FindingDerivatives
Occasionallyyou willneed to compute the derivative of aquotient witha constant
numerator,like10=x
2
. Ofcourseyoucanusethequotientrule,butitisusuallynotthe
easiestmethod. Ifwedouseithere,weget
d
dx
10
x2
=
x
2
0 102x
x4
=
20
x3
;
sincethederivativeof10is0.Butitissimplertodothis:
d
dx
10
x2
=
d
dx
10x
2
= 20x
3
:
Admittedly,x
2
isaparticularlysimpledenominator,butwewillseethatasimilarcalcu-
lationisusuallypossible. Anotherapproachistorememberthat
d
dx
1
g(x)
=
g
0
(x)
g(x)2
;
butthisrequiresextramemorization. Usingthisformula,
d
dx
10
x2
=10
2x
x4
:
Notethatwerstuselinearityofthederivativetopullthe10outinfront.
Exercises 3.4.
Findthederivativesofthefunctionsin1{4usingthequotientrule.
1.
x
3
x 5x+10
)
2.
x
2
+5x 3
x 6x+3x 7x+1
)
3.
p
x
p
625 x2
)
4.
p
625 x2
x20
)
5. Findanequationforthetangentlinetof(x)=(x
2
4)=(5 x)atx=3. )
6. Findanequationforthetangentlinetof(x)=(x 2)=(x
3
+4x 1)atx=1.)
7. LetP P beapolynomialofdegreenandletQbeapolynomialofdegreem(withQnotthe
zeropolynomial).Usingsigmanotationwecanwrite
P=
Xn
k=0
a
k
x
k
;
Q=
Xm
k=0
b
k
x
k
:
UsesigmanotationtowritethederivativeoftherationalfunctionP=Q.
8. Thecurvey=1=(1+x
2
)isanexampleofaclassofcurveseachofwhichiscalledawitch
of Agnesi. . Sketchthe e curve andndthetangent linetothecurve at x=5. . (Theword
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3.5 The e ChainRule
65
witch hereisamistranslationoftheoriginalItalian,asdescribedat
http://mathworld.wolfram.com/WitchofAgnesi.html
and
http://instructional1.calstatela.edu/sgray/Agnesi/
WitchHistory/Historynamewitch.html.)
)
9. Iff
0
(4)=5,g
0
(4)=12,(fg)(4)=f(4)g(4)=2,andg(4)=6,computef(4)and
d
dx
f
g
at4.
)
So far we have seen how w to o compute the e derivative e of f a a function n built t up p from m other
functions by addition, , subtraction, , multiplication and division. There e is another very
important way y that t we combine simple functions s to o make more complicated functions:
functioncomposition,asdiscussedinsection2.3. Forexample,consider
p
625 x2.This
functionhasmanysimplercomponents,like625andx
2
,andthenthereisthatsquareroot
symbol,sothesquarerootfunction
p
x=x
1=2
isinvolved. Theobviousquestionis: : can
wecomputethederivativeusingthederivativesoftheconstituents625 x
2
and
p
x?We
canindeed. Ingeneral,iff(x)andg(x)arefunctions,wecancomputethederivativesof
f(g(x))andg(f(x))intermsoff
0
(x)andg
0
(x).
EXAMPLE 3.5.1
Form the e two o possible e compositions of f(x) ) =
p
x and d g(x) ) =
625 x
2
and compute the derivatives. First, , f(g(x)) ) =
p
625 x2,and d thederivative
is x=
p
625 xaswehaveseen.Second,g(f(x))=625 (
p
x)
2
=625 xwithderiva-
tive  1. Of f course, , these e calculations s do o not use e anything g new, , and in particularthe
derivativeoff(g(x))wassomewhattedioustocomputefromthedenition.
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
0
(x)whenwetakethelimit.
Therstfractionismorecomplicated,butit toolookssomethinglikeaderivative. The
denominator,g(x+x)) g(x),isa a change in the value of g,so let’sabbreviate it as
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66
Chapter 3 3 Rulesfor r FindingDerivatives
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
:
Nowthislooksexactlylikeaderivative,namelyf
0
(g(x)),thatis,thefunctionf
0
(x)with
xreplacedbyg(x).Ifthisallwithstandsscrutiny,wethenget
d
dx
f(g(x))=f
0
(g(x))g
0
(x):
Unfortunately,thereisasmall awintheargument.Recallthatwhatwemeanbylim
x!0
involveswhathappenswhenxiscloseto0butnotequalto0. Thequalicationisvery
important,sincewemustbeabletodividebyx. Butwhenxiscloseto0butnotequal
to0,g=g(x+x)) g(x)iscloseto0andpossiblyequalto0. . Thismeansitdoesn’t
reallymakesensetodivide byg. Fortunately,it t ispossible torecasttheargument to
avoidthisdiculty,but it isa bit tricky; ; wewillnot t includethe details,whichcan be
found inmanycalculusbooks. Note e that manyfunctions s g do o have the property y that
g(x+x) g(x)6=0whenxissmall,andforthese e functionsthe argumentaboveis
ne.
ThechainrulehasaparticularlysimpleexpressionifweusetheLeibniznotationfor
thederivative. Thequantityf0(g(x))isthederivativeoff withxreplacedbyg;thiscan
bewrittendf=dg.Asusual,g
0
(x)=dg=dx. Thenthechainrulebecomes
df
dx
=
df
dg
dg
dx
:
Thislooksliketrivialarithmetic,butitisnot:dg=dxisnotafraction,thatis,notliteral
division,butasinglesymbolthatmeansg
0
(x).Nevertheless,itturnsoutthatwhatlooks
liketrivialarithmetic,andisthereforeeasytoremember,isreallytrue.
It willtakeabit ofpracticetomake theuseofthe chainrulecomenaturally|itis
morecomplicatedthantheearlierdierentiationruleswehaveseen.
EXAMPLE 3.5.2
Compute thederivativeof
p
625 x2. Wealreadyknowthat t the
answeris x=
p
625 x2,computeddirectlyfromthelimit. Inthecontext t ofthechain
rule, we e have f(x) ) =
p
x, g(x) ) = 625 x
2
. We e know that f
0
(x) = (1=2)x
1=2
, so
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3.5 The e ChainRule
67
f
0
(g(x))=(1=2)(625 x
2
)
1=2
. Notethatthisisatwostepcomputation: rstcompute
f
0
(x),thenreplacexbyg(x). Sinceg
0
(x)= 2xwehave
f
0
(g(x))g
0
(x)=
1
2
p
625 x2
( 2x)=
x
p
625 x2
:
EXAMPLE3.5.3
Computethederivativeof1=
p
625 x2. Thisisaquotientwitha
constantnumerator,sowecouldusethequotientrule,butitissimplertousethechain
rule.Thefunctionis(625 x
2
)
1=2
,thecompositionoff(x)=x
1=2
andg(x)=625 x
2
.
Wecomputef
0
(x)=( 1=2)x
3=2
usingthepowerrule,andthen
f
0
(g(x))g
0
(x)=
1
2(625 x2)3=2
( 2x)=
x
(625 x2)3=2
:
Inpractice,ofcourse,youwillneedtousemorethanoneoftheruleswehavedeveloped
tocomputethederivativeofacomplicatedfunction.
EXAMPLE3.5.4
Computethederivativeof
f(x)=
x
2
1
x
p
x2+1
:
The\last"operationhere isdivision,sotoget startedweneedtousethe quotient rule
rst.Thisgives
f
0
(x)=
(x2 1)0x
p
x2+1 (x2 1)(x
p
x2+1)0
x2(x2+1)
=
2x2
p
x2+1 (x2 1)(x
p
x2+1)0
x2(x2+1)
:
Nowweneed tocomputethe derivative ofx
p
x2+1. Thisisaproduct,so o weusethe
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
p
x2+1
:
68
Chapter 3 3 Rulesfor r FindingDerivatives
Andputtingitalltogether:
f
0
(x)=
2x
2
p
x2+1 (x
2
1)(x
p
x2+1)
0
x2(x2+1)
:
=
2x
2
p
x2+1 (x
2
1)
x
x
p
x2+1
+
p
x2+1
x2(x2+1)
:
Thiscanbesimpliedofcourse,butwehavedoneallthecalculus,sothatonlyalgebrais
left.
EXAMPLE 3.5.5
Compute the derivative of
r
1+
q
1+
p
x. Herewe e have amore
complicatedchainofcompositions,soweusethechainruletwice.Attheoutermost\layer"
wehavethefunctiong(x)=1+
q
1+
p
xpluggedintof(x)=
p
x,soapplyingthechain
ruleoncegives
d
dx
r
1+
q
1+
p
x=
1
2
1+
q
1+
p
x
1=2
d
dx
1+
q
1+
p
x
:
Nowweneedthederivativeof
q
1+
p
x. Usingthechainruleagain:
d
dx
q
1+
p
x=
1
2
1+
p
x
1=2
1
2
x
1=2
:
Sotheoriginalderivativeis
d
dx
r
1+
q
1+
p
x=
1
2
1+
q
1+
p
x
1=2
1
2
1+
p
x
1=2
1
2
x
1=2
:
=
1
8
p
x
p
1+
p
x
q
1+
p
1+
p
x
Usingthechainrule,thepowerrule,andtheproductrule,itispossibletoavoidusing
thequotientruleentirely.
3.5 TheChainRule
69
EXAMPLE3.5.6 Computethederivativeoff(x)=
x
3
x2+1
.Writef(x)=x
3
(x
2
+1)
1
,
then
f
0
(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
Notethatwealreadyhadthederivativeonthesecondline;alltherestissimplication.It
iseasiertogettothisanswerbyusingthequotientrule,sothere’satradeo:morework
forfewermemorizedformulas.
Exercises3.5.
Findthederivativesofthefunctions. Forextrapractice,andtocheckyouranswers,dosomeof
theseinmorethanonewayifpossible.
1. x
4
3x
3
+(1=2)x
2
+7x )
2. x
3
2x
2
+4
p
x)
3. (x
2
+1)
3
)
4. x
p
169 x)
5. (x
2
4x+5)
p
25 x)
6.
p
r x2,risaconstant)
7.
p
1+x)
8.
1
p
p
x
)
9. (1+3x)
2
)
10.
(x
2
+x+1)
(1 x)
)
11.
p
25 x2
x
)
12.
r
169
x
x)
13.
p
x x (1=x))
14. 100=(100 x
2
)
3=2
)
15.
3
p
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))
23. 4(2x
2
x+3)
2
)
24.
1
1+1=x
)
25.
3
4x 2x+1
)
26. (x
2
+1)(5 2x)=2)
70
Chapter 3 3 Rulesfor r FindingDerivatives
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  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)
p
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;
q
4+
p
5). )
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