lesson 7: the derivative
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. . . . . .
Section2.1–2TheDerivativeandRatesofChange
TheDerivativeasaFunction
V63.0121.006/016, CalculusI
February9, 2010
Announcements
I OfficeHoursthisweek: Monday, Wednesdays1:30–2:30,Thursdays9–10am(WWH 726)
I PleasecompletetheGet-to-know-yousurveyanduploadaphoto.
. . . . . .
CommonErrorsonHomework#2
I Rememberthat limx→a−
f(x) meansthelimitas x approaches a
fromtheleft. Thisisdifferentfrom limx→−a
f(x).
I Likewise, limx→a+
f(x) meansthelimitas x approaches a from
theright.I Rememberthatthelimitlawsworkonlyiftheindividuallimitsexist.
. . . . . .
Problem1.4.32
ProblemFind lim
x→0+
√x[1+ sin2(2π/x)
]
. . . . . .
Problem1.4.32
ProblemFind lim
x→0+
√x[1+ sin2(2π/x)
]WrongSolution#1
limx→0+
√x[1+ sin2(2π/x)
]=
(limx→0+
√x)
limx→0+
[1+ sin2(2π/x)
]= 0 · lim
x→0+
[1+ sin2(2π/x)
]= 0
Youhavenotcheckedthat limx→0+
[1+ sin2(2π/x)
]exists.
. . . . . .
Problem1.4.32
ProblemFind lim
x→0+
√x[1+ sin2(2π/x)
]WrongSolution#1
limx→0+
√x[1+ sin2(2π/x)
]=
(limx→0+
√x)
limx→0+
[1+ sin2(2π/x)
]= 0 · lim
x→0+
[1+ sin2(2π/x)
]= 0
Youhavenotcheckedthat limx→0+
[1+ sin2(2π/x)
]exists.
. . . . . .
Problem1.4.32
ProblemFind lim
x→0+
√x[1+ sin2(2π/x)
]WrongSolution#2
limx→0+
√x[1+ sin2(2π/x)
]=
(limx→0+
√x)
limx→0+
[1+ sin2(2π/x)
]= 0 · lim
x→0+
[1+ sin2(2π/x)
]= 0 ·
[1+ lim
x→0+sin2(2π/x)
]= 0 · [1+ 0] = 0
Unfortunately, limx→0+
sin2(2π/x) doesnotexist.
. . . . . .
Problem1.4.32
ProblemFind lim
x→0+
√x[1+ sin2(2π/x)
]WrongSolution#2
limx→0+
√x[1+ sin2(2π/x)
]=
(limx→0+
√x)
limx→0+
[1+ sin2(2π/x)
]= 0 · lim
x→0+
[1+ sin2(2π/x)
]= 0 ·
[1+ lim
x→0+sin2(2π/x)
]= 0 · [1+ 0] = 0
Unfortunately, limx→0+
sin2(2π/x) doesnotexist.
. . . . . .
Problem1.4.32
ProblemFind lim
x→0+
√x[1+ sin2(2π/x)
]WrongSolution#3
limx→0+
√x[1+ sin2(2π/x)
]=
(limx→0+
√x)
limx→0+
[1+ sin2(2π/x)
]= 0 · lim
x→0+
[1+ sin2(2π/x)
]= 0 · [1+DNE]
= 0 · [1+DNE] = DNE
Remember“DNE” isnotanumberandlimitslawsonlyworkonfunctionsthathavelimits.
. . . . . .
Problem1.4.32
ProblemFind lim
x→0+
√x[1+ sin2(2π/x)
]WrongSolution#3
limx→0+
√x[1+ sin2(2π/x)
]=
(limx→0+
√x)
limx→0+
[1+ sin2(2π/x)
]= 0 · lim
x→0+
[1+ sin2(2π/x)
]= 0 · [1+DNE]
= 0 · [1+DNE] = DNE
Remember“DNE” isnotanumberandlimitslawsonlyworkonfunctionsthathavelimits.
. . . . . .
Problem1.4.32
ProblemFind lim
x→0+
√x[1+ sin2(2π/x)
]SolutionSince −1 ≤ sin(2π/x) ≤ 1 forall x > 0, wehave
0 ≤ sin2(2π/x) ≤ 1
=⇒ 1 ≤ 1+ sin2(2π/x) ≤ 2
=⇒√x ≤
√x[1+ sin2(2π/x)
]≤ 2
√x
Since limx→0+
√x = 0 = lim
x→0+2√x, bytheSqueezeTheoremitmust
alsobetruethat
limx→0+
√x[1+ sin2(2π/x)
]= 0
. . . . . .
WolframAlphaoutput
plot sqrt�x��1�sin^2�2pi�x�� from 0 to 1
Input interpretation:
plot x 1� sin22 Πx
x � 0 to 1
Plot:
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
Generated by Wolfram|Alpha (www.wolframalpha.com) on February 9, 2010 from Champaign, IL.© Wolfram Alpha LLC—A Wolfram Research Company
1
. . . . . .
Formatofwrittenwork
Please:I Usescratchpaperandcopyyourfinalworkontofreshpaper.
I Useloose-leafpaper(nottornfromanotebook).
I Writeyourname,lecturesection,assignmentnumber, anddateatthetop.
I Stapleyourhomeworktogether.
Seethewebsiteformoreinformation.
. . . . . .
Explanations
Fromthesyllabus:
Graderswillbeexpectingyoutoexpressyourideasclearly, legibly, andcompletely, oftenrequiringcompleteEnglishsentencesratherthanmerelyjustalongstringofequationsorunconnectedmathematicalexpressions. Thismeansyoucouldlosepointsforunexplainedanswers.
. . . . . .
Rubric
Points DescriptionofWork3 Work is completely accurate and essentially perfect.
Workisthoroughlydeveloped, neat, andeasytoread.Completesentencesareused.
2 Work is good, but incompletely developed, hard toread, unexplained, or jumbled. Answers which arenotexplained, evenifcorrect, willgenerallyreceive2points. Workcontains“rightidea”butisflawed.
1 Workissketchy. Thereissomecorrectwork, butmostofworkisincorrect.
0 Workminimalornon-existent. Solutioniscompletelyincorrect.
. . . . . .
Quiz1
Thefirstquizwillbeinrecitationthisweek.I ThequizcoversSections1.1–1.4.I Thequizcoversbothsidesofasheetofpaper: twoquestionswithparts.
I Thequizstartspromptlyatthebeginningofrecitationandlasts15minutes.
I Youmustgotothesectionyouareregisteredfor.
. . . . . .
Policyonout-of-sequenceexamsandquizzesFromthesyllabus
I Wemayapproveout-of-sequenceexamsandquizzesinthefollowingcases:1. A documentedmedicalexcuse.2. A Universitysponsoredeventsuchasanathletictournament,
aplay, oramusicalperformance.3. A religiousholiday.4. Extremehardshipsuchasafamilyemergency.
I Wewillnotbeabletoaccommodateout-of-sequenceexams, quizzes, andfinalsforpurposesofmoreconvenienttravel
I Wedropthelowestquiztogiveyouonefreepass.I IfyourequireadditionalaccommodationsasdeterminedbytheCenterforStudentDisabilities, pleaseletmeknowASAP.
. . . . . .
Outline
RatesofChangeTangentLinesVelocityPopulationgrowthMarginalcosts
Thederivative, definedDerivativesof(some)powerfunctionsWhatdoes f tellyouabout f′?
Howcanafunctionfailtobedifferentiable?
Othernotations
Thesecondderivative
. . . . . .
Thetangentproblem
ProblemGivenacurveandapointonthecurve, findtheslopeofthelinetangenttothecurveatthatpoint.
ExampleFindtheslopeofthelinetangenttothecurve y = x2 atthepoint(2, 4).
UpshotIfthecurveisgivenby y = f(x), andthepointonthecurveis(a, f(a)), thentheslopeofthetangentlineisgivenby
mtangent = limx→a
f(x)− f(a)x− a
. . . . . .
Thetangentproblem
ProblemGivenacurveandapointonthecurve, findtheslopeofthelinetangenttothecurveatthatpoint.
ExampleFindtheslopeofthelinetangenttothecurve y = x2 atthepoint(2, 4).
UpshotIfthecurveisgivenby y = f(x), andthepointonthecurveis(a, f(a)), thentheslopeofthetangentlineisgivenby
mtangent = limx→a
f(x)− f(a)x− a
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
x m =x2 − 22
x− 2
3 52.5 4.52.1 4.12.01 4.01limit 41.99 3.991.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
.
..3
..9
x m =x2 − 22
x− 23
52.5 4.52.1 4.12.01 4.01limit 41.99 3.991.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
.
..3
..9
x m =x2 − 22
x− 23 5
2.5 4.52.1 4.12.01 4.01limit 41.99 3.991.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
.
..2.5
..6.25
x m =x2 − 22
x− 23 52.5
4.52.1 4.12.01 4.01limit 41.99 3.991.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
.
..2.5
..6.25
x m =x2 − 22
x− 23 52.5 4.5
2.1 4.12.01 4.01limit 41.99 3.991.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 ..
..2.1
..4.41
x m =x2 − 22
x− 23 52.5 4.52.1
4.12.01 4.01limit 41.99 3.991.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 ..
..2.1
..4.41
x m =x2 − 22
x− 23 52.5 4.52.1 4.1
2.01 4.01limit 41.99 3.991.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 ..
..2.01
..4.0401
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01
4.01limit 41.99 3.991.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 ..
..2.01
..4.0401
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01 4.01
limit 41.99 3.991.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
.
..1
..1
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01 4.01
limit 41.99 3.991.9 3.91.5 3.5
1
3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
.
..1
..1
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01 4.01
limit 41.99 3.991.9 3.91.5 3.5
1 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
.
..1.5
..2.25
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01 4.01
limit 41.99 3.991.9 3.9
1.5
3.5
1 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
.
..1.5
..2.25
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01 4.01
limit 41.99 3.991.9 3.9
1.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 ..
..1.9
..3.61
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01 4.01
limit 41.99 3.99
1.9
3.9
1.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 ..
..1.9
..3.61.
..1.99
..3.9601
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01 4.01
limit 41.99 3.99
1.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 ..
..1.99
..3.9601
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01 4.01
limit 4
1.99
3.99
1.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01 4.01
limit 4
1.99 3.991.9 3.91.5 3.51 3
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
.
..3
..9
.
..2.5
..6.25
.
..2.1
..4.41 .
..2.01
..4.0401
.
..1
..1
.
..1.5
..2.25
.
..1.9
..3.61.
..1.99
..3.9601
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01 4.01limit 41.99 3.991.9 3.91.5 3.51 3
. . . . . .
Thetangentproblem
ProblemGivenacurveandapointonthecurve, findtheslopeofthelinetangenttothecurveatthatpoint.
ExampleFindtheslopeofthelinetangenttothecurve y = x2 atthepoint(2, 4).
UpshotIfthecurveisgivenby y = f(x), andthepointonthecurveis(a, f(a)), thentheslopeofthetangentlineisgivenby
mtangent = limx→a
f(x)− f(a)x− a
. . . . . .
VelocityProblemGiventhepositionfunctionofamovingobject, findthevelocityoftheobjectatacertaininstantintime.
ExampleDropaballofftheroofoftheSilverCentersothatitsheightcanbedescribedby
h(t) = 50− 5t2
where t issecondsafterdroppingitand h ismetersabovetheground. Howfastisitfallingonesecondafterwedropit?
SolutionTheansweris
v = limt→1
(50− 5t2)− 45t− 1
= limt→1
5− 5t2
t− 1= lim
t→1
5(1− t)(1+ t)t− 1
= (−5) limt→1
(1+ t) = −5 · 2 = −10
. . . . . .
Numericalevidence
t vave =h(t)− h(1)
t− 12 − 15
1.5 − 12.51.1 − 10.51.01 − 10.051.001 − 10.005
. . . . . .
Numericalevidence
t vave =h(t)− h(1)
t− 12 − 151.5
− 12.51.1 − 10.51.01 − 10.051.001 − 10.005
. . . . . .
Numericalevidence
t vave =h(t)− h(1)
t− 12 − 151.5 − 12.5
1.1 − 10.51.01 − 10.051.001 − 10.005
. . . . . .
Numericalevidence
t vave =h(t)− h(1)
t− 12 − 151.5 − 12.51.1
− 10.51.01 − 10.051.001 − 10.005
. . . . . .
Numericalevidence
t vave =h(t)− h(1)
t− 12 − 151.5 − 12.51.1 − 10.5
1.01 − 10.051.001 − 10.005
. . . . . .
Numericalevidence
t vave =h(t)− h(1)
t− 12 − 151.5 − 12.51.1 − 10.51.01
− 10.051.001 − 10.005
. . . . . .
Numericalevidence
t vave =h(t)− h(1)
t− 12 − 151.5 − 12.51.1 − 10.51.01 − 10.05
1.001 − 10.005
. . . . . .
Numericalevidence
t vave =h(t)− h(1)
t− 12 − 151.5 − 12.51.1 − 10.51.01 − 10.051.001
− 10.005
. . . . . .
Numericalevidence
t vave =h(t)− h(1)
t− 12 − 151.5 − 12.51.1 − 10.51.01 − 10.051.001 − 10.005
. . . . . .
VelocityProblemGiventhepositionfunctionofamovingobject, findthevelocityoftheobjectatacertaininstantintime.
ExampleDropaballofftheroofoftheSilverCentersothatitsheightcanbedescribedby
h(t) = 50− 5t2
where t issecondsafterdroppingitand h ismetersabovetheground. Howfastisitfallingonesecondafterwedropit?
SolutionTheansweris
v = limt→1
(50− 5t2)− 45t− 1
= limt→1
5− 5t2
t− 1= lim
t→1
5(1− t)(1+ t)t− 1
= (−5) limt→1
(1+ t) = −5 · 2 = −10
. . . . . .
UpshotIftheheightfunctionisgivenby h(t), theinstantaneousvelocityattime t0 isgivenby
v = limt→t0
h(t)− h(t0)t− t0
= lim∆t→0
h(t0 +∆t)− h(t0)∆t
. .t
.y = h(t).
.
..t0
..t
.∆t
. . . . . .
Populationgrowth
ProblemGiventhepopulationfunctionofagroupoforganisms, findtherateofgrowthofthepopulationataparticularinstant.
ExampleSupposethepopulationoffishintheEastRiverisgivenbythefunction
P(t) =3et
1+ et
where t isinyearssince2000and P isinmillionsoffish. Isthefishpopulationgrowingfastestin1990, 2000, or2010? (Estimatenumerically)
SolutionWeestimatestheratesofgrowthtobe 0.000143229, 0.749376,and 0.0001296. Sothepopulationisgrowingfastestin2000.
. . . . . .
Populationgrowth
ProblemGiventhepopulationfunctionofagroupoforganisms, findtherateofgrowthofthepopulationataparticularinstant.
ExampleSupposethepopulationoffishintheEastRiverisgivenbythefunction
P(t) =3et
1+ et
where t isinyearssince2000and P isinmillionsoffish. Isthefishpopulationgrowingfastestin1990, 2000, or2010? (Estimatenumerically)
SolutionWeestimatestheratesofgrowthtobe 0.000143229, 0.749376,and 0.0001296. Sothepopulationisgrowingfastestin2000.
. . . . . .
Derivation
Let ∆t beanincrementintimeand ∆P thecorrespondingchangeinpopulation:
∆P = P(t+∆t)− P(t)
Thisdependson ∆t, sowewant
lim∆t→0
∆P∆t
= lim∆t→0
1∆t
(3et+∆t
1+ et+∆t −3et
1+ et
)
Toohard! Tryasmall ∆t toapproximate.
. . . . . .
Derivation
Let ∆t beanincrementintimeand ∆P thecorrespondingchangeinpopulation:
∆P = P(t+∆t)− P(t)
Thisdependson ∆t, sowewant
lim∆t→0
∆P∆t
= lim∆t→0
1∆t
(3et+∆t
1+ et+∆t −3et
1+ et
)Toohard! Tryasmall ∆t toapproximate.
. . . . . .
Numericalevidence
r1990 ≈ P(−10+ 0.1)− P(−10)0.1
=
0.000143229
r2000 ≈ P(0.1)− P(0)0.1
= 0.749376
r2010 ≈ P(10+ 0.1)− P(10)0.1
= 0.0001296
. . . . . .
Numericalevidence
r1990 ≈ P(−10+ 0.1)− P(−10)0.1
= 0.000143229
r2000 ≈ P(0.1)− P(0)0.1
= 0.749376
r2010 ≈ P(10+ 0.1)− P(10)0.1
= 0.0001296
. . . . . .
Numericalevidence
r1990 ≈ P(−10+ 0.1)− P(−10)0.1
= 0.000143229
r2000 ≈ P(0.1)− P(0)0.1
=
0.749376
r2010 ≈ P(10+ 0.1)− P(10)0.1
= 0.0001296
. . . . . .
Numericalevidence
r1990 ≈ P(−10+ 0.1)− P(−10)0.1
= 0.000143229
r2000 ≈ P(0.1)− P(0)0.1
= 0.749376
r2010 ≈ P(10+ 0.1)− P(10)0.1
= 0.0001296
. . . . . .
Numericalevidence
r1990 ≈ P(−10+ 0.1)− P(−10)0.1
= 0.000143229
r2000 ≈ P(0.1)− P(0)0.1
= 0.749376
r2010 ≈ P(10+ 0.1)− P(10)0.1
=
0.0001296
. . . . . .
Numericalevidence
r1990 ≈ P(−10+ 0.1)− P(−10)0.1
= 0.000143229
r2000 ≈ P(0.1)− P(0)0.1
= 0.749376
r2010 ≈ P(10+ 0.1)− P(10)0.1
= 0.0001296
. . . . . .
Populationgrowth
ProblemGiventhepopulationfunctionofagroupoforganisms, findtherateofgrowthofthepopulationataparticularinstant.
ExampleSupposethepopulationoffishintheEastRiverisgivenbythefunction
P(t) =3et
1+ et
where t isinyearssince2000and P isinmillionsoffish. Isthefishpopulationgrowingfastestin1990, 2000, or2010? (Estimatenumerically)
SolutionWeestimatestheratesofgrowthtobe 0.000143229, 0.749376,and 0.0001296. Sothepopulationisgrowingfastestin2000.
. . . . . .
UpshotTheinstantaneouspopulationgrowthisgivenby
lim∆t→0
P(t+∆t)− P(t)∆t
. . . . . .
Marginalcosts
ProblemGiventheproductioncostofagood, findthemarginalcostofproductionafterhavingproducedacertainquantity.
ExampleSupposethecostofproducing q tonsofriceonourpaddyinayearis
C(q) = q3 − 12q2 + 60q
Wearecurrentlyproducing 5 tonsayear. Shouldwechangethat?
ExampleIf q = 5, then C = 125, ∆C = 19, while AC = 25. Soweshouldproducemoretoloweraveragecosts.
. . . . . .
Marginalcosts
ProblemGiventheproductioncostofagood, findthemarginalcostofproductionafterhavingproducedacertainquantity.
ExampleSupposethecostofproducing q tonsofriceonourpaddyinayearis
C(q) = q3 − 12q2 + 60q
Wearecurrentlyproducing 5 tonsayear. Shouldwechangethat?
ExampleIf q = 5, then C = 125, ∆C = 19, while AC = 25. Soweshouldproducemoretoloweraveragecosts.
. . . . . .
Comparisons
q C(q)
AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)
4
112 28 13
5
125 25 19
6
144 24 31
. . . . . .
Comparisons
q C(q)
AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)
4 112
28 13
5
125 25 19
6
144 24 31
. . . . . .
Comparisons
q C(q)
AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)
4 112
28 13
5 125
25 19
6
144 24 31
. . . . . .
Comparisons
q C(q)
AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)
4 112
28 13
5 125
25 19
6 144
24 31
. . . . . .
Comparisons
q C(q) AC(q) = C(q)/q
∆C = C(q+ 1)− C(q)
4 112
28 13
5 125
25 19
6 144
24 31
. . . . . .
Comparisons
q C(q) AC(q) = C(q)/q
∆C = C(q+ 1)− C(q)
4 112 28
13
5 125
25 19
6 144
24 31
. . . . . .
Comparisons
q C(q) AC(q) = C(q)/q
∆C = C(q+ 1)− C(q)
4 112 28
13
5 125 25
19
6 144
24 31
. . . . . .
Comparisons
q C(q) AC(q) = C(q)/q
∆C = C(q+ 1)− C(q)
4 112 28
13
5 125 25
19
6 144 24
31
. . . . . .
Comparisons
q C(q) AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)4 112 28
13
5 125 25
19
6 144 24
31
. . . . . .
Comparisons
q C(q) AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)4 112 28 135 125 25
19
6 144 24
31
. . . . . .
Comparisons
q C(q) AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)4 112 28 135 125 25 196 144 24
31
. . . . . .
Comparisons
q C(q) AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)4 112 28 135 125 25 196 144 24 31
. . . . . .
Marginalcosts
ProblemGiventheproductioncostofagood, findthemarginalcostofproductionafterhavingproducedacertainquantity.
ExampleSupposethecostofproducing q tonsofriceonourpaddyinayearis
C(q) = q3 − 12q2 + 60q
Wearecurrentlyproducing 5 tonsayear. Shouldwechangethat?
ExampleIf q = 5, then C = 125, ∆C = 19, while AC = 25. Soweshouldproducemoretoloweraveragecosts.
. . . . . .
Upshot
I Theincrementalcost
∆C = C(q+ 1)− C(q)
isuseful, butdependsonunits.
I Themarginalcostafterproducing q givenby
MC = lim∆q→0
C(q+∆q)− C(q)∆q
ismoreusefulsinceit’sunit-independent.
. . . . . .
Upshot
I Theincrementalcost
∆C = C(q+ 1)− C(q)
isuseful, butdependsonunits.I Themarginalcostafterproducing q givenby
MC = lim∆q→0
C(q+∆q)− C(q)∆q
ismoreusefulsinceit’sunit-independent.
. . . . . .
Outline
RatesofChangeTangentLinesVelocityPopulationgrowthMarginalcosts
Thederivative, definedDerivativesof(some)powerfunctionsWhatdoes f tellyouabout f′?
Howcanafunctionfailtobedifferentiable?
Othernotations
Thesecondderivative
. . . . . .
Thedefinition
Alloftheseratesofchangearefoundthesameway!
DefinitionLet f beafunctionand a apointinthedomainof f. Ifthelimit
f′(a) = limh→0
f(a+ h)− f(a)h
= limx→a
f(x)− f(a)x− a
exists, thefunctionissaidtobe differentiableat a and f′(a) isthederivativeof f at a.
. . . . . .
Thedefinition
Alloftheseratesofchangearefoundthesameway!
DefinitionLet f beafunctionand a apointinthedomainof f. Ifthelimit
f′(a) = limh→0
f(a+ h)− f(a)h
= limx→a
f(x)− f(a)x− a
exists, thefunctionissaidtobe differentiableat a and f′(a) isthederivativeof f at a.
. . . . . .
Derivativeofthesquaringfunction
ExampleSuppose f(x) = x2. Usethedefinitionofderivativetofind f′(a).
Solution
f′(a) = limh→0
f(a+ h)− f(a)h
= limh→0
(a+ h)2 − a2
h
= limh→0
(a2 + 2ah+ h2)− a2
h= lim
h→0
2ah+ h2
h= lim
h→0(2a+ h) = 2a.
. . . . . .
Derivativeofthesquaringfunction
ExampleSuppose f(x) = x2. Usethedefinitionofderivativetofind f′(a).
Solution
f′(a) = limh→0
f(a+ h)− f(a)h
= limh→0
(a+ h)2 − a2
h
= limh→0
(a2 + 2ah+ h2)− a2
h= lim
h→0
2ah+ h2
h= lim
h→0(2a+ h) = 2a.
. . . . . .
Derivativeofthereciprocalfunction
Example
Suppose f(x) =1x. Usethe
definitionofthederivativetofind f′(2).
Solution
f′(2) = limx→2
1/x− 1/2x− 2
= limx→2
2− x2x(x− 2)
= limx→2
−12x
= −14
. .x
.x
.
. . . . . .
Derivativeofthereciprocalfunction
Example
Suppose f(x) =1x. Usethe
definitionofthederivativetofind f′(2).
Solution
f′(2) = limx→2
1/x− 1/2x− 2
= limx→2
2− x2x(x− 2)
= limx→2
−12x
= −14
. .x
.x
.
. . . . . .
TheSure-FireSallyRule(SFSR) foraddingFractionsInanticipationofthequestion, “Howdidyougetthat?”
ab± c
d=
ad± bcbd
So
1x− 1
2x− 2
=
2− x2x
x− 2
=2− x
2x(x− 2)
. . . . . .
TheSure-FireSallyRule(SFSR) foraddingFractionsInanticipationofthequestion, “Howdidyougetthat?”
ab± c
d=
ad± bcbd
So
1x− 1
2x− 2
=
2− x2x
x− 2
=2− x
2x(x− 2)
. . . . . .
Whatdoes f tellyouabout f′?
I If f isafunction, wecancomputethederivative f′(x) ateachpoint x where f isdifferentiable, andcomeupwithanotherfunction, thederivativefunction.
I Whatcanwesayaboutthisfunction f′?
I If f isdecreasingonaninterval, f′ isnegative(technically,nonpositive)onthatinterval
I If f isincreasingonaninterval, f′ ispositive(technically,nonnegative)onthatinterval
. . . . . .
Whatdoes f tellyouabout f′?
I If f isafunction, wecancomputethederivative f′(x) ateachpoint x where f isdifferentiable, andcomeupwithanotherfunction, thederivativefunction.
I Whatcanwesayaboutthisfunction f′?I If f isdecreasingonaninterval, f′ isnegative(technically,nonpositive)onthatinterval
I If f isincreasingonaninterval, f′ ispositive(technically,nonnegative)onthatinterval
. . . . . .
Derivativeofthereciprocalfunction
Example
Suppose f(x) =1x. Usethe
definitionofthederivativetofind f′(2).
Solution
f′(2) = limx→2
1/x− 1/2x− 2
= limx→2
2− x2x(x− 2)
= limx→2
−12x
= −14
. .x
.x
.
. . . . . .
Whatdoes f tellyouabout f′?
I If f isafunction, wecancomputethederivative f′(x) ateachpoint x where f isdifferentiable, andcomeupwithanotherfunction, thederivativefunction.
I Whatcanwesayaboutthisfunction f′?I If f isdecreasingonaninterval, f′ isnegative(technically,nonpositive)onthatinterval
I If f isincreasingonaninterval, f′ ispositive(technically,nonnegative)onthatinterval
. . . . . .
Graphicallyandnumerically
. .x
.y
..2
..4 .
.
..1
..1
x m =x2 − 22
x− 23 52.5 4.52.1 4.12.01 4.01
limit 41.99 3.991.9 3.91.5 3.5
1
3
. . . . . .
Whatdoes f tellyouabout f′?FactIf f isdecreasingon (a,b), then f′ ≤ 0 on (a,b).
Proof.If f isdecreasingon (a,b), and ∆x > 0, then
f(x+∆x) < f(x) =⇒ f(x+∆x)− f(x)∆x
< 0
Butif ∆x < 0, then x+∆x < x, and
f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x
< 0
still! Eitherway,f(x+∆x)− f(x)
∆x< 0, so
f′(x) = lim∆x→0
f(x+∆x)− f(x)∆x
≤ 0
. . . . . .
Whatdoes f tellyouabout f′?FactIf f isdecreasingon (a,b), then f′ ≤ 0 on (a,b).
Proof.If f isdecreasingon (a,b), and ∆x > 0, then
f(x+∆x) < f(x) =⇒ f(x+∆x)− f(x)∆x
< 0
Butif ∆x < 0, then x+∆x < x, and
f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x
< 0
still!
Eitherway,f(x+∆x)− f(x)
∆x< 0, so
f′(x) = lim∆x→0
f(x+∆x)− f(x)∆x
≤ 0
. . . . . .
Whatdoes f tellyouabout f′?FactIf f isdecreasingon (a,b), then f′ ≤ 0 on (a,b).
Proof.If f isdecreasingon (a,b), and ∆x > 0, then
f(x+∆x) < f(x) =⇒ f(x+∆x)− f(x)∆x
< 0
Butif ∆x < 0, then x+∆x < x, and
f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x
< 0
still! Eitherway,f(x+∆x)− f(x)
∆x< 0, so
f′(x) = lim∆x→0
f(x+∆x)− f(x)∆x
≤ 0
. . . . . .
Outline
RatesofChangeTangentLinesVelocityPopulationgrowthMarginalcosts
Thederivative, definedDerivativesof(some)powerfunctionsWhatdoes f tellyouabout f′?
Howcanafunctionfailtobedifferentiable?
Othernotations
Thesecondderivative
. . . . . .
Differentiabilityissuper-continuity
TheoremIf f isdifferentiableat a, then f iscontinuousat a.
Proof.Wehave
limx→a
(f(x)− f(a)) = limx→a
f(x)− f(a)x− a
· (x− a)
= limx→a
f(x)− f(a)x− a
· limx→a
(x− a)
= f′(a) · 0 = 0
Notetheproperuseofthelimitlaw: if thefactorseachhavealimitat a, thelimitoftheproductistheproductofthelimits.
. . . . . .
Differentiabilityissuper-continuity
TheoremIf f isdifferentiableat a, then f iscontinuousat a.
Proof.Wehave
limx→a
(f(x)− f(a)) = limx→a
f(x)− f(a)x− a
· (x− a)
= limx→a
f(x)− f(a)x− a
· limx→a
(x− a)
= f′(a) · 0 = 0
Notetheproperuseofthelimitlaw: if thefactorseachhavealimitat a, thelimitoftheproductistheproductofthelimits.
. . . . . .
Differentiabilityissuper-continuity
TheoremIf f isdifferentiableat a, then f iscontinuousat a.
Proof.Wehave
limx→a
(f(x)− f(a)) = limx→a
f(x)− f(a)x− a
· (x− a)
= limx→a
f(x)− f(a)x− a
· limx→a
(x− a)
= f′(a) · 0 = 0
Notetheproperuseofthelimitlaw: if thefactorseachhavealimitat a, thelimitoftheproductistheproductofthelimits.
. . . . . .
DifferentiabilityFAILKinks
. .x
.f(x)
. .x
.f′(x)
.
.
. . . . . .
DifferentiabilityFAILKinks
. .x
.f(x)
. .x
.f′(x)
.
.
. . . . . .
DifferentiabilityFAILKinks
. .x
.f(x)
. .x
.f′(x)
.
.
. . . . . .
DifferentiabilityFAILCusps
. .x
.f(x)
. .x
.f′(x)
. . . . . .
DifferentiabilityFAILCusps
. .x
.f(x)
. .x
.f′(x)
. . . . . .
DifferentiabilityFAILCusps
. .x
.f(x)
. .x
.f′(x)
. . . . . .
DifferentiabilityFAILVerticalTangents
. .x
.f(x)
. .x
.f′(x)
. . . . . .
DifferentiabilityFAILVerticalTangents
. .x
.f(x)
. .x
.f′(x)
. . . . . .
DifferentiabilityFAILVerticalTangents
. .x
.f(x)
. .x
.f′(x)
. . . . . .
DifferentiabilityFAILWeird, Wild, Stuff
. .x
.f(x)
Thisfunctionisdifferentiableat 0.
. .x
.f′(x)
Butthederivativeisnotcontinuousat 0!
. . . . . .
DifferentiabilityFAILWeird, Wild, Stuff
. .x
.f(x)
Thisfunctionisdifferentiableat 0.
. .x
.f′(x)
Butthederivativeisnotcontinuousat 0!
. . . . . .
Outline
RatesofChangeTangentLinesVelocityPopulationgrowthMarginalcosts
Thederivative, definedDerivativesof(some)powerfunctionsWhatdoes f tellyouabout f′?
Howcanafunctionfailtobedifferentiable?
Othernotations
Thesecondderivative
. . . . . .
Notation
I Newtoniannotation
f′(x) y′(x) y′
I Leibniziannotation
dydx
ddx
f(x)dfdx
Theseallmeanthesamething.
. . . . . .
MeettheMathematician: IsaacNewton
I English, 1643–1727I ProfessoratCambridge(England)
I PhilosophiaeNaturalisPrincipiaMathematicapublished1687
. . . . . .
MeettheMathematician: GottfriedLeibniz
I German, 1646–1716I Eminentphilosopheraswellasmathematician
I Contemporarilydisgracedbythecalculusprioritydispute
. . . . . .
Outline
RatesofChangeTangentLinesVelocityPopulationgrowthMarginalcosts
Thederivative, definedDerivativesof(some)powerfunctionsWhatdoes f tellyouabout f′?
Howcanafunctionfailtobedifferentiable?
Othernotations
Thesecondderivative
. . . . . .
Thesecondderivative
If f isafunction, sois f′, andwecanseekitsderivative.
f′′ = (f′)′
Itmeasurestherateofchangeoftherateofchange!
Leibniziannotation:
d2ydx2
d2
dx2f(x)
d2fdx2
. . . . . .
Thesecondderivative
If f isafunction, sois f′, andwecanseekitsderivative.
f′′ = (f′)′
Itmeasurestherateofchangeoftherateofchange! Leibniziannotation:
d2ydx2
d2
dx2f(x)
d2fdx2
. . . . . .
function, derivative, secondderivative
. .x
.y
.f(x) = x2
.f′(x) = 2x
.f′′(x) = 2
. . . . . .
Whathavewelearnedtoday?
I ThederivativemeasuresinstantaneousrateofchangeI Thederivativehasmanyinterpretations: slopeofthetangentline, velocity, marginalquantities, etc.
I Thederivativereflectsthemonotonicity(increasingordecreasing)ofthegraph
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