lesson 7: the derivative

. . . . . .

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#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)

]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

. . . . . .

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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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

. . . . . .


. .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− 12 − 151.5

− 12.51.1 − 10.51.01 − 10.051.001 − 10.005

. . . . . .

Numericalevidence


t− 12 − 151.5 − 12.5

1.1 − 10.51.01 − 10.051.001 − 10.005

. . . . . .

Numericalevidence


t− 12 − 151.5 − 12.51.1

− 10.51.01 − 10.051.001 − 10.005

. . . . . .

Numericalevidence


t− 12 − 151.5 − 12.51.1 − 10.5

1.01 − 10.051.001 − 10.005

. . . . . .

Numericalevidence


t− 12 − 151.5 − 12.51.1 − 10.51.01

− 10.051.001 − 10.005

. . . . . .

Numericalevidence


t− 12 − 151.5 − 12.51.1 − 10.51.01 − 10.05

1.001 − 10.005

. . . . . .

Numericalevidence


t− 12 − 151.5 − 12.51.1 − 10.51.01 − 10.051.001

− 10.005

. . . . . .

Numericalevidence


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.

. . . . . .

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.

. . . . . .

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


∆C = C(q+ 1)− C(q)

4 112 28

13

5 125

25 19

6 144

24 31

. . . . . .

Comparisons


∆C = C(q+ 1)− C(q)

4 112 28

13

5 125 25

19

6 144

24 31

. . . . . .

Comparisons


∆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




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.

. . . . . .

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

.

. . . . . .

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

. . . . . .



I Whatcanwesayaboutthisfunction f′?I If f isdecreasingonaninterval, f′ isnegative(technically,nonpositive)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

.

. . . . . .



I Whatcanwesayaboutthisfunction f′?I If f isdecreasingonaninterval, f′ isnegative(technically,nonpositive)onthatinterval


. . . . . .


. .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

. . . . . .



f(x+∆x) < f(x) =⇒ f(x+∆x)− f(x)∆x

< 0


f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x

< 0

still!

Eitherway,f(x+∆x)− f(x)

∆x< 0, so



≤ 0

. . . . . .



f(x+∆x) < f(x) =⇒ f(x+∆x)− f(x)∆x

< 0


f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x

< 0

still! Eitherway,f(x+∆x)− f(x)

∆x< 0, so



≤ 0

. . . . . .

Outline




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.

. . . . . .

DifferentiabilityFAILKinks

. .x

.f(x)

. .x

.f′(x)

.

.

. . . . . .

DifferentiabilityFAILCusps

. .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!

. . . . . .

Outline




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




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

lesson 7: the derivative

Education