Lesson 7: The Derivative

. . . . . .

Section2.1TheDerivativeandRatesofChange

V63.0121.027, CalculusI

September24, 2009

Announcements

I WebAssignmentsdueTuesday.I OfficeHourstoday3-4. SeeSectionCalendarforup-to-dateOH.

. . . . . .

RegardingWebAssignWefeelyourpain

. . . . . .

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.

. . . . . .

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




mtangent = limx→a

f(x) − f(a)x− a

. . . . . .

Graphicallyandnumerically

. .x

.y

..2

..4 .

x m

3 52.5 4.252.1 4.12.01 4.01limit 41.99 3.991.9 3.91.5 3.51 3

. . . . . .


. .x

.y

..2

..4 .

.

..3

..9

x m3 5

2.5 4.252.1 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 m3 52.5 4.25

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 m3 52.5 4.252.1 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 m3 52.5 4.252.1 4.12.01 4.01

limit 41.99 3.991.9 3.91.5 3.51 3

. . . . . .


. .x

.y

..2

..4 .

.

..1

..1

x m3 52.5 4.252.1 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 m3 52.5 4.252.1 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 m3 52.5 4.252.1 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 m3 52.5 4.252.1 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 m3 52.5 4.252.1 4.12.01 4.01limit 41.99 3.991.9 3.91.5 3.51 3

. . . . . .

Thetangentproblem




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)?

SolutionTheestimatedratesofgrowthare 0.000136, 0.75, and 0.000136.

. . . . . .

Populationgrowth



P(t) =3et

1 + et



. . . . . .

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≈

0.000136

r2000 ≈ P(0.1) − P(0)

0.1≈ 0.75

r2010 ≈ P(10 + 0.1) − P(10)

0.1≈ 0.000136

. . . . . .

Numericalevidence

r1990 ≈ P(−10 + 0.1) − P(−10)

0.1≈ 0.000136

r2000 ≈ P(0.1) − P(0)

0.1≈ 0.75

r2010 ≈ P(10 + 0.1) − P(10)

0.1≈ 0.000136

. . . . . .

Numericalevidence

r1990 ≈ P(−10 + 0.1) − P(−10)

0.1≈ 0.000136

r2000 ≈ P(0.1) − P(0)

0.1≈

0.75

r2010 ≈ P(10 + 0.1) − P(10)

0.1≈ 0.000136

. . . . . .

Numericalevidence

r1990 ≈ P(−10 + 0.1) − P(−10)

0.1≈ 0.000136

r2000 ≈ P(0.1) − P(0)

0.1≈ 0.75

r2010 ≈ P(10 + 0.1) − P(10)

0.1≈ 0.000136

. . . . . .

Numericalevidence

r1990 ≈ P(−10 + 0.1) − P(−10)

0.1≈ 0.000136

r2000 ≈ P(0.1) − P(0)

0.1≈ 0.75

r2010 ≈ P(10 + 0.1) − P(10)

0.1≈

0.000136

. . . . . .

Numericalevidence

r1990 ≈ P(−10 + 0.1) − P(−10)

0.1≈ 0.000136

r2000 ≈ P(0.1) − P(0)

0.1≈ 0.75

r2010 ≈ P(10 + 0.1) − P(10)

0.1≈ 0.000136

. . . . . .

Populationgrowth



P(t) =3et

1 + et



. . . . . .

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



C(q) = q3 − 12q2 + 60q



. . . . . .

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



C(q) = q3 − 12q2 + 60q



. . . . . .

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

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

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

.

. . . . . .


Example



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(well,nonpositive)onthatinterval

I If f isincreasingonaninterval, f′ ispositive(well,nonnegative)onthatinterval

. . . . . .



I Whatcanwesayaboutthisfunction f′?I If f isdecreasingonaninterval, f′ isnegative(well,

nonpositive)onthatinterval

I If f isincreasingonaninterval, f′ ispositive(well,nonnegative)onthatinterval

. . . . . .


Example



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(well,

nonpositive)onthatintervalI If f isincreasingonaninterval, f′ ispositive(well,

nonnegative)onthatinterval

. . . . . .


. .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 m3 52.5 4.252.1 4.12.01 4.01limit 41.99 3.991.9 3.91.5 3.51 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


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


f(x + ∆x) − f(x)∆x

≤ 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.

. . . . . .



Proof.Wehave

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


. . . . . .



Proof.Wehave

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


. . . . . .

Howcanafunctionfailtobedifferentiable?Kinks

. .x

.f(x)

. .x

.f′(x)

.

.

. . . . . .


. .x

.f(x)

. .x

.f′(x)

.

.

. . . . . .


. .x

.f(x)

. .x

.f′(x)

.

.

. . . . . .

Howcanafunctionfailtobedifferentiable?Cusps

. .x

.f(x)

. .x

.f′(x)

. . . . . .


. .x

.f(x)

. .x

.f′(x)

. . . . . .


. .x

.f(x)

. .x

.f′(x)

. . . . . .

Howcanafunctionfailtobedifferentiable?VerticalTangents

. .x

.f(x)

. .x

.f′(x)

. . . . . .


. .x

.f(x)

. .x

.f′(x)

. . . . . .


. .x

.f(x)

. .x

.f′(x)

. . . . . .

Howcanafunctionfailtobedifferentiable?Weird, Wild, Stuff

. .x

.f(x)

Thisfunctionisdifferentiableat 0.

. .x

.f′(x)

Butthederivativeisnotcontinuousat 0!

. . . . . .

Howcanafunctionfailtobedifferentiable?Weird, 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

Education

Lesson 7: The Derivative