lesson 7: the derivative (section 41 slides)

Section 2.1–2.2The Derivative and Rates of Change

The Derivative as a Function

V63.0121.041, Calculus I

New York University

September 27, 2010

Announcements

I Quiz this week in recitation on §§1.1–1.4I Get-to-know-you/photo due Friday October 1

. . . . . .

Announcements

I Quiz this week in recitationon §§1.1–1.4

I Get-to-know-you/photodue Friday October 1

V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 2 / 46

. . . . . .

Format of written work

Please:I Use scratch paper andcopy your final work ontofresh paper.

I Use loose-leaf paper (nottorn from a notebook).

I Write your name, lecturesection, assignmentnumber, recitation, anddate at the top.

I Staple your homeworktogether.

See the website for more information.

. . . . . .

Objectives for Section 2.1

I Understand and state thedefinition of the derivativeof a function at a point.

I Given a function and apoint in its domain, decideif the function isdifferentiable at the pointand find the value of thederivative at that point.

I Understand and giveseveral examples ofderivatives modeling ratesof change in science.

. . . . . .

Objectives for Section 2.2

I Given a function f, use thedefinition of the derivativeto find the derivativefunction f’.

I Given a function, find itssecond derivative.

I Given the graph of afunction, sketch the graphof its derivative.

. . . . . .

Outline

Rates of ChangeTangent LinesVelocityPopulation growthMarginal costs

The derivative, definedDerivatives of (some) power functionsWhat does f tell you about f′?

How can a function fail to be differentiable?

Other notations

The second derivative

. . . . . .

The tangent problem

ProblemGiven a curve and a point on the curve, find the slope of the linetangent to the curve at that point.

Example

Find the slope of the line tangent to the curve y = x2 at the point (2,4).

Upshot

If the curve is given by y = f(x), and the point on the curve is (a, f(a)),then the slope of the tangent line is given by

mtangent = limx→a

f(x)− f(a)x− a

. . . . . .

The tangent problem

Example

Upshot

mtangent = limx→a

f(x)− f(a)x− a

. . . . . .

Graphically and numerically

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 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 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

. . . . . .

..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

. . . . . .

..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

. . . . . .

..4 ..

..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

. . . . . .

..4 ..

..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

. . . . . .

..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

. . . . . .

..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 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

. . . . . .

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

. . . . . .

..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

. . . . . .

..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

. . . . . .

..4 ..

..3.61

x m =x2 − 22

x− 23 52.5 4.52.1 4.12.01 4.01

limit 41.99 3.99

1.5 3.51 3

. . . . . .

..4 ..

..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 3.91.5 3.51 3

. . . . . .

..4 ..

..1.99

..3.9601

x m =x2 − 22

x− 23 52.5 4.52.1 4.12.01 4.01

limit 4

1.9 3.91.5 3.51 3

. . . . . .

..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 3.991.9 3.91.5 3.51 3

. . . . . .

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

. . . . . .

..6.25

..4.41 .

..2.01

..4.0401

..2.25

..3.61.

..1.99

..3.9601

x m =x2 − 22

x− 23 52.5 4.52.1 4.12.01 4.01limit

1.99 3.991.9 3.91.5 3.51 3

. . . . . .

x m =x2 − 22

x− 23 52.5 4.52.1 4.12.01 4.01

41.99 3.991.9 3.91.5 3.51 3

. . . . . .

The tangent problem

Example

Upshot

mtangent = limx→a

f(x)− f(a)x− a

. . . . . .

Velocity.

ProblemGiven the position function of a moving object, find the velocity of the object ata certain instant in time.

Example

Drop a ball off the roof of the Silver Center so that its height can be describedby

h(t) = 50− 5t2

where t is seconds after dropping it and h is meters above the ground. Howfast is it falling one second after we drop it?

SolutionThe answer is

v = limt→1

(50− 5t2)− 45t− 1

= limt→1

5− 5t2

t− 1= lim

5(1− t)(1+ t)t− 1

= (−5) limt→1

(1+ t) = −5 · 2 = −10

. . . . . .

Numerical evidence

h(t) = 50− 5t2

Fill in the table:t vave =

h(t)− h(1)t− 1

2 − 15

1.5 − 12.51.1 − 10.51.01 − 10.051.001 − 10.005

. . . . . .

Numerical evidence

h(t) = 50− 5t2

h(t)− h(1)t− 1

2 − 151.5

− 12.51.1 − 10.51.01 − 10.051.001 − 10.005

. . . . . .

Numerical evidence

h(t) = 50− 5t2

h(t)− h(1)t− 1

2 − 151.5 − 12.5

1.1 − 10.51.01 − 10.051.001 − 10.005

. . . . . .

Numerical evidence

h(t) = 50− 5t2

h(t)− h(1)t− 1

2 − 151.5 − 12.51.1

− 10.51.01 − 10.051.001 − 10.005

. . . . . .

Numerical evidence

h(t) = 50− 5t2

h(t)− h(1)t− 1

2 − 151.5 − 12.51.1 − 10.5

1.01 − 10.051.001 − 10.005

. . . . . .

Numerical evidence

h(t) = 50− 5t2

h(t)− h(1)t− 1

2 − 151.5 − 12.51.1 − 10.51.01

− 10.051.001 − 10.005

. . . . . .

Numerical evidence

h(t) = 50− 5t2

h(t)− h(1)t− 1

2 − 151.5 − 12.51.1 − 10.51.01 − 10.05

1.001 − 10.005

. . . . . .

Numerical evidence

h(t) = 50− 5t2

h(t)− h(1)t− 1

2 − 151.5 − 12.51.1 − 10.51.01 − 10.051.001

− 10.005

. . . . . .

Numerical evidence

h(t) = 50− 5t2

h(t)− h(1)t− 1

2 − 151.5 − 12.51.1 − 10.51.01 − 10.051.001 − 10.005

. . . . . .

Velocity.

ProblemGiven the position function of a moving object, find the velocity of the object ata certain instant in time.

Example

Drop a ball off the roof of the Silver Center so that its height can be describedby

h(t) = 50− 5t2

where t is seconds after dropping it and h is meters above the ground. Howfast is it falling one second after we drop it?

SolutionThe answer is

v = limt→1

(50− 5t2)− 45t− 1

= limt→1

5− 5t2

t− 1= lim

5(1− t)(1+ t)t− 1

= (−5) limt→1

(1+ t) = −5 · 2 = −10

. . . . . .

Velocity in general

Upshot

If the height function is given byh(t), the instantaneous velocityat time t0 is given by

v = limt→t0

h(t)− h(t0)t− t0

= lim∆t→0

h(t0 +∆t)− h(t0)∆t

..y = h(t).

..h(t0)

..h(t0 +∆t)

. . . . . .

Population growth

ProblemGiven the population function of a group of organisms, find the rate ofgrowth of the population at a particular instant.

Example

Suppose the population of fish in the East River is given by the function

P(t) =3et

where t is in years since 2000 and P is in millions of fish. Is the fishpopulation growing fastest in 1990, 2000, or 2010? (Estimatenumerically)

AnswerWe estimate the rates of growth to be 0.000143229, 0.749376, and0.0001296. So the population is growing fastest in 2000.

. . . . . .

Population growth

ProblemGiven the population function of a group of organisms, find the rate ofgrowth of the population at a particular instant.

Example

P(t) =3et

where t is in years since 2000 and P is in millions of fish. Is the fishpopulation growing fastest in 1990, 2000, or 2010? (Estimatenumerically)

AnswerWe estimate the rates of growth to be 0.000143229, 0.749376, and0.0001296. So the population is growing fastest in 2000.

. . . . . .

Derivation

SolutionLet ∆t be an increment in time and ∆P the corresponding change inpopulation:

∆P = P(t+∆t)− P(t)

This depends on ∆t, so ideally we would want

lim∆t→0

∆P∆t

= lim∆t→0

(3et+∆t

1+ et+∆t −3et

But rather than compute a complicated limit analytically, let usapproximate numerically. We will try a small ∆t, for instance 0.1.

. . . . . .

Derivation

SolutionLet ∆t be an increment in time and ∆P the corresponding change inpopulation:

∆P = P(t+∆t)− P(t)

This depends on ∆t, so ideally we would want

lim∆t→0

∆P∆t

= lim∆t→0

(3et+∆t

1+ et+∆t −3et

)But rather than compute a complicated limit analytically, let usapproximate numerically. We will try a small ∆t, for instance 0.1.

. . . . . .

Numerical evidence

Solution (Continued)

To approximate the population change in year n, use the difference

quotientP(t+∆t)− P(t)

∆t, where ∆t = 0.1 and t = n− 2000.

≈ P(−10+ 0.1)− P(−10)0.1

(3e−9.9

1+ e−9.9 − 3e−10

1+ e−10

)= 0.000143229

≈ P(0.1)− P(0)0.1

(3e0.1

1+ e0.1− 3e0

)= 0.749376

≈ P(10+ 0.1)− P(10)0.1

(3e10.1

1+ e10.1− 3e10

1+ e10

)= 0.0001296

. . . . . .

Numerical evidence

∆t, where ∆t = 0.1 and t = n− 2000.

r1990 ≈ P(−10+ 0.1)− P(−10)0.1

(3e−9.9

1+ e−9.9 − 3e−10

1+ e−10

)= 0.000143229

r2000 ≈ P(0.1)− P(0)0.1

(3e0.1

1+ e0.1− 3e0

)= 0.749376

r2010 ≈ P(10+ 0.1)− P(10)0.1

(3e10.1

1+ e10.1− 3e10

1+ e10

)= 0.0001296

. . . . . .

Numerical evidence

∆t, where ∆t = 0.1 and t = n− 2000.

r1990 ≈ P(−10+ 0.1)− P(−10)0.1

(3e−9.9

1+ e−9.9 − 3e−10

1+ e−10

= 0.000143229

r2000 ≈ P(0.1)− P(0)0.1

(3e0.1

1+ e0.1− 3e0

= 0.749376

r2010 ≈ P(10+ 0.1)− P(10)0.1

(3e10.1

1+ e10.1− 3e10

1+ e10

= 0.0001296

. . . . . .

Numerical evidence

∆t, where ∆t = 0.1 and t = n− 2000.

r1990 ≈ P(−10+ 0.1)− P(−10)0.1

(3e−9.9

1+ e−9.9 − 3e−10

1+ e−10

)= 0.000143229

r2000 ≈ P(0.1)− P(0)0.1

(3e0.1

1+ e0.1− 3e0

= 0.749376

r2010 ≈ P(10+ 0.1)− P(10)0.1

(3e10.1

1+ e10.1− 3e10

1+ e10

= 0.0001296

. . . . . .

Numerical evidence

∆t, where ∆t = 0.1 and t = n− 2000.

r1990 ≈ P(−10+ 0.1)− P(−10)0.1

(3e−9.9

1+ e−9.9 − 3e−10

1+ e−10

)= 0.000143229

r2000 ≈ P(0.1)− P(0)0.1

(3e0.1

1+ e0.1− 3e0

)= 0.749376

r2010 ≈ P(10+ 0.1)− P(10)0.1

(3e10.1

1+ e10.1− 3e10

1+ e10

= 0.0001296

. . . . . .

Numerical evidence

∆t, where ∆t = 0.1 and t = n− 2000.

r1990 ≈ P(−10+ 0.1)− P(−10)0.1

(3e−9.9

1+ e−9.9 − 3e−10

1+ e−10

)= 0.000143229

r2000 ≈ P(0.1)− P(0)0.1

(3e0.1

1+ e0.1− 3e0

)= 0.749376

r2010 ≈ P(10+ 0.1)− P(10)0.1

(3e10.1

1+ e10.1− 3e10

1+ e10

)= 0.0001296

. . . . . .

Population growth.

ProblemGiven the population function of a group of organisms, find the rate of growthof the population at a particular instant.

Example

P(t) =3et

where t is in years since 2000 and P is in millions of fish. Is the fish populationgrowing fastest in 1990, 2000, or 2010? (Estimate numerically)

AnswerWe estimate the rates of growth to be 0.000143229, 0.749376, and 0.0001296.So the population is growing fastest in 2000.

. . . . . .

Population growth in general

Upshot

The instantaneous population growth is given by

lim∆t→0

P(t+∆t)− P(t)∆t

. . . . . .

Marginal costs

ProblemGiven the production cost of a good, find the marginal cost ofproduction after having produced a certain quantity.

Example

Suppose the cost of producing q tons of rice on our paddy in a year is

C(q) = q3 − 12q2 + 60q

We are currently producing 5 tons a year. Should we change that?

Answer

If q = 5, then C = 125, ∆C = 19, while AC = 25. So we shouldproduce more to lower average costs.

. . . . . .

Marginal costs

Example

C(q) = q3 − 12q2 + 60q

Answer

If q = 5, then C = 125, ∆C = 19, while AC = 25. So we shouldproduce more to lower average costs.

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

q C(q)

AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)

112 28 13

125 25 19

144 24 31

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

q C(q)

AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)

125 25 19

144 24 31

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

q C(q)

AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)

144 24 31

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

q C(q)

AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

q C(q) AC(q) = C(q)/q

∆C = C(q+ 1)− C(q)

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

∆C = C(q+ 1)− C(q)

4 112 28

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

∆C = C(q+ 1)− C(q)

4 112 28

5 125 25

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

∆C = C(q+ 1)− C(q)

4 112 28

5 125 25

6 144 24

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

q C(q) AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)4 112 28

5 125 25

6 144 24

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

q C(q) AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)4 112 28 135 125 25

6 144 24

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

q C(q) AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)4 112 28 135 125 25 196 144 24

. . . . . .

Comparisons

Solution

C(q) = q3 − 12q2 + 60q

Fill in the table:

q C(q) AC(q) = C(q)/q ∆C = C(q+ 1)− C(q)4 112 28 135 125 25 196 144 24 31

. . . . . .

Marginal costs

Example

C(q) = q3 − 12q2 + 60q

AnswerIf q = 5, then C = 125, ∆C = 19, while AC = 25. So we shouldproduce more to lower average costs.

. . . . . .

Marginal Cost in General

Upshot

I The incremental cost

∆C = C(q+ 1)− C(q)

is useful, but is still only an average rate of change.

I The marginal cost after producing q given by

MC = lim∆q→0

C(q+∆q)− C(q)∆q

is more useful since it’s an instantaneous rate of change.

. . . . . .

Marginal Cost in General

Upshot

I The incremental cost

∆C = C(q+ 1)− C(q)

is useful, but is still only an average rate of change.I The marginal cost after producing q given by

MC = lim∆q→0

C(q+∆q)− C(q)∆q

is more useful since it’s an instantaneous rate of change.

. . . . . .

Outline

Other notations

. . . . . .

The definition

All of these rates of change are found the same way!

DefinitionLet f be a function and a a point in the domain of f. If the limit

f′(a) = limh→0

f(a+ h)− f(a)h

= limx→a

f(x)− f(a)x− a

exists, the function is said to be differentiable at a and f′(a) is thederivative of f at a.

. . . . . .

The definition

All of these rates of change are found the same way!

DefinitionLet f be a function and a a point in the domain of f. If the limit

f′(a) = limh→0

f(a+ h)− f(a)h

= limx→a

f(x)− f(a)x− a

exists, the function is said to be differentiable at a and f′(a) is thederivative of f at a.

. . . . . .

Derivative of the squaring function

Example

Suppose f(x) = x2. Use the definition of derivative to find f′(a).

Solution

f′(a) = limh→0

f(a+ h)− f(a)h

= limh→0

(a+ h)2 − a2

= limh→0

(a2 + 2ah+ h2)− a2

h= lim

2ah+ h2

h= lim

h→0(2a+ h) = 2a.

. . . . . .

Derivative of the squaring function

Example

Suppose f(x) = x2. Use the definition of derivative to find f′(a).

Solution

f′(a) = limh→0

f(a+ h)− f(a)h

= limh→0

(a+ h)2 − a2

= limh→0

(a2 + 2ah+ h2)− a2

h= lim

2ah+ h2

h= lim

h→0(2a+ h) = 2a.

. . . . . .

Derivative of the reciprocal function

Example

Suppose f(x) =1x. Use the

definition of the derivative tofind f′(2).

Solution

f′(2) = limx→2

1/x− 1/2x− 2

= limx→2

2− x2x(x− 2)

= limx→2

−12x

= −14

. . . . . .

Example

Solution

f′(2) = limx→2

1/x− 1/2x− 2

= limx→2

2− x2x(x− 2)

= limx→2

−12x

= −14

. . . . . .

The Sure-Fire Sally Rule (SFSR) for adding FractionsIn anticipation of the question, “How did you get that?"

ab± c

ad± bcbd

1x− 1

2x− 2

2− x2x

x− 2

=2− x

2x(x− 2)

. . . . . .

The Sure-Fire Sally Rule (SFSR) for adding FractionsIn anticipation of the question, “How did you get that?"

ab± c

ad± bcbd

1x− 1

2x− 2

2− x2x

x− 2

=2− x

2x(x− 2)

. . . . . .

What does f tell you about f′?

I If f is a function, we can compute the derivative f′(x) at each pointx where f is differentiable, and come up with another function, thederivative function.

I What can we say about this function f′?

I If f is decreasing on an interval, f′ is negative (technically,nonpositive) on that interval

I If f is increasing on an interval, f′ is positive (technically,nonnegative) on that interval

. . . . . .

I What can we say about this function f′?I If f is decreasing on an interval, f′ is negative (technically,

nonpositive) on that interval

I If f is increasing on an interval, f′ is positive (technically,nonnegative) on that interval

. . . . . .

Example

Solution

f′(2) = limx→2

1/x− 1/2x− 2

= limx→2

2− x2x(x− 2)

= limx→2

−12x

= −14

. . . . . .

I What can we say about this function f′?I If f is decreasing on an interval, f′ is negative (technically,

nonpositive) on that intervalI If f is increasing on an interval, f′ is positive (technically,

nonnegative) on that interval

. . . . . .

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

. . . . . .

What does f tell you about f′?.

FactIf f is decreasing on (a,b), then f′ ≤ 0 on (a,b).

Proof.If f is decreasing on (a,b), and ∆x > 0, then

f(x+∆x) < f(x) =⇒ f(x+∆x)− f(x)∆x

But if ∆x < 0, then x+∆x < x, and

f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x

still! Either way,f(x+∆x)− f(x)

∆x< 0, so

f′(x) = lim∆x→0

f(x+∆x)− f(x)∆x

. . . . . .

f(x+∆x) < f(x) =⇒ f(x+∆x)− f(x)∆x

f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x

∆x< 0, so

. . . . . .

f(x+∆x) < f(x) =⇒ f(x+∆x)− f(x)∆x

f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x

still!

Either way,f(x+∆x)− f(x)

∆x< 0, so

. . . . . .

f(x+∆x) < f(x) =⇒ f(x+∆x)− f(x)∆x

f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x

∆x< 0, so

. . . . . .

Outline

Other notations

. . . . . .

Differentiability is super-continuity

TheoremIf f is differentiable at a, then f is continuous at a.

Proof.We have

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

Note the proper use of the limit law: if the factors each have a limit ata, the limit of the product is the product of the limits.

. . . . . .

Proof.We have

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.We have

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

. . . . . .

Differentiability FAILKinks

Example

Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.

.f′(x)

. . . . . .

Example

.f′(x)

. . . . . .

Example

.f′(x)

. . . . . .

Differentiability FAILCusps

Example

.f′(x)

. . . . . .

Example

.f′(x)

. . . . . .

Example

.f′(x)

. . . . . .

Example

.f′(x)

. . . . . .

Differentiability FAILVertical Tangents

Example

.f′(x)

. . . . . .

Example

.f′(x)

. . . . . .

Example

.f′(x)

. . . . . .

Example

.f′(x)

. . . . . .

Differentiability FAILWeird, Wild, Stuff

Example

This function is differentiableat 0.

.f′(x)

But the derivative is notcontinuous at 0!

. . . . . .

Example

.f′(x)

. . . . . .

Example

.f′(x)

. . . . . .

Example

.f′(x)

. . . . . .

Outline

Other notations

. . . . . .

Notation

I Newtonian notation

f′(x) y′(x) y′

I Leibnizian notation

f(x)dfdx

These all mean the same thing.

. . . . . .

Meet the Mathematician: Isaac Newton

I English, 1643–1727I Professor at Cambridge(England)

I Philosophiae NaturalisPrincipia Mathematicapublished 1687

. . . . . .

Meet the Mathematician: Gottfried Leibniz

I German, 1646–1716I Eminent philosopher aswell as mathematician

I Contemporarily disgracedby the calculus prioritydispute

. . . . . .

Outline

Other notations

. . . . . .

If f is a function, so is f′, and we can seek its derivative.

f′′ = (f′)′

It measures the rate of change of the rate of change!

Leibniziannotation:

d2ydx2

dx2f(x)

d2fdx2

. . . . . .

If f is a function, so is f′, and we can seek its derivative.

f′′ = (f′)′

It measures the rate of change of the rate of change! Leibniziannotation:

d2ydx2

dx2f(x)

d2fdx2

. . . . . .

function, derivative, second derivative

.y.f(x) = x2

.f′(x) = 2x

.f′′(x) = 2

. . . . . .

What have we learned today?

I The derivative measures instantaneous rate of changeI The derivative has many interpretations: slope of the tangent line,velocity, marginal quantities, etc.

I The derivative reflects the monotonicity (increasing or decreasing)of the graph

lesson 7: the derivative (section 41 slides)

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