lesson 7: the derivative (slides)

.... NYUMathematics

Sec on 2.1–2.2The Deriva ve

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

February 14, 2011

Announcements

I Quiz this week onSec ons 1.1–1.4

I No class Monday,February 21

ObjectivesThe Derivative

I Understand and state the defini on ofthe deriva ve of a func on at a point.

I Given a func on and a point in itsdomain, decide if the func on isdifferen able at the point and find thevalue of the deriva ve at that point.

I Understand and give several examplesof deriva ves modeling rates of changein science.

ObjectivesThe Derivative as a Function

I Given a func on f, use the defini on ofthe deriva ve to find the deriva vefunc on f’.

I Given a func on, find its secondderiva ve.

I Given the graph of a func on, sketchthe graph of its deriva ve.

OutlineRates of Change

Tangent LinesVelocityPopula on growthMarginal costs

The deriva ve, definedDeriva ves of (some) power func onsWhat does f tell you about f′?

How can a func on fail to be differen able?Other nota onsThe second deriva ve

The tangent problemA geometric rate of change

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

Solu onIf 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

A tangent problem

Example

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

Graphically and numerically

.. 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

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

....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.

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

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

The tangent problemA geometric rate of change

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

Solu onIf 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 velocity problemKinematics—Physical rates of change

ProblemGiven the posi on func on of a moving object, find the velocity ofthe object at a certain instant in me.

A velocity problemExample

Drop a ball off the roof of theSilver Center so that its height canbe described by

h(t) = 50− 5t2

where t is seconds a er droppingit and h is meters above theground. How fast is it falling onesecond a er we drop it?

Solu onThe answer is

v = limt→1

(50− 5t2)− 45t− 1

= limt→1

5− 5t2

t− 1

= limt→1

5(1− t)(1+ t)t− 1

= (−5) limt→1

(1+ t)

= −5 · 2 = −10

Numerical evidenceh(t) = 50− 5t2

Fill in the table:

t vave =h(t)− h(1)

t− 12 − 15

1.5 − 12.51.1 − 10.51.01 − 10.051.001 − 10.005


Fill in the table:


t− 12 − 151.5

− 12.51.1 − 10.51.01 − 10.051.001 − 10.005


Fill in the table:


t− 12 − 151.5 − 12.5

1.1 − 10.51.01 − 10.051.001 − 10.005


Fill in the table:


t− 12 − 151.5 − 12.51.1

− 10.51.01 − 10.051.001 − 10.005


Fill in the table:


t− 12 − 151.5 − 12.51.1 − 10.5

1.01 − 10.051.001 − 10.005


Fill in the table:


t− 12 − 151.5 − 12.51.1 − 10.51.01

− 10.051.001 − 10.005


Fill in the table:


t− 12 − 151.5 − 12.51.1 − 10.51.01 − 10.05

1.001 − 10.005


Fill in the table:


t− 12 − 151.5 − 12.51.1 − 10.51.01 − 10.051.001

− 10.005


Fill in the table:


t− 12 − 151.5 − 12.51.1 − 10.51.01 − 10.051.001 − 10.005

A velocity problemExample

Drop a ball off the roof of theSilver Center so that its height canbe described by

h(t) = 50− 5t2

where t is seconds a er droppingit and h is meters above theground. How fast is it falling onesecond a er we drop it?

Solu onThe answer is

v = limt→1

(50− 5t2)− 45t− 1

= limt→1

5− 5t2

t− 1

= limt→1

5(1− t)(1+ t)t− 1

= (−5) limt→1

(1+ t)

= −5 · 2 = −10

Velocity in generalUpshot

If the height func on is givenby h(t), the instantaneousvelocity at me t0 is given by

v = limt→t0

h(t)− h(t0)t− t0

= lim∆t→0

h(t0 +∆t)− h(t0)∆t ... t..

y = h(t)

....t0

..t

..

h(t0)

..

h(t0 +∆t)

. ∆t.

∆h

Population growthBiological Rates of Change

ProblemGiven the popula on func on of a group of organisms, find the rateof growth of the popula on at a par cular instant.

Solu onThe instantaneous popula on growth is given by

lim∆t→0

P(t+∆t)− P(t)∆t

Population growth exampleExample

Suppose the popula on of fish in the East River is given by thefunc on

P(t) =3et

1+ etwhere t is in years since 2000 and P is in millions of fish. Is the fishpopula on growing fastest in 1990, 2000, or 2010? (Es matenumerically)

AnswerWe es mate the rates of growth to be 0.000143229, 0.749376, and0.0001296. So the popula on is growing fastest in 2000.

DerivationSolu onLet∆t be an increment in me and∆P the corresponding change inpopula on:

∆P = P(t+∆t)− P(t)

This depends on∆t, so ideally we would want

lim∆t→0

∆P∆t

= lim∆t→0

1∆t

(3et+∆t

1+ et+∆t −3et

1+ et

)

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

DerivationSolu onLet∆t be an increment in me and∆P the corresponding change inpopula on:

∆P = P(t+∆t)− P(t)

This depends on∆t, so ideally we would want

lim∆t→0

∆P∆t

= lim∆t→0

1∆t

(3et+∆t

1+ et+∆t −3et

1+ et

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

Numerical evidenceSolu on (Con nued)

To approximate the popula on change in year n, use the difference

quo entP(t+∆t)− P(t)

∆t, where∆t = 0.1 and t = n− 2000.

r1990

≈ P(−10+ 0.1)− P(−10)0.1

=10.1

(3e−9.9

1+ e−9.9 −3e−10

1+ e−10

)= 0.000143229

r2000

≈ P(0.1)− P(0)0.1

=10.1

(3e0.1

1+ e0.1− 3e0

1+ e0

)= 0.749376




∆t, where∆t = 0.1 and t = n− 2000.

r1990 ≈P(−10+ 0.1)− P(−10)

0.1

=10.1

(3e−9.9

1+ e−9.9 −3e−10

1+ e−10

)= 0.000143229

r2000 ≈P(0.1)− P(0)

0.1

=10.1

(3e0.1

1+ e0.1− 3e0

1+ e0

)= 0.749376




∆t, where∆t = 0.1 and t = n− 2000.

r1990 ≈P(−10+ 0.1)− P(−10)

0.1=

10.1

(3e−9.9

1+ e−9.9 −3e−10

1+ e−10

)

= 0.000143229

r2000 ≈P(0.1)− P(0)

0.1=

10.1

(3e0.1

1+ e0.1− 3e0

1+ e0

)

= 0.749376




∆t, where∆t = 0.1 and t = n− 2000.

r1990 ≈P(−10+ 0.1)− P(−10)

0.1=

10.1

(3e−9.9

1+ e−9.9 −3e−10

1+ e−10

)= 0.000143229

r2000 ≈P(0.1)− P(0)

0.1=

10.1

(3e0.1

1+ e0.1− 3e0

1+ e0

)

= 0.749376




∆t, where∆t = 0.1 and t = n− 2000.

r1990 ≈P(−10+ 0.1)− P(−10)

0.1=

10.1

(3e−9.9

1+ e−9.9 −3e−10

1+ e−10

)= 0.000143229

r2000 ≈P(0.1)− P(0)

0.1=

10.1

(3e0.1

1+ e0.1− 3e0

1+ e0

)= 0.749376

Solu on (Con nued)

r2010

≈ P(10+ 0.1)− P(10)0.1

=10.1

(3e10.1

1+ e10.1− 3e10

1+ e10

)= 0.0001296

Solu on (Con nued)

r2010 ≈P(10+ 0.1)− P(10)

0.1

=10.1

(3e10.1

1+ e10.1− 3e10

1+ e10

)= 0.0001296

Solu on (Con nued)

r2010 ≈P(10+ 0.1)− P(10)

0.1=

10.1

(3e10.1

1+ e10.1− 3e10

1+ e10

)

= 0.0001296

Solu on (Con nued)

r2010 ≈P(10+ 0.1)− P(10)

0.1=

10.1

(3e10.1

1+ e10.1− 3e10

1+ e10

)= 0.0001296

Population growth exampleExample

Suppose the popula on of fish in the East River is given by thefunc on

P(t) =3et

1+ etwhere t is in years since 2000 and P is in millions of fish. Is the fishpopula on growing fastest in 1990, 2000, or 2010? (Es matenumerically)

AnswerWe es mate the rates of growth to be 0.000143229, 0.749376, and0.0001296. So the popula on is growing fastest in 2000.

Population growthBiological Rates of Change

ProblemGiven the popula on func on of a group of organisms, find the rateof growth of the popula on at a par cular instant.

Solu onThe instantaneous popula on growth is given by

lim∆t→0

P(t+∆t)− P(t)∆t

Marginal costsRates of change in economics

ProblemGiven the produc on cost of a good, find the marginal cost ofproduc on a er having produced a certain quan ty.

Solu onThe marginal cost a er producing q is given by

MC = lim∆q→0

C(q+∆q)− C(q)∆q

Marginal Cost ExampleExample

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?

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

ComparisonsSolu on

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 13

5

125 25 19

6

144 24 31

ComparisonsSolu on

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 13

5

125 25 19

6

144 24 31

ComparisonsSolu on

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 13

5 125

25 19

6

144 24 31

ComparisonsSolu on

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 13

5 125

25 19

6 144

24 31

ComparisonsSolu on

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 13

5 125

25 19

6 144

24 31

ComparisonsSolu on

C(q) = q3 − 12q2 + 60q

Fill in the table:


∆C = C(q+ 1)− C(q)

4 112 28

13

5 125

25 19

6 144

24 31

ComparisonsSolu on

C(q) = q3 − 12q2 + 60q

Fill in the table:


∆C = C(q+ 1)− C(q)

4 112 28

13

5 125 25

19

6 144

24 31

ComparisonsSolu on

C(q) = q3 − 12q2 + 60q

Fill in the table:


∆C = C(q+ 1)− C(q)

4 112 28

13

5 125 25

19

6 144 24

31

ComparisonsSolu on

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

13

5 125 25

19

6 144 24

31

ComparisonsSolu on

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

19

6 144 24

31

ComparisonsSolu on

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

ComparisonsSolu on

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 Cost ExampleExample

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?

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

Marginal costsRates of change in economics

ProblemGiven the produc on cost of a good, find the marginal cost ofproduc on a er having produced a certain quan ty.

Solu onThe marginal cost a er producing q is given by

MC = lim∆q→0

C(q+∆q)− C(q)∆q

The definitionAll of these rates of change are found the same way!

Defini onLet f be a func on 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 func on is said to be differen able at a and f′(a) is thederiva ve of f at a.

Derivative of the squaring functionExample

Suppose f(x) = x2. Use the defini on of deriva ve to find f′(a).

Solu on

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



Solu on

f′(a) = limh→0

f(a+ h)− f(a)h

= limh→0

(a+ h)2 − a2

h

= limh→0

(a2 + 2ah+ h2)− a2

h

= limh→0

2ah+ h2

h= lim

h→0(2a+ h) = 2a



Solu on

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

= limh→0

(2a+ h) = 2a



Solu on

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



Solu on

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

Derivative of the reciprocalExample

Suppose f(x) =1x. Use the defini on of the deriva ve to find f′(2).

Solu on

f′(2) = limx→2

1/x− 1/2x− 2

= limx→2

2− x2x(x− 2)

= limx→2

−12x

= −14

..x

.

y

.

“Can you do it the other way?”Same limit, different form

Solu on

f′(2) = limh→0

f(2+ h)− f(2)h

= limh→0

12+h −

12

h

= limh→0

2− (2+ h)2h(2+ h)

= limh→0

−h2h(2+ h)

= limh→0

−12(2+ h)

= −14

“How did you get that?”The Sure-Fire Sally Rule (SFSR) for adding fractions

Fact

ab± c

d=

ad± bcbd

1x− 1

2x− 2

=

2− x2x

x− 2=

2− x2x(x− 2)

Paul Sally

“How did you get that?”The Sure-Fire Sally Rule (SFSR) for adding fractions

Fact

ab± c

d=

ad± bcbd

1x− 1

2x− 2

=

2− x2x

x− 2=

2− x2x(x− 2) Paul Sally

What does f tell you about f′?

I If f is a func on, we can compute the deriva ve f′(x) at eachpoint x where f is differen able, and come up with anotherfunc on, the deriva ve func on.

I What can we say about this func on f′?

I If f is decreasing on an interval, f′ is nega ve (technically, nonposi ve)on that interval

I If f is increasing on an interval, f′ is posi ve (technically, nonnega ve)on that interval



I What can we say about this func on f′?I If f is decreasing on an interval, f′ is nega ve (technically, nonposi ve)

on that interval

I If f is increasing on an interval, f′ is posi ve (technically, nonnega ve)on that interval

Derivative of the reciprocalExample

Suppose f(x) =1x. Use the defini on of the deriva ve to find f′(2).

Solu on

f′(2) = limx→2

1/x− 1/2x− 2

= limx→2

2− x2x(x− 2)

= limx→2

−12x

= −14

..x

.

y

.



I What can we say about this func on f′?I If f is decreasing on an interval, f′ is nega ve (technically, nonposi ve)

on that intervalI If f is increasing on an interval, f′ is posi ve (technically, nonnega ve)

on that interval


.. 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

What does f tell you about f′?FactIf f is decreasing on the open interval (a, b), then f′ ≤ 0 on (a, b).

Picture Proof.

If f is decreasing, then all secant linespoint downward, hence havenega ve slope. The deriva ve is alimit of slopes of secant lines, whichare all nega ve, so the limit must be≤ 0. ..

x.

y

......

What does f tell you about f′?FactIf f is decreasing on on the open interval (a, b), then f′ ≤ 0 on (a, b).

The Real Proof.

I If∆x > 0, then

f(x+∆x) < f(x) =⇒ f(x+∆x)− f(x)∆x

< 0


The Real Proof.

I If∆x > 0, then

f(x+∆x) < f(x) =⇒ f(x+∆x)− f(x)∆x

< 0

I If∆x < 0, then x+∆x < x, and

f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x

< 0


The Real Proof.

I Either way,f(x+∆x)− f(x)

∆x< 0, so

f′(x) = lim∆x→0

f(x+∆x)− f(x)∆x

≤ 0

Going the Other Way?

Ques on

If a func on has a nega ve deriva ve on an interval, must it bedecreasing on that interval?

AnswerMaybe.

Differentiability is super-continuityTheoremIf f is differen able at a, then f is con nuous 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


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 le -hand side below. Sketch the graph ofthe deriva ve f′.

.. x.

f(x)

.. x.

f′(x)

.

.

Differentiability FAILKinks

Example


.. x.

f(x)

.. x.

f′(x)

..

Differentiability FAILCusps

Example


.. x.

f(x)

.. x.

f′(x)

Differentiability FAILVertical Tangents

Example


.. x.

f(x)

.. x.

f′(x)

Differentiability FAILWeird, Wild, Stuff

Example

.. x.

f(x)

This func on is differen ableat 0.

.. x.

f′(x)

.

But the deriva ve is notcon nuous at 0!

Notation

I Newtonian nota on

f′(x) y′(x) y′

I Leibnizian nota on

dydx

ddx

f(x)dfdx

These all mean the same thing.

Meet the MathematicianIsaac Newton

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

I Philosophiae NaturalisPrincipia Mathema capublished 1687

Meet the MathematicianGottfried Leibniz

I German, 1646–1716I Eminent philosopher aswell as mathema cian

I Contemporarily disgracedby the calculus prioritydispute

The second derivative

If f is a func on, so is f′, and we can seek its deriva ve.

f′′ = (f′)′

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

Leibniziannota on:

d2ydx2

d2

dx2f(x)

d2fdx2

The second derivative

If f is a func on, so is f′, and we can seek its deriva ve.

f′′ = (f′)′

It measures the rate of change of the rate of change! Leibniziannota on:

d2ydx2

d2

dx2f(x)

d2fdx2

Function, derivative, second derivative

.. x.

y

.

f(x) = x2

.

f′(x) = 2x

.f′′(x) = 2

SummaryWhat have we learned today?

I The deriva ve measures instantaneous rate of changeI The deriva ve has many interpreta ons: slope of the tangentline, velocity, marginal quan es, etc.

I The deriva ve reflects the monotonicity (increasing-ness ordecreasing-ness) of the graph

lesson 7: the derivative (slides)

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