Lesson 7: The Derivative (Section 21 slide)

Section 2.1–2.2
The Derivative and Rates of Change
The Derivative as a Function

V63.0121.021, Calculus I

New York University

September 28, 2010

Announcements

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

. . . . . .

Announcements

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

. . . . . .

V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 2 / 49

Format of written work

Please:
Use scratch paper and
copy your final work onto
fresh paper.
Use loose-leaf paper (not
torn from a notebook).
Write your name, lecture
section, assignment
number, recitation, and
date at the top.
Staple your homework
together.
See the website for more information.

. . . . . .


Objectives for Section 2.1

Understand and state the
definition of the derivative
of a function at a point.
Given a function and a
point in its domain, decide
if the function is
differentiable at the point
and find the value of the
derivative at that point.
Understand and give
several examples of
derivatives modeling rates
of change in science.

. . . . . .


Objectives for Section 2.2

Given a function f, use the
definition of the derivative
to find the derivative
function f’.
Given a function, find its
second derivative.
Given the graph of a
function, sketch the graph
of its derivative.

. . . . . .


Outline

Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs

The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′ ?

How can a function fail to be differentiable?

Other notations

The second derivative

. . . . . .


The tangent problem

Problem
Given a curve and a point on the curve, find the slope of the line
tangent to the curve at that point.

. . . . . .


The tangent problem

Problem

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

. . . . . .


Graphically and numerically

y
.
x2 − 22
x m=
x−2

. .
4 .

. . x
.
2
.
. . . . . .



y
.
x2 − 22
x m=
x−2
3
. .
9 .

. .
4 .

. . . x
.
2
. 3
.
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
. .
9 .

. .
4 .

. . . x
.
2
. 3
.
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5

. .25 .
6 .

. .
4 .

. . . x
.
22
. . .5
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5

. .25 .
6 .

. .
4 .

. . . x
.
22
. . .5
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1

. .41 .
4 .
. .
4 .

. .. x
.
2
.. .1
2
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1

. .41 .
4 .
. .
4 .

. .. x
.
2
.. .1
2
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01

. .0401 .
4 4
. .

. . x
.
2.
. .01
2
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01

. .0401 .
4 4
. .

. . x
.
2.
. .01
2
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01

. .
4 .

1
. .
1 .
. . . x
.
1
. 2
.
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01

. .
4 .

1 3
. .
1 .
. . . x
.
1
. 2
.
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01

. .
4 .
1.5
. .25 .
2 .
1 3
. . . x
.
1 2
. .5 .
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01

. .
4 .
1.5 3.5
. .25 .
2 .
1 3
. . . x
.
1 2
. .5 .
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01

. .
4 .
. .61 .
3 . 1.9
1.5 3.5
1 3
. .. x
.
1.
. .9
2
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01

. .
4 .
. .61 .
3 . 1.9 3.9
1.5 3.5
1 3
. .. x
.
1.
. .9
2
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01

1.99
. .9601 .
3 4
. .
1.9 3.9
1.5 3.5
1 3
. . x
.
1.
. .99
2
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01

1.99 3.99
. .9601 .
3 4
. .
1.9 3.9
1.5 3.5
1 3
. . x
.
1.
. .99
2
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01

1.99 3.99
. .
4 .
1.9 3.9
1.5 3.5
1 3
. . x
.
2
.
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
. .
9 .
2.5 4.5
2.1 4.1
2.01 4.01
. .25 .
6 .
limit
. .41 .
4 . 1.99 3.99
. .0401 .
4.9601 .
3 . .61
3
4
. .. 1.9 3.9
1.5 3.5
. .25 .
2 .
1 3
. .
1 .
. . . ... . . x
.
1 1 . .. .1 3
. . .5 .99 .5 .
12 .
2.9 2
2
.01
. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
. .
9 .
2.5 4.5
2.1 4.1
2.01 4.01
. .25 .
6 .
limit 4
. .41 .
4 . 1.99 3.99
. .0401 .
4.9601 .
3 . .61
3
4
. .. 1.9 3.9
1.5 3.5
. .25 .
2 .
1 3
. .
1 .
. . . ... . . x
.
1 1 . .. .1 3
. . .5 .99 .5 .
12 .
2.9 2
2
.01
. . . . . .


The tangent problem

Problem

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

f(x) − f(a)
mtangent = lim
x→a x−a

. . . . . .


Velocity
.
Problem
Given the position function of a moving object, find the velocity of the object at
a certain instant in time.

Example
Drop a ball off the roof of the Silver Center so that its height can be described
by
h(t) = 50 − 5t2
where t is seconds after dropping it and h is meters above the ground. How
fast is it falling one second after we drop it?

. . . . . . .


Numerical evidence

h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15

. . . . . .


Numerical evidence

h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5

. . . . . .


Numerical evidence

h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5

. . . . . .


Numerical evidence

h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1

. . . . . .


Numerical evidence

h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5

. . . . . .


Numerical evidence

h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01

. . . . . .


Numerical evidence

h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05

. . . . . .


Numerical evidence

h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001

. . . . . .


Numerical evidence

h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001 − 10.005

. . . . . .


Velocity
.
Problem

Example
by
h(t) = 50 − 5t2

Solution
The answer is

(50 − 5t2 ) − 45
v = lim
t→1 t−1

. . . . . . .


Velocity
.
Problem

Example
by
h(t) = 50 − 5t2

Solution
The answer is

(50 − 5t2 ) − 45 5 − 5t2
v = lim = lim
t→1 t−1 t→1 t − 1

. . . . . . .


Velocity
.
Problem

Example
by
h(t) = 50 − 5t2

Solution
The answer is

(50 − 5t2 ) − 45 5 − 5t2 5(1 − t)(1 + t)
v = lim = lim = lim
t→1 t−1 t→1 t − 1 t→1 t−1

. . . . . . .


Velocity
.
Problem

Example
by
h(t) = 50 − 5t2

Solution
The answer is

(50 − 5t2 ) − 45 5 − 5t2 5(1 − t)(1 + t)
v = lim = lim = lim
t→1 t−1 t→1 t − 1 t→1 t−1
= (−5) lim (1 + t)
t→1

. . . . . . .


Velocity
.
Problem

Example
by
h(t) = 50 − 5t2

Solution
The answer is

(50 − 5t2 ) − 45 5 − 5t2 5(1 − t)(1 + t)
v = lim = lim = lim
t→1 t−1 t→1 t − 1 t→1 t−1
= (−5) lim (1 + t) = −5 · 2 = −10
t→1

. . . . . . .


Velocity in general

. . = h(t)
y
Upshot
. (t0 ) .
h .
If the height function is given by
h(t), the instantaneous velocity . h
∆
at time t0 is given by
. (t0 + ∆t) .
h .
h(t) − h(t0 )
v = lim
t→t0 t − t0
h(t0 + ∆t) − h(t0 )
= lim
∆t→0 ∆t
. . . t .
∆ ..
t
t
.0 t
.

. . . . . .


Population growth
Problem
Given the population function of a group of organisms, find the rate of
growth of the population at a particular instant.

. . . . . .


Population growth
Problem
Given the population function of a group of organisms, find the rate of
growth of the population at a particular instant.

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

3et
P(t) =
1 + et
where t is in years since 2000 and P is in millions of fish. Is the fish
population growing fastest in 1990, 2000, or 2010? (Estimate
numerically)

. . . . . .


Derivation

Solution
Let ∆t be an increment in time and ∆P the corresponding change in
population:
∆P = P(t + ∆t) − P(t)
This depends on ∆t, so ideally we would want
( )
∆P 1 3et+∆t 3et
lim = lim −
∆t→0 ∆t ∆t→0 ∆t 1 + et+∆t 1 + et

. . . . . .


Derivation

Solution
Let ∆t be an increment in time and ∆P the corresponding change in
population:
∆P = P(t + ∆t) − P(t)
This depends on ∆t, so ideally we would want
( )
∆P 1 3et+∆t 3et
lim = lim −
∆t→0 ∆t ∆t→0 ∆t 1 + et+∆t 1 + et

But rather than compute a complicated limit analytically, let us
approximate 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
P(t + ∆t) − P(t)
quotient , where ∆t = 0.1 and t = n − 2000.
∆t

r1990

r2000

r2010

. . . . . .


Numerical evidence
P(t + ∆t) − P(t)
∆t

P(−10 + 0.1) − P(−10)
r1990 ≈
0.1

P(0.1) − P(0)
r2000 ≈
0.1

P(10 + 0.1) − P(10)
r2010 ≈
0.1

. . . . . .


Numerical evidence
P(t + ∆t) − P(t)
∆t
( )
P(−10 + 0.1) − P(−10) 1 3e−9.9 3e−10
r1990 ≈ = −
0.1 0.1 1 + e−9.9 1 + e−10

( )
P(0.1) − P(0) 1 3e0.1 3e0
r2000 ≈ = −
0.1 0.1 1 + e0.1 1 + e0

( )
P(10 + 0.1) − P(10) 1 3e10.1 3e10
r2010 ≈ = −
0.1 0.1 1 + e10.1 1 + e10

. . . . . .


Numerical evidence
P(t + ∆t) − P(t)
∆t
( )
P(−10 + 0.1) − P(−10) 1 3e−9.9 3e−10
r1990 ≈ = −
0.1 0.1 1 + e−9.9 1 + e−10
= 0.000143229
( )
P(0.1) − P(0) 1 3e0.1 3e0
r2000 ≈ = −
0.1 0.1 1 + e0.1 1 + e0

( )
P(10 + 0.1) − P(10) 1 3e10.1 3e10
r2010 ≈ = −
0.1 0.1 1 + e10.1 1 + e10

. . . . . .


Numerical evidence
P(t + ∆t) − P(t)
∆t
( )
P(−10 + 0.1) − P(−10) 1 3e−9.9 3e−10
r1990 ≈ = −
0.1 0.1 1 + e−9.9 1 + e−10
= 0.000143229
( )
P(0.1) − P(0) 1 3e0.1 3e0
r2000 ≈ = −
0.1 0.1 1 + e0.1 1 + e0
= 0.749376
( )
P(10 + 0.1) − P(10) 1 3e10.1 3e10
r2010 ≈ = −
0.1 0.1 1 + e10.1 1 + e10

. . . . . .


Numerical evidence
P(t + ∆t) − P(t)
∆t
( )
P(−10 + 0.1) − P(−10) 1 3e−9.9 3e−10
r1990 ≈ = −
0.1 0.1 1 + e−9.9 1 + e−10
= 0.000143229
( )
P(0.1) − P(0) 1 3e0.1 3e0
r2000 ≈ = −
0.1 0.1 1 + e0.1 1 + e0
= 0.749376
( )
P(10 + 0.1) − P(10) 1 3e10.1 3e10
r2010 ≈ = −
0.1 0.1 1 + e10.1 1 + e10
= 0.0001296
. . . . . .


Population growth
.
Problem
Given the population function of a group of organisms, find the rate of growth
of the population at a particular instant.

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

3et
P(t) =
1 + et
where t is in years since 2000 and P is in millions of fish. Is the fish population
growing fastest in 1990, 2000, or 2010? (Estimate numerically)

Answer
We 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

P(t + ∆t) − P(t)
lim
∆t→0 ∆t

. . . . . .


Marginal costs

Problem
Given the production cost of a good, find the marginal cost of
production after having produced a certain quantity.

. . . . . .


Marginal costs

Problem

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?

. . . . . .


Comparisons

Solution

C(q) = q3 − 12q2 + 60q
Fill in the table:

q C(q)
4
5
6

. . . . . .


Comparisons

Solution

C(q) = q3 − 12q2 + 60q
Fill in the table:

q C(q)
4 112
5
6

. . . . . .


Comparisons

Solution

C(q) = q3 − 12q2 + 60q
Fill in the table:

q C(q)
4 112
5 125
6

. . . . . .


Comparisons

Solution

C(q) = q3 − 12q2 + 60q
Fill in the table:

q C(q)
4 112
5 125
6 144

. . . . . .


Comparisons

Solution

C(q) = q3 − 12q2 + 60q
Fill in the table:

q C(q) AC(q) = C(q)/q
4 112
5 125
6 144

. . . . . .


Comparisons

Solution

C(q) = q3 − 12q2 + 60q
Fill in the table:

4 112 28
5 125
6 144

. . . . . .


Comparisons

Solution

C(q) = q3 − 12q2 + 60q
Fill in the table:

4 112 28
5 125 25
6 144

. . . . . .


Comparisons

Solution

C(q) = q3 − 12q2 + 60q
Fill in the table:

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 13
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 13
5 125 25 19
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 13
5 125 25 19
6 144 24 31

. . . . . .


Marginal costs

Problem

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 should
produce more to lower average costs.
. . . . . .


Marginal Cost in General

Upshot

The incremental cost

∆C = C(q + 1) − C(q)

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

. . . . . .


Marginal Cost in General

Upshot

The incremental cost

∆C = C(q + 1) − C(q)

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

C(q + ∆q) − C(q)
MC = lim
∆q→0 ∆q

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

. . . . . .


Outline

Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs



Other notations


. . . . . .


The definition

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

. . . . . .


The definition

All of these rates of change are found the same way!
Definition
Let f be a function and a a point in the domain of f. If the limit

f(a + h) − f(a) f(x) − f(a)
f′ (a) = lim = lim
h→0 h x→a x−a

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

. . . . . .


Derivative of the squaring function

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

. . . . . .



Example

Solution

f(a + h) − f(a)
f′ (a) = lim
h→0 h

. . . . . .



Example

Solution

f(a + h) − f(a) (a + h)2 − a2
h→0 h h→0 h

. . . . . .



Example

Solution

f(a + h) − f(a) (a + h)2 − a2
h→0 h h→0 h
(a2 + 2ah + h2 ) − a2
= lim
h→0 h

. . . . . .



Example

Solution

f(a + h) − f(a) (a + h)2 − a2
h→0 h h→0 h
(a2 + 2ah + h2 ) − a2 2ah + h2
= lim = lim
h→0 h h→0 h

. . . . . .



Example

Solution

f(a + h) − f(a) (a + h)2 − a2
h→0 h h→0 h
(a2 + 2ah + h2 ) − a2 2ah + h2
= lim = lim
h→0 h h→0 h
= lim (2a + h)
h→0

. . . . . .



Example

Solution

f(a + h) − f(a) (a + h)2 − a2
h→0 h h→0 h
(a2 + 2ah + h2 ) − a2 2ah + h2
= lim = lim
h→0 h h→0 h
= lim (2a + h) = 2a
h→0

. . . . . .


Derivative of the reciprocal function

Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
find f′ (2). x
.

.
. x
.

. . . . . .



Example
1
x
find f′ (2). x
.

Solution

1/x − 1/2
f′ (2) = lim .
x→2 x−2 . x
.

. . . . . .



Example
1
x
find f′ (2). x
.

Solution

1/x − 1/2
f′ (2) = lim .
x→2 x−2 . x
.
2−x
= lim
x→2 2x(x − 2)

. . . . . .



Example
1
x
find f′ (2). x
.

Solution

1/x − 1/2
f′ (2) = lim .
x→2 x−2 . x
.
2−x
= lim
x→2 2x(x − 2)
−1
= lim
x→2 2x

. . . . . .



Example
1
x
find f′ (2). x
.

Solution

1/x − 1/2
f′ (2) = lim .
x→2 x−2 . x
.
2−x
= lim
x→2 2x(x − 2)
−1 1
= lim =−
x→2 2x 4
. . . . . .


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

Solution

f(2 + h) − f(2)
f′ (2) = lim
h→0 h
1
−1
= lim 2+h 2
h→0 h

. . . . . .



Solution

f(2 + h) − f(2)
f′ (2) = lim
h→0 h
1
−1
= lim 2+h 2
h→0 h
2 − (2 + h)
= lim
h→0 2h(2 + h)

. . . . . .



Solution

f(2 + h) − f(2)
f′ (2) = lim
h→0 h
1
−1
= lim 2+h 2
h→0 h
2 − (2 + h) −h
= lim = lim
h→0 2h(2 + h) h→0 2h(2 + h)

. . . . . .



Solution

f(2 + h) − f(2)
f′ (2) = lim
h→0 h
1
−1
= lim 2+h 2
h→0 h
2 − (2 + h) −h
= lim = lim
h→0 2h(2 + h) h→0 2h(2 + h)
−1
= lim
h→0 2(2 + h)

. . . . . .



Solution

f(2 + h) − f(2)
f′ (2) = lim
h→0 h
1
−1
= lim 2+h 2
h→0 h
2 − (2 + h) −h
= lim = lim
h→0 2h(2 + h) h→0 2h(2 + h)
−1 1
= lim =−
h→0 2(2 + h) 4

. . . . . .


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

Fact

a c ad ± bc
± =
b d bd

So
1 1 2−x
−
x 2 = 2x
x−2 x−2
2−x
=
2x(x − 2)

. . . . . .



If f is a function, we can compute the derivative f′ (x) at each point
x where f is differentiable, and come up with another function, the
derivative function.
What can we say about this function f′ ?

. . . . . .



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

. . . . . .



Example
1
x
find f′ (2). x
.

Solution

1/x − 1/2
f′ (2) = lim .
x→2 x−2 . x
.
2−x
= lim
x→2 2x(x − 2)
−1 1
= lim =−
x→2 2x 4
. . . . . .



If f is decreasing on an interval, f′ is negative (technically,
nonpositive) on that interval
If f is increasing on an interval, f′ is positive (technically,
nonnegative) on that interval

. . . . . .



y
.
x2 − 22
x m=
x−2
3 5
. .
9 .
2.5 4.5
2.1 4.1
2.01 4.01
. .25 .
6 .
limit 4
. .41 .
4 . 1.99 3.99
. .0401 .
4.9601 .
3 . .61
3
4
. .. 1.9 3.9
1.5 3.5
. .25 .
2 .
1 3
. .
1 .
. . . ... . . x
.
1 1 . .. .1 3
. . .5 .99 .5 .
12 .
2.9 2
2
.01
. . . . . .



Fact
If f is decreasing on the open interval (a, b), then f′ ≤ 0 on (a, b).

. . . . . .



Fact
If f is decreasing on the open interval (a, b), then f′ ≤ 0 on (a, b).

Picture Proof. .

y
.
If f is decreasing, then all
secant lines point downward,
hence have negative slope.
The derivative is a limit of
slopes of secant lines, which
are all negative, so the limit .
must be ≤ 0. . .
.
.
x
.

. . . . . .


.
Fact
If f is decreasing on on the open interval (a, b), then f′ ≤ 0 on (a, b).

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

f(x + ∆x) − f(x)
f(x + ∆x) < f(x) =⇒ <0
∆x

. . . . . . .


.
Fact

The Real Proof.

f(x + ∆x) − f(x)
f(x + ∆x) < f(x) =⇒ <0
∆x
But if ∆x < 0, then x + ∆x < x, and
f(x + ∆x) − f(x)
f(x + ∆x) > f(x) =⇒ <0
∆x

still!

. . . . . . .


.
Fact

The Real Proof.

f(x + ∆x) − f(x)
f(x + ∆x) < f(x) =⇒ <0
∆x
But if ∆x < 0, then x + ∆x < x, and
f(x + ∆x) − f(x)
f(x + ∆x) > f(x) =⇒ <0
∆x
f(x + ∆x) − f(x)
still! Either way, < 0, so
∆x
f(x + ∆x) − f(x)
f′ (x) = lim ≤0
∆x→0 ∆x

. . . . . . .


Going the Other Way?

Question
If a function has a negative derivative on an interval, must it be
decreasing on that interval?

. . . . . .


Going the Other Way?

Question
If a function has a negative derivative on an interval, must it be
decreasing on that interval?

Answer
Maybe.

. . . . . .


Outline

Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs



Other notations


. . . . . .


Differentiability is super-continuity

Theorem
If f is differentiable at a, then f is continuous at a.

. . . . . .



Theorem

Proof.
We have
f(x) − f(a)
lim (f(x) − f(a)) = lim · (x − a)
x→a x→a x−a
f(x) − f(a)
= lim · lim (x − a)
x→a x−a x→a
′
= f (a) · 0 = 0

. . . . . .



Theorem

Proof.
We have
f(x) − f(a)
lim (f(x) − f(a)) = lim · (x − a)
x→a x→a x−a
f(x) − f(a)
= lim · lim (x − a)
x→a x−a x→a
′
= f (a) · 0 = 0

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


Differentiability FAIL
Kinks

Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x)

. x
.

. . . . . .


Kinks

Example
f
.(x) .′ (x)
f

. x
. . x
.

. . . . . .


Kinks

Example
f
.(x) .′ (x)
f

.

. x
. . x
.

. . . . . .


Cusps

Example
f
.(x)

. x
.

. . . . . .


Cusps

Example
f
.(x) .′ (x)
f

. x
. . x
.

. . . . . .


Vertical Tangents

Example
f
.(x)

. x
.

. . . . . .


Vertical Tangents

Example
f
.(x) .′ (x)
f

. x
. . x
.

. . . . . .


Weird, Wild, Stuff

Example

f
.(x)

. x
.

This function is differentiable
at 0.
. . . . . .


Lesson 7: The Derivative (Section 21 slide)

Lesson 7: The Derivative (Section 21 slide)

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Lesson 7: The Derivative (Section 21 slide)