Lesson 6: The Derivative

Section 2.1
The Derivative and Rates of Change

V63.0121.002.2010Su, Calculus I

New York University

May 20, 2010

Announcements

Written Assignment 1 is on the website

Announcements

Written Assignment 1 is on
the website

V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 2 / 28

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,
assignment number, and
date at the top.
Staple your homework
together.
See the website for more information.


Objectives

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.


Outline

Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs

The derivative, deﬁned
Derivatives of (some) power functions


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 = x 2 at the point (2, 4).


Graphically and numerically

y
x 2 − 22
x m=
x −2

4

x
2


y
x 2 − 22
x m=
x −2
3
9

4

x
2 3


y
x 2 − 22
x m=
x −2
3 5
9

4

x
2 3


y
x 2 − 22
x m=
x −2
3 5
2.5

6.25

4

x
2 2.5


y
x 2 − 22
x m=
x −2
3 5
2.5 4.5

6.25

4

x
2 2.5


y
x 2 − 22
x m=
x −2
3 5
2.5 4.5
2.1

4.41
4

x
2.1
2


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

4.41
4

x
2.1
2


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

4.0401
4

x
2.01
2


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

4.0401
4

x
2.01
2


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

4
1.5
2.25
1 3

x
1.5 2


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

4
1.5 3.5
2.25
1 3

x
1.5 2


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

4 1.9
3.61
1.5 3.5
1 3

x
1.9
2


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

4
3.9601 1.9 3.9
3.61
1.5 3.5
1 3

x
1.99
1.9
2


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

1.99
4
3.9601 1.9 3.9
1.5 3.5
1 3

x
1.99
2


y
x 2 − 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
x 2 − 22
x m=
x −2
3 5
9 2.5 4.5
2.1 4.1
2.01 4.01
6.25
limit 4
4.41 1.99 3.99
4.0401
4
3.9601 1.9 3.9
3.61
1.5 3.5
2.25
1 3
1
x
1 1.5 2.1 3
1.99
1.9 2.5
2.01
2

The tangent problem

Problem

Example
Find the slope of the line tangent to the curve y = x 2 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 − 5t 2
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) − h(1)
t vave =
t −1
2 − 15


Numerical evidence

h(t) − h(1)
t vave =
t −1
2 − 15
1.5


Numerical evidence

h(t) − h(1)
t vave =
t −1
2 − 15
1.5 − 12.5


Numerical evidence

h(t) − h(1)
t vave =
t −1
2 − 15
1.5 − 12.5
1.1


Numerical evidence

h(t) − h(1)
t vave =
t −1
2 − 15
1.5 − 12.5
1.1 − 10.5


Numerical evidence

h(t) − h(1)
t vave =
t −1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01


Numerical evidence

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) − 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) − 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
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 − 5t 2
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?

Solution
The answer is
(50 − 5t 2 ) − 45 5 − 5t 2 5(1 − t)(1 + t)
v = lim = lim = lim
t→1 t −1 t→1 t − 1 t→1 t −1
V63.0121.002.2010Su, Calculus I (NYU) t) = Section 2.1 = −10
= (−5) lim (1 + −5 · 2 The Derivative May 20, 2010 11 / 28

y = h(t)
Upshot
If the height function is given by
h(t), the instantaneous velocity
at time t0 is given by

h(t) − h(t0 )
v = lim
t→t0 t − t0
h(t0 + ∆t) − h(t0 )
= lim
∆t→0 ∆t
∆t
t
t0 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

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

3e t
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

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 we want

∆P 1 3e t+∆t 3e t
lim = lim −
∆t→0 ∆t ∆t→0 ∆t 1 + e t+∆t 1 + et


Derivation

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 we want

∆P 1 3e t+∆t 3e t
lim = lim −
∆t→0 ∆t ∆t→0 ∆t 1 + e t+∆t 1 + et

Too hard! Try a small ∆t to approximate.


Numerical evidence

∆P ∆P
Use ∆t = 0.1, and use to approximate lim .
∆t ∆t→0 ∆t

P(−10 + 0.1) − P(−10)
r1990 ≈ =
0.1


Numerical evidence

∆P ∆P
∆t ∆t→0 ∆t

P(−10 + 0.1) − P(−10)
r1990 ≈ = 0.000143229
0.1


Numerical evidence

∆P ∆P
∆t ∆t→0 ∆t

P(−10 + 0.1) − P(−10)
r1990 ≈ = 0.000143229
0.1

P(0.1) − P(0)
r2000 ≈ =
0.1


Numerical evidence

∆P ∆P
∆t ∆t→0 ∆t

P(−10 + 0.1) − P(−10)
r1990 ≈ = 0.000143229
0.1

P(0.1) − P(0)
r2000 ≈ = 0.749376
0.1


Numerical evidence

∆P ∆P
∆t ∆t→0 ∆t

P(−10 + 0.1) − P(−10)
r1990 ≈ = 0.000143229
0.1

P(0.1) − P(0)
r2000 ≈ = 0.749376
0.1
P(10 + 0.1) − P(10)
r2010 ≈ =
0.1


Numerical evidence

∆P ∆P
∆t ∆t→0 ∆t

P(−10 + 0.1) − P(−10)
r1990 ≈ = 0.000143229
0.1

P(0.1) − P(0)
r2000 ≈ = 0.749376
0.1
P(10 + 0.1) − P(10)
r2010 ≈ = 0.0001296
0.1


Population growth
Problem

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

3e t
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)

Solution
We estimates the rates of growth to be 0.000143229, 0.749376, and
0.0001296. So the population is growing fastest in 2000.

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) = q 3 − 12q 2 + 60q

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


Comparisons

q C (q)
4
5
6


Comparisons

q C (q)
4 112
5
6


Comparisons

q C (q)
4 112
5 125
6


Comparisons

q C (q)
4 112
5 125
6 144


Comparisons

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


Comparisons

q C (q) AC (q) = C (q)/q
4 112 28
5 125
6 144


Comparisons

q C (q) AC (q) = C (q)/q
4 112 28
5 125 25
6 144


Comparisons

q C (q) AC (q) = C (q)/q
4 112 28
5 125 25
6 144 24


Comparisons

q C (q) AC (q) = C (q)/q ∆C = C (q + 1) − C (q)
4 112 28
5 125 25
6 144 24


Comparisons

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

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

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) = q 3 − 12q 2 + 60q

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

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


Upshot

The incremental cost

∆C = C (q + 1) − C (q)

is useful, but depends on units.


Upshot

The incremental cost

∆C = C (q + 1) − C (q)

is useful, but depends on units.
The marginal cost after producing q given by

C (q + ∆q) − C (q)
MC = lim
∆q→0 ∆q

is more useful since it’s unit-independent.


Outline

Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs

The derivative, deﬁned
Derivatives of (some) power functions


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) = x 2 . Use the deﬁnition of derivative to ﬁnd f (a).


Derivative of the squaring function

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

Solution

f (a + h) − f (a) (a + h)2 − a2
f (a) = lim = lim
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).


Derivative of the reciprocal function

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

Solution

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


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

a c ad ± bc
± =
b d bd
So
1 1 2−x
−
x 2 = 2x
x −2 x −2
2−x
=
2x(x − 2)


Worksheet


Summary

The derivative measures instantaneous rate of change
The derivative has many interpretations: slope of the tangent line,
velocity, marginal quantities, etc.
The derivative is calculated with a limit.


Lesson 6: The Derivative

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