1.
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
. . . . . .
2.
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
3.
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 3 / 49
4.
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 4 / 49
5.
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 5 / 49
6.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 6 / 49
7.
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 7 / 49
8.
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.
Example
Find the slope of the line tangent to the curve y = x2 at the point (2, 4).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 7 / 49
9.
Graphically and numerically
y
.
x2 − 22
x m=
x−2
. .
4 .
. . x
.
2
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
10.
Graphically and numerically
y
.
x2 − 22
x m=
x−2
3
. .
9 .
. .
4 .
. . . x
.
2
. 3
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
11.
Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
. .
9 .
. .
4 .
. . . x
.
2
. 3
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
12.
Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5
. .25 .
6 .
. .
4 .
. . . x
.
22
. . .5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
13.
Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
. .25 .
6 .
. .
4 .
. . . x
.
22
. . .5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
14.
Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1
. .41 .
4 .
. .
4 .
. .. x
.
2
.. .1
2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
15.
Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
. .41 .
4 .
. .
4 .
. .. x
.
2
.. .1
2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
16.
Graphically and numerically
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
17.
Graphically and numerically
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
18.
Graphically and numerically
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
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
19.
Graphically and numerically
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
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
20.
Graphically and numerically
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 .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
21.
Graphically and numerically
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 .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
22.
Graphically and numerically
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
23.
Graphically and numerically
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
24.
Graphically and numerically
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
25.
Graphically and numerically
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
26.
Graphically and numerically
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
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
29.
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.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 9 / 49
30.
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?
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 10 / 49
31.
Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
32.
Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
33.
Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
34.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
35.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
36.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
37.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
38.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
39.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
40.
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?
Solution
The answer is
(50 − 5t2 ) − 45
v = lim
t→1 t−1
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
41.
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?
Solution
The answer is
(50 − 5t2 ) − 45 5 − 5t2
v = lim = lim
t→1 t−1 t→1 t − 1
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
42.
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?
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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
43.
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?
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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
44.
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?
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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
45.
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
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 13 / 49
46.
Population growth
Problem
Given the population function of a group of organisms, find the rate of
growth of the population at a particular instant.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 14 / 49
47.
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)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 14 / 49
48.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 15 / 49
49.
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 15 / 49
50.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
51.
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
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
52.
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
( )
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
53.
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
( )
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
54.
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
( )
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
55.
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
( )
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
56.
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.
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 17 / 49
57.
Population growth in general
Upshot
The instantaneous population growth is given by
P(t + ∆t) − P(t)
lim
∆t→0 ∆t
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 18 / 49
58.
Marginal costs
Problem
Given the production cost of a good, find the marginal cost of
production after having produced a certain quantity.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 19 / 49
59.
Marginal costs
Problem
Given the production cost of a good, find the marginal cost of
production 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?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 19 / 49
60.
Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q)
4
5
6
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
61.
Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q)
4 112
5
6
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
62.
Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q)
4 112
5 125
6
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
63.
Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q)
4 112
5 125
6 144
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
64.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
65.
Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q) AC(q) = C(q)/q
4 112 28
5 125
6 144
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
66.
Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q) AC(q) = C(q)/q
4 112 28
5 125 25
6 144
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
67.
Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q) AC(q) = C(q)/q
4 112 28
5 125 25
6 144 24
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
68.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
69.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
70.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
71.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
72.
Marginal costs
Problem
Given the production cost of a good, find the marginal cost of
production 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 should
produce more to lower average costs.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 21 / 49
73.
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 22 / 49
74.
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 22 / 49
75.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 23 / 49
76.
The definition
All of these rates of change are found the same way!
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 24 / 49
77.
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 24 / 49
78.
Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (a).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
79.
Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (a).
Solution
f(a + h) − f(a)
f′ (a) = lim
h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
80.
Derivative of the squaring function
Example
Suppose f(x) = x2 . 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
81.
Derivative of the squaring function
Example
Suppose f(x) = x2 . 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
= lim
h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
82.
Derivative of the squaring function
Example
Suppose f(x) = x2 . 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
83.
Derivative of the squaring function
Example
Suppose f(x) = x2 . 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)
h→0
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
84.
Derivative of the squaring function
Example
Suppose f(x) = x2 . 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
85.
Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
find f′ (2). x
.
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
86.
Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
find f′ (2). x
.
Solution
1/x − 1/2
f′ (2) = lim .
x→2 x−2 . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
87.
Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
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)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
88.
Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
89.
Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
90.
“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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
91.
“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
2 − (2 + h)
= lim
h→0 2h(2 + h)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
92.
“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
2 − (2 + h) −h
= lim = lim
h→0 2h(2 + h) h→0 2h(2 + h)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
93.
“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
2 − (2 + h) −h
= lim = lim
h→0 2h(2 + h) h→0 2h(2 + h)
−1
= lim
h→0 2(2 + h)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
94.
“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
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
95.
“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)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 28 / 49
96.
“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)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 28 / 49
97.
What does f tell you about f′ ?
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′ ?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 29 / 49
98.
What does f tell you about f′ ?
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 29 / 49
99.
Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 30 / 49
100.
What does f tell you about f′ ?
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
If f is increasing on an interval, f′ is positive (technically,
nonnegative) on that interval
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 31 / 49
102.
What does f tell you about f′ ?
Fact
If f is decreasing on the open interval (a, b), then f′ ≤ 0 on (a, b).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 33 / 49
103.
What does f tell you about f′ ?
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
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 33 / 49
104.
What does f tell you about f′ ?
.
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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 34 / 49
105.
What does f tell you about f′ ?
.
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
But if ∆x < 0, then x + ∆x < x, and
f(x + ∆x) − f(x)
f(x + ∆x) > f(x) =⇒ <0
∆x
still!
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 34 / 49
106.
What does f tell you about f′ ?
.
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
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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 34 / 49
107.
Going the Other Way?
Question
If a function has a negative derivative on an interval, must it be
decreasing on that interval?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 35 / 49
108.
Going the Other Way?
Question
If a function has a negative derivative on an interval, must it be
decreasing on that interval?
Answer
Maybe.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 35 / 49
109.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 36 / 49
110.
Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 37 / 49
111.
Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 37 / 49
112.
Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 37 / 49
113.
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
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 38 / 49
114.
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)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 38 / 49
115.
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)
f
.
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 38 / 49
116.
Differentiability FAIL
Cusps
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x)
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
117.
Differentiability FAIL
Cusps
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
118.
Differentiability FAIL
Cusps
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
119.
Differentiability FAIL
Cusps
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
120.
Differentiability FAIL
Vertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x)
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 40 / 49
121.
Differentiability FAIL
Vertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 40 / 49
122.
Differentiability FAIL
Vertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 40 / 49
123.
Differentiability FAIL
Vertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 40 / 49
124.
Differentiability FAIL
Weird, Wild, Stuff
Example
f
.(x)
. x
.
This function is differentiable
at 0.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 41 / 49
125.
Differentiability FAIL
Weird, Wild, Stuff
Example
f
.(x) .′ (x)
f
. x
. . x
.
This function is differentiable
at 0.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 41 / 49
126.
Differentiability FAIL
Weird, Wild, Stuff
Example
f
.(x) .′ (x)
f
. x
. . x
.
This function is differentiable
at 0.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 41 / 49
127.
Differentiability FAIL
Weird, Wild, Stuff
Example
f
.(x) .′ (x)
f
. x
. . x
.
This function is differentiable But the derivative is not
at 0. continuous at 0!
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 41 / 49
128.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 42 / 49
129.
Notation
Newtonian notation
f′ (x) y′ (x) y′
Leibnizian notation
dy d df
f(x)
dx dx dx
These all mean the same thing.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 43 / 49
130.
Meet the Mathematician: Isaac Newton
English, 1643–1727
Professor at Cambridge
(England)
Philosophiae Naturalis
Principia Mathematica
published 1687
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 44 / 49
131.
Meet the Mathematician: Gottfried Leibniz
German, 1646–1716
Eminent philosopher as
well as mathematician
Contemporarily disgraced
by the calculus priority
dispute
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 45 / 49
132.
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 46 / 49
133.
The second derivative
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!
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 47 / 49
134.
The second derivative
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! Leibnizian
notation:
d2 y d2 d2 f
f(x)
dx2 dx2 dx2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 47 / 49
135.
function, derivative, second derivative
y
.
.(x) = x2
f
.′ (x) = 2x
f
.′′ (x) = 2
f
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 48 / 49
136.
What have we learned today?
The derivative measures instantaneous rate of change
The derivative has many interpretations: slope of the tangent line,
velocity, marginal quantities, etc.
The derivative reflects the monotonicity (increasing or decreasing)
of the graph
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 49 / 49
Be the first to comment