lesson 7: the derivative (section 41 slides)
DESCRIPTION
The derivative is the instantaneous rate of change of a function.TRANSCRIPT
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 3 / 46
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 4 / 46
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 5 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 6 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 7 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 7 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
Graphically and numerically
. .x
.y
..2
..4 .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 8 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 9 / 46
. . . . . .
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
t→1
5(1− t)(1+ t)t− 1
= (−5) limt→1
(1+ t) = −5 · 2 = −10
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 10 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 11 / 46
. . . . . .
Numerical evidence
h(t) = 50− 5t2
Fill in the table:t vave =
h(t)− h(1)t− 1
2 − 151.5
− 12.51.1 − 10.51.01 − 10.051.001 − 10.005
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 11 / 46
. . . . . .
Numerical evidence
h(t) = 50− 5t2
Fill in the table:t vave =
h(t)− h(1)t− 1
2 − 151.5 − 12.5
1.1 − 10.51.01 − 10.051.001 − 10.005
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 11 / 46
. . . . . .
Numerical evidence
h(t) = 50− 5t2
Fill in the table:t vave =
h(t)− h(1)t− 1
2 − 151.5 − 12.51.1
− 10.51.01 − 10.051.001 − 10.005
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 11 / 46
. . . . . .
Numerical evidence
h(t) = 50− 5t2
Fill in the table:t vave =
h(t)− h(1)t− 1
2 − 151.5 − 12.51.1 − 10.5
1.01 − 10.051.001 − 10.005
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 11 / 46
. . . . . .
Numerical evidence
h(t) = 50− 5t2
Fill in the table:t vave =
h(t)− h(1)t− 1
2 − 151.5 − 12.51.1 − 10.51.01
− 10.051.001 − 10.005
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 11 / 46
. . . . . .
Numerical evidence
h(t) = 50− 5t2
Fill in the table:t vave =
h(t)− h(1)t− 1
2 − 151.5 − 12.51.1 − 10.51.01 − 10.05
1.001 − 10.005
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 11 / 46
. . . . . .
Numerical evidence
h(t) = 50− 5t2
Fill in the table:t vave =
h(t)− h(1)t− 1
2 − 151.5 − 12.51.1 − 10.51.01 − 10.051.001
− 10.005
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 11 / 46
. . . . . .
Numerical evidence
h(t) = 50− 5t2
Fill in the table:t vave =
h(t)− h(1)t− 1
2 − 151.5 − 12.51.1 − 10.51.01 − 10.051.001 − 10.005
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 11 / 46
. . . . . .
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
t→1
5(1− t)(1+ t)t− 1
= (−5) limt→1
(1+ t) = −5 · 2 = −10
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 12 / 46
. . . . . .
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
. ..t
..y = h(t).
.
..t0
..t
..h(t0)
..h(t0 +∆t)
.∆t
.∆h
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 13 / 46
. . . . . .
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
1+ et
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 14 / 46
. . . . . .
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
1+ et
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 14 / 46
. . . . . .
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
1∆t
(3et+∆t
1+ et+∆t −3et
1+ et
)
But rather than compute a complicated limit analytically, let usapproximate numerically. We will try a small ∆t, for instance 0.1.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 15 / 46
. . . . . .
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
1∆t
(3et+∆t
1+ et+∆t −3et
1+ et
)But rather than compute a complicated limit analytically, let usapproximate numerically. We will try a small ∆t, for instance 0.1.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 15 / 46
. . . . . .
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.
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
r2010
≈ P(10+ 0.1)− P(10)0.1
=10.1
(3e10.1
1+ e10.1− 3e10
1+ e10
)= 0.0001296
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 16 / 46
. . . . . .
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.
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
r2010 ≈ P(10+ 0.1)− P(10)0.1
=10.1
(3e10.1
1+ e10.1− 3e10
1+ e10
)= 0.0001296
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 16 / 46
. . . . . .
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.
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
r2010 ≈ P(10+ 0.1)− P(10)0.1
=10.1
(3e10.1
1+ e10.1− 3e10
1+ e10
)
= 0.0001296
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 16 / 46
. . . . . .
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.
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
r2010 ≈ P(10+ 0.1)− P(10)0.1
=10.1
(3e10.1
1+ e10.1− 3e10
1+ e10
)
= 0.0001296
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 16 / 46
. . . . . .
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.
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
r2010 ≈ P(10+ 0.1)− P(10)0.1
=10.1
(3e10.1
1+ e10.1− 3e10
1+ e10
)
= 0.0001296
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 16 / 46
. . . . . .
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.
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
r2010 ≈ P(10+ 0.1)− P(10)0.1
=10.1
(3e10.1
1+ e10.1− 3e10
1+ e10
)= 0.0001296
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 16 / 46
. . . . . .
Population growth.
.
ProblemGiven the population function of a group of organisms, find the rate of growthof the population at a particular instant.
Example
Suppose the population of fish in the East River is given by the function
P(t) =3et
1+ et
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 17 / 46
. . . . . .
Population growth in general
Upshot
The instantaneous population growth is given by
lim∆t→0
P(t+∆t)− P(t)∆t
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 18 / 46
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 19 / 46
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 19 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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
19
6 144 24
31
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 20 / 46
. . . . . .
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?
AnswerIf q = 5, then C = 125, ∆C = 19, while AC = 25. So we shouldproduce more to lower average costs.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 21 / 46
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 22 / 46
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 22 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 23 / 46
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 24 / 46
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 24 / 46
. . . . . .
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
h
= limh→0
(a2 + 2ah+ h2)− a2
h= lim
h→0
2ah+ h2
h= lim
h→0(2a+ h) = 2a.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 25 / 46
. . . . . .
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
h
= limh→0
(a2 + 2ah+ h2)− a2
h= lim
h→0
2ah+ h2
h= lim
h→0(2a+ h) = 2a.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 25 / 46
. . . . . .
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
. .x
.x
.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 26 / 46
. . . . . .
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
. .x
.x
.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 26 / 46
. . . . . .
The Sure-Fire Sally Rule (SFSR) for adding FractionsIn anticipation of the question, “How did you get that?"
ab± c
d=
ad± bcbd
So
1x− 1
2x− 2
=
2− x2x
x− 2
=2− x
2x(x− 2)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 27 / 46
. . . . . .
The Sure-Fire Sally Rule (SFSR) for adding FractionsIn anticipation of the question, “How did you get that?"
ab± c
d=
ad± bcbd
So
1x− 1
2x− 2
=
2− x2x
x− 2
=2− x
2x(x− 2)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 27 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 28 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 28 / 46
. . . . . .
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
. .x
.x
.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 29 / 46
. . . . . .
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 intervalI If f is increasing on an interval, f′ is positive (technically,
nonnegative) on that interval
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 30 / 46
. . . . . .
Graphically and numerically
. .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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 31 / 46
. . . . . .
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
< 0
But if ∆x < 0, then x+∆x < x, and
f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x
< 0
still! Either way,f(x+∆x)− f(x)
∆x< 0, so
f′(x) = lim∆x→0
f(x+∆x)− f(x)∆x
≤ 0
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 32 / 46
. . . . . .
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
< 0
But if ∆x < 0, then x+∆x < x, and
f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x
< 0
still! Either way,f(x+∆x)− f(x)
∆x< 0, so
f′(x) = lim∆x→0
f(x+∆x)− f(x)∆x
≤ 0
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 32 / 46
. . . . . .
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
< 0
But if ∆x < 0, then x+∆x < x, and
f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x
< 0
still!
Either way,f(x+∆x)− f(x)
∆x< 0, so
f′(x) = lim∆x→0
f(x+∆x)− f(x)∆x
≤ 0
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 32 / 46
. . . . . .
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
< 0
But if ∆x < 0, then x+∆x < x, and
f(x+∆x) > f(x) =⇒ f(x+∆x)− f(x)∆x
< 0
still! Either way,f(x+∆x)− f(x)
∆x< 0, so
f′(x) = lim∆x→0
f(x+∆x)− f(x)∆x
≤ 0
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 32 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 33 / 46
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 34 / 46
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 34 / 46
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 34 / 46
. . . . . .
Differentiability FAILKinks
Example
Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.
. .x
.f(x)
. .x
.f′(x)
.
.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 35 / 46
. . . . . .
Differentiability FAILKinks
Example
Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.
. .x
.f(x)
. .x
.f′(x)
.
.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 35 / 46
. . . . . .
Differentiability FAILKinks
Example
Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.
. .x
.f(x)
. .x
.f′(x)
.
.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 35 / 46
. . . . . .
Differentiability FAILCusps
Example
Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.
. .x
.f(x)
. .x
.f′(x)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 36 / 46
. . . . . .
Differentiability FAILCusps
Example
Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.
. .x
.f(x)
. .x
.f′(x)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 36 / 46
. . . . . .
Differentiability FAILCusps
Example
Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.
. .x
.f(x)
. .x
.f′(x)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 36 / 46
. . . . . .
Differentiability FAILCusps
Example
Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.
. .x
.f(x)
. .x
.f′(x)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 36 / 46
. . . . . .
Differentiability FAILVertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.
. .x
.f(x)
. .x
.f′(x)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 37 / 46
. . . . . .
Differentiability FAILVertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.
. .x
.f(x)
. .x
.f′(x)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 37 / 46
. . . . . .
Differentiability FAILVertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.
. .x
.f(x)
. .x
.f′(x)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 37 / 46
. . . . . .
Differentiability FAILVertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′.
. .x
.f(x)
. .x
.f′(x)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 37 / 46
. . . . . .
Differentiability FAILWeird, Wild, Stuff
Example
. .x
.f(x)
This function is differentiableat 0.
. .x
.f′(x)
.
But the derivative is notcontinuous at 0!
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 38 / 46
. . . . . .
Differentiability FAILWeird, Wild, Stuff
Example
. .x
.f(x)
This function is differentiableat 0.
. .x
.f′(x)
.
But the derivative is notcontinuous at 0!
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 38 / 46
. . . . . .
Differentiability FAILWeird, Wild, Stuff
Example
. .x
.f(x)
This function is differentiableat 0.
. .x
.f′(x)
.
But the derivative is notcontinuous at 0!
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 38 / 46
. . . . . .
Differentiability FAILWeird, Wild, Stuff
Example
. .x
.f(x)
This function is differentiableat 0.
. .x
.f′(x)
.
But the derivative is notcontinuous at 0!
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 38 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 39 / 46
. . . . . .
Notation
I Newtonian notation
f′(x) y′(x) y′
I Leibnizian notation
dydx
ddx
f(x)dfdx
These all mean the same thing.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 40 / 46
. . . . . .
Meet the Mathematician: Isaac Newton
I English, 1643–1727I Professor at Cambridge(England)
I Philosophiae NaturalisPrincipia Mathematicapublished 1687
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 41 / 46
. . . . . .
Meet the Mathematician: Gottfried Leibniz
I German, 1646–1716I Eminent philosopher aswell as mathematician
I Contemporarily disgracedby the calculus prioritydispute
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 42 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 43 / 46
. . . . . .
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!
Leibniziannotation:
d2ydx2
d2
dx2f(x)
d2fdx2
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 44 / 46
. . . . . .
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! Leibniziannotation:
d2ydx2
d2
dx2f(x)
d2fdx2
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 44 / 46
. . . . . .
function, derivative, second derivative
. .x
.y.f(x) = x2
.f′(x) = 2x
.f′′(x) = 2
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 45 / 46
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 27, 2010 46 / 46