lesson 6: the derivative
TRANSCRIPT
Section 2.1The Derivative and Rates of Change
V63.0121.002.2010Su, Calculus I
New York University
May 20, 2010
Announcements
I Written Assignment 1 is on the website
Announcements
I Written Assignment 1 is onthe website
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 2 / 28
Format of written work
Please:
I Use scratch paper and copyyour final work onto freshpaper.
I Use loose-leaf paper (nottorn from a notebook).
I Write your name,assignment number, anddate at the top.
I Staple your homeworktogether.
See the website for more information.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 3 / 28
Objectives
I Understand and state thedefinition of the derivative ofa function at a point.
I Given a function and a pointin its domain, decide if thefunction is differentiable atthe point and find the valueof the derivative at thatpoint.
I Understand and give severalexamples of derivativesmodeling rates of change inscience.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 4 / 28
Outline
Rates of ChangeTangent LinesVelocityPopulation growthMarginal costs
The derivative, definedDerivatives of (some) power functions
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 5 / 28
The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line tangentto 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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 6 / 28
The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line tangentto 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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 6 / 28
Graphically and numerically
x
y
2
4
x m =x2 − 22
x − 2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
3
9
x m =x2 − 22
x − 23
5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
3
9
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
2.5
6.25
x m =x2 − 22
x − 23 5
2.5
4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
2.5
6.25
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
2.1
4.41
x m =x2 − 22
x − 23 5
2.5 4.5
2.1
4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
2.1
4.41
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
2.01
4.0401
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01
4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
2.01
4.0401
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
1
1
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1
3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
1
1
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
1.5
2.25
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5
3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
1.5
2.25
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
1.9
3.61
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9
3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
1.9
3.61
1.99
3.9601
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
1.99
3.9601
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99
3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
Graphically and numerically
x
y
2
4
x m =x2 − 22
x − 23 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
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 5
2.5 4.5
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 7 / 28
The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line tangentto 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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 8 / 28
Velocity
Problem
Given the position function of a moving object, find the velocity of theobject at a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can bedescribed 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
v = limt→1
(50 − 5t2) − 45
t − 1= lim
t→1
5 − 5t2
t − 1= lim
t→1
5(1 − t)(1 + t)
t − 1
= (−5) limt→1
(1 + t) = −5 · 2 = −10
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 9 / 28
Numerical evidence
t vave =h(t) − h(1)
t − 12 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001 − 10.005
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
Numerical evidence
t vave =h(t) − h(1)
t − 12 − 15
1.5
− 12.5
1.1 − 10.5
1.01 − 10.05
1.001 − 10.005
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
Numerical evidence
t vave =h(t) − h(1)
t − 12 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001 − 10.005
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
Numerical evidence
t vave =h(t) − h(1)
t − 12 − 15
1.5 − 12.5
1.1
− 10.5
1.01 − 10.05
1.001 − 10.005
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
Numerical evidence
t vave =h(t) − h(1)
t − 12 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001 − 10.005
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
Numerical evidence
t vave =h(t) − h(1)
t − 12 − 15
1.5 − 12.5
1.1 − 10.5
1.01
− 10.05
1.001 − 10.005
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
Numerical evidence
t vave =h(t) − h(1)
t − 12 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001 − 10.005
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
Numerical evidence
t vave =h(t) − h(1)
t − 12 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001
− 10.005
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
Numerical evidence
t vave =h(t) − h(1)
t − 12 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001 − 10.005
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 10 / 28
Velocity
Problem
Given the position function of a moving object, find the velocity of theobject at a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can bedescribed 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
v = limt→1
(50 − 5t2) − 45
t − 1= lim
t→1
5 − 5t2
t − 1= lim
t→1
5(1 − t)(1 + t)
t − 1
= (−5) limt→1
(1 + t) = −5 · 2 = −10V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 11 / 28
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
∆t
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 12 / 28
Population growth
Problem
Given 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? (Estimate numerically)
Solution
We estimates the rates of growth to be 0.000143229, 0.749376, and0.0001296. So the population is growing fastest in 2000.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 13 / 28
Population growth
Problem
Given 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? (Estimate numerically)
Solution
We estimates the rates of growth to be 0.000143229, 0.749376, and0.0001296. So the population is growing fastest in 2000.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 13 / 28
Derivation
Let ∆t be an increment in time and ∆P the corresponding change inpopulation:
∆P = P(t + ∆t) − P(t)
This depends on ∆t, so we want
lim∆t→0
∆P
∆t= lim
∆t→0
1
∆t
(3et+∆t
1 + et+∆t− 3et
1 + et
)
Too hard! Try a small ∆t to approximate.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 14 / 28
Derivation
Let ∆t be an increment in time and ∆P the corresponding change inpopulation:
∆P = P(t + ∆t) − P(t)
This depends on ∆t, so we want
lim∆t→0
∆P
∆t= lim
∆t→0
1
∆t
(3et+∆t
1 + et+∆t− 3et
1 + et
)Too hard! Try a small ∆t to approximate.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 14 / 28
Numerical evidence
Use ∆t = 0.1, and use∆P
∆tto approximate lim
∆t→0
∆P
∆t.
r1990 ≈ P(−10 + 0.1) − P(−10)
0.1=
0.000143229
r2000 ≈ P(0.1) − P(0)
0.1= 0.749376
r2010 ≈ P(10 + 0.1) − P(10)
0.1= 0.0001296
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
Numerical evidence
Use ∆t = 0.1, and use∆P
∆tto approximate lim
∆t→0
∆P
∆t.
r1990 ≈ P(−10 + 0.1) − P(−10)
0.1= 0.000143229
r2000 ≈ P(0.1) − P(0)
0.1= 0.749376
r2010 ≈ P(10 + 0.1) − P(10)
0.1= 0.0001296
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
Numerical evidence
Use ∆t = 0.1, and use∆P
∆tto approximate lim
∆t→0
∆P
∆t.
r1990 ≈ P(−10 + 0.1) − P(−10)
0.1= 0.000143229
r2000 ≈ P(0.1) − P(0)
0.1=
0.749376
r2010 ≈ P(10 + 0.1) − P(10)
0.1= 0.0001296
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
Numerical evidence
Use ∆t = 0.1, and use∆P
∆tto approximate lim
∆t→0
∆P
∆t.
r1990 ≈ P(−10 + 0.1) − P(−10)
0.1= 0.000143229
r2000 ≈ P(0.1) − P(0)
0.1= 0.749376
r2010 ≈ P(10 + 0.1) − P(10)
0.1= 0.0001296
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
Numerical evidence
Use ∆t = 0.1, and use∆P
∆tto approximate lim
∆t→0
∆P
∆t.
r1990 ≈ P(−10 + 0.1) − P(−10)
0.1= 0.000143229
r2000 ≈ P(0.1) − P(0)
0.1= 0.749376
r2010 ≈ P(10 + 0.1) − P(10)
0.1=
0.0001296
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
Numerical evidence
Use ∆t = 0.1, and use∆P
∆tto approximate lim
∆t→0
∆P
∆t.
r1990 ≈ P(−10 + 0.1) − P(−10)
0.1= 0.000143229
r2000 ≈ P(0.1) − P(0)
0.1= 0.749376
r2010 ≈ P(10 + 0.1) − P(10)
0.1= 0.0001296
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 15 / 28
Population growth
Problem
Given 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? (Estimate numerically)
Solution
We estimates the rates of growth to be 0.000143229, 0.749376, and0.0001296. So the population is growing fastest in 2000.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 16 / 28
Upshot
The instantaneous population growth is given by
lim∆t→0
P(t + ∆t) − P(t)
∆t
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 17 / 28
Marginal costs
Problem
Given the production cost of a good, find the marginal cost of productionafter 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?
Example
If q = 5, then C = 125, ∆C = 19, while AC = 25. So we should producemore to lower average costs.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 18 / 28
Marginal costs
Problem
Given the production cost of a good, find the marginal cost of productionafter 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?
Example
If q = 5, then C = 125, ∆C = 19, while AC = 25. So we should producemore to lower average costs.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 18 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 19 / 28
Marginal costs
Problem
Given the production cost of a good, find the marginal cost of productionafter 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?
Example
If q = 5, then C = 125, ∆C = 19, while AC = 25. So we should producemore to lower average costs.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 20 / 28
Upshot
I The incremental cost
∆C = C (q + 1) − C (q)
is useful, but depends on units.
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 unit-independent.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 21 / 28
Upshot
I The incremental cost
∆C = C (q + 1) − C (q)
is useful, but depends on units.
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 unit-independent.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 21 / 28
Outline
Rates of ChangeTangent LinesVelocityPopulation growthMarginal costs
The derivative, definedDerivatives of (some) power functions
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 22 / 28
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) = limh→0
f (a + h) − f (a)
h= lim
x→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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 23 / 28
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) = limh→0
f (a + h) − f (a)
h= lim
x→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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 23 / 28
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= lim
h→0
(a + h)2 − a2
h
= limh→0
(a2 + 2ah + h2) − a2
h= lim
h→0
2ah + h2
h
= limh→0
(2a + h) = 2a.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 24 / 28
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= lim
h→0
(a + h)2 − a2
h
= limh→0
(a2 + 2ah + h2) − a2
h= lim
h→0
2ah + h2
h
= limh→0
(2a + h) = 2a.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 24 / 28
Derivative of the reciprocal function
Example
Suppose f (x) =1
x. Use the
definition of the derivative to findf ′(2).
Solution
f ′(2) = limx→2
1/x − 1/2
x − 2= lim
x→2
2 − x
2x(x − 2)
= limx→2
−1
2x= −1
4
x
x
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 25 / 28
Derivative of the reciprocal function
Example
Suppose f (x) =1
x. Use the
definition of the derivative to findf ′(2).
Solution
f ′(2) = limx→2
1/x − 1/2
x − 2= lim
x→2
2 − x
2x(x − 2)
= limx→2
−1
2x= −1
4
x
x
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 25 / 28
The Sure-Fire Sally Rule (SFSR) for addingFractionsIn anticipation of the question, “How did you get that?”
a
b± c
d=
ad ± bc
bd
So
1
x− 1
2x − 2
=
2 − x
2xx − 2
=2 − x
2x(x − 2)
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 26 / 28
The Sure-Fire Sally Rule (SFSR) for addingFractionsIn anticipation of the question, “How did you get that?”
a
b± c
d=
ad ± bc
bd
So
1
x− 1
2x − 2
=
2 − x
2xx − 2
=2 − x
2x(x − 2)
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 26 / 28
Worksheet
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 27 / 28
Summary
I The derivative measures instantaneous rate of change
I The derivative has many interpretations: slope of the tangent line,velocity, marginal quantities, etc.
I The derivative is calculated with a limit.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.1 The Derivative May 20, 2010 28 / 28