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Section 2.1–2.2The Derivative and Rates of Change
The Derivative as a Function
V63.0121.041, Calculus I
New York University
September 26, 2010
Announcements
I Quiz this week in recitation on §§1.1–1.4
I 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/photo dueFriday October 1
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 2 / 46
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, lecturesection, assignment number,recitation, and date at thetop.
I Staple your homeworktogether.
See the website for more information.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 3 / 46
Notes
Notes
Notes
1
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
Objectives for Section 2.1
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 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 graph ofits derivative.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 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 26, 2010 6 / 46
Notes
Notes
Notes
2
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 7 / 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 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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 8 / 46
Velocity
Problem
Given the position function of a moving object, find the velocity of the object at acertain 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 isit 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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 10 / 46
Notes
Notes
Notes
3
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
Numerical evidence
h(t) = 50− 5t2
Fill in the table:
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
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 11 / 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 26, 2010 13 / 46
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 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 26, 2010 14 / 46
Notes
Notes
Notes
4
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
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 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 26, 2010 15 / 46
Numerical evidence
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=
1
0.1
(3e−9.9
1 + e−9.9− 3e−10
1 + e−10
)=
0.000143229
r2000 ≈P(0.1)− P(0)
0.1=
1
0.1
(3e0.1
1 + e0.1− 3e0
1 + e0
)=
0.749376
r2010 ≈P(10 + 0.1)− P(10)
0.1=
1
0.1
(3e10.1
1 + e10.1− 3e10
1 + e10
)=
0.0001296
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 16 / 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 26, 2010 18 / 46
Notes
Notes
Notes
5
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
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?
Answer
If q = 5, then C = 125, ∆C = 19, while AC = 25. So we should producemore to lower average costs.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 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 26, 2010 20 / 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 26, 2010 22 / 46
Notes
Notes
Notes
6
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
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 26, 2010 23 / 46
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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 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= 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.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 25 / 46
Notes
Notes
Notes
7
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
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
= limx→2
2− x
2x(x − 2)
= limx→2
−1
2x= −1
4
x
x
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 26 / 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 26, 2010 28 / 46
What does f tell you about f ′?
Fact
If 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 26, 2010 32 / 46
Notes
Notes
Notes
8
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
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 26, 2010 33 / 46
Differentiability is super-continuity
Theorem
If 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 at a,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 26, 2010 34 / 46
Differentiability FAILKinks
x
f (x)
x
f ′(x)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 35 / 46
Notes
Notes
Notes
9
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
Differentiability FAILCusps
x
f (x)
x
f ′(x)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 36 / 46
Differentiability FAILVertical Tangents
x
f (x)
x
f ′(x)
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 37 / 46
Differentiability FAILWeird, Wild, Stuff
x
f (x)
This function is differentiable at0.
x
f ′(x)
But the derivative is notcontinuous at 0!
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 38 / 46
Notes
Notes
Notes
10
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
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 26, 2010 39 / 46
Notation
I Newtonian notation
f ′(x) y ′(x) y ′
I Leibnizian notation
dy
dx
d
dxf (x)
df
dx
These all mean the same thing.
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 40 / 46
Meet the Mathematician: Isaac Newton
I English, 1643–1727
I Professor at Cambridge(England)
I Philosophiae NaturalisPrincipia Mathematicapublished 1687
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 41 / 46
Notes
Notes
Notes
11
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
Meet the Mathematician: Gottfried Leibniz
I German, 1646–1716
I Eminent philosopher as wellas mathematician
I Contemporarily disgraced bythe calculus priority dispute
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 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 26, 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! Leibnizian notation:
d2y
dx2
d2
dx2f (x)
d2f
dx2
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 44 / 46
Notes
Notes
Notes
12
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010
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 26, 2010 45 / 46
What have we learned today?
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 reflects the monotonicity (increasing or decreasing) ofthe graph
V63.0121.041, Calculus I (NYU) Section 2.1–2.2 The Derivative September 26, 2010 46 / 46
Notes
Notes
Notes
13
Section 2.1–2.2 : The DerivativeV63.0121.041, Calculus I September 26, 2010