Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function
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Transcript of Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function
![Page 1: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/1.jpg)
Sections 2.1–2.2Derivatives and Rates of Changes
The Derivative as a Function
V63.0121, Calculus I
February 9–12, 2009
Announcements
I Quiz 2 is next week: Covers up through 1.6
I Midterm is March 4/5: Covers up to 2.4 (next T/W)
![Page 2: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/2.jpg)
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
![Page 3: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/3.jpg)
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
![Page 4: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/4.jpg)
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
![Page 5: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/5.jpg)
Graphically and numerically
x
y
2
4
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
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Graphically and numerically
x
y
2
4
3
9
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
![Page 7: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/7.jpg)
Graphically and numerically
x
y
2
4
2.5
6.25
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
![Page 8: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/8.jpg)
Graphically and numerically
x
y
2
4
2.1
4.41
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
![Page 9: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/9.jpg)
Graphically and numerically
x
y
2
4
2.01
4.0401
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
![Page 10: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/10.jpg)
Graphically and numerically
x
y
2
4
1
1
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
![Page 11: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/11.jpg)
Graphically and numerically
x
y
2
4
1.5
2.25
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
![Page 12: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/12.jpg)
Graphically and numerically
x
y
2
4
1.9
3.61
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
![Page 13: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/13.jpg)
Graphically and numerically
x
y
2
4
1.99
3.9601
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
![Page 14: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/14.jpg)
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
3 5
2.5 4.25
2.1 4.1
2.01 4.01
limit 4
1.99 3.99
1.9 3.9
1.5 3.5
1 3
![Page 15: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/15.jpg)
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
![Page 16: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/16.jpg)
Velocity
ProblemGiven the position function of a moving object, find the velocity ofthe object at a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height canbe described by
h(t) = 50− 10t2
where t is seconds after dropping it and h is meters above theground. How fast is it falling one second after we drop it?
SolutionThe answer is
limt→1
(50− 10t2)− 40
t − 1= −20.
![Page 17: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/17.jpg)
Velocity
ProblemGiven the position function of a moving object, find the velocity ofthe object at a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height canbe described by
h(t) = 50− 10t2
where t is seconds after dropping it and h is meters above theground. How fast is it falling one second after we drop it?
SolutionThe answer is
limt→1
(50− 10t2)− 40
t − 1= −20.
![Page 18: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/18.jpg)
Numerical evidence
t vave =h(t)− h(1)
t − 12 −30
1.5 −25
1.1 −21
1.01 −20.01
1.001 −20.001
![Page 19: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/19.jpg)
Velocity
ProblemGiven the position function of a moving object, find the velocity ofthe object at a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height canbe described by
h(t) = 50− 10t2
where t is seconds after dropping it and h is meters above theground. How fast is it falling one second after we drop it?
SolutionThe answer is
limt→1
(50− 10t2)− 40
t − 1= −20.
![Page 20: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/20.jpg)
Upshot
If the height function is given by h(t), the instantaneous velocityat time t is given by
v = lim∆t→0
h(t + ∆t)− h(t)
∆t
![Page 21: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/21.jpg)
Population growth
ProblemGiven the population function of a group of organisms, find therate of growth of the population at a particular instant.
Example
Suppose the population of fish in the East River is given by thefunction
P(t) =3et
1 + et
where t is in years since 2000 and P is in millions of fish. Is thefish population growing fastest in 1990, 2000, or 2010? (Estimatenumerically)?
SolutionThe estimated rates of growth are 0.000136, 0.75, and 0.000136.
![Page 22: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/22.jpg)
Population growth
ProblemGiven the population function of a group of organisms, find therate of growth of the population at a particular instant.
Example
Suppose the population of fish in the East River is given by thefunction
P(t) =3et
1 + et
where t is in years since 2000 and P is in millions of fish. Is thefish population growing fastest in 1990, 2000, or 2010? (Estimatenumerically)?
SolutionThe estimated rates of growth are 0.000136, 0.75, and 0.000136.
![Page 23: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/23.jpg)
Numerical evidence
r1990 ≈P(−10 + 0.1)− P(−10)
0.1≈ 0.000136
r2000 ≈P(0.1)− P(0)
0.1≈ 0.75
r2010 ≈P(10 + 0.1)− P(10)
0.1≈ 0.000136
![Page 24: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/24.jpg)
Numerical evidence
r1990 ≈P(−10 + 0.1)− P(−10)
0.1≈ 0.000136
r2000 ≈P(0.1)− P(0)
0.1≈ 0.75
r2010 ≈P(10 + 0.1)− P(10)
0.1≈ 0.000136
![Page 25: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/25.jpg)
Numerical evidence
r1990 ≈P(−10 + 0.1)− P(−10)
0.1≈ 0.000136
r2000 ≈P(0.1)− P(0)
0.1≈ 0.75
r2010 ≈P(10 + 0.1)− P(10)
0.1≈ 0.000136
![Page 26: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/26.jpg)
Population growth
ProblemGiven the population function of a group of organisms, find therate of growth of the population at a particular instant.
Example
Suppose the population of fish in the East River is given by thefunction
P(t) =3et
1 + et
where t is in years since 2000 and P is in millions of fish. Is thefish population growing fastest in 1990, 2000, or 2010? (Estimatenumerically)?
SolutionThe estimated rates of growth are 0.000136, 0.75, and 0.000136.
![Page 27: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/27.jpg)
Upshot
The instantaneous population growth is given by
lim∆t→0
P(t + ∆t)− P(t)
∆t
![Page 28: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/28.jpg)
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 ayear 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 shouldproduce more to lower average costs.
![Page 29: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/29.jpg)
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 ayear 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 shouldproduce more to lower average costs.
![Page 30: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/30.jpg)
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
![Page 31: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/31.jpg)
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 ayear 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 shouldproduce more to lower average costs.
![Page 32: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/32.jpg)
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.
![Page 33: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/33.jpg)
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.
![Page 34: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/34.jpg)
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
![Page 35: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/35.jpg)
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
exists, the function is said to be differentiable at a and f ′(a) isthe derivative of f at a.
![Page 36: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/36.jpg)
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
exists, the function is said to be differentiable at a and f ′(a) isthe derivative of f at a.
![Page 37: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/37.jpg)
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.
![Page 38: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/38.jpg)
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.
![Page 39: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/39.jpg)
What does f tell you about f ′?
I If f is a function, we can compute the derivative f ′(x) at eachpoint x where f is differentiable, and come up with anotherfunction, the derivative function.
I What can we say about this function f ′?I If f is decreasing on an interval, f ′ is negative (well,
nonpositive) on that intervalI If f is increasing on an interval, f ′ is positive (well,
nonnegative) on that interval
![Page 40: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/40.jpg)
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
![Page 41: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/41.jpg)
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 alimit at a, the limit of the product is the product of the limits.
![Page 42: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/42.jpg)
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 alimit at a, the limit of the product is the product of the limits.
![Page 43: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/43.jpg)
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 alimit at a, the limit of the product is the product of the limits.
![Page 44: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/44.jpg)
How can a function fail to be differentiable?Kinks
x
f (x)
x
f ′(x)
![Page 45: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/45.jpg)
How can a function fail to be differentiable?Kinks
x
f (x)
x
f ′(x)
![Page 46: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/46.jpg)
How can a function fail to be differentiable?Kinks
x
f (x)
x
f ′(x)
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How can a function fail to be differentiable?Cusps
x
f (x)
x
f ′(x)
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How can a function fail to be differentiable?Cusps
x
f (x)
x
f ′(x)
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How can a function fail to be differentiable?Cusps
x
f (x)
x
f ′(x)
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How can a function fail to be differentiable?Vertical Tangents
x
f (x)
x
f ′(x)
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How can a function fail to be differentiable?Vertical Tangents
x
f (x)
x
f ′(x)
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How can a function fail to be differentiable?Vertical Tangents
x
f (x)
x
f ′(x)
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How can a function fail to be differentiable?Weird, Wild, Stuff
x
f (x)
x
f ′(x)
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How can a function fail to be differentiable?Weird, Wild, Stuff
x
f (x)
x
f ′(x)
![Page 55: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/55.jpg)
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
![Page 56: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/56.jpg)
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.
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Meet the Mathematician: Isaac Newton
I English, 1643–1727
I Professor at Cambridge(England)
I Philosophiae NaturalisPrincipia Mathematicapublished 1687
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Meet the Mathematician: Gottfried Leibniz
I German, 1646–1716
I Eminent philosopher aswell as mathematician
I Contemporarily disgracedby the calculus prioritydispute
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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
![Page 60: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/60.jpg)
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:
d2y
dx2
d2
dx2f (x)
d2f
dx2
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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:
d2y
dx2
d2
dx2f (x)
d2f
dx2
![Page 62: Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function](https://reader033.fdocuments.in/reader033/viewer/2022052411/5562b2bcd8b42a15548b54df/html5/thumbnails/62.jpg)
function, derivative, second derivative
x
y
f (x) = x2
f ′(x) = 2x
f ′′(x) = 2