Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function

transcript

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)

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

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

The tangent problem

Example

Upshot

mtangent = limx→a

f (x)− f (a)

x − a

Graphically and numerically

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

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

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

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

4.0401

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

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

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

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

3.9601

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

4.0401

3.9601

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

The tangent problem

Example

Upshot

mtangent = limx→a

f (x)− f (a)

x − a

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.

Velocity

Example

h(t) = 50− 10t2

limt→1

(50− 10t2)− 40

t − 1= −20.

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

Velocity

Example

h(t) = 50− 10t2

limt→1

(50− 10t2)− 40

t − 1= −20.

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)

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.

Population growth

Example

P(t) =3et

1 + et

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

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

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

Population growth

Example

P(t) =3et

1 + et

Upshot

The instantaneous population growth is given by

lim∆t→0

P(t + ∆t)− P(t)

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.

Marginal costs

Example

C (q) = q3 − 12q2 + 60q

Example

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

Marginal costs

Example

C (q) = q3 − 12q2 + 60q

Example

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)

is more useful since it’s unit-independent.

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)

is more useful since it’s unit-independent.

Outline

Other notations

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)

exists, the function is said to be differentiable at a and f ′(a) isthe derivative of f at a.

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)

exists, the function is said to be differentiable at a and f ′(a) isthe derivative of f at a.

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

(a + h)2 − a2

= limh→0

(a2 + 2ah + h2)− a2

h= lim

2ah + h2

= limh→0

(2a + h) = 2a.

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

(a + h)2 − a2

= limh→0

(a2 + 2ah + h2)− a2

h= lim

2ah + h2

= limh→0

(2a + h) = 2a.

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

Outline

Other notations

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.

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

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

How can a function fail to be differentiable?Kinks

f ′(x)

How can a function fail to be differentiable?Cusps

f ′(x)

How can a function fail to be differentiable?Vertical Tangents

f ′(x)

How can a function fail to be differentiable?Weird, Wild, Stuff

f ′(x)

How can a function fail to be differentiable?Weird, Wild, Stuff

f ′(x)

Outline

Other notations

Notation

I Newtonian notation

f ′(x) y ′(x) y ′

I Leibnizian notation

dxf (x)

These all mean the same thing.

Meet the Mathematician: Isaac Newton

I English, 1643–1727

I Professor at Cambridge(England)

I Philosophiae NaturalisPrincipia Mathematicapublished 1687

Meet the Mathematician: Gottfried Leibniz

I German, 1646–1716

I Eminent philosopher aswell as mathematician

I Contemporarily disgracedby the calculus prioritydispute

Outline

Other notations

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:

dx2f (x)

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:

dx2f (x)

function, derivative, second derivative

f (x) = x2

f ′(x) = 2x

f ′′(x) = 2

Lesson 7-8: Derivatives and Rates of Change, The Derivative as a function

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