lesson 6: the derivative

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.


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.


Outline

Rates of ChangeTangent LinesVelocityPopulation growthMarginal costs

The derivative, definedDerivatives of (some) power functions


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


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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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


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


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


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


Numerical evidence


t − 12 − 15

1.5

− 12.5

1.1 − 10.5

1.01 − 10.05

1.001 − 10.005


Numerical evidence


t − 12 − 15

1.5 − 12.5

1.1 − 10.5

1.01 − 10.05

1.001 − 10.005


Numerical evidence


t − 12 − 15

1.5 − 12.5

1.1

− 10.5

1.01 − 10.05

1.001 − 10.005


Numerical evidence


t − 12 − 15

1.5 − 12.5

1.1 − 10.5

1.01 − 10.05

1.001 − 10.005


Numerical evidence


t − 12 − 15

1.5 − 12.5

1.1 − 10.5

1.01

− 10.05

1.001 − 10.005


Numerical evidence


t − 12 − 15

1.5 − 12.5

1.1 − 10.5

1.01 − 10.05

1.001 − 10.005


Numerical evidence


t − 12 − 15

1.5 − 12.5

1.1 − 10.5

1.01 − 10.05

1.001

− 10.005


Numerical evidence


t − 12 − 15

1.5 − 12.5

1.1 − 10.5

1.01 − 10.05

1.001 − 10.005


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


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.


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.


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.


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


Numerical evidence



∆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


Numerical evidence



∆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


Numerical evidence



∆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


Numerical evidence



∆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


Numerical evidence



∆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


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.


Upshot

The instantaneous population growth is given by

lim∆t→0

P(t + ∆t) − P(t)

∆t


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.


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


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


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


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


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


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


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


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


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


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


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


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

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.


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.


Outline

Rates of ChangeTangent LinesVelocityPopulation growthMarginal costs

The derivative, definedDerivatives of (some) power functions


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.


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.


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


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)


Worksheet


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.


lesson 6: the derivative

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