H.Melikian1 § 10.5 Limits and the Derivative Derivatives of Constants, Power Forms, and Sums Dr.Hayk Melikyan Departmen of Mathematics and CS [email protected]

H.Melikian 1

§ 10.5 Limits and the Derivative

Derivatives of Constants, Power Forms, and Sums

Dr .Hayk MelikyanDepartmen of Mathematics and CS

[email protected]

The student will learn about:

the derivative of a constant function, the power rule,

a constant times f (x),

derivatives of sums and differences, and

an application.

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WARM UP EXERCISE

The ozone level (in parts per billion) on a summer day at R University is given by

P(x) = 80 + 12x – x 2

Where t is hours after 9 am.

1. Use the two-step process to find P’(x).

2. Find P(3) and P’(3).

3. Write an interpretation.

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Objectives for Section 10.5 Power Rule and Differentiation Properties

■ The student will be able to calculate the derivative of a constant function.

■ The student will be able to apply the power rule.

■ The student will be able to apply the constant multiple and sum and difference properties.

■ The student will be able to solve applications.

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Derivative Notation

In the preceding section we defined the derivative of a function. There are several widely used symbols to represent the derivative. Given y = f (x), the derivative of f at x may be represented by any of the following:

■ f (x)

■ y

■ dy/dx

H.Melikian 5Barnett/Ziegler/Byleen College Mathematics 12e

Example 1

What is the slope of a constant function?

H.Melikian 6Barnett/Ziegler/Byleen College Mathematics 12e

Example 1(continued)

What is the slope of a constant function?

The graph of f (x) = C is a horizontal line with slope 0, so we would expect f ’(x) = 0.

Theorem 1. Let y = f (x) = C be a constant function, then

y = f (x) = 0.

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Power Rule

A function of the form f (x) = xn is called a power function. This includes f (x) = x (where n = 1) and radical functions (fractional n).

Theorem 2. (Power Rule) Let y = xn be a power function, then

y = f (x) = dy/dx = n xn – 1.

THEOREM 2 IS VERY IMPORTANT. IT WILL BE USED A LOT!

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Example 2

Differentiate f (x) = x5.

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Example 2

Differentiate f (x) = x5.

Solution:

By the power rule, the derivative of xn is n xn–1.

In our case n = 5, so we get f (x) = 5 x4.

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Differentiate

Example 3

.)( 3 xxf

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Differentiate

Solution:

Rewrite f (x) as a power function, and apply the power rule:

Example 3

3/1)( xxf

.)( 3 xxf

f (x)

1

3x 2/3

1

3 x23

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Constant Multiple Property

Theorem 3. Let y = f (x) = k u(x) be a constant k times a function u(x). Then

y = f (x) = k u (x).

In words: The derivative of a constant times a function is the constant times the derivative of the function.

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Example 4

Differentiate f (x) = 7x4.

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Example 4

Differentiate f (x) = 7x4.

Solution:

Apply the constant multiple property and the power rule.

f (x) = 7(4x3) = 28 x3.

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Sum and Difference Properties

Theorem 5. If y = f (x) = u(x) ± v(x),

then y = f (x) = u(x) ± v(x).

In words:■ The derivative of the sum of two differentiable functions is the sum of the derivatives.■ The derivative of the difference of two differentiable functions is the difference of the derivatives.

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Differentiate f (x) = 3x5 + x4 – 2x3 + 5x2 – 7x + 4.

Example 5

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Differentiate f (x) = 3x5 + x4 – 2x3 + 5x2 – 7x + 4.

Solution:

Apply the sum and difference rules, as well as the constant multiple property and the power rule.

f (x) = 15x4 + 4x3 – 6x2 + 10x – 7.

Example 5

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Applications

Remember that the derivative gives the instantaneous rate of change of the function with respect to x. That might be:

■ Instantaneous velocity.

■ Tangent line slope at a point on the curve of the function.

■ Marginal Cost. If C(x) is the cost function, that is, the total cost of producing x items, then C(x) approximates the cost of producing one more item at a production level of x items. C(x) is called the marginal cost.

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Tangent Line Example

Let f (x) = x4 – 6x2 + 10.

(a) Find f (x)

(b) Find the equation of the tangent line at x = 1

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Tangent Line Example(continued)

Let f (x) = x4 – 6x2 + 10.

(a) Find f (x)

(b) Find the equation of the tangent line at x = 1

Solution:

(a) f (x) = 4x3 - 12x

(b) Slope: f (1) = 4(13) – 12(1) = -8.Point: If x = 1, then y = f (1) = 1 – 6 + 10 = 5. Point-slope form: y – y1 = m(x – x1)

y – 5 = –8(x –1) y = –8x + 13

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Application Example

The total cost (in dollars) of producing x portable radios per day is

C(x) = 1000 + 100x – 0.5x2

for 0 ≤ x ≤ 100.

1. Find the marginal cost at a production level of x radios.

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Example(continued)

The total cost (in dollars) of producing x portable radios per day is

C(x) = 1000 + 100x – 0.5x2

for 0 ≤ x ≤ 100.

1. Find the marginal cost at a production level of x radios.

Solution: The marginal cost will be

C(x) = 100 – x.

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Example(continued)

2. Find the marginal cost at a production level of 80 radios and interpret the result.

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Example(continued)


Solution: C(80) = 100 – 80 = 20.

It will cost approximately $20 to produce the 81st radio.

3. Find the actual cost of producing the 81st radio and compare this with the marginal cost.

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Example(continued)


Solution: C(80) = 100 – 80 = 20.

It will cost approximately $20 to produce the 81st radio.

3. Find the actual cost of producing the 81st radio and compare this with the marginal cost.

Solution: The actual cost of the 81st radio will be

C(81) – C(80) = $5819.50 – $5800 = $19.50.

This is approximately equal to the marginal cost.

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Summary

If f (x) = C, then f (x) = 0

If f (x) = xn, then f (x) = n xn-1

If f (x) = ku(x), then f (x) = ku(x)

If f (x) = u(x) ± v(x), then f (x) =

u(x) ± v(x).

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Practice Problems

§3.4

1, 5, 9, 13, 17, 19, 23, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69, 73, 77, 81, 83, 87, 89.

H.Melikian 28Hayk Melikyan/ MATH2000/ 28

Learning Objectives for Section 10.6 Differentials

The student will be able to apply the concept of increments.

The student will be able to compute differentials.

The student will be able to calculate approximations using differentials.

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Increments

In a previous section we defined the derivative of f at x as the limit of the difference quotient:

Increment notation will enable us to interpret the numerator and the denominator of the difference quotient separately.

h

xfhxfxf

h

)()(lim)('

0

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Example

Let y = f (x) = x3. If x changes from 2 to 2.1, then y will change from y = f (2) = 8 to y = f (2.1) = 9.261.

We can write this using increment notation. The change in x is called the increment in x and is denoted by x. is the Greek letter “delta”, which often stands for a difference or change. Similarly, the change in y is called the increment in y and is denoted by y.

In our example,

x = 2.1 – 2 = 0.1

y = f (2.1) – f (2) = 9.261 – 8 = 1.261.

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Graphical Illustration of Increments

For y = f (x)

x = x2 - x1 y = y2 - y1

x2 = x1 + x = f (x2) – f (x1) = f (x1 + x) – f (x1)

(x1, f (x1))

(x2, f (x2))

x1 x2

x

y■ y represents the

change in y corresponding to a x change in x.

■ x can be either positive or negative.

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Assume that the limit exists.

For small x,

Multiplying both sides of this equation by x gives us

y f ’(x) x.

Here the increments x and y represent the actual changes in x and y.

Differentials

x

yxf

x

0

lim)('

x

yxf

)('

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One of the notations for the derivative is

If we pretend that dx and dy are actual quantities, we get

We treat this equation as a definition, and call dx and dy differentials.

Differentials(continued)

dx

dyxf )('

dxxfdy )('

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x and dx are the same, and represent the change in x.

The increment y stands for the actual change in y resulting from the change in x.

The differential dy stands for the approximate change in y, estimated by using derivatives.

In applications, we use dy (which is easy to calculate) to estimate y (which is what we want).

Interpretation of Differentials

dxxfdyy )('

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Example 1

Find dy for f (x) = x2 + 3x and evaluate dy for x = 2 and dx = 0.1.

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Example 1

Find dy for f (x) = x2 + 3x and evaluate dy for x = 2 and dx = 0.1.

Solution:

dy = f ’(x) dx = (2x + 3) dx

When x = 2 and dx = 0.1, dy = [2(2) + 3] 0.1 = 0.7.

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Example 2 Cost-Revenue

A company manufactures and sells x transistor radios per week. If the weekly cost and revenue equations are

find the approximate changes in revenue and profit if production is increased from 2,000 to 2,010 units/week.

000,80

000,110)(

2000,5)(2

x

xxxR

xxC

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The profit is

We will approximate R and P with dR and dP, respectively, using x = 2,000 and dx = 2,010 – 2,000 = 10.

Example 2 Solution

000,5000,1

8)()()(2

x

xxCxRxP

per week60$10)500

000,210(

)500

10()('

dxx

dxxRdR

per week40$10)500

000,28(

)500

8()('

dxx

dxxPdP

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Objectives for Section-10.7 Marginal Analysis

The student will be able to compute:

■ Marginal cost, revenue and profit

■ Marginal average cost, revenue and profit

■ The student will be able to solve applications

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Marginal Cost

Remember that marginal refers to an instantaneous rate of change, that is, a derivative.

Definition:

If x is the number of units of a product produced in some time interval, then

Total cost = C(x)

Marginal cost = C’(x)

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Marginal Revenue and Marginal Profit

Definition:If x is the number of units of a product sold in some time interval, then

Total revenue = R(x) Marginal revenue = R’(x)

If x is the number of units of a product produced and sold in some time interval, then

Total profit = P(x) = R(x) – C(x)Marginal profit = P’(x) = R’(x) – C’(x)

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Marginal Cost and Exact Cost

Assume C(x) is the total cost of producing x items. Then the exact cost of producing the (x + 1)st item is

C(x + 1) – C(x).

The marginal cost is an approximation of the exact cost.

C’(x) ≈ C(x + 1) – C(x).

Similar statements are true for revenue and profit.

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Example 1

The total cost of producing x electric guitars is C(x) = 1,000 + 100x – 0.25x2.

1. Find the exact cost of producing the 51st guitar.

2. Use the marginal cost to approximate the cost of producing the 51st guitar.

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The total cost of producing x electric guitars is C(x) = 1,000 + 100x – 0.25x2.

1. Find the exact cost of producing the 51st guitar.

The exact cost is C(x + 1) – C(x).

C(51) – C(50) = 5,449.75 – 5375 = $74.75.

2. Use the marginal cost to approximate the cost of producing the 51st guitar.

The marginal cost is C’(x) = 100 – 0.5x

C’(50) = $75.

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Marginal Average Cost

Definition:

If x is the number of units of a product produced in some time interval, then

Average cost per unit

Marginal average cost

x

xCxC

)()(

)()(' xCdx

dxC

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If x is the number of units of a product sold in some time interval, then

Average revenue per unit

Marginal average revenue

If x is the number of units of a product produced and sold in some time interval, then

Average profit per unit

Marginal average profit

Marginal Average Revenue Marginal Average Profit

x

xRxR

)()(

)()(' xRdx

dxR

x

xPxP

)()(

)()(' xPdx

dxP

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Warning!

To calculate the marginal averages you must calculate the average first (divide by x), and then the derivative. If you change this order you will get no useful economic interpretations.

STOP

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Example 2

The total cost of printing x dictionaries is

C(x) = 20,000 + 10x

1. Find the average cost per unit if 1,000 dictionaries are produced.

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The total cost of printing x dictionaries is

C(x) = 20,000 + 10x

1. Find the average cost per unit if 1,000 dictionaries are produced.

= $30

x

xCxC

)()(

)000,1(C000,1

000,10000,20

x

x10000,20

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2. Find the marginal average cost at a production level of 1,000 dictionaries, and interpret the results.

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2. Find the marginal average cost at a production level of 1,000 dictionaries, and interpret the results.

Marginal average cost =

)()(' xCdx

dxC

x

x

dx

dxC

1020000)('

21000

20000)1000('C

2

20000

x

02.0

This means that if you raise production from 1,000 to 1,001 dictionaries, the price per book will fall approximately 2 cents.

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3. Use the results from above to estimate the average cost per dictionary if 1,001 dictionaries are produced.

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3. Use the results from above to estimate the average cost per dictionary if 1,001 dictionaries are produced.

Average cost for 1000 dictionaries = $30.00Marginal average cost = - 0.02

The average cost per dictionary for 1001 dictionaries would be the average for 1000, plus the marginal average cost, or

$30.00 + $(- 0.02) = $29.98

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The price-demand equation and the cost function for the production of television sets are given by

where x is the number of sets that can be sold at a price of $p per set, and C(x) is the total cost of producing x sets.

1. Find the marginal cost.

Example 3

xxCx

xp 30000,150)(and30

300)(

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The price-demand equation and the cost function for the production of television sets are given by

where x is the number of sets that can be sold at a price of $p per set, and C(x) is the total cost of producing x sets.

1. Find the marginal cost.

Solution: The marginal cost is C’(x) = $30.


xxCx

xp 30000,150)(and30

300)(

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2. Find the revenue function in terms of x.


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The revenue function is

3. Find the marginal revenue.


30300)()(

2xxxpxxR

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The marginal revenue is

4. Find R’(1500) and interpret the results.


30300)()(

2xxxpxxR

15300)('

xxR

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The marginal revenue is

4. Find R’(1500) and interpret the results.

At a production rate of 1,500, each additional set increases revenue by approximately $200.


30300)()(

2xxxpxxR

15300)('

xxR

200$15

1500300)1500(' R

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5. Graph the cost function and the revenue function on the same coordinate. Find the break-even point.

0 < y < 700,000

0 < x < 9,000

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5. Graph the cost function and the revenue function on the same coordinate. Find the break-even point.

0 < y < 700,000

0 < x < 9,000

(600,168,000)(7500, 375,000)

Solution: There are two break-even points.

C(x)

R(x)

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6. Find the profit function in terms of x.


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The profit is revenue minus cost, so

7. Find the marginal profit.


15000027030

)(2

xx

xP

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8. Find P’(1500) and interpret the results.


15000027030

)(2

xx

xP

15270)('

xxP

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8. Find P’(1500) and interpret the results.

At a production level of 1500 sets, profit is increasing at a rate of about $170 per set.


15000027030

)(2

xx

xP

15270)('

xxP

17015

1500270)1500(' P

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QUIZZ Time

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§ 9.5 Derivates of Products and Quotients

The student will learn about:

the derivative of a product of two functions, and

the derivative of a quotient of two functions.

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Derivates of Products

The derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.

Theorem 1 - Product Rule

If f (x) = F (x) • S (x),

Then f’(x) = F (x)’ • S (x) + F (x) • S’ (x), OR

f’(x) = F + Sdx

dS

dx

dF

H.Melikian 69

Example

Find the derivative of y = 5x2(x3 + 2).

Product Rule

Let F (x) = 5x2 then F ‘ (x) =

Let S (x) = x3 + 2 then S ‘ (x) =

f ‘ (x) = 5x2 • 3x2 + (x3 + 2) • 10x= 15x4 + 10x4 + 20x = 25x4 + 20x

If f (x) = F (x) • S (x),

Then f ’ (x) = F (x)’ • S (x) + F (x) • S’ (x).

10x

3x2, and

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Derivatives of Quotients

The derivative of the quotient of two functions is the bottom function times the derivative of the top function minus the top function times the derivative of the bottom function, all over the bottom function squared.

Theorem 2. Quotient Rule:

If y = f (x) = T (x) / B (x), Then

2])x(B[

)x('B)x(T)x('T)x(B)x('f

H.Melikian 71

Derivatives of Quotients

May also be expressed as -

2BdxdB

TdxdT

B

dx

dy

2])x(B[


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Example

Let T (x) = 3x and then T ‘ (x) =

Find the derivative of .5x2

x3y

2])x(B[


Let B (x) = 2x + 5 and then B ‘ (x) =

2)5x2(

2x33)5x2()x('f 2)5x2(

15

3.2.

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Summary.

Product Rule. If F (x) and S (x), then

F (x) • S‘(x) + S (x) • F‘(x). )x(S)x(Fdx

d

Quotient Rule. If T(x) and B(x), then

2BdxdB

TdxdT

B

)x(B

)x(T

dx

d

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Practice Problems

§9.5

1, 5, 9, 13, 17, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 65, 69, 71.

Documents

H.Melikian1 § 10.5 Limits and the Derivative Derivatives of Constants, Power Forms, and Sums Dr.Hayk Melikyan Departmen of Mathematics and CS [email protected]