h.melikian1 § 10.5 limits and the derivative derivatives of constants, power forms, and sums...
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H.Melikian 1
§ 10.5 Limits and the Derivative
Derivatives of Constants, Power Forms, and Sums
Dr .Hayk MelikyanDepartmen of Mathematics and CS
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)
2. Find the marginal cost at a production level of 80 radios and interpret the result.
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)
2. Find the marginal cost at a production level of 80 radios and interpret the result.
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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Example 1(continued)
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
H.Melikian 4646
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
H.Melikian 4747
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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Example 2(continued)
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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Example 2(continued)
2. Find the marginal average cost at a production level of 1,000 dictionaries, and interpret the results.
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Example 2(continued)
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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Example 2(continued)
3. Use the results from above to estimate the average cost per dictionary if 1,001 dictionaries are produced.
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Example 2(continued)
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
H.Melikian 5454
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)(
H.Melikian 5555
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.
Example 3(continued)
xxCx
xp 30000,150)(and30
300)(
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2. Find the revenue function in terms of x.
Example 3(continued)
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2. Find the revenue function in terms of x.
The revenue function is
3. Find the marginal revenue.
Example 3(continued)
30300)()(
2xxxpxxR
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2. Find the revenue function in terms of x.
The revenue function is
3. Find the marginal revenue.
The marginal revenue is
4. Find R’(1500) and interpret the results.
Example 3(continued)
30300)()(
2xxxpxxR
15300)('
xxR
H.Melikian 5959
2. Find the revenue function in terms of x.
The revenue function is
3. Find the marginal revenue.
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.
Example 3(continued)
30300)()(
2xxxpxxR
15300)('
xxR
200$15
1500300)1500(' R
H.Melikian 6060
Example 3(continued)
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
H.Melikian 6161
Example 3(continued)
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.
Example 3(continued)
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6. Find the profit function in terms of x.
The profit is revenue minus cost, so
7. Find the marginal profit.
Example 3(continued)
15000027030
)(2
xx
xP
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6. Find the profit function in terms of x.
The profit is revenue minus cost, so
7. Find the marginal profit.
8. Find P’(1500) and interpret the results.
Example 3(continued)
15000027030
)(2
xx
xP
15270)('
xxP
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6. Find the profit function in terms of x.
The profit is revenue minus cost, so
7. Find the marginal profit.
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.
Example 3(continued)
15000027030
)(2
xx
xP
15270)('
xxP
17015
1500270)1500(' P
H.Melikian 66
QUIZZ Time
H.Melikian 67
§ 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.
H.Melikian 68
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[
)x('B)x(T)x('T)x(B)x('f
H.Melikian 72
Example
Let T (x) = 3x and then T ‘ (x) =
Find the derivative of .5x2
x3y
2])x(B[
)x('B)x(T)x('T)x(B)x('f
Let B (x) = 2x + 5 and then B ‘ (x) =
2)5x2(
2x33)5x2()x('f 2)5x2(
15
3.2.
H.Melikian 73
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
H.Melikian 74
Practice Problems
§9.5
1, 5, 9, 13, 17, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 65, 69, 71.