h.melikian1 chain rule: power form marginal analysis in business and economics dr.hayk melikyan...
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H.Melikian 1
Chain Rule: Power FormMarginal Analysis in Business
and Economics
Dr .Hayk MelikyanDepartmen of Mathematics and CS
The student will learn about:the chain rule, combining different rules of derivation, and an application.
Marginal cost, revenue, and profit as well as,
applications, andmarginal average cost,
revenue and profit.
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Chain Rule: Power Rule. We have already made extensive use of the power rule with xn,
We wish to generalize this rule to cover [u (x)]n.
1nn xnxdx
d
That is, we already know how to find the derivative of
f (x) = x 5
We now want to find the derivative of
f (x) = (3x 2 + 2x + 1) 5
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Chain Rule: Power Rule.
General Power Rule. [Chain Rule]
Theorem 1. If u (x) is a differential function, n is any real number, and
y = f (x) = [u (x)]n
then
f ’ (x) = n[ u (x)]n – 1 u’ (x) = n un – 1u’or
dx
duunu
dx
d 1nn
* * * * * VERY IMPORTANT * * * * *
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Example 1Find the derivative of y = (x3 + 2) 5.
Let u (x) = x3 + 2, then y = u 5 and u ‘ (x) = 3x2
53 )2x(dx
d5 (x3 + 2) 3x24
= 15x2(x3 + 2)4
Chain Rule
dx
duunu
dx
d 1nn
NOTE: If we let u = x 3 + 2, then y = u 5.
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Example 2Find the derivative of y =
Rewrite as y = (x 3 + 3) 1/2
= 3/2 x2 (x3 + 3) –1/2
3x 3
Then y’ = 1/2Then y’ = 1/2 (x 3 + 3) – 1/2Then y’ = 1/2 (x 3 + 3) – 1/2 (3x2)
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Combining Rules of Differentiation
The chain rule just developed may be used in combination with the previous rules for taking derivatives
Some examples follow.
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Example 3
Find f ’ (x) if f (x) = .)8x3(
x2
4
We will use a combination of the quotient rule and the chain rule.
Let the top be t (x) = x4, then t ‘ (x) = 4x3
Let the bottom be b (x) = (3x – 8)2, then using the chain rule b ‘ (x) = 2 (3x – 8) 3 = 6 (3x – 8)
22
432
))8x3((
)8x3(6x)x4()8x3()x('f
4
432
)8x3(
)8x3(x6)x4()8x3()x('f
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Example 4Find f ’ (x) and find the equation of the line tangent to the graph of f at the indicated value of x.
f (x) = x2 (1 – x)4; at x = 2.
We will use the point-slope form. The point will come from (2, f(2)) and the slope from f ‘ (2).
Point - When x = 2, f (x) = 22 (1 – 2)4 = (4) (1) = 4
Hence the tangent goes through the point (2,4).
f ‘ (x) = x2 4 (1 – x)3 (-1) + (1 – x)4 2x
= - 4x2 (1 – x)3 + 2x(1 – x)4 and
f ‘ (2) = (- 4) (4) (-1)3 + (2) (2) (-1)4 = 16 + 4 = 20 = slope
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Example 4 continuedFind f ’ (x) and find the equation of the line tangent to the graph of f at the indicated value of x.
f (x) = x2 (1 – x)4; at x = 2.
We will use the point-slope form. The point is (2, 4) and the slope is 20.
y – 4 = 20 (x – 2) = 20x – 40 or
y = 20x - 36.
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Example 4 continuedFind f ’ (x) and find the equation of the line tangent to the graph of f at the indicated value of x.
f (x) = x2 (1 – x)4; at x = 2.
By graphing calculator!Graph the function and use “Math”, “tangent”.
-2 ≤ x ≤ 3
-1 ≤ y ≤ 20
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Application P. 202, #78. The number x of stereo speakers people are willing to buy per week at a price of $p is given by
x = 1,000 - 60 25p for 20 ≤ p ≤ 100
1. Find dx/dp. f ‘ (p) =
25p
30
- (60) (1/2) (p + 25)-1/2 (1)
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Application continuedThe number x of stereo speakers people are willing to buy per week at a price of $p is given by
x = 1,000 - 60 25p for 20 ≤ p ≤ 100
2. Find the demand and the instantaneous rate of change of demand with respect to price when the price is $75.
That is, find f (75) and f ‘ (75).
2575
30
f ‘ (75) = -30/10 = - 3
f (75) = 1,000 – 60 2575 = 1000 – 600 = 400
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Application continuedThe number x of stereo speakers people are willing to buy per week at a price of $p is given by
x = 1,000 - 60 25p for 20 ≤ p ≤ 100
3. Give a verbal interpretation of these results.
With f (75) = 400 and f ‘ (75) = - 3 that means
that the demand at a price of $75 is 400 speakers and
each time the price is raised $1, three fewer speakers are purchased.
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Summary.
Ify = f (x) = [u (x)]n
then
dx
duunu
dx
d 1nn
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Marginal Cost, Revenue, and Profit
Remember that margin refers to an instantaneous rate of change, that is, a derivative.
Marginal Cost
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 Cost, Revenue, and Profit
Marginal Revenue
If x is the number of units of a product sold in some time interval, then
Total revenue = R (x)
Marginal revenue = R’ (x)
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Marginal Cost, Revenue, and Profit
Marginal Profit
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. Theorem 1. C (x) is the total cost of producing x items and
C (x + 1) is the cost of producing x + 1 items. Then the exact cost of producing the x + 1st item is
C (x + 1) – C (x)The marginal cost is an approximation of the exact cost. Hence,
C ’ (x) ≈ C (x + 1) – C (x).
The same is true for revenue and profit.
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Example 1
P. 210, #2. The total cost of producing x electric guitars is
C (x) = 1,000 + 100 x – 0.25 x2
1. Find the exact cost of producing the 51st guitar.
Exact cost is C (x + 1) – C (x)
C (51) =
C (50) =
Exact cost = $5449.75 - $5375 = $74.75
$5449.75
$5375.00
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Example 1 continued
The total cost of producing x electric guitars is
C (x) = 1,000 + 100 x – 0.25 x2
2. Use the marginal cost to approximate the cost of producing the 51st guitar.
C ‘ (x) =
C ‘ (50) =
Exact cost = $5449.75 - $5375 = $74.75
The marginal cost is C ‘ (x)
100 – 0.5x
$75.00
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Marginal Average Cost
If x is the number of units of a product produced in some time interval, then
Average cost per unit = x
)x(C)x(C
Marginal average cost = )x(Cdx
d)x('C
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Marginal Average Revenue
If x is the number of units of a product sold in some time interval, then
Average revenue per unit =
Marginal average revenue =
x
)x(R)x(R
)x(Rdx
d)x('R
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Marginal Average Profit.
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 =
x
)x(P)x(P
)x(Pdx
d)x('P
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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 2P. 210, # 4. 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
)x(C)x(C
)1000(C1000
000,10000,20
x
x1020000
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Example 2 continuedThe total cost of printing x dictionaries is
C (x) = 20,000 + 10x
2. Find the marginal average cost at a production level of 1,000 dictionaries, and interpret the results.
Marginal average cost = )x(Cdx
d)x('C
x
x1020000
dx
d)x('C
21000
20000)1000('C
2x
20000
02.0What does this mean?
2x
20000
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Example 2 continuedThe total cost of printing x dictionaries is
C (x) = 20,000 + 10x
3. Use the results from above to estimate the average cost per dictionary if 1,001 dictionaries are produced.
Average cost = $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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Example 3 P. 211, #14. The price-demand equation and the cost function for the production of television sets are given, respectively by
p (x) = 300 - and C (x) = 150,000 + 30x
The marginal cost is C ‘ (x) so
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.
C ‘ (x) = $30.
30
x
What does this mean?
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Example 3 continuedThe price-demand equation and the cost function for the production of television sets are given, respectively by
p (x) = 300 - and C (x) = 150,000 + 30x
The revenue function is R (x) = x · p (x), so
2. Find the revenue function in terms of x.
30
x
)x(R30
xx300
2
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Example 3 continuedThe price-demand equation and the cost function for the production of television sets are given, respectively by
and C (x) = 150,000 + 30x
The marginal revenue is R ‘ (x), so
3. Find the marginal revenue.
)x('R
30
xx300)x(R
2
15
x300
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Example 3 continuedThe price-demand equation and the cost function for the production of television sets are given, respectively by
and C (x) = 150,000 + 30x
4. Find R’ (1,500) and interpret the results.
)1500('R
At a production rate of 1,500 sets, revenue is increasing at the rate of about $200 per set.
15
x300)x('R
15
1500300 200$ What does
this mean?
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Example 3 continuedThe price-demand equation and the cost function for the production of television sets are given, respectively by
and C (x) = 150,000 + 30x
5. Graph the cost function and the revenue function on the same coordinate. Find the break-even point.
30
xx300)x(R
2
0 ≤ y ≤ 700,0000 ≤ x ≤ 9,000
(600,168000) (7500, 375000)
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Example 3 continuedThe price-demand equation and the cost function for the production of television sets are given, respectively by
and C (x) = 150,000 + 30x
6. Find the profit function in terms of x.
30
xx300)x(R
2
The profit is revenue minus cost, so
)x(P
150000x27030
x)x(P
2
x3015000030
xx300
2
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Example 3 continuedThe price-demand equation and the cost function for the production of television sets are given, respectively by
7. Find the marginal profit.
The marginal profit is P ‘ (x), so
)x('P
150000x27030
x)x(P
2
15
x270
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Example 3 continuedThe price-demand equation and the cost function for the production of television sets are given, respectively by
7. Find P’ (1,500) and interpret the results.
At a production level of 1500 sets, profit is increasing at a rate of about $170 per set.
)1500('P
15
x270)x('P
17015
1500270 What does
this mean?
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Summary.
In business the instantaneous rate of change, the derivative, is referred to as the margin.
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