h.melikian1 chain rule: power form marginal analysis in business and economics dr.hayk melikyan...

H.Melikian 1

Chain Rule: Power FormMarginal Analysis in Business

and Economics

Dr .Hayk MelikyanDepartmen of Mathematics and CS

melikyan@nccu.edu

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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