Chain Rule: Power Form Marginal Analysis in Business and Economics 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. The slides for this text are organized into chapters. This lecture covers Chapter 1. Chapter 1: Introduction to Database Systems Chapter 2: The Entity-Relationship Model Chapter 3: The Relational Model Chapter 4 (Part A): Relational Algebra Chapter 4 (Part B): Relational Calculus Chapter 5: SQL: Queries, Programming, Triggers Chapter 6: Query-by-Example (QBE) Chapter 7: Storing Data: Disks and Files Chapter 8: File Organizations and Indexing Chapter 9: Tree-Structured Indexing Chapter 10: Hash-Based Indexing Chapter 11: External Sorting Chapter 12 (Part A): Evaluation of Relational Operators Chapter 12 (Part B): Evaluation of Relational Operators: Other Techniques Chapter 13: Introduction to Query Optimization Chapter 14: A Typical Relational Optimizer Chapter 15: Schema Refinement and Normal Forms Chapter 16 (Part A): Physical Database Design Chapter 16 (Part B): Database Tuning Chapter 17: Security Chapter 18: Transaction Management Overview Chapter 19: Concurrency Control Chapter 20: Crash Recovery Chapter 21: Parallel and Distributed Databases Chapter 22: Internet Databases Chapter 23: Decision Support Chapter 24: Data Mining Chapter 25: Object-Database Systems Chapter 26: Spatial Data Management Chapter 27: Deductive Databases Chapter 28: Additional Topics Dr .Hayk Melikyan Departmen of Mathematics and CS melikyan@nccu.edu

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. 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

* * * * * VERY IMPORTANT * * * * * 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 * * * * * VERY IMPORTANT * * * * *

Find the derivative of y = (x3 + 2) 5. Example 1 Find the derivative of y = (x3 + 2) 5. Chain Rule NOTE: If we let u = x 3 + 2, then y = u 5. Let u (x) = x3 + 2, then y = u 5 and u ‘ (x) = 3x2 5 (x3 + 2) 4 3x2 = 15x2(x3 + 2)4

Find the derivative of y = Example 2 Find the derivative of y = Rewrite as y = (x 3 + 3) 1/2 Then y’ = 1/2 (x 3 + 3) – 1/2 Then y’ = 1/2 Then y’ = 1/2 (x 3 + 3) – 1/2 (3x2) = 3/2 x2 (x3 + 3) –1/2

Combining Rules of Differentiation The chain rule just developed may be used in combination with the previous rules for taking derivatives Some examples follow.

Example 3 Find f ’ (x) if f (x) = 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)

Example 4 Find 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

Example 4 continued Find 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.

By graphing calculator! Example 4 continued Find 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

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 for 20 ≤ p ≤ 100 Find dx/dp. f ‘ (p) = - (60) (1/2) (p + 25)-1/2 (1)

Application continued 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 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). f (75) = 1,000 – 60 = 1000 – 600 = 400 f ‘ (75) = -30/10 = - 3

Application continued 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 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.

Summary. If y = f (x) = [u (x)]n then

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)

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)

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)

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.

Example 1 P. 210, #2. The total cost of producing x electric guitars is C (x) = 1,000 + 100 x – 0.25 x2 Find the exact cost of producing the 51st guitar. Exact cost is C (x + 1) – C (x) C (51) = $5449.75 C (50) = $5375.00 Exact cost = $5449.75 - $5375 = $74.75

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. The marginal cost is C ‘ (x) C ‘ (x) = 100 – 0.5x C ‘ (50) = $75.00 Exact cost = $5449.75 - $5375 = $74.75

Marginal Average Cost If x is the number of units of a product produced in some time interval, then Average cost per unit = Marginal average cost =

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 =

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 =

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

Example 2 P. 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

Example 2 continued Marginal average cost = The 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 = What does this mean?

Example 2 continued Average cost = $30.00 The 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 = $30.00 Marginal 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

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 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. Find the marginal cost. The marginal cost is C ‘ (x) so What does this mean? C ‘ (x) = $30.

Example 3 continued 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 2. Find the revenue function in terms of x. The revenue function is R (x) = x · p (x), so

Example 3 continued The price-demand equation and the cost function for the production of television sets are given, respectively by and C (x) = 150,000 + 30x 3. Find the marginal revenue. The marginal revenue is R ‘ (x), so

Example 3 continued The 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. What does this mean? At a production rate of 1,500 sets, revenue is increasing at the rate of about $200 per set.

Example 3 continued The price-demand equation and the cost function for the production of television sets are given, respectively by C (x) = 150,000 + 30x and 5. Graph the cost function and the revenue function on the same coordinate. Find the break-even point. 0 ≤ x ≤ 9,000 0 ≤ y ≤ 700,000 (600,168000) (7500, 375000)

Example 3 continued The price-demand equation and the cost function for the production of television sets are given, respectively by C (x) = 150,000 + 30x and 6. Find the profit function in terms of x. The profit is revenue minus cost, so

Example 3 continued The 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

Example 3 continued The 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. What does this mean? At a production level of 1500 sets, profit is increasing at a rate of about $170 per set.

Summary. In business the instantaneous rate of change, the derivative, is referred to as the margin.