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1 Lecture 8 §4.1 The Product and Quotient Rules Product Rule or, first.(der. of sec) + sec.(der. of first) Quotient Rule or, low.(d. high) – high.(d. low)

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Presentation on theme: "1 Lecture 8 §4.1 The Product and Quotient Rules Product Rule or, first.(der. of sec) + sec.(der. of first) Quotient Rule or, low.(d. high) – high.(d. low)"— Presentation transcript:

1 1 Lecture 8 §4.1 The Product and Quotient Rules Product Rule or, first.(der. of sec) + sec.(der. of first) Quotient Rule or, low.(d. high) – high.(d. low) draw the line & square below Chapter 4

2 2 Techniques of Differentiation Chapter 4 The Product and Quotient Rules (Lecture 8) The Chain Rule (Lectures 9 & 10) Derivatives of Logarithmic and Exponential Functions (Lecture 11)

3 3 The Quotient Rule The Product Rule

4 4 Ex. Derivative of first Derivative of Second The Product Rule

5 5 The Quotient Rule Ex. Derivative of numerator Derivative of denominator

6 6 Compute the Derivative Ex. = –10

7 7 Calculation Thought Experiment Given an expression, consider the steps you would use in computing its value. If the last operation is multiplication, treat the expression as a product; if the last operation is division, treat the expression as a quotient; and so on.

8 8 Ex. To compute a value first you would evaluate the parentheses then multiply the results, so this can be treated as a product. Ex. To compute a value the last operation would be to subtract, so this can be treated as a difference. Calculation Thought Experiment

9 9 Proof of Product Rule If you add a quantity, then subtract the same quantity, the value is the same. We could – f(x)g(x + h), then + f(x)g(x + h) to numerator Now factor out the common term from the first 2 parts, then factor out the common term from the last 2 parts

10 10 Proof of Product Rule (continued)

11 11 Given 3 factors y = (x)(x + 1)(x – 3), we could change this given to y = (x 2 + x)(x – 3), or treat the problem as y = [(x)(x + 1)][x – 3] Example:

12 12 68. Bus TravelThoroughbred Bus Company finds that its monthly cost for one particular year were given C(t) = 100 + t 2 dollars after t months. After t months the company had P(t) = 1000 +t 2 passengers per month. How fast is its cost per passenger changing after 6 months? §4.1 HW Problem: Solution: The cost per passenger is increasing at a rate of ?$ Per month.

13 13 74. Military Spending in the 1990sThe annual cost per active-duty armed service member in the United States increased from $80,000 in 1995 to 90,000 in 2000, In 1990 there were 2 million armed service personnel, and this number decreased to 1.5 million in 2000. Use linear models for annual cost and personnel to estimate to the nearest $10 million the rate of change of total military personnel cost in 1995. §4.1 HW Problem: Solution: Let t = 0 represent 1990. The annual cost is the line through (5, 80,000) and (10, 90,000), which is C(t) = ?

14 14 §4.1 HW Problem 74 (continued) : Hence, the total personnel cost in 1995 (t = 5) was decreasing at a rate of $? million per year. The number of personnel is the line through (0, 2) and (10, 1.5), which is P(t) = ? The total personnel cost is then C(t)P(t) and its rate of change is


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