1 / 14

Lecture 8

Chapter 4. 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. Techniques of Differentiation. Chapter 4.

jeremy-good
Télécharger la présentation

Lecture 8

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter4 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

  2. Techniques of Differentiation Chapter4 • The Product and Quotient Rules (Lecture 8) • The Chain Rule (Lectures 9 & 10) • Derivatives of Logarithmic and Exponential Functions (Lecture 11)

  3. The Quotient Rule The Product Rule

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

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

  6. Compute the Derivative Ex. = –10

  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. Calculation Thought Experiment 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.

  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. Proof of Product Rule (continued)

  11. Example: Given 3 factors y = (x)(x + 1)(x – 3), we could change this given to y = (x2 + x)(x – 3),or treat the problem as y = [(x)(x + 1)][x – 3]

  12. §4.1HW Problem: 68.Bus Travel Thoroughbred Bus Company finds that its monthly cost for one particular year were given C(t) = 100 + t2 dollars after tmonths. After t months the company had P(t) = 1000 +t2passengers per month. How fast is its cost per passenger changing after 6 months? Solution: The cost per passenger is increasing at a rate of ?$ Per month.

  13. §4.1HW Problem: 74.Military Spending in the 1990s The 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. 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. §4.1HW Problem 74(continued): 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 Hence, the total personnel cost in 1995 (t = 5) was decreasing at a rate of $? million per year.

More Related