1 / 9

§ 1.6

§ 1.6. Some Rules for Differentiation. Section Outline. Some Rules of Differentiation Differentiation The Derivative as a Rate of Change. Rules of Differentiation. Differentiation. EXAMPLE. Differentiate. SOLUTION. This is the given function. We begin to differentiate.

mhorner
Download Presentation

§ 1.6

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. §1.6 Some Rules for Differentiation

  2. Section Outline • Some Rules of Differentiation • Differentiation • The Derivative as a Rate of Change

  3. Rules of Differentiation

  4. Differentiation EXAMPLE Differentiate SOLUTION This is the given function. We begin to differentiate. Rewrite the rational expression with a negative exponent. Use the General Power Rule taking x3 + x + 1 to be g(x). Use the Sum Rule.

  5. Differentiation CONTINUED Differentiate. Simplify. Simplify. Simplify.

  6. Differentiation EXAMPLE Differentiate SOLUTION This is the given function. We begin to differentiate. Rewrite the rational expression with a negative exponent. Use the Constant Multiple Rule. Use the General Power Rule taking to be g(x).

  7. Differentiation CONTINUED Use the Sum Rule and rewrite as x1/2. Differentiate. Simplify. Simplify. Simplify.

  8. The Derivative as a Rate of Change

  9. The Derivative as a Rate of Change EXAMPLE Let S(x) represent the total sales (in thousands of dollars) for month x in the year 2005 at a certain department store. Represent each statement below by an equation involving S or . (a) The sales at the end of January reached $120,560 and were rising at the rate of $1500 per month. (b) At the end of March, the sales for this month dropped to $80,000 and were falling by about $200 a day (Use 1 month = 30 days). SOLUTION • Since the sales at the end of January (the first month, so x = 1) reached $120,560 and S(x) represents the amount of sales for a given month, we have: S(1) = 120,560. Further, since the rate of change of sales (rate of change means we will use the derivative of S(x)) for the month of January is a positive $1500 per month, we have: (b) At the end of March (the third month, so x = 3), the sales dropped to $80,000. Therefore, sales for the month of March was $80,000. That is: S(3) = 80,000. Additionally, since sales were dropping by $200 per day during March, this means that the rate of change of the function S(x) was (30 days) x (-200 dollars) = -6000 dollars per month. Therefore, we have:

More Related