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§ 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.

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§ 1.6

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  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:

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