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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 33 Chapter 3 Techniques of Differentiation.

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Presentation on theme: "Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 33 Chapter 3 Techniques of Differentiation."— Presentation transcript:

1 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 33 Chapter 3 Techniques of Differentiation

2 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 33  The Product and Quotient Rules  The Chain Rule and the General Power Rule  Implicit Differentiation and Related Rates Chapter Outline

3 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 33 § 3.1 The Product and Quotient Rules

4 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 33  The Product Rule  The Quotient Rule  Rate of Change Section Outline

5 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 33 The Product Rule

6 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 33 The Product RuleEXAMPLE SOLUTION Differentiate the function. Let and. Then, using the product rule, and the general power rule to compute g΄(x),

7 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 33 The Quotient Rule

8 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 of 33 The Quotient RuleEXAMPLE SOLUTION Differentiate. Let and. Then, using the quotient rule Now simplify.

9 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 33 The Quotient Rule Now let’s differentiate again, but first simplify the expression. Now we can differentiate the function in its new form. CONTINUED Notice that the same answer was acquired both ways.

10 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 10 of 33 Rate of ChangeEXAMPLE SOLUTION (Rate of Change) The width of a rectangle is increasing at a rate of 3 inches per second and its length is increasing at the rate of 4 inches per second. At what rate is the area of the rectangle increasing when its width is 5 inches and its length is 6 inches? [Hint: Let W(t) and L(t) be the widths and lengths, respectively, at time t.] Since we are looking for the rate at which the area of the rectangle is changing, we will need to evaluate the derivative of an area function, A(x) for those given values (and to simplify, let’s say that this is happening at time t = t 0 ). Thus This is the area function. Differentiate using the product rule.

11 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 11 of 33 Rate of Change Now, since the width of the rectangle is increasing at a rate of 3 inches per second, we know W΄(t) = 3. And since the length is increasing at a rate of 4 inches per second, we know L΄(t) = 4. Now we substitute into the derivative of A. This is the derivative function. CONTINUED W΄(t) = 3, L΄(t) = 4, W(t) = 5, and L(t) = 6. Simplify. Now, we are determining the rate at which the area of the rectangle is increasing when its width is 5 inches (W(t) = 5) and its length is 6 inches (L(t) = 6).

12 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 12 of 33 The Product Rule & Quotient Rule Another way to order terms in the product and quotient rules, for the purpose of memorizing them more easily, is PRODUCT RULE QUOTIENT RULE


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