Objectives for Section 11.3 Derivatives of Products and Quotients

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11.3:Derivatives of Products and Quotients

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Presentation transcript:

Objectives for Section 11.3 Derivatives of Products and Quotients The student will be able to calculate: the derivative of a product of two functions, and the derivative of a quotient of two functions. Barnett/Ziegler/Byleen Business Calculus 11e

Derivatives of Products Theorem 1 (Product Rule) If f (x) = F(x)  S(x), and if F ’(x) and S ’(x) exist, then f ’ (x) = F(x)  S ’(x) + F ’(x)  S(x), or In words: The derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. Barnett/Ziegler/Byleen Business Calculus 11e

Example Find the derivative of y = 5x2(x3 + 2). Barnett/Ziegler/Byleen Business Calculus 11e

Example Find the derivative of y = 5x2(x3 + 2). Solution: Let F(x) = 5x2, so F ’(x) = 10x Let S(x) = x3 + 2, so S ’(x) = 3x2. Then f ’ (x) = F(x)  S ’(x) + F ’(x)  S(x) = 5x2  3x2 + 10x  (x3 + 2) = 15x4 + 10x4 + 20x = 25x4 + 20x. Barnett/Ziegler/Byleen Business Calculus 11e

Derivatives of Quotients Theorem 2 (Quotient Rule) If f (x) = T (x) / B(x), and if T ’(x) and B ’(x) exist, then or In words: The derivative of the quotient of two functions is the bottom function times the derivative of the top function minus the top function times the derivative of the bottom function, all over the bottom function squared. Barnett/Ziegler/Byleen Business Calculus 11e

Example Find the derivative of y = 3x / (2x + 5). Barnett/Ziegler/Byleen Business Calculus 11e

Example Find the derivative of y = 3x / (2x + 5). Solution: Let T(x) = 3x, so T ’(x) = 3 Let B(x) = 2x + 5, so B ’(x) = 2. Then Barnett/Ziegler/Byleen Business Calculus 11e

Tangent Lines Let f (x) = (2x - 9)(x2 + 6). Find the equation of the line tangent to the graph of f (x) at x = 3. Barnett/Ziegler/Byleen Business Calculus 11e

Tangent Lines Let f (x) = (2x - 9)(x2 + 6). Find the equation of the line tangent to the graph of f (x) at x = 3. Solution: First, find f ’(x): f ’(x) = (2x - 9) (2x) + (2) (x2 + 6) Then find f (3) and f ’(3): f (3) = -45 f ’(3) = 12 The tangent has slope 12 and goes through the point (3, -45). Using the point-slope form y - y1 = m(x - x1), we get y – (-45) = 12(x - 3) or y = 12x - 81 Barnett/Ziegler/Byleen Business Calculus 11e

Summary Product Rule: Quotient Rule: Barnett/Ziegler/Byleen Business Calculus 11e