§11.3 Derivatives of Products and Quotients $11.4 The Chain Rule

1. §11.3 Derivatives of Products and Quotients $11.4 The Chain Rule • the derivative of a product of two functions, and • the derivative of a quotient of two functions. • The student will be able to • form the composition of two functions • apply the general power rule. • the chain rule.

2. 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.

3. Example Find the derivative of y = 5x2(x3 + 2).

4. Example Find the derivative of y = 5x2(x3 + 2). Solution: Let F(x) = 5x2, so F ’(x) = 10xLet 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.

5. 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.

6. Example Find the derivative of y = 3x / (2x + 5).

7. Example Find the derivative of y = 3x / (2x + 5). Solution: Let T(x) = 3x, so T ’(x) = 3Let B(x) = 2x + 5, so B ’(x) = 2. Then

8. 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.

9. 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

10. Summary Product Rule: Quotient Rule:

11. Composite Functions Definition: A function m is a composite of functions f and g if m(x) = f [g(x)] The domain of m is the set of all numbers x such that x is in the domain of g and g(x) is in the domain of f.

12. General Power Rule We have already made extensive use of the power rule: Now we want to generalize this rule so that we can differentiate composite functions of the form [u(x)]n, where u(x) is a differentiable function. Is the power rule still valid if we replace x with a function u(x)?

13. Example Let u(x) = 2x2 and f (x) = [u(x)]3 = 8x6. Which of the following is f ’(x)? (a) 3[u(x)]2 (b) 3[u’(x)]2 (c) 3[u(x)]2 u’(x)

14. Example Let u(x) = 2x2 and f (x) = [u(x)]3 = 8x6. Which of the following is f ’(x)? (a) 3[u(x)]2 (b) 3[u’(x)]2 (c) 3[u(x)]2 u’(x) We know that f ’(x) = 48x5. (a)3[u(x)]2 = 3(2x2)2 = 3(4x4) = 12 x4. This is not correct. (b) 3[u’(x)]2 = 3(4x)2 = 3(16x2) = 48x2. This is not correct. (c) 3[u(x)]2 u’(x) = 3[2x2]2(4x) = 3(4x4)(4x) = 48x5. This is the correct choice.

15. Generalized Power Rule What we have seen is an example of the generalized power rule: If u is a function of x, then For example,

16. Chain Rule We have used the generalized power rule to find derivatives of composite functions of the form f (g(x)) where f (u) = un is a power function. But what if f is not a power function? It is a more general rule, the chain rule, that enables us to compute the derivatives of many composite functions of the form f(g(x)). Chain Rule: If y = f (u) and u = g(x) define the composite function y = f (u) = f [g(x)], then

17. Generalized Derivative Rules If y = u n , then y’= nu n - 1du/dx 1. If y = ln u, then y’= 1/u du/dx 2. If y = e u, then y’ = e udu/dx 3.

18. Examples for the Power Rule Chain rule terms are marked:

19. Examples for Exponential Derivatives

20. Examples for Logarithmic Derivatives Barnett/Ziegler/Bylee Business Calculus 11e