2.4 The Product and Quotient Rules

Transcription

1 Hartfield MATH 040 Unit Page 1.4 The Product and Quotient Rules For functions which are the result of multiplying or dividing epressions, special rules apply which involve multiple steps. E. 1: Find the derivative using Product Rule. f ( ) 1 4 Rule 5: Product Rule d d f g. f g f g Rule 6: Quotient Rule d f g f f g d g g.

2 Hartfield MATH 040 Unit Page E. : Find the derivative using Product Rule. E. 1: Find the derivative using Quotient Rule. f ( ) 5 3 f ( ) 1 1

3 Hartfield MATH 040 Unit Page 3 E. : Find the derivative using Quotient Rule. E. 3: Find the derivative using Quotient Rule. f ( ) 1 f ( ) 1 1

4 Hartfield MATH 040 Unit Page 4 Marginal Average Revenue/Cost/Profit It may be helpful to know the average cost per unit which is simply the result of dividing the total cost by the number of units. Definition: Let AR() be the average revenue function, AC() be the average cost function, and AP() be the average profit function. Then, AR() = AC() = AP() = R( ), C ( ), P( ). and By taking the derivative of any of these average functions, you get a marginal average function which would be the change in the average function when producing one more unit. Definition: Let MAR() be the marginal average revenue function, MAC() be the marginal average cost function, and MAP() be the marginal average profit function. Then, MAR() = MAC() = MAP() = d d d d d d R ( ) C ( ),, P ( ). and

5 Hartfield MATH 040 Unit Page 5 Applications E. 1: To create a new line of pottery it takes an artisan $30 for the materials to make a jug along with $000 in startup costs. Thus, the cost function for making the jugs is B. Find the average cost function. C ( ) A. Evaluate C(50) and interpret the result. C. Evaluate AC(50) and interpret the result.

6 Hartfield MATH 040 Unit Page 6 D. Find the marginal average cost function. E. A Porsche s fuel economy (in miles per gallon) is given as a function E() of speed in mph, where E (Source: Finite Mathematics & Calculus Applied to the Real World (1996) p. 76, modified #55) E. Evaluate MAC(50) and interpret the result. A. Evaluate E(50) and E(70) and interpret the results.

7 Hartfield MATH 040 Unit Page 7 B. Find E (). C. Evaluate E (50) and E (70) and interpret the results. D. Using the evaluations from part C, what does the sign of each value tell you?

8 Hartfield MATH 040 Unit Page 8.6 Chain Rule and Generalized Power Rule Recall from algebra that composed functions consist of one function inside a second function. For eample, 3 1 is considered to be a composite function because the function ² 1 eists within a cubing function (that is, the output of ² 1 is the input to the cube). We can decompose 3 1 by defining the two functions that are brought together to make the new function. E.: Decompose each of the following functions so that f(g()) returns the original function. A f g ( ) with f ( ) 3 g ( ) 1 B. 7 We tend to call g the inside function and f the outside function.

9 Hartfield MATH 040 Unit Page 9 Chain Rule: Differentiating Compositions To differentiate a composition, it is important to recognize the roles that both the inside and outside functions play. Rule 7: Chain Rule E.: Find the derivative by applying Chain Rule. y 3 1 d f g ( ) f g ( ) g ( ) d This rule says to multiply the derivative of the outside function (while holding the inside function fied) to the derivative of the inside function. This derivative is conditioned on the assumption that the inside derivative eists. Restate the composition y f g ( ) as y = f(u) and u = g(). Their respective derivatives dy would be f ( u) du du and g( ) d. Then be appropriate substitution, we can present Chain Rule using Leibniz s Notation: d y d y d u d d u d.

10 Hartfield MATH 040 Unit Page 10 Generalized Power Rule Chain Rule can be applied to any composition. A special case of the Chain Rule is useful since frequently our outside function is a power function. E. 1: Find the derivative using Generalized Power Rule. 3 6 y Rule 7a: Generalized Power Rule d d n n 1 g ( ) n g ( ) g ( ) As we have run into previously, some functions require some etra attention when using Power Rule (or in this case Generalized P.R.)

11 Hartfield MATH 040 Unit Page 11 E. : Find the derivative using Generalized Power Rule. E. 3: Find the derivative using Generalized Power Rule. y 7 y 1 3 You may find it necessary in some cases to apply the Generalized Power Rule more than once.

12 Hartfield MATH 040 Unit Page 1 E. 4: Find the derivative using Generalized Power Rule. 3 y E. 5: Suppose that for a group of 10,000 people, the number who survive to age is N ( ) (Source: 4 th edition p. 167, #96) Find N (3 6 ) and interpret your result.

13 Hartfield MATH 040 Unit Page 13 Generalized Power Rule and Product and/or Quotient Rules An unavoidable challenge is that some compositions may involve products or quotients. Application of the Generalized Power Rule (or Chain Rule) with Product Rule or Quotient Rule requires careful application of each rule involved. E. 1: Find the derivative y

14 Hartfield MATH 040 Unit Page 14 E. : Find the derivative. y 1 3

15 Hartfield MATH 040 Unit Page 15.7 Nondifferentiable Functions Some functions, which are otherwise differentiable, will have points for which a derivative does not eist. If a function is either undefined or discontinuous at an -value, it follows that no derivative would eist for that -value. What we want to pay special attention to are situations where a function is defined and continuous and yet lacks a derivative. Consider the absolute value function: E.: Show that f() = is not differentiable at = 0.

16 Reasons a function may not be differentiable at Besides the possibility that a function is undefined or discontinuous at, there are two main reasons a function may lack a derivative: A. If f has a corner point at = c, f will not be differentiable at c. E. Hartfield MATH 040 Unit Page 16 Using the graph below, identify the values for where f is not differentiable. B. If f has a vertical tangent at = c, f will not be differentiable at c.

17 Hartfield MATH 040 Unit Page a Optimization, part 1 Definition: An absolute maimum value is a function value which is greater than or equal to any other value within the domain of the function. Definition: An absolute maimum value is a function value which is less than or equal to any other value within the domain of the function. If the domain of a function is limited and the function is continuous over that domain, then the function must have an absolute maimum value and an absolute minimum value. Continuous functions whose domains are not unlimited may have an absolute maimum value and/or an absolute minimum value but may have neither or just one. Function has an absolute maimum value but no absolute minimum. Function has neither an absolute maimum or absolute minimum value. Function has both an absolute maimum value and an absolute minimum value.

18 Hartfield MATH 040 Unit Page 18 Given a continuous function f on a closed interval [a, b], the absolute maimum value and the absolute minimum value of f will either be found at the -value endpoints of the domain or at -values within the domain where the derivative of the function equals zero (these are called critical numbers and will be more fully defined in the net unit). To find each value, evaluate at all relevant -values identified previously; the greatest value found will be the absolute maimum value and the least value found will be the absolute minimum value. E. 1: Find the absolute etreme values of the function on the given closed interval. 3 on [0, 6] f ( ) Collectively we call absolute maimum points and absolute minimum points together absolute etreme points.

19 Hartfield MATH 040 Unit Page 19 E. : Find the absolute etreme values of the function on the given closed interval. f ( ) 6 3 on [, 4] E. 3.: The fuel economy (in miles per gallon) on an average American compact car is E() = , where is the driving speed in miles per hour and 0 < < 60. At what speed is fuel economy greatest? (Source: 4 th edition p. 05, #6)

20 E. 4: A copier company finds that copiers that are years old require on average R() = 1.² repairs annually for the first 5 years after it is produced (thus, 0 < < 5). Find the year that requires the least repairs, rounding to nearest year. (Source: 4 th edition p. 05, #8) Hartfield MATH 040 Unit Page 0