## Xem mẫu

16.2 The Derivative of where is any Real Number 521 22. 32 23. 2 12 24. 3 2 1 25. 33 26. Prove that if ’3 & where ’, & 0 and ’ &, then has a point of inßection at ’. 27. (a) Which of the following are equal to ? Identify all correct answers. i. ii. iii. iv. v. 2 vi. 2 vii. 2 viii. 2 (b) Differentiate 2. 28. Graph 2 and answer the following questions. (a) What is the domain of ? The range? (b) Is an even function, an odd function, or neither? (c) For what values of is increasing? Decreasing? (d) Find all relative maximum and minimum points. (e) Does haveanabsolutemaximumvalue?Anabsoluteminimumvalue?Agreatest lower bound? If any of these exist, identify them. (f) Find the -coordinates of the points of inßection. (Note: This function is a good one to keep in mind. It is extremely useful in both probability and statistics because, with some minor adjustments, it gives us a normal distribution curve.) 16.2 THE DERIVATIVE OF WHERE IS ANY REAL NUMBER Atthispointwehaveproventhat 1 foranyinteger.Infact,thisformulaholds if is any real number. We can use the Chain Rule to prove this fact. Suppose that is any real number and we want to Þnd . The key is to rewrite so we can use derivative formulas that we already know. ln ln and ln are inverses, so ln ( (. Use log rule (iii). Now we take the derivative. 522 CHAPTER 16 Taking the Derivative of Composite Functions ln ln ln ln 1 Use the Chain Rule: mess mess mess . Keep in mind that is a constant. Rewrite ln as . We now can take the derivative of where is any real number. 1 EXERCISE 16.1 Find if 22.1 . Answer 2[1 ln 2.122.11.1] P R O B L E M S F O R S E C T I O N 1 6 . 2 For Problems 1 through 3, Þnd . 1. 2 2 2. 2 1 3. 321 4. Differentiate the following, simplifying the expression Þrst if useful. (a) 32 (d) ln 2.6 (b) ln 1 (e) ln 21.5 (c) 2 4 (f) 3 ln 1 5. Find , simplifying the expression Þrst where useful. (a) (d) ln 1 3.5 (g) ln 1 (b) 1 (e) ln 1 (h) 5 6 (c) (f) 1 ln 54 6. Identify and classify all critical points of the function 2 41 for 0. 16.3 Using the Chain Rule 523 16.3 USING THE CHAIN RULE EXAMPLE 16.6 SOLUTION In this section, we take a second look at applying the Chain Rule. Once we know the Chain Rule, there is no problem if we forget the Quotient Rule, because can be written as [ ]1. Derive the Quotient Rule from the Product Rule. [ ]1 [ 1] [ 1] 1[ ]2 [ ]1 [ ]2 [ ]2 [ ]2 [ ]2 Rewrite the quotient as a product. Use the Product Rule. Use the Chain Rule to Þnd [ 1] 1[ ]2 . Rewrite negative exponents as fractions. Get a common denominator of [ ]2. Combine into one fraction. EXAMPLE 16.7 SOLUTION We can apply the Chain Rule to relate rates of change; as illustrated in the following example. In Nepal the daily kerosene consumption in restaurants and guesthouses throughout the country can be modeled by the function ), where is the number of tourists in the country. Tourism is seasonal; let be the number of tourists at time . In our model, ) and are continuous and differentiable. At a certain time, there are 5,000 tourists in Nepal and the number is decreasing at a rate of 40 per day. Write an expression for the rate at which kerosene consumption is changing with time at this moment. WeÕre looking for . We know that , and 40. Therefore, ) 50004040) 5000, where ) 5000 ) . 5000 Before applying the Chain Rule to more complicated expressions and situations, weÕll summarize our knowledge of differentiation, using the Chain Rule to express familiar differentiation formulas in a more general way. Derivatives: A Summary GraphicalInterpretation:If ,thenthederivativeof istheslopefunction,giving the slope of the tangent line to at the point , . For instance, 2 gives the slope of at 2. Instantaneous Rate of Change: The derivative evaluated at 2 gives the instantaneous rate of change of with respect to at 2. Notation: , , , , , all mean the same thing. is an operator that means Òtake the derivative of what follows.Ó 524 CHAPTER 16 Taking the Derivative of Composite Functions Limit DeÞnition: The following expressions are equivalent and are equal to . lim ! lim " lim " !0 "0 "0 D i f f e r e n t i a t i o n F o r m u l a s 1. 2. constant Sum Rule Constant Multiple Rule 3. 4. 5. [ ] constant [ ]1 Product Rule For the product of two functions Quotient Rule Generalized Power Rule (Variable in base) 6. & b constant & ln & Exponential Function (Variable in exponent) 7. log& ln[ ] 8. ln & Logarithmic Function Chain Rule The world of functions we can differentiate has broadened immensely! If faced with a function we cannot differentiate using these shortcuts,2 we can return to the limit deÞnition of derivative. If we cannot make headway obtaining an exact value for a derivative and the functionweÕreworkingwithisreasonablywellbehaved,wecanstillusenumericalmethods to approximate the derivative at a point, and we can obtain qualitative information using a graphical approach. CAUTION Be sure you know the difference between an exponential function like 12 and a generalized power function like 2 2. If the variable is in the base, then we differentiate using the generalized power rule; if the variable is in the exponent, we use the generalized rule for exponential functions. Feel free to think of some of these differentiation formulas more informally. For instance, the derivative of the derivative of the derivative of (mess) ln(mess) mess is mess1 mess , is mess mess , and is mess mess . 2 This will be the case when we Þrst run into trigonometric functions. 16.3 Using the Chain Rule 525 These rules are all obtained by applying the Chain Rule to the differentiation rules we have been using. Then when you look at a complicated expression, you can categorize it. For instance, the expression 1 ln 1 is basically mess12, meaning were we to construct the expression assembly-line style, the last worker on the line would take the mess coming along the line and raise it to the 12. The derivative of mess12 is 1mess32 mess . WhatÕs the mess? Basically, it is lnstuff, so mess stuff stuff . Putting this together, 1 ln2 1 1 ln2 1 2 ln2 13 2 1 2. What weÕve doneÑalbeit informallyÑis to decompose ln21 into !, where 12, So ln , and !2 1. 2 1 ln2 1 ln2 1 ! ! !! . A WORD OF ADVICE Many students, having learned shortcuts to differentiation, approach problems by charging them, sleeves rolled up, brutally hacking away. There is no virtue in doing a problem in the most difÞcult way possible; it does not put you on higher moral ground. Rather, you increase the likelihood of making an error. Instead, when presented with a function to differentiate, take a moment to consider how to prepare the function for differentiation. Sometimes, a bit of thoughtful reformulation will save you time and energy. (Of course, there are times when it is necessary to dig in and get your hands all dirty. If it is necessary, do it with relish!) The meaning of this advice is clariÞed in the next example. EXAMPLE 16.8 SOLUTION Differentiate ln 13 . If youÕre enthusiastically charging the problem, you might informally say that basically this is lnmess, where mess is stuff12 and to differentiate stuff, the Quotient Rule and generalized power rule come into play. If youÕre formally charging headlong into this, you might valiantly decompose the function as follows: ... - --nqh--