Exploratory Problems for Chapter 16 531 (c) Find d dx f g x=3 . (d) Let y = f(x) 2 g(x 2 ). What is dy dx x=2 ? (e) Let y = f(x 2 ).Find y (2). 9. Below is a graph of f(x)on the interval [−2, 3]. j(x)is given by j(x)=ln f(x) on the domain [−2, 3]. 1 1 –1–2 2 2 3 f x (a) How many zeros does j(x) have? (b) Find j (x). (c) Approximate the critical points of j. (d) Identify the local extrema of j. (Estimate the positions.) (e) Where does j attain its absolute maximum value? Its absolute minimum value? (f) Which function has a higher absolute maximum value, f or j? Which function has a lower absolute minimum value, f or j? Explain. 10. g(x) is a continuous function with exactly two zeros, one at x = 1 and the other at x = 4. g(x) has a local minimum at x = 3 and a local maximum at x = 7. These are the only local extrema of g. Let f(x)= g(x) 4 . (a) Find f (x) in terms of g and its derivatives. (b) Can we determine (definitively) whether g has an absolute minimum value on (−∞, ∞)? If we can, where is that absolute minimum value attained? Can we determine (definitively) whether g has an absolute maximum value? If we can, where is that absolute maximum value attained? (c) What are the critical points of f ? (d) On what intervals is the graph of f increasing? On what intervals is it decreasing? (e) Identify the local maximum and minimum points of f . (f) Can we determine (definitively) whether f has an absolute minimum value? If so, can we determine what that value is? If you haven’t already done so, step back, take a good look at the problem (a bird’s-eye view) and make sure your answers make sense. 11. Assume that f , g, and h are differentiable. Differentiate p(x) where (a) p(x) = f (x)g(x)h(x).(Hint: Use the Product Rule twice.) (b) p(x) = √ g(x) + ln f(x). 532 CHAPTER 16 Taking the Derivative of Composite Functions 12. Let f(x)be the function whose graph is drawn on the axes below. 1 1–1 –1 –2 –2 –3–4–5–62345 2 3 4 f(x) (–5, –1) (3, –1) Let a(x) =2f(x), b(x) = f(x+2), and c(x) = f(2x). (a) On three separate sets of axes, draw the graphs of the function a(x) =2f(x), b(x) =f(x +2), and c(x) = f(2x), labeling the x-intercepts, the y-intercept, and the x- and y-coordinates of the local extrema. (b) Suppose we know that f (−4) = 1/2, f (−2) = 3/2, f (0) = 0. Find the following, explaining your reasoning briefly. i. a (−2) = ii. b (−2) = iii. c (−2) = In Problems 13 through 18, find h (x). Assume that f and g are differentiable on (−∞, ∞). Your answers may be in terms of f , g, f , and g . 13. h(x) = f(ln x) −ln(f (x)) 14. h(x) = √ f (x)g(x) 15. h(x) = 1 √ f (g(x)) 16. h(x) = f(x 2 )e g(x) 17. h(x) = 1 [f(x)] 2 +f 1 x 2 18. h(x) = [f(x)] 3 g(2x) In Problems 19 through 22, find dy dx . Take the time to prepare the expression so that it is as simple as possible to differentiate. 19. y = 3ln x 2 −1 x+2 Exploratory Problems for Chapter 16 533 20. y = (x 2 + 3) 5 21. y = 5x π −x 3 −1 x 2 22. y = 5ln 3x x 2 +1 In Problems 23 through 29, differentiate. In Problems 23 through 25, assume f is differentiable. Your answers may be in terms of f and f . 23. y = ln x f(x 2 ) 24. y = [f(x)] 2 +2 f(x) 25. y = ln x·f(x) √ 3x 3 +2 26. f(x)=ln(e (x+5) 2 ) 27. f(x)=e (g(x)) 2 28. f(x)=(x 3 + e) π 29. Let f(x)=x x . (a) Use numerical methods to approximate f (2). (b) Refer to your answer to part (a) to show that f (x) = x · x x−1 . What is it about f that makes it not a power function? (c) Refer to your answer to part (a) to show that f (x) = ln x ·x x . What is it about f that makes it not an exponential function? (d) Challenge: Figure out how to rewrite x x so you can use the Chain Rule to differ- entiate it. 30. Consider the function f(x)=e −x 2 (−x 2 +1).You must give exact answers for all of the following questions. Show your work. Your work must stand independent of your calculator. (a) Find all the x-intercepts. (b) Identify the local extrema of f(x). (c) Sketch a graph of f , labeling the x-coordinates of all local and global extrema. (d) Now consider the function g(x) =|f(x)|. i. What are the critical points of g? ii. Classify the critical points of g(x). 31. A craftsman is making a mobile consisting of hanging circles each with an inscribed triangle of stained glass. Each piece of stained glass will be an isosceles triangle. Show that if she wants to maximize the amount of stained glass used, the glass triangles should be equilateral. In other words, show that the isosceles triangle of maximum area that can be inscribed in a circle of radius R is an equilateral triangle. 534 CHAPTER 16 Taking the Derivative of Composite Functions 32. A holiday ornament is being constructed by inscribing a right circular cone of brightly colored material in a transparent spherical ball of radius 2 inches. What is the maximum possible volume of such an inscribed cone? 33. A craftsman is making a ribbon ornament by inscribing an open hollow cylinder of colored ribbon in a transparent spherical ball of radius R. What is the maximum surface area of such a cylinder? 34. Phone cables are to be run from an island to a town on the shore. The island is 5 miles from the shore and 13 miles from the town. The cable will be run in a straight line from the island to the shore and then in another straight line along the shoreline. If it costs 60% more to run the cable under water than it does to run it under the ground, how far should the cable be run along the shore? 35. The volume of a cylindrical tree trunk varies with time. Let r(t) give the radius of the trunk at time t and let h(t) give the height of time t. (a) Express the rate of change of A, the cross-sectional area, with respect to time in terms of r and r . (b) Express the rate of change of volume with respect to time in terms of r, r , h, and h . 36. Suppose the amount of power generated by an energy generating system is a function of v, the volume of water flowing through the system. The function is given by P = P(v). The volume of water in the sytem is determined by r, the radius of an adjustable valve; v = v(r). The radius varies with time: r = r(t). (a) Express dP dr , the rate of change of the power with respect to a change in the valve’s radius, in terms of the functions P(v) and v(r) and their derivatives. (b) Express dP dt , the rate of change of the power with respect to time, in terms of the functions P(v),v(r), and r(t) and their derivatives. 17 CHAPTER Implicit Differentiation and its Applications 17.1 INTRODUCTORY EXAMPLE In this section you will add sophistication to your differentiation skills by applying two fundamental principles. Principle (i): If two expressions are equal, their derivatives are equal. Principle (ii): Whether y is given implicitly or explicitly in terms of x, 1 we can differentiate an expression in y with respect to x by treating y the same way we would treat g(x) and using the Chain Rule. Try Example 17.1 below. Do the problem on your own, writing down your solution so you can compare it with the discussion that follows. ◆ EXAMPLE 17.1 Let f(x)=x x+1 ,where x>0. (a) Approximate f (2) numerically. (b) Using appropriate rules of differentiation, find f (2) exactly. (c) Compare your answers to parts (a) and (b). If they are not very close to one another, identify your error. SOLUTION (a) To approximate the slope of the tangent line to y = f(x) at x = 2, we find the slope of the secant line through (2, f(2)) and a nearby point on the graph of f , say (2.0001, f(2.0001)). 1 For example, xy 3 = 1 is an equation that gives y implicitly in terms of x; y = x 2 + 3x is an equation that gives y explicitly in terms of x. 535 536 CHAPTER 17 Implicit Differentiation and its Applications f (2) ≈ f(2.0001) − f(2) 2.0001 − 2 (Equivalently, we could say f (2) = lim h→0 f(2+h)−f(2) h ,sof (2)≈ f(2+h)−f(2) h for h very small. Then we choose a very small value for h, say h = 0.0001.) f (2) ≈ (2.0001) 2.0001+1 − 2 2+1 0.0001 f (2) ≈ (2.0001) 3.0001 − 2 3 0.0001 ≈ 17.547 REMARK When using a calculator to evaluate this, do not round off the numerator before dividing by 0.0001. To divide by 0.0001 is to multiply by 10,000, so any roundoff error in the numerator will be multiplied ten thousand-fold. (b) There are two popular but fatally flawed approaches to differentiating x x+1 .We’ll set them up and knock them down before proceeding. Fatally flawed approach #1: We know that d dx x n = nx n−1 for any constant n. It is easy to succumb to temptation and propose (x + 1)x x as the derivative of x x+1 . But this differentiation rule cannot be applied because (x + 1) is not a constant; x x+1 is not a power function! (If you apply it anyway, comparing your answers to parts (a) and (b) (17.547 versus 24) ought to alert you to the error.) Fatally flawed approach #2: We know that d dx b g(x) = ln b · b g(x) · g (x) for any positive constant b. Trying to apply this rule to x x+1 by letting b = x leads to a derivative of ln x · x x+1 . The error is similar to that in the previous approach; b must be a constant, and x is not constant. Again, comparing your answers in parts (a) and (b) (17.547 versus 5.545) ought to alert you to an error if you use this approach. A correct approach: In order to apply a differentiation rule we know, we must alter the form of f(x). The problem is that x is in both the base and the exponent. A flash of inspiration can lead us to express f(x)as follows. f(x)=x x+1 =e ln x x+1 = e (x+1) ln x Now we have an expression of the form e g(x) and can use the fact that d dx e g(x) = e g(x) · g (x) to obtain f (x) = e (x+1) ln x d dx [(x + 1) ln x] = e (x+1) ln x x + 1 x + 1 · ln x = x x+1 x + 1 x + ln x . Then f (2) = 2 3 3 2 + ln 2 = 12 + 8 ln 2. This is 17.54517 , quite close to the approximation found in part (a). From what source could this flash of inspiration, expressing x x+1 as e (x+1) ln x , have arisen? We know that e ln A = A for any positive A, but what could make us think 17.1 Introductory Example 537 of applying that knowledge in this problem? Recall that in proving d dx x n = nx n−1 for any real number n we used the same idea. 2 We rewrote x n as e ln x n and then as e n ln x . (This is why we needed the Chain Rule before we could establish the proof.) An alternative correct approach: In the approach taken above, to differentiate f(x)= x x+1 we changed its form, converting the expression x x+1 to something we could more easily differentiate. Another approach is to deal instead with the entire equation f(x)= x x+1 .Ifwecan bring the (x + 1) down from the exponent before we differentiate, we’ll be in good shape. Taking logs of both sides of the equation f(x)=x x+1 accomplishes this while still preserving the balance of the equation. Keep in mind that taking the log of an expression changes it into a new and different expression; we are not proposing to take the logarithm of an expression, but rather of both sides of an equation. f(x)=x x+1 ln f(x)=ln(x x+1 ) Take ln of each side. (If A = B and A, B>0, then ln A = ln B.) ln f(x)=(x + 1)(ln x) Use ln a b = b ln a to bring down the exponent. Now we can differentiate both sides using principle (i), which says that if g(x) = h(x), then d dx g(x) = d dx h(x). Remember to use the Chain Rule on the left side (where we have ln(f (x))) and the Product Rule on the right. d dx [ln f(x)]= d dx [(x + 1)(ln x)] Differentiate each side. 1 f(x) f (x) = (1)(ln x) + (x + 1) 1 x f (x) = ln x + x + 1 x f(x) Solve for f (x). f (x) = ln x + x + 1 x x x+1 Replace f(x)by x x+1 . This gives us an expression for f (x), as desired. ◆ Expressions Versus Equations The distinction between an expression and an equation is critical to understanding the two approaches discussed above, so we’ll reiterate the distinction. The governing criterion for working with an expression is that its value remain un- changed; only its form may vary. Zero can be added to an expression; an expression can be multiplied by 1; an expression can be factored. An expression A in the domain of f can be written as f −1 (f (A)). For example, g(x) = ln(e g(x) ) and if g(x) > 0, then g(x) = e ln g(x) . 2 Sometimes establishing proofs and doing practical problem solving seem to be disparate activities. Often, however, they are more intimately intertwined; a proof can provide problem-solving inspiration, and problem solving might suggest a more general theorem and sometimes even a proof. 538 CHAPTER 17 Implicit Differentiation and its Applications An equation, on the other hand, establishes an equality of two expressions; we can manipulate an equation in any way that does not alter the equality of the two sides. The governing criterion for working with equations is that balance be maintained. Adding/subtracting the same thing to/from both sides of the equation maintains the balance. Multiplying/dividing both sides of the equation by the same nonzero quantity maintains the balance. If A = C, where A and C are both positive, then A n = C n . For example, squaring both sides of an equation (n = 2) maintains balance (although it may introduce extrane- ous roots). Similarly, taking reciprocals of both sides of an equation (n =−1) main- tains balance. If J + K = C, then (J + K) n = C n . In particular, if J + K = C, then 1 J +K = 1 C . If A = C, where A and C are both positive, then ln A = ln C. For example, if J + K = C, then ln(J + K) = ln C. If A = C, then b A = b C for any positive constant b. For example, if J + K = C then e J +K = e C . We have presented two different successful approaches to tackling the problem pre- sented in Example 17.1(b). The first dealt with the expression x x+1 ; we used the fact that e ln A = A for any positive A to convert the expression x x+1 to the equivalent expression e (x+1) ln x . The second approach dealt with the equation f(x)=x x+1 .Wetook the natural logarithm of both sides of the equation to obtain an equivalent equation, differentiated both sides of the equation, and solved for f (x). This latter technique of differentiation is called logarithmic differentiation. PROBLEMS FOR SECTION 17.1 Differentiate the following. 1. (a) y = 3 x (b) y = x 3 (c) y = x x , where x>0. 2. y = (x + 1) (x+1) , where x>−1. 3. y = (3x 2 + 2) x 4. y = x x 2 , where x>0. 17.2 LOGARITHMIC DIFFERENTIATION Logarithmic differentiation deals with the task of differentiating a positive function f(x) by working with both sides of the equation y = f(x)as follows. Using Logarithmic Differentiation to Find y 1. Begin with an equation y = f(x),where f(x)>0.Takethe natural logarithm of both sides of the equation. 17.2 Logarithmic Differentiation 539 2. Use log rules to bring down exponents and/or simplify expressions. 3. Differentiate both sides of the equation. y is a function of x so the Chain Rule must be applied to differentiate ln y or ln f(x). d dx [ln f(x)]= 1 f(x) f (x) or, equivalently, d dx [ln y] = 1 y dy dx . 4. Solve for dy dx or f (x). 5. To express y in terms of x, replace y or f(x)with the equivalent expression in terms of x. When to Use Logarithmic Differentiation i. The technique is useful in differentiating a function that has the variable in both the base and the exponent. ii. We may choose to use logarithmic differentiation to make the differentiation of quo- tients or products more palatable, provided that we only take the log of positive quan- tities. ◆ EXAMPLE 17.2 Let y = 2x e x , where x>0.Find dy dx . SOLUTION The variable is in the base and the exponent; logarithmic differentiation enables us to bring down the expression in the exponent. y = 2x e x ln y = ln 2x e x Take the natural logarithm of each side. ln y = ln 2 + e x ln x Use log rules (i) and (iii) to bring down the exponent. d dx [ln y] = d dx [ln 2 + e x ln x] Differentiate each side. ln 2 is a constant. 1 y dy dx = (e x )(ln x) + (e x ) 1 x Use the Chain Rule on left because y is a function of x. dy dx = e x ln x + 1 x y Solve for dy dx . dy dx = e x ln x + 1 x 2x e x Replace y by its expression in x. ◆ ◆ EXAMPLE 17.3 Differentiate y = (x + 3) 5 (x 2 + 7x) 8 x(x 2 + 5) 3 , where x>0. 3 3 The condition x>0assures that (x + 3), (x 2 + 7x), and x are all positive. 540 CHAPTER 17 Implicit Differentiation and its Applications y = (x + 3) 5 (x 2 + 7x) 8 x(x 2 +5) 3 ln y = ln (x + 3) 5 (x 2 + 7x) 8 x(x 2 +5) 3 Take the natural logarithm of each side. ln y = ln[(x + 3) 5 (x 2 + 7x) 8 ] −ln[x(x 2 +5) 3 ] Use the fact that ln a b = ln a − ln b. ln y = 5ln(x + 3) + 8ln(x 2 + 7x) −ln x − 3ln(x 2 + 5) Use ln(ab) = ln a + ln b and ln a b = b ln a. d dx [ln y] = d dx [5 ln(x + 3) +8ln(x 2 + 7x) −ln x − 3ln(x 2 + 5)] Differentiate each side. 1 y dy dx = 5 x + 3 + 8(2x + 7) x 2 + 7x − 1 x − 3(2x) x 2 + 5 Apply the Chain Rule. dy dx = 5 x + 3 + 16x + 56 x 2 + 7x − 1 x − 6x x 2 + 5 y Solve for dy dx . dy dx = 5 x + 3 + 16x + 56 x 2 + 7x − 1 x − 6x x 2 + 5 (x + 3) 5 (x 2 + 7x) 8 x(x 2 +5) 3 Replace y by its expression in x. Ugly, but not as painful as differentiating this using the Quotient and Product Rules. 4 ◆ PROBLEMS FOR SECTION 17.2 1. Find f (x). (a) f(x)=2x x ,where x>0 (b) f(x)=5(x 2 + 1) x (c) f(x)=(2x 4 +5) 3x+1 2. Find f (x). (a) f(x)=3·2 x +2·x 3 +3·x 2x+3 ,where x>0 (b) f(x)=x(2x 3 +1) x +5, where x>0 3. Find g (t). (a) g(t) = 2 t t 2t , where t>0 (b) g(t) = ln(t + 1) t 2 +1 , where t>−1 4. Find dy dx using logarithmic differentiation. You need not simplify. (a) y = x ln √ x , where x>0 (b) y = xe 5x (x+1) 2 √ x−2 , where x>0 (c) y = (e 2x )(x 2 + 3) 5 (2x 2 + 1) 3 5. Suppose y = f (x)g(x), where f(x) and g(x) are positive for all x. Use logarithmic differentiation to find dy dx . Verify that your result is simply the Product Rule. 4 If you ever forget the Product or Quotient Rules but remember the derivative of ln x, you can use logarithmic differentiation to reconstruct the other rules for yourself. . respect to a change in the valve’s radius, in terms of the functions P(v) and v(r) and their derivatives. (b) Express dP dt , the rate of change of the power with respect to time, in terms of the functions. of time t. (a) Express the rate of change of A, the cross-sectional area, with respect to time in terms of r and r . (b) Express the rate of change of volume with respect to time in terms of. spherical ball of radius R. What is the maximum surface area of such a cylinder? 34. Phone cables are to be run from an island to a town on the shore. The island is 5 miles from the shore and 13 miles