Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 50 pot

Exploratory Problem for Chapter 14 471 y = log b x ⇐⇒ b y = x ln b y = ln x y ln b = ln x y = ln x ln b log b x = ln x ln b = 1 ln b ln x 1 ln b is simply a constant. We know that the derivative of a constant times f(x)is that constant times f  (x),so d dx  1 ln b ln x  = 1 ln b d dx (ln x) = 1 ln b 1 x . Conclusion: d dx log b x = 1 ln b 1 x . The derivative of log b x is just a constant times 1 x . For example, d dx log x = 1 ln 10 1 x , because log x = 1 ln 10 ln x. Notice that there is no escaping the natural logarithm. It pops its head 3 right up into the derivative of the log base b of x no matter what the value of b! ◆ EXAMPLE 14.1 Find the equation of the line tangent to f(x)=log 3 27x 2 at the point (1, 3). SOLUTION The slope of the tangent line at x = 1isgivenbyf  (1). The equation of the line is y − y 1 = m(x−x 1 ),ory−3=f  (1)(x−1). f(x)=log 3 27 + log 3 x 2 = 3 + log 3 x 2 . For positive x, f(x)=3+2log 3 x. We need to differentiate this and evaluate the derivative at x = 1. Recall: d dx (3 + 2 log 3 x)     x=1 says “take the derivative of 3 + 2 log 3 x and evaluate it at x = 1.” d dx  3 + 2log 3 x       x=1 =  0 + 2  1 ln 3 1 x       x=1 = 2 ln 3 . Therefore, the equation of the tangent line is y − 3 = 2 ln 3 (x − 1) or y = 2 ln 3 x − 2 ln 3 + 3. This is a linear equation; 2 ln 3 is a constant. ◆ PROBLEMS FOR SECTION 14.1 In Problems 1 through 7, find y  . 3 Of course we could write log b log e in place of ln b. In particular, ln 10 = log 10 log e = 1 log e .So d dx log x = 1 ln 10 1 x = (log e) 1 x ;bute is involved however we present the derivative. 472 CHAPTER 14 Differentiating Logarithmic and Exponential Functions 1. y = 2ln5x 2. y = π ln √ x 3. y = ln 3x 5 4. y = x ln x 5. y = ln √ 2x x 6. y = 3 log x 7. y = log 2 x 3 8. Show that f(x)= ln √ 3x 2 + 3 is invertible. Find f −1 (x). 9. Show that g(x) =π log 2 (πx) − π 2 is invertible. Find g −1 (x). 10. Find and classify the critical points of f(x)=xln x. 11. What is the lowest value taken on by the function g(x) =x 2 ln x? Is there a highest value? Explain. 12. Use a tangent line approximation of ln x at x = 1 to approximate: (a) ln(0.9). (b) ln(1.1). 13. Graph f(x)= √ x −ln x, indicating all local maxima, minima, and points of inflection. Do this without your graphing calculator. (You can use your calculator to check your answer.) To aid in doing the graphing, do the following. (a) On a number line, indicate the sign of f  . Above this number line draw arrows indicating whether f is increasing or decreasing. (b) On a number line indicate the sign of f  . Above this number line indicate the concavity of f . (c) Find lim x→0 + f(x)and lim x→∞ f(x)using all tools available to you. You should be able to give a strong argument supporting your answer to the former. The latter requires a bit more ingenuity, but you can do it. 14. Let f(x)=ln x − x. (a) What is the domain of this function? (b) Find all the critical points of f . (The critical points must be in the domain of f .) (c) By looking at the sign of f  , find all local maxima and minima. Give both the x- and y-coordinates of the extrema. (d) Find f  . Where is f concave up and where is f concave down? (e) Sketch the graph of ln x − x without using a calculator (except possibly to check your work). 14.2 The Derivative of b x Revisited 473 14.2 THE DERIVATIVE OF b x REVISITED EXERCISE 14.3 Label the y-coordinates of the three points indicated in Figure 14.4 below. Use numerical approximations, rounding off your answers to three decimal places. Do these numbers look familiar? y 12345 678910 x y = ln x (2, __) (3, __) (10, __) Figure 14.4 When we first investigated the derivative of b x , we showed that d dx b x = kb x , where k is a constant that depends on the base b. We know that k is the slope of b x at x = 0, and we approximated k for various values of b. We found d dx 2 x ≈ (0.693)2 x d dx 3 x ≈ (1.099)3 x d dx 10 x ≈ (2.303)10 x , and as a result of work on Exploratory Problems you may also know that d dx 2.7 x ≈ (0.993)2.7 x and d dx 2.8 x ≈ (1.030)2.8 x . None of this is completely satisfying. These results drive us to search for some structure to these constants. That search, or the exercise at the beginning of this section, might lead us to conjecture that d dx b x = ln b · b x . This conjecture we have is wonderful and delightful. If it is true, those constants that you’ve been memorizing or looking up in order to approximate the derivatives of 2 x , 3 x , and 10 x (the proportionality constants for the rates of growth of these functions), are just ln 2, ln 3, and ln 10, respectively. The natural logarithm function just pops up—oh so naturally! We will show that the conjecture d dx b x = ln b · b x holds. Our argument will rest on the fact that log b x and b x are inverse functions. For each point (c, d) on the graph of b x , the point (d, c) lies on the graph of log b x. Notice that if (c, d) lies on the graph of b x , then d = b c . 474 CHAPTER 14 Differentiating Logarithmic and Exponential Functions y 1 x L 1 L 1 L 2 L 2 y = log b x y = b x (c, b c ) = (c, d) (d, c) Figure 14.5 The slope of L 1 , the line tangent to the graph of b x at x = c,is 1 the slope of L 2 , where L 2 is the line tangent to the graph of log b x at the point (d, c). The slope of L 2 at the point (d, c) = d dx log b x     x=d = 1 ln b 1 x     x=d = 1 ln b 1 d = 1 d ln b . Therefore, the slope of L 1 = 1 1 d ln b = d ln b = ln b · d. But (c, d) lies on the graph of b x so d = b c . Consequently, the slope of L 1 = ln b · b c . We have shown that the derivative of b x at x = c is ln b · b c . Conclusion: d dx b x = ln b · b x An Alternative Approach to Finding the Derivative of b x We know that the derivative of b x is b x · (the slope of the tangent to b x at x = 0). We can concentrate our energy simply on finding the slope of the tangent line to b x at (0,1). It is the reciprocal of the slope of log b x at the point (1,0). 14.2 The Derivative of b x Revisited 475 y (1, 0) (0, 1) x y = log b x y = b x Figure 14.6 The slope of the line tangent to log b x at x = 1is d dx log b x    x=1 = 1 ln b 1 x    x=1 = 1 ln b . Therefore, the slope of the tangent line to b x at x = 0islnb. Weconclude that d dx b x =ln b · b x , as before. EXERCISE 14.4 Differentiate f(x)=3·2 x . Answer f  (x) = 3ln2·2 x PROBLEMS FOR SECTION 14.2 For Problems 1 through 3, find dy dx . 1. y = x 2 · 2 x 2. y = 5·2 x 3 3. y = x 5 5 x 5 4. Find the absolute minimum value of f(x)=x ·2 x . 5. Find f  (x) if (a) f(x)=x 2 +e x +x e +e 2 . (b) f(x)=πe x − 6e x √ 29 . (c) f(x)=3e x+3 .(Hint: Break this up into the product of e x and a constant.) 6. Using the definition of derivative, show that the derivative of a x is a x times the slope of the tangent to a x at x = 0. (We’ve done this, but refresh your memory.) 7. If your last name begins with A–J: Approximate the derivative of (2.7) x using the results of Problem 6 and h = 0.000001. If your last name begins with K–Z: Approximate the derivative of (2.8) x using the results of Problem 6 and h = 0.000001. 476 CHAPTER 14 Differentiating Logarithmic and Exponential Functions 8. Graph f(x)=e x −x. Only use a calculator to check your work after working on your own. (a) Find f  (x). Draw a number line and indicate where f  is positive, zero, and negative. (b) Label the x- and y-coordinates of any local extrema (local maxima or minima). (c) Using your picture, determine how many solutions there are to the following equations. i. f(x)=5 ii. f(x)=0.5 Notice that these equations are “intractable”— try to solve e x − x = 5 alge- braically to see what this means. If we want to estimate the solutions, we can do so using a graphing calculator. At this point, we should know how many solutions to expect. 9. Find the quantity indicated. (a) y = ln 5x + ln x 5 + ln 5 x i. Find y  . ii. Find the slope of the graph of the function at x = 1. (b) f(x)=log 10 x i. Find f  (x). ii. Find f  (100). (c) P(x)=7 x i. Find P  (x). ii. Find the instantaneous rate of change of P with respect to x when x = 0. (d) y = e 3x i. Find y  . ii. Find the slope of the graph of the function at x = 0. (e) f(x)=14e x/2 i. Find f  . ii. Find f  (ln 9) and simplify your answer. 10. Differentiate y = e 3x ln( 1 √ 5x ). 14.3 WORKED EXAMPLES INVOLVING DIFFERENTIATION In Sections 14.1 and 14.2 we found some important derivative formulas. We’ve shown the following. i. d dx ln x = 1 x ii. d dx log b x = d dx  ln x ln b  = d dx  1 ln b ln x  = 1 ln b · 1 x = 1 x ln b iii. d dx b x =ln b · b x In this section we will work examples using a combination of these differentiation formulas and the laws of logs and exponents. The main piece of advice is to spend time 14.3 Worked Examples Involving Differentiation 477 putting the expression to be differentiated into a form that makes it simple to apply the differentiation formulas given above. Be sure to distinguish between constants and variables. Worked Examples In each of the following problems, find dy dx . ◆ EXAMPLE 14.2 y = π ln  7x 6 8  SOLUTION Rewrite: y = π(ln 7x 6 − ln 8) = π ln 7 + π ln x 6 − π ln 8 = π ln 7 + 6π ln x − π ln 8 (First and third terms are constant.) Differentiate: dy dx = 6π x ◆ ◆ EXAMPLE 14.3 y = 10 log(bx 3 ) √ π SOLUTION Rewrite: y = 10 √ π log  bx 3  = 10 √ π [log b + 3 log x] = 10 √ π log b + 30 √ π ln x ln 10 Notice that 10 √ π log b is constant, as is 30 √ π 1 ln 10 . Differentiate: dy dx = 30 √ π 1 x ln 10 ◆ ◆ EXAMPLE 14.4 y = 17(3) t/5 SOLUTION Rewrite: y = 17  3 1/5  t We’ ll treat 3 1/5 as the base of y = b x . Differentiate: dy dx =17  ln 3 1/5  3 1/5  t = 17 5 ( ln 3 )  3 t/5  ◆ 478 CHAPTER 14 Differentiating Logarithmic and Exponential Functions ◆ EXAMPLE 14.5 y = 5 2x+3 SOLUTION Rewrite: y = 5 2x · 5 3 =  5 3  5 2  x =5 3  25 x  Notice that 5 3 is a constant. We’ ll treat 5 2 , or 25, as the base of y = b x . Differentiate: dy dx = 5 3 ln  5 2  · 25 x = 125 ln 5 2 · 5 2x = 250 ln 5 · 5 2x ◆ In the following examples, we generalize what we have done. Try them on your own before looking at the answers. ◆ EXAMPLE 14.6 y = A ln  Bx C  , where A, B, and C are constants and B>0. SOLUTION Rewrite: y = A[ln B + C ln x] = A ln B + AC ln x Differentiate: dy dx = AC 1 x = AC x ◆ ◆ EXAMPLE 14.7 y = Ab kx , where A, b, and k are constants and b>0. SOLUTION Rewrite: y = A  b k  x Differentiate: dy dx = A ln  b k  b k  x =Ak ln b · b kx ◆ In Chapter 16 we will acquire alternative means of finding these derivatives. ◆ EXAMPLE 14.8 Approximate ln 1.1 with the help of the first derivative. SOLUTION This problem is like the tangent-line approximations we discussed in Chapter 8. We begin by sketching ln x and getting an off-the-cuff approximation of ln 1.1. 14.3 Worked Examples Involving Differentiation 479 x y y = lnx y = lnx 1 1 1.1 1.1 enlarged tangent line at x = 1 slope = 1 tangent line ∆y ∆x Figure 14.7 ln 1.1 is a bit larger than ln 1; how much larger can be approximated by looking at the rate at which ln x is increasing near x = 1. The derivative of ln x at x = 1 gives the rate of increase. d dx ln x| x=1 = 1 x     x=1 = 1. Knowing the rate of increase of ln x at x = 1 enables us to judge how to adjust our approximation (in this case, 0) to fit a nearby value of x. dy dx ≈ y x for x small (because dy dx = lim x→0 y x ). In this problem, x = 0.1 y = lnx 1 1.1 tangent line approx. actual y-value Figure 14.8 dy dx = 1 ≈ y .1 ⇒ y ≈ 0.1 Therefore, y ≈ 0.1. We obtain the approximation ln 1.1 ≈ 0 + 0.1 = 0.1. Is the approximation ln 1.1 ≈ 0.1 an overestimate, or an underestimate? We know that ln x is concave down. d 2 dx 2 ln x = d dx  1 x  =− 1 x 2 < 0 Therefore the tangent line lies above the graph of the function, and this approximation is a bit too big. Comparing our approximation, ln 1.1 ≈ 0.1, with the calculator approximation, ln 1.1 ≈ 0.0953, we see that the approximation is fairly good. 480 CHAPTER 14 Differentiating Logarithmic and Exponential Functions REMARK If we wanted to further refine our approximation, we could use information supplied by the second derivative to make the required adjustment. 4 This is equivalent to approximating the curve y = ln x at x = 1 by a parabola instead of a line. Further successive refinements involve higher order derivatives and give us what is called a Taylor polynomial approximation to the function at x = 1. ◆ EXERCISE 14.5 Approximate ln 0.9 using a tangent-line approximation. Answer ln 0.9 ≈−0.1 Logarithms: A Summary log b x = y is equivalent to x = b y . This equivalence follows from the definition of log b x. log b x is the number we must raise b to in order to obtain x. log b x and b x are inverse functions, so log b b  =  and b log b  = . log x is the way we write log 10 x (log base 10 of x, or the common log). ln x is the way we write log e x (log base e of x, or the natural log). Thus, ln x = y is equivalent to x = e y . It follows that e ln  =  and ln e  = . y 1 1 x y = ln x y = x y = e x Figure 14.9 4 The larger the value of the second derivative at x = 1 the larger the rate of change of the slope near x = 1. . in the domain of f .) (c) By looking at the sign of f  , find all local maxima and minima. Give both the x- and y-coordinates of the extrema. (d) Find f  . Where is f concave up and where is. Derivative of b x We know that the derivative of b x is b x · (the slope of the tangent to b x at x = 0). We can concentrate our energy simply on finding the slope of the tangent line to b x at. to solve e x − x = 5 alge- braically to see what this means. If we want to estimate the solutions, we can do so using a graphing calculator. At this point, we should know how many solutions to