DEFINITION Derivative Function. The derivative of the function ƒ(x) with respect to the variable x is the function ƒ whose value at x is.

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1 Chapter 3 DIFFERENTIATION OVERVIEW In Chapter 2, we efine the slope of a curve at a point as the limit of secant slopes. This limit, calle a erivative, measures the rate at which a function changes, an it is one of the most important ieas in calculus. Derivatives are use to calculate velocit an acceleration, to estimate the rate of sprea of a isease, to set levels of prouction so as to maimize efficienc, to fin the best imensions of a clinrical can, to fin the age of a prehistoric artifact, an for man other applications. In this chapter, we evelop techniques to calculate erivatives easil an learn how to use erivatives to approimate complicate functions. 3. HISTORICAL ESSAY The Derivative The Derivative as a Function At the en of Chapter 2, we efine the slope of a curve = to be lim h: ƒs + h - ƒs. h = ƒs at the point where We calle this limit, when it eiste, the erivative of ƒ at. We now investigate the erivative as a function erive from ƒ b consiering the limit at each point of the omain of ƒ. DEFINITION Derivative Function The erivative of the function ƒ() with respect to the variable is the function ƒ whose value at is ƒs + h - ƒs ƒ s = lim, h: h provie the limit eists. 47

2 48 Chapter 3: Differentiation P(, f()) Q(z, f(z)) f() f(z) f() Secant slope is f(z) f() z We use the notation ƒ() rather than simpl ƒ in the efinition to emphasize the inepenent variable, which we are ifferentiating with respect to. The omain of ƒ is the set of points in the omain of ƒ for which the limit eists, an the omain ma be the same or smaller than the omain of ƒ. If ƒ eists at a particular, we sa that ƒ is ifferentiable (has a erivative) at. If ƒ eists at ever point in the omain of ƒ, we call ƒ ifferentiable. If we write z = + h, then h = z - an h approaches if an onl if z approaches. Therefore, an equivalent efinition of the erivative is as follows (see Figure 3.). h z Derivative of f at is f'() lim h lim z z f( h) f() h f(z) f() z FIGURE 3. The wa we write the ifference quotient for the erivative of a function ƒ epens on how we label the points involve. Alternative Formula for the Derivative ƒsz - ƒs ƒ s = lim z: z -. Calculating Derivatives from the Definition The process of calculating a erivative is calle ifferentiation. To emphasize the iea that ifferentiation is an operation performe on a function = ƒs, we use the notation as another wa to enote the erivative ƒ s. Eamples 2 an 3 of Section 2.7 illustrate the ifferentiation process for the functions = m + b an = >. Eample 2 shows that For instance, In Eample 3, we see that Here are two more eamples. ƒs sm + b = m. a3 2-4b = 3 2. a b =- 2. EXAMPLE Differentiate ƒs = Appling the Definition -. Solution Here we have ƒs = -

3 3. The Derivative as a Function 49 an = lim h: ƒs + h = ƒs + h - ƒs ƒ s = lim h: h + h + h - - s + h s + h -, so = lim # s + hs - - s + h - h: h s + h - s - = lim # -h h: h s + h - s - h - = lim h: - s + h - s - = - s - 2. a b - c a - cb = b EXAMPLE 2 Derivative of the Square Root Function (a) Fin the erivative of = for 7. (b) Fin the tangent line to the curve = at = 4. You will often nee to know the erivative of for 7 : = 2. (4, 2) 4 4 FIGURE 3.2 The curve = an its tangent at (4, 2). The tangent s slope is foun b evaluating the erivative at = 4 (Eample 2). Solution (a) We use the equivalent form to calculate ƒ : ƒsz - ƒs ƒ s = lim z: z - (b) The slope of the curve at = 4 is z - = lim z: z - z - = lim z: A z - BAz + B = lim z: z + = 2. ƒ s4 = The tangent is the line through the point (4, 2) with slope >4 (Figure 3.2): = 2 + s = = 4. We consier the erivative of = when = in Eample 6.

4 5 Chapter 3: Differentiation Notations There are man was to enote the erivative of a function = ƒs, where the inepenent variable is an the epenent variable is. Some common alternative notations for the erivative are ƒ s = = = ƒ = ƒs = Dsƒs = D ƒs. The smbols > an D inicate the operation of ifferentiation an are calle ifferentiation operators. We rea > as the erivative of with respect to, an ƒ> an ( >)ƒ() as the erivative of ƒ with respect to. The prime notations an ƒ come from notations that Newton use for erivatives. The > notations are similar to those use b Leibniz. The smbol > shoul not be regare as a ratio (until we introuce the iea of ifferentials in Section 3.8). Be careful not to confuse the notation D(ƒ) as meaning the omain of the function ƒ instea of the erivative function ƒ. The istinction shoul be clear from the contet. To inicate the value of a erivative at a specifie number = a, we use the notation ƒ sa = ` = a For instance, in Eample 2b we coul write = f ` = a = ƒs ` = a. ƒ s4 = ` = 4 = 2 ` = 4 To evaluate an epression, we sometimes use the right bracket ] in place of the vertical bar = 224 = 4. ƒ. Graphing the Derivative We can often make a reasonable plot of the erivative of = ƒs b estimating the slopes on the graph of ƒ. That is, we plot the points s, ƒ s in the -plane an connect them with a smooth curve, which represents = ƒ s. EXAMPLE 3 Graphing a Derivative Graph the erivative of the function = ƒs in Figure 3.3a. On April 23, 988, the human-powere airplane Daealus flew a recor-breaking 9 km from Crete to the islan of Santorini in the Aegean Sea, southeast of mainlan Greece. Dur- Solution We sketch the tangents to the graph of ƒ at frequent intervals an use their slopes to estimate the values of ƒ s at these points. We plot the corresponing s, ƒ s pairs an connect them with a smooth curve as sketche in Figure 3.3b. What can we learn from the graph of = ƒ s? At a glance we can see. where the rate of change of ƒ is positive, negative, or zero; 2. the rough size of the growth rate at an an its size in relation to the size of ƒ(); 3. where the rate of change itself is increasing or ecreasing. Here s another eample. EXAMPLE 4 Concentration of Bloo Sugar

5 3. The Derivative as a Function 5 Slope A 5 B Slope C Slope units 5 5 (a) D Slope f() E Slope 8 -units units/-unit Slope f '() E' A' 2 D' 5 5 C' B' Vertical coorinate (b) FIGURE 3.3 We mae the graph of = ƒ s in (b) b plotting slopes from the graph of = ƒs in (a). The vertical coorinate of B is the slope at B an so on. The graph of ƒ is a visual recor of how the slope of ƒ changes with. ing the 6-hour enurance tests before the flight, researchers monitore the prospective pilots bloo-sugar concentrations. The concentration graph for one of the athlete-pilots is shown in Figure 3.4a, where the concentration in milligrams> eciliter is plotte against time in hours. The graph consists of line segments connecting ata points. The constant slope of each segment gives an estimate of the erivative of the concentration between measurements. We calculate the slope of each segment from the coorinate gri an plotte the erivative as a step function in Figure 3.4b. To make the plot for the first hour, for instance, we observe that the concentration increase from about 79 mg> L to 93 mg> L. The net increase was = = 4 mg>l. Diviing this b t = hour gave the rate of change as t = 4 = 4 mg>l per hour. Notice that we can make no estimate of the concentration s rate of change at times t =, 2, Á, 5, where the graph we have rawn for the concentration has a corner an no slope. The erivative step function is not efine at these times.

6 52 Chapter 3: Differentiation Concentration, mg/l Rate of change of concentration, mg/l h Time (h) (a) 5 5 ' Time (h) (b) Slope f(a h) f(a) lim h h f() Slope f(b h) f(b) lim h h t t Differentiable on an Interval; One-Sie Derivatives A function = ƒs is ifferentiable on an open interval (finite or infinite) if it has a erivative at each point of the interval. It is ifferentiable on a close interval [a, b] if it is ifferentiable on the interior (a, b) an if the limits Right-han erivative at a ƒsb + h - ƒsb lim Left-han erivative at b h: - h eist at the enpoints (Figure 3.5). Right-han an left-han erivatives ma be efine at an point of a function s omain. The usual relation between one-sie an two-sie limits hols for these erivatives. Because of Theorem 6, Section 2.4, a function has a erivative at a point if an onl if it has left-han an right-han erivatives there, an these one-sie erivatives are equal. EXAMPLE 5 FIGURE 3.4 (a) Graph of the sugar concentration in the bloo of a Daealus pilot uring a 6-hour preflight enurance test. (b) The erivative of the pilot s bloo-sugar concentration shows how rapil the concentration rose an fell uring various portions of the test. GREECE Meiterranean Sea Athens ƒsa + h - ƒsa lim h: + h SANTORINI CRETE Aegean Sea Sea of Crete Heraklion Daealus's flight path on April 23, 988 = ƒ ƒ Is Not Differentiable at the Origin TURKEY RHODES 5 5 km Show that the function = ƒ ƒ is ifferentiable on s - q, an s, q but has no erivative at =. a a h h b h h FIGURE 3.5 Derivatives at enpoints are one-sie limits. b Solution To the left, To the right of the origin, s ƒ ƒ = s = s # =. s ƒ ƒ = s - = s - # = - sm + b = m, ƒ ƒ = ƒ ƒ = -

7 3. The Derivative as a Function 53 ' ' ' not efine at : right-han erivative left-han erivative FIGURE 3.6 The function = ƒ ƒ is not ifferentiable at the origin where the graph has a corner. (Figure 3.6). There can be no erivative at the origin because the one-sie erivatives iffer there: EXAMPLE 6 Right-han erivative of ƒ ƒ at zero = Left-han erivative of ƒ ƒ at zero = = In Eample 2 we foun that for 7, Is Not Differentiable at = = 2. We appl the efinition to eamine if the erivative eists at = : 2 + h - 2 lim h: + h ƒ + h ƒ - ƒ ƒ lim h: + h = lim h = lim h: + = h: + h ƒ + h ƒ - ƒ ƒ lim h: - h ƒ ƒ = lim h h = -h when h 6. = lim - -. = h: - h: - -h = lim h: + h = q. ƒ h ƒ = lim h: + h ƒ h ƒ = h when h 7. ƒ h ƒ = lim h: - h Since the (right-han) limit is not finite, there is no erivative at =. Since the slopes of the secant lines joining the origin to the points Ah, hb on a graph of = approach q, the graph has a vertical tangent at the origin. When Does a Function Not Have a Derivative at a Point? A function has a erivative at a point if the slopes of the secant lines through Ps, ƒs an a nearb point Q on the graph approach a limit as Q approaches P. Whenever the secants fail to take up a limiting position or become vertical as Q approaches P, the erivative oes not eist. Thus ifferentiabilit is a smoothness conition on the graph of ƒ. A function whose graph is otherwise smooth will fail to have a erivative at a point for several reasons, such as at points where the graph has. a corner, where the one-sie 2. a cusp, where the slope of PQ erivatives iffer. approaches q from one sie an - q from the other. P P Q Q Q Q

8 54 Chapter 3: Differentiation 3. a vertical tangent, where the slope of PQ approaches q from both sies or approaches - q from both sies (here, - q). Q P Q 4. a iscontinuit. P P Q Q Q Q Differentiable Functions Are Continuous A function is continuous at ever point where it has a erivative. THEOREM Differentiabilit Implies Continuit If ƒ has a erivative at = c, then ƒ is continuous at = c. Proof Given that ƒ sc eists, we must show that lim :c ƒs = ƒsc, or equivalentl, that lim h: ƒsc + h = ƒsc. If h Z, then ƒsc + h = ƒsc + sƒsc + h - ƒsc ƒsc + h - ƒsc = ƒsc + # h. h

9 3. The Derivative as a Function 55 Now take limits as h :. B Theorem of Section 2.2, ƒsc + h - ƒsc lim ƒsc + h = lim ƒsc + lim # lim h h: h: h: h h: = ƒsc + ƒ sc # = ƒsc + = ƒsc. Similar arguments with one-sie limits show that if ƒ has a erivative from one sie (right or left) at = c then ƒ is continuous from that sie at = c. Theorem on page 54 sas that if a function has a iscontinuit at a point (for instance, a jump iscontinuit), then it cannot be ifferentiable there. The greatest integer function = :; = int fails to be ifferentiable at ever integer = n (Eample 4, Section 2.6). CAUTION The converse of Theorem is false. A function nee not have a erivative at a point where it is continuous, as we saw in Eample 5. U() The Intermeiate Value Propert of Derivatives Not ever function can be some function s erivative, as we see from the following theorem. FIGURE 3.7 The unit step function oes not have the Intermeiate Value Propert an cannot be the erivative of a function on the real line. THEOREM 2 If a an b are an two points in an interval on which ƒ is ifferentiable, then ƒ takes on ever value between ƒ sa an ƒ sb. Theorem 2 (which we will not prove) sas that a function cannot be a erivative on an interval unless it has the Intermeiate Value Propert there. For eample, the unit step function in Figure 3.7 cannot be the erivative of an real-value function on the real line. In Chapter 5 we will see that ever continuous function is a erivative of some function. In Section 4.4, we invoke Theorem 2 to analze what happens at a point on the graph of a twice-ifferentiable function where it changes its bening behavior.

10 EXERCISES 3. Fining Derivative Functions an Values Using the efinition, calculate the erivatives of the functions in Eercises 6. Then fin the values of the erivatives as specifie.. 2. ƒs = 4-2 ; ƒ s -3, ƒ s, ƒ s Fs = s ; F s -, F s, F s2 3. gst = t 2 ; g s -, g s2, g A 23B 3. The Derivative as a Function ksz = - z 2z ; k s -, k s, k A 22B psu = 23u ; p s, p s3, p s2>3 rss = 22s + ; r s, r s, r s>2 In Eercises 7 2, fin the inicate erivatives. r 7. if = if r = s3 s 2 +

11 56 Chapter 3: Differentiation Slopes an Tangent Lines In Eercises 3 6, ifferentiate the functions an fin the slope of the tangent line at the given value of the inepenent variable. 3. s t t p q if s = if = t - t if p = t 2t + z w if z = 23w - 2 ƒs = + 9, = -3 2q + Graphs Match the functions graphe in Eercises 27 3 with the erivatives graphe in the accompaning figures (a) (). ' (a) ' ' (b) ' s = t 3 - t 2, t = - 6. = s + 3, = -2 In Eercises 7 8, ifferentiate the functions. Then fin an equation of the tangent line at the inicate point on the graph of the function ks = = ƒs = 2 +, = 2 8, s, = s6, w = gsz = z, sz, w = s3, 2 In Eercises 9 22, fin the values of the erivatives. (c) f () () f 2 () s t ` t =- ` = 23 r u ` u = w z ` z = 4 if s = - 3t 2 if = - if r = u if w = z + z f 3 () f 4 () Using the Alternative Formula for Derivatives Use the formula ƒsz - ƒs ƒ s = lim z: z - to fin the erivative of the functions in Eercises ƒs = ƒs = s gs = gs = + 3. a. The graph in the accompaning figure is mae of line segments joine en to en. At which points of the interval [-4, 6] is ƒ not efine? Give reasons for our answer. (, 2) (6, 2) f() ( 4, ) 6 (, 2) (4, 2)

12 3. The Derivative as a Function 57 b. Graph the erivative of ƒ. The graph shoul show a step function. 32. Recovering a function from its erivative a. Use the following information to graph the function ƒ over the close interval [-2, 5]. i) The graph of ƒ is mae of close line segments joine en to en. ii) The graph starts at the point s -2, 3. iii) The erivative of ƒ is the step function in the figure shown here. ' ' f'() b. Repeat part (a) assuming that the graph starts at s -2, instea of s -2, Growth in the econom The graph in the accompaning figure shows the average annual percentage change = ƒst in the U.S. gross national prouct (GNP) for the ears Graph >t (where efine). (Source: Statistical Abstracts of the Unite States, th Eition, U.S. Department of Commerce, p. 427.) 7% Fruit flies (Continuation of Eample 3, Section 2..) Populations starting out in close environments grow slowl at first, when there are relativel few members, then more rapil as the number of reproucing iniviuals increases an resources are still abunant, then slowl again as the population reaches the carring capacit of the environment. a. Use the graphical technique of Eample 3 to graph the erivative of the fruit fl population introuce in Section 2.. The graph of the population is reprouce here p b. During what as oes the population seem to be increasing fastest? Slowest? One-Sie Derivatives Compare the right-han an left-han erivatives to show that the functions in Eercises are not ifferentiable at the point P f() 2 f() P(, 2) P(, ) f() P(, ) Time (as) 2 Differentiabilit an Continuit on an Interval Each figure in Eercises shows the graph of a function over a close interval D. At what omain points oes the function appear to be a. ifferentiable? b. continuous but not ifferentiable? c. neither continuous nor ifferentiable? P(, ) f() t

13 58 Chapter 3: Differentiation Give reasons for our answers f() D: f() D: f() D: Theor an Eamples f() D: 3 3 In Eercises 45 48, a. Fin the erivative ƒ s of the given function = ƒs. b. Graph = ƒs an = ƒ s sie b sie using separate sets of coorinate aes, an answer the following questions. c. For what values of, if an, is ƒ positive? Zero? Negative?. Over what intervals of -values, if an, oes the function = ƒs increase as increases? Decrease as increases? How is this relate to what ou foun in part (c)? (We will sa more about this relationship in Chapter 4.) 45. = = -> 47. = 3 >3 48. = 4 >4 49. Does the curve = 3 ever have a negative slope? If so, where? Give reasons for our answer. 5. Does the curve = 2 have an horizontal tangents? If so, where? Give reasons for our answer f() D: f() D: Tangent to a parabola Does the parabola = have a tangent whose slope is -? If so, fin an equation for the line an the point of tangenc. If not, wh not? 52. Tangent to Does an tangent to the curve = cross the -ais at = -? If so, fin an equation for the line an the point of tangenc. If not, wh not? 53. Greatest integer in Does an function ifferentiable on s - q, q have = int, the greatest integer in (see Figure 2.55), as its erivative? Give reasons for our answer. 54. Derivative of ƒ ƒ Graph the erivative of ƒs = ƒ ƒ. Then graph = sƒ ƒ - >s - = ƒ ƒ >. What can ou conclue? 55. Derivative of ƒ Does knowing that a function ƒ() is ifferentiable at = tell ou anthing about the ifferentiabilit of the function -ƒ at =? Give reasons for our answer. 56. Derivative of multiples Does knowing that a function g(t) is ifferentiable at t = 7 tell ou anthing about the ifferentiabilit of the function 3g at t = 7? Give reasons for our answer. 57. Limit of a quotient Suppose that functions g(t) an h(t) are efine for all values of t an gs = hs =. Can lim t: sgst>shst eist? If it oes eist, must it equal zero? Give reasons for our answers. 58. a. Let ƒ() be a function satisfing ƒ ƒs ƒ 2 for -. Show that ƒ is ifferentiable at = an fin ƒ s. b. Show that 2 sin ƒs =, Z L, = is ifferentiable at = an fin ƒ s. T 59. Graph = > A2B in a winow that has 2. Then, on the same screen, graph = + h - h for h =,.5,.. Then tr h = -, -.5, -.. Eplain what is going on. T 6. Graph = 3 2 in a winow that has -2 2, 3. Then, on the same screen, graph = s + h3-3 h for h = 2,,.2. Then tr h = -2, -, -.2. Eplain what is going on. T 6. Weierstrass s nowhere ifferentiable continuous function The sum of the first eight terms of the Weierstrass function ƒ() = g n q = s2>3 n cos s9 n p is gs = cos sp + s2>3 cos s9p + s2>3 2 cos s9 2 p + s2>3 3 cos s9 3 p + Á + s2>3 7 cos s9 7 p. Graph this sum. Zoom in several times. How wiggl an bump is this graph? Specif a viewing winow in which the isplae portion of the graph is smooth.

14 3. The Derivative as a Function 59 COMPUTER EXPLORATIONS Use a CAS to perform the following steps for the functions in Eercises a. Plot = ƒs to see that function s global behavior. b. Define the ifference quotient q at a general point, with general step size h. c. Take the limit as h :. What formula oes this give?. Substitute the value = an plot the function = ƒs together with its tangent line at that point. e. Substitute various values for larger an smaller than into the formula obtaine in part (c). Do the numbers make sense with our picture? f. Graph the formula obtaine in part (c). What oes it mean when its values are negative? Zero? Positive? Does this make sense with our plot from part (a)? Give reasons for our answer. 62. ƒs = , = 63. ƒs = >3 + 2>3, = ƒs = , = 2 ƒs = , = ƒs = sin 2, 67. ƒs = 2 = p>2 cos, = p>4

15 3.2 Differentiation Rules Differentiation Rules This section introuces a few rules that allow us to ifferentiate a great variet of functions. B proving these rules here, we can ifferentiate functions without having to appl the efinition of the erivative each time. Powers, Multiples, Sums, an Differences The first rule of ifferentiation is that the erivative of ever constant function is zero. RULE Derivative of a Constant Function If ƒ has the constant value ƒs = c, then ƒ = sc =. EXAMPLE If ƒ has the constant value ƒs = 8, then c (, c) ( h, c) c f = s8 =. Similarl, h h a- p 2 b = an a23b =. FIGURE 3.8 The rule s>sc = is another wa to sa that the values of constant functions never change an that the slope of a horizontal line is zero at ever point. Proof of Rule We appl the efinition of erivative to ƒs = c, the function whose outputs have the constant value c (Figure 3.8). At ever value of, we fin that ƒ s = lim h: ƒs + h - ƒs h = lim h: c - c h = lim =. h:

16 6 Chapter 3: Differentiation The secon rule tells how to ifferentiate n if n is a positive integer. RULE 2 Power Rule for Positive Integers If n is a positive integer, then n = n n -. To appl the Power Rule, we subtract from the original eponent (n) an multipl the result b n. EXAMPLE 2 Interpreting Rule 2 ƒ ƒ Á Á HISTORICAL BIOGRAPHY Richar Courant ( ) First Proof of Rule 2 The formula z n - n = sz - sz n - + z n Á + z n n - can be verifie b multipling out the right-han sie. Then from the alternative form for the efinition of the erivative, ƒsz - ƒs ƒ s = lim z: z - = n n - z = lim n - n z: z - = lim z: sz n - + z n Á + z n n - Secon Proof of Rule 2 If ƒs = n, then ƒs + h = s + h n. Since n is a positive integer, we can epan s + h n b the Binomial Theorem to get ƒs + h - ƒs ƒ s = lim h: h = lim h: = lim h: nn - h + = lim h: cn n - + = n n - c n + n n - h + nsn - 2 nsn - 2 s + h n - n = lim h: h nsn - 2 n - 2 h 2 + Á + nh n - + h n h n - 2 h 2 + Á + nh n - + h n - n n - 2 h + Á + nh n h n - The thir rule sas that when a ifferentiable function is multiplie b a constant, its erivative is multiplie b the same constant. h

17 3.2 Differentiation Rules 6 RULE 3 Constant Multiple Rule If u is a ifferentiable function of, an c is a constant, then u scu = c. 3 2 In particular, if n is a positive integer, then scn = cn n -. Slope 3(2) 6 3 (, 3) Slope 6() Slope 2 Slope 2() 2 (, ) FIGURE 3.9 The graphs of = 2 an = 3 2. Tripling the -coorinates triples the slope (Eample 3). Denoting Functions b u an Y The functions we are working with when we nee a ifferentiation formula are likel to be enote b letters like ƒ an g. When we appl the formula, we o not want to fin it using these same letters in some other wa. To guar against this problem, we enote the functions in ifferentiation rules b letters like u an that are not likel to be alrea in use. 2 EXAMPLE 3 (a) The erivative formula sas that if we rescale the graph of = 2 b multipling each -coorinate b 3, then we multipl the slope at each point b 3 (Figure 3.9). (b) A useful special case The erivative of the negative of a ifferentiable function u is the negative of the function s erivative. Rule 3 with c = - gives Proof of Rule 3 s32 = 3 # 2 = 6 s -u = s - # u = - # su =-u. cus + h - cus cu = lim h: h us + h - us = c lim Limit propert h: h = c u u is ifferentiable. The net rule sas that the erivative of the sum of two ifferentiable functions is the sum of their erivatives. RULE 4 Derivative Sum Rule If u an are ifferentiable functions of, then their sum u + is ifferentiable at ever point where u an are both ifferentiable. At such points, u su + = +. Derivative efinition with ƒs = cus

18 62 Chapter 3: Differentiation EXAMPLE 4 Proof of Rule 4 Derivative of a Sum = = s4 + s2 = We appl the efinition of erivative to ƒs = us + s: [us + h + s + h] - [us + s] [us + s] = lim h: h us + h - us = lim c h: h us + h - us = lim h: h Combining the Sum Rule with the Constant Multiple Rule gives the Difference Rule, which sas that the erivative of a ifference of ifferentiable functions is the ifference of their erivatives. su - = u [u + s -] = The Sum Rule also etens to sums of more than two functions, as long as there are onl finitel man functions in the sum. If are ifferentiable at, then so is u + u 2 + Á u, u 2, Á, u n + u n, an su + u 2 + Á + u n = u + u 2 + Á + u n. + s + h - s h s + h - s + lim h: h = u +. + s - = u - EXAMPLE 5 Derivative of a Polnomial = = 3 + a4 3 2 b - s5 + s = # = Notice that we can ifferentiate an polnomial term b term, the wa we ifferentiate the polnomial in Eample 5. All polnomials are ifferentiable everwhere. Proof of the Sum Rule for Sums of More Than Two Functions We prove the statement su + u 2 + Á + u n = u + u 2 + Á + u n b mathematical inuction (see Appeni ). The statement is true for n = 2, as was just prove. This is Step of the inuction proof.

19 3.2 Differentiation Rules 63 Step 2 is to show that if the statement is true for an positive integer n = k, where k Ú n = 2, then it is also true for n = k +. So suppose that Then su + u 2 + Á + u k = u + u 2 + Á + u k. (u + u 2 + Á + u k + u k + ) (++++)++++* ()* Call the function Call this efine b this sum u. function. () = su + u 2 + Á + u k + u k + Rule 4 for su Eq. () With these steps verifie, the mathematical inuction principle now guarantees the Sum Rule for ever integer n Ú 2. EXAMPLE 6 = u + u 2 + Á + u k + u k +. Fining Horizontal Tangents Does the curve = have an horizontal tangents? If so, where? (, ) (, 2) (, ) FIGURE 3. The curve = an its horizontal tangents (Eample 6). Solution The horizontal tangents, if an, occur where the slope > is zero. We have, Now solve the equation = s = = for : = 4s 2 - = =,, -. The curve = has horizontal tangents at =,, an -. The corresponing points on the curve are (, 2), (, ) an s -,. See Figure 3.. Proucts an Quotients While the erivative of the sum of two functions is the sum of their erivatives, the erivative of the prouct of two functions is not the prouct of their erivatives. For instance, s # = s2 = 2, while s # s = # =. The erivative of a prouct of two functions is the sum of two proucts, as we now eplain. RULE 5 Derivative Prouct Rule If u an are ifferentiable at, then so is their prouct u, an su = u + u.

20 64 Chapter 3: Differentiation Picturing the Prouct Rule If u() an () are positive an increase when increases, an if h 7, ( h) () u() u()() u () u u() u u( h) then the total shae area in the picture is us + hs + h - uss = us + h + s + h u - u. Diviing both sies of this equation b h gives As h : +, leaving us + hs + h - uss h = us + h h - u h. + s + h u h u # h : # =, su = u + u. The erivative of the prouct u is u times the erivative of plus times the erivative of u. In prime notation, su =u +u. In function notation, EXAMPLE 7 Fin the erivative of Using the Prouct Rule Solution We appl the Prouct Rule with u = > an = 2 + s>: c a2 + b = a2-2 b + a2 + ba- 2 b Proof of Rule 5 [ƒsgs] = ƒsg s + gsƒ s. = = = a2 + b. us + hs + h - uss su = lim h: h To change this fraction into an equivalent one that contains ifference quotients for the erivatives of u an, we subtract an a us + hs in the numerator: us + hs + h - us + hs + us + hs - uss su = lim h: h s + h - s = lim cus + h h: h = lim us + h # s + h - s lim h: h: h + s As h approaches zero, us + h approaches u() because u, being ifferentiable at, is continuous at. The two fractions approach the values of > at an u> at. In short, su = u + u. su = u + u, an a b =- b 2 Eample 3, Section 2.7. us + h - us h + s # us + h - us lim. h: h In the following eample, we have onl numerical values with which to work. EXAMPLE 8 Derivative from Numerical Values Let = u be the prouct of the functions u an. Fin s2 if us2 = 3, u s2 = -4, s2 =, an s2 = 2. Solution From the Prouct Rule, in the form =su =u +u,

21 3.2 Differentiation Rules 65 we have s2 = us2 s2 + s2u s2 = s3s2 + ss -4 = 6-4 = 2. EXAMPLE 9 Differentiating a Prouct in Two Was Fin the erivative of = s 2 + s Solution (a) From the Prouct Rule with u = 2 + an = 3 + 3, we fin CA2 + BA 3 + 3BD = s 2 + s3 2 + s 3 + 3s2 = = (b) This particular prouct can be ifferentiate as well (perhaps better) b multipling out the original epression for an ifferentiating the resulting polnomial: = s 2 + s = = This is in agreement with our first calculation. Just as the erivative of the prouct of two ifferentiable functions is not the prouct of their erivatives, the erivative of the quotient of two functions is not the quotient of their erivatives. What happens instea is the Quotient Rule. RULE 6 Derivative Quotient Rule If u an are ifferentiable at an if s Z, then the quotient u> is ifferentiable at, an u au b = - u 2. In function notation, c ƒs gsƒ s - ƒsg s = gs g 2. s EXAMPLE Fin the erivative of Using the Quotient Rule = t 2 - t 2 +.

22 66 Chapter 3: Differentiation Solution We appl the Quotient Rule with Proof of Rule 6 t u = t 2 - an = t 2 + : = st 2 + # 2t - st 2 - # 2t st = 2t 3 + 2t - 2t 3 + 2t st = 4t st us + h au b = lim s + h - us s h: h = lim h: sus + h - uss + h hs + hs To change the last fraction into an equivalent one that contains the ifference quotients for the erivatives of u an, we subtract an a ()u() in the numerator. We then get au b = lim sus + h - sus + sus - uss + h h: hs + hs us + h - us s h = lim h: - us s + hs t au su>t - us>t b = 2 s + h - s h Taking the limit in the numerator an enominator now gives the Quotient Rule.. Negative Integer Powers of The Power Rule for negative integers is the same as the rule for positive integers. RULE 7 Power Rule for Negative Integers If n is a negative integer an Z, then sn = n n -. EXAMPLE (a) Agrees with Eample 3, Section 2.7 a b = s - = s - -2 =- 2 (b) a 4 3 b = 4 s -3 = 4s -3-4 =- 2 4

23 3.2 Differentiation Rules 67 Proof of Rule 7 The proof uses the Quotient Rule. If n is a negative integer, then n = -m, where m is a positive integer. Hence, n = -m = > m, an sn = a m b (, 3) 2 4 FIGURE 3. The tangent to the curve = + s2> at (, 3) in Eample 2. The curve has a thir-quarant portion not shown here. We see how to graph functions like this one in Chapter 4. 3 EXAMPLE 2 Tangent to a Curve Fin an equation for the tangent to the curve at the point (, 3) (Figure 3.). Solution The slope at = is = - mm - 2m = -m -m - = n n -. The slope of the curve is ` = = + 2 = c = The line through (, 3) with slope m = - is = m # AB - # Am B s m 2-3 = s -s - = = Quotient Rule with u = an = m Since m 7, sm = m m - Since -m = n = s + 2 a b = + 2 a- 2 b = = - 2 = -. Point-slope equation The choice of which rules to use in solving a ifferentiation problem can make a ifference in how much work ou have to o. Here is an eample. EXAMPLE 3 Choosing Which Rule to Use Rather than using the Quotient Rule to fin the erivative of epan the numerator an ivie b 4 : = s - s2-2 4 = s - s2-2 4, = =

24 68 Chapter 3: Differentiation Then use the Sum an Power Rules: = s s -3-4 = Secon- an Higher-Orer Derivatives If = ƒs is a ifferentiable function, then its erivative ƒ s is also a function. If ƒ is also ifferentiable, then we can ifferentiate ƒ to get a new function of enote b ƒ. So ƒ =sƒ. The function ƒ is calle the secon erivative of ƒ because it is the erivative of the first erivative. Notationall, ƒ s = 2 2 = a b = = =D 2 sƒs = D 2 ƒs. The smbol means the operation of ifferentiation is performe twice. If = 6, then =6 5 an we have D 2 = = A65 B = 3 4. How to Rea the Smbols for Derivatives prime ouble prime 2 square square 2 triple prime sn super n n to the n of b to the n n D n D to the n Thus D 2 A 6 B = 3 4. If is ifferentiable, its erivative, = > = 3 > 3 is the thir erivative of with respect to. The names continue as ou imagine, with sn = sn - = n n = D n enoting the nth erivative of with respect to for an positive integer n. We can interpret the secon erivative as the rate of change of the slope of the tangent to the graph of = ƒs at each point. You will see in the net chapter that the secon erivative reveals whether the graph bens upwar or ownwar from the tangent line as we move off the point of tangenc. In the net section, we interpret both the secon an thir erivatives in terms of motion along a straight line. EXAMPLE 4 Fining Higher Derivatives The first four erivatives of = are First erivative: Secon erivative: Thir erivative: Fourth erivative: =3 2-6 =6-6 =6 s4 =. The function has erivatives of all orers, the fifth an later erivatives all being zero.

25 3.2 Differentiation Rules 69 EXERCISES 3.2 Derivative Calculations In Eercises 2, fin the first an secon erivatives.. = = s = 5t 3-3t 5 4. w = 3z 7-7z 3 + 2z 2 5. = w = 3z -2 - z 8. s = -2t t 2 9. = = r = 3s 2-5 2s 2. r = 2 u - 4 u 3 + u 4 In Eercises 3 6, fin (a) b appling the Prouct Rule an (b) b multipling the factors to prouce a sum of simpler terms to ifferentiate. 3. = s3-2 s = s - s = 6. = a + ba - s2 + a b + b Fin the erivatives of the functions in Eercises = z = t 2. ƒst = - gs = t 2 + t = s - ts + t w = s2-7 - s ƒss = s u = 5 + s = r = 2 a 2u + 2ub 27. s + s + 2 = 28. = s 2 - s s - s - 2 Fin the erivatives of all orers of the functions in Eercises 29 an = 4 3. = Fin the first an secon erivatives of the functions in Eercises = r = su - su2 + u u 3 = s = t 2 + 5t - t 2 u = s2 + s w = a + 3z bs3 - z 36. w = sz + sz - sz 2 + 3z q p = a q p = 2q baq4 q 3 b sq sq + 3 Using Numerical Values 39. Suppose u an are functions of that are ifferentiable at = an that us = 5, u s = -3, s = -, s = 2. Fin the values of the following erivatives at =. a. b. c.. s7-2u su au b a u b 4. Suppose u an are ifferentiable functions of an that us = 2, u s =, s = 5, s = -. Fin the values of the following erivatives at =. a. b. c.. s7-2u su au b a u b Slopes an Tangents 4. a. Normal to a curve Fin an equation for the line perpenicular to the tangent to the curve = at the point (2, ). b. Smallest slope What is the smallest slope on the curve? At what point on the curve oes the curve have this slope? c. Tangents having specifie slope Fin equations for the tangents to the curve at the points where the slope of the curve is a. Horizontal tangents Fin equations for the horizontal tangents to the curve = Also fin equations for the lines that are perpenicular to these tangents at the points of tangenc. b. Smallest slope What is the smallest slope on the curve? At what point on the curve oes the curve have this slope? Fin an equation for the line that is perpenicular to the curve s tangent at this point. 43. Fin the tangents to Newton s serpentine (graphe here) at the origin an the point (, 2) (, 2) 2 3 4

26 7 Chapter 3: Differentiation 44. Fin the tangent to the Witch of Agnesi (graphe here) at the point (2, ). 45. Quaratic tangent to ientit function The curve = a 2 + b + c passes through the point (, 2) an is tangent to the line = at the origin. Fin a, b, an c. 46. Quaratics having a common tangent The curves = 2 + a + b an = c - 2 have a common tangent line at the point (, ). Fin a, b, an c. 47. a. Fin an equation for the line that is tangent to the curve = 3 - at the point s -,. T b. Graph the curve an tangent line together. The tangent intersects the curve at another point. Use Zoom an Trace to estimate the point s coorinates. T c. Confirm our estimates of the coorinates of the secon intersection point b solving the equations for the curve an tangent simultaneousl (Solver ke). 48. a. Fin an equation for the line that is tangent to the curve = at the origin. T b. Graph the curve an tangent together. The tangent intersects the curve at another point. Use Zoom an Trace to estimate the point s coorinates. T c. Confirm our estimates of the coorinates of the secon intersection point b solving the equations for the curve an tangent simultaneousl (Solver ke). Theor an Eamples 49. The general polnomial of egree n has the form (2, ) Ps = a n n + a n - n - + Á + a a + a where a n Z. Fin P s. 5. The bo s reaction to meicine The reaction of the bo to a ose of meicine can sometimes be represente b an equation of the form R = M 2 a C 2 - M 3 b, where C is a positive constant an M is the amount of meicine absorbe in the bloo. If the reaction is a change in bloo pressure, R is measure in millimeters of mercur. If the reaction is a change in temperature, R is measure in egrees, an so on. Fin R>M. This erivative, as a function of M, is calle the sensitivit of the bo to the meicine. In Section 4.5, we will see 3 how to fin the amount of meicine to which the bo is most sensitive. 5. Suppose that the function in the Prouct Rule has a constant value c. What oes the Prouct Rule then sa? What oes this sa about the Constant Multiple Rule? 52. The Reciprocal Rule a. The Reciprocal Rule sas that at an point where the function () is ifferentiable an ifferent from zero, Show that the Reciprocal Rule is a special case of the Quotient Rule. b. Show that the Reciprocal Rule an the Prouct Rule together impl the Quotient Rule. 53. Generalizing the Prouct Rule The Prouct Rule gives the formula su = u + u for the erivative of the prouct u of two ifferentiable functions of. a. What is the analogous formula for the erivative of the prouct uw of three ifferentiable functions of? b. What is the formula for the erivative of the prouct u u 2 u 3 u 4 of four ifferentiable functions of? c. What is the formula for the erivative of a prouct u u 2 u 3 Á u n of a finite number n of ifferentiable functions of? 54. Rational Powers a. Fin b writing as # >2 3>2 an using the Prouct A3>2 B Rule. Epress our answer as a rational number times a rational power of. Work parts (b) an (c) b a similar metho. b. Fin s5>2. c. Fin s7>2.. What patterns o ou see in our answers to parts (a), (b), an (c)? Rational powers are one of the topics in Section Cliner pressure If gas in a cliner is maintaine at a constant temperature T, the pressure P is relate to the volume V b a formula of the form P = a b =- 2. nrt V - nb - an2 V 2, in which a, b, n, an R are constants. Fin P>V. (See accompaning figure.)

27 3.2 Differentiation Rules The best quantit to orer One of the formulas for inventor management sas that the average weekl cost of orering, paing for, an holing merchanise is Asq = km q hq + cm + 2, where q is the quantit ou orer when things run low (shoes, raios, brooms, or whatever the item might be); k is the cost of placing an orer (the same, no matter how often ou orer); c is the cost of one item (a constant); m is the number of items sol each week (a constant); an h is the weekl holing cost per item (a constant that takes into account things such as space, utilities, insurance, an securit). Fin A>q an 2 A>q 2.

28 The Derivative as a Rate of Change 7 The Derivative as a Rate of Change In Section 2., we initiate the stu of average an instantaneous rates of change. In this section, we continue our investigations of applications in which erivatives are use to moel the rates at which things change in the worl aroun us. We revisit the stu of motion along a line an eamine other applications. It is natural to think of change as change with respect to time, but other variables can be treate in the same wa. For eample, a phsician ma want to know how change in osage affects the bo s response to a rug. An economist ma want to stu how the cost of proucing steel varies with the number of tons prouce. Instantaneous Rates of Change If we interpret the ifference quotient sƒs + h - ƒs>h as the average rate of change in ƒ over the interval from to + h, we can interpret its limit as h : as the rate at which ƒ is changing at the point. DEFINITION Instantaneous Rate of Change The instantaneous rate of change of ƒ with respect to at is the erivative ƒ s = lim h: ƒs + h - ƒs h, provie the limit eists. Thus, instantaneous rates are limits of average rates. It is conventional to use the wor instantaneous even when oes not represent time. The wor is, however, frequentl omitte. When we sa rate of change, we mean instantaneous rate of change.

29 72 Chapter 3: Differentiation EXAMPLE How a Circle s Area Changes with Its Diameter The area A of a circle is relate to its iameter b the equation A = p 4 D 2. How fast oes the area change with respect to the iameter when the iameter is m? Solution The rate of change of the area with respect to the iameter is A D = p 4 # 2D = pd 2. When D = m, the area is changing at rate sp>2 = 5p m 2 >m. Position at time t an at time t Δt Δs s f(t) s Δs f(t Δt) FIGURE 3.2 The positions of a bo moving along a coorinate line at time t an shortl later at time t + t. s Motion Along a Line: Displacement, Velocit, Spee, Acceleration, an Jerk Suppose that an object is moving along a coorinate line (sa an s-ais) so that we know its position s on that line as a function of time t: The isplacement of the object over the time interval from t to t + t(figure 3.2) is an the average velocit of the object over that time interval is a = isplacement travel time s = ƒst. s = ƒst + t - ƒst, = s t ƒst + t-ƒst =. t To fin the bo s velocit at the eact instant t, we take the limit of the average velocit over the interval from t to t + tas t shrinks to zero. This limit is the erivative of ƒ with respect to t. DEFINITION Velocit Velocit (instantaneous velocit) is the erivative of position with respect to time. If a bo s position at time t is s = ƒst, then the bo s velocit at time t is st = s t = lim t: ƒst + t-ƒst. t EXAMPLE 2 Fining the Velocit of a Race Car Figure 3.3 shows the time-to-istance graph of a 996 Rile & Scott Mk III-Ols WSC race car. The slope of the secant PQ is the average velocit for the 3-sec interval from t = 2 to t = 5 sec; in this case, it is about ft> sec or 68 mph. The slope of the tangent at P is the speeometer reaing at t = 2 sec, about 57 ft> sec or 39 mph. The acceleration for the perio shown is a nearl constant 28.5 ft>sec 2 uring

30 3.3 The Derivative as a Rate of Change 73 s Distance (ft) 5 Secant slope is average velocit 4 for interval from Q t 2 to t 5. Tangent slope 3 is speeometer 2 reaing at t 2 (instantaneous velocit). P t Elapse time (sec) FIGURE 3.3 The time-to-istance graph for Eample 2. The slope of the tangent line at P is the instantaneous velocit at t = 2 sec. each secon, which is about.89g, where g is the acceleration ue to gravit. The race car s top spee is an estimate 9 mph. (Source: Roa an Track, March 997.) Besies telling how fast an object is moving, its velocit tells the irection of motion. When the object is moving forwar (s increasing), the velocit is positive; when the bo is moving backwar (s ecreasing), the velocit is negative (Figure 3.4). s s s f(t) s f(t) s t s t s increasing: positive slope so moving forwar t s ecreasing: negative slope so moving backwar t FIGURE 3.4 For motion s = ƒst along a straight line, = s/t is positive when s increases an negative when s ecreases. If we rive to a frien s house an back at 3 mph, sa, the speeometer will show 3 on the wa over but it will not show -3 on the wa back, even though our istance from home is ecreasing. The speeometer alwas shows spee, which is the absolute value of velocit. Spee measures the rate of progress regarless of irection.

31 74 Chapter 3: Differentiation DEFINITION Spee Spee is the absolute value of velocit. s Spee = ƒ st ƒ = ` t ` EXAMPLE 3 Horizontal Motion Figure 3.5 shows the velocit = ƒ st of a particle moving on a coorinate line. The particle moves forwar for the first 3 sec, moves backwar for the net 2 sec, stans still for a secon, an moves forwar again. The particle achieves its greatest spee at time t = 4, while moving backwar. MOVES FORWARD ( ) f'(t) FORWARD AGAIN ( ) Spees up Stea ( const) Slows own Spees up Stans still ( ) t (sec) Greatest spee Spees up Slows own MOVES BACKWARD ( ) FIGURE 3.5 The velocit graph for Eample 3. HISTORICAL BIOGRAPHY Bernar Bolzano (78 848) The rate at which a bo s velocit changes is the bo s acceleration. The acceleration measures how quickl the bo picks up or loses spee. A suen change in acceleration is calle a jerk. When a rie in a car or a bus is jerk, it is not that the accelerations involve are necessaril large but that the changes in acceleration are abrupt.

32 3.3 The Derivative as a Rate of Change 75 DEFINITIONS Acceleration, Jerk Acceleration is the erivative of velocit with respect to time. If a bo s position at time t is s = ƒst, then the bo s acceleration at time t is ast = = 2 s t t 2. Jerk is the erivative of acceleration with respect to time: jst = a t = 3 s t 3. t (secons) t t t 2 t 3 s (meters) Near the surface of the Earth all boies fall with the same constant acceleration. Galileo s eperiments with free fall (Eample, Section 2.) lea to the equation where s is istance an g is the acceleration ue to Earth s gravit. This equation hols in a vacuum, where there is no air resistance, an closel moels the fall of ense, heav objects, such as rocks or steel tools, for the first few secons of their fall, before air resistance starts to slow them own. The value of g in the equation s = s>2gt 2 epens on the units use to measure t an s. With t in secons (the usual unit), the value of g etermine b measurement at sea level is approimatel 32 ft>sec 2 (feet per secon square) in English units, an g = 9.8 m>sec 2 (meters per secon square) in metric units. (These gravitational constants epen on the istance from Earth s center of mass, an are slightl lower on top of Mt. Everest, for eample.) The jerk of the constant acceleration of gravit sg = 32 ft>sec 2 is zero: An object oes not ehibit jerkiness uring free fall. EXAMPLE 4 Moeling Free Fall Figure 3.6 shows the free fall of a heav ball bearing release from rest at time t = sec. (a) How man meters oes the ball fall in the first 2 sec? (b) What is its velocit, spee, an acceleration then? Solution s = 2 gt 2, j = sg =. t (a) The metric free-fall equation is s = 4.9t 2. During the first 2 sec, the ball falls ss2 = 4.9s2 2 = 9.6 m. (b) At an time t, velocit is the erivative of position: FIGURE 3.6 A ball bearing falling from rest (Eample 4). st = s st = t s4.9t 2 = 9.8t.

33 76 Chapter 3: Differentiation Height (ft) s s ma 256 t? At t = 2, the velocit is s2 = 9.6 m>sec in the ownwar (increasing s) irection. The spee at t = 2 is Spee = ƒ s2 ƒ = 9.6 m>sec. The acceleration at an time t is ast = st = s st = 9.8 m>sec 2. At t = 2, the acceleration is 9.8 m>sec 2. EXAMPLE 5 Moeling Vertical Motion 4 s, s (a) s 6t 6t 2 A namite blast blows a heav rock straight up with a launch velocit of 6 ft> sec (about 9 mph) (Figure 3.7a). It reaches a height of s = 6t - 6t 2 ft after t sec. (a) How high oes the rock go? (b) What are the velocit an spee of the rock when it is 256 ft above the groun on the wa up? On the wa own? (c) What is the acceleration of the rock at an time t uring its flight (after the blast)? () When oes the rock hit the groun again? 6 6 s t t (b) FIGURE 3.7 (a) The rock in Eample 5. (b) The graphs of s an as functions of time; s is largest when = s/t =. The graph of s is not the path of the rock: It is a plot of height versus time. The slope of the plot is the rock s velocit, graphe here as a straight line. t Solution (a) In the coorinate sstem we have chosen, s measures height from the groun up, so the velocit is positive on the wa up an negative on the wa own. The instant the rock is at its highest point is the one instant uring the flight when the velocit is. To fin the maimum height, all we nee to o is to fin when = an evaluate s at this time. At an time t, the velocit is The velocit is zero when The rock s height at t = 5 sec is s ma = ss5 = 6s5-6s5 2 = 8-4 = 4 ft. See Figure 3.7b. (b) To fin the rock s velocit at 256 ft on the wa up an again on the wa own, we first fin the two values of t for which To solve this equation, we write = s t = t s6t - 6t 2 = 6-32t ft>sec. 6-32t = or t = 5 sec. sst = 6t - 6t 2 = t 2-6t = 6st 2 - t + 6 = st - 2st - 8 = t = 2 sec, t = 8 sec.

34 3.3 The Derivative as a Rate of Change 77 (ollars) Slope marginal cost (tons/week) h c() FIGURE 3.8 Weekl steel prouction: c() is the cost of proucing tons per week. The cost of proucing an aitional h tons is cs + h - cs. The rock is 256 ft above the groun 2 sec after the eplosion an again 8 sec after the eplosion. The rock s velocities at these times are s2 = 6-32s2 = 6-64 = 96 ft>sec. s8 = 6-32s8 = = -96 ft>sec. At both instants, the rock s spee is 96 ft> sec. Since s2 7, the rock is moving upwar (s is increasing) at t = 2 sec; it is moving ownwar (s is ecreasing) at t = 8 because s8 6. (c) At an time uring its flight following the eplosion, the rock s acceleration is a constant a = t = t s6-32t = -32 ft>sec2. The acceleration is alwas ownwar. As the rock rises, it slows own; as it falls, it spees up. () The rock hits the groun at the positive time t for which s =. The equation 6t - 6t 2 = factors to give 6ts - t =, so it has solutions t = an t =. At t =, the blast occurre an the rock was thrown upwar. It returne to the groun sec later. c c() c FIGURE 3.9 The marginal cost c> is approimatel the etra cost c of proucing = more unit. Derivatives in Economics Engineers use the terms velocit an acceleration to refer to the erivatives of functions escribing motion. Economists, too, have a specialize vocabular for rates of change an erivatives. The call them marginals. In a manufacturing operation, the cost of prouction c() is a function of, the number of units prouce. The marginal cost of prouction is the rate of change of cost with respect to level of prouction, so it is c>. Suppose that c() represents the ollars neee to prouce tons of steel in one week. It costs more to prouce + h units per week, an the cost ifference, ivie b h, is the average cost of proucing each aitional ton: cs + h - cs h = The limit of this ratio as h : is the marginal cost of proucing more steel per week when the current weekl prouction is tons (Figure 3.8). c = lim cs + h - cs h: h Sometimes the marginal cost of prouction is loosel efine to be the etra cost of proucing one unit: c average cost of each of the aitional h tons of steel prouce. = marginal cost of prouction. cs + - cs =, which is approimate b the value of c> at. This approimation is acceptable if the slope of the graph of c oes not change quickl near. Then the ifference quotient will be close to its limit c>, which is the rise in the tangent line if = (Figure 3.9). The approimation works best for large values of.

35 78 Chapter 3: Differentiation Economists often represent a total cost function b a cubic polnomial cs = a 3 + b 2 + g + where represents fie costs such as rent, heat, equipment capitalization, an management costs. The other terms represent variable costs such as the costs of raw materials, taes, an labor. Fie costs are inepenent of the number of units prouce, whereas variable costs epen on the quantit prouce. A cubic polnomial is usuall complicate enough to capture the cost behavior on a relevant quantit interval. EXAMPLE 6 Suppose that it costs Marginal Cost an Marginal Revenue cs = ollars to prouce raiators when 8 to 3 raiators are prouce an that rs = gives the ollar revenue from selling raiators. Your shop currentl prouces raiators a a. About how much etra will it cost to prouce one more raiator a a, an what is our estimate increase in revenue for selling raiators a a? Solution c s: The cost of proucing one more raiator a a when are prouce is about c s = A B = c s = 3s - 2s + 5 = 95. The aitional cost will be about $95. The marginal revenue is r s = A B = The marginal revenue function estimates the increase in revenue that will result from selling one aitional unit. If ou currentl sell raiators a a, ou can epect our revenue to increase b about if ou increase sales to raiators a a. r s = 3s - 6s + 2 = $252 EXAMPLE 7 Marginal Ta Rate To get some feel for the language of marginal rates, consier marginal ta rates. If our marginal income ta rate is 28% an our income increases b $, ou can epect to pa an etra $28 in taes. This oes not mean that ou pa 28% of our entire income in taes. It just means that at our current income level I, the rate of increase of taes T with respect to income is T>I =.28. You will pa $.28 out of ever etra ollar ou earn in taes. Of course, if ou earn a lot more, ou ma lan in a higher ta bracket an our marginal rate will increase.

36 3.3 The Derivative as a Rate of Change 79 Sensitivit to Change When a small change in prouces a large change in the value of a function ƒ(), we sa that the function is relativel sensitive to changes in. The erivative ƒ s is a measure of this sensitivit. EXAMPLE 8 Genetic Data an Sensitivit to Change The Austrian monk Gregor Johann Menel ( ), working with garen peas an other plants, provie the first scientific eplanation of hbriization. His careful recors showe that if p (a number between an ) is the frequenc of the gene for smooth skin in peas (ominant) an s - p is the frequenc of the gene for wrinkle skin in peas, then the proportion of smooth-skinne peas in the net generation will be = 2ps - p + p 2 = 2p - p 2. The graph of versus p in Figure 3.2a suggests that the value of is more sensitive to a change in p when p is small than when p is large. Inee, this fact is borne out b the erivative graph in Figure 3.2b, which shows that >p is close to 2 when p is near an close to when p is near. /p 2 p 2 2p 2p p 2 p p (a) (b) FIGURE 3.2 (a) The graph of = 2p - p 2, escribing the proportion of smooth-skinne peas. (b) The graph of >p (Eample 8). The implication for genetics is that introucing a few more ominant genes into a highl recessive population (where the frequenc of wrinkle skin peas is small) will have a more ramatic effect on later generations than will a similar increase in a highl ominant population.

37 EXERCISES 3.3 Motion Along a Coorinate Line Eercises 6 give the positions s = ƒst of a bo moving on a coorinate line, with s in meters an t in secons. a. Fin the bo s isplacement an average velocit for the given time interval. 3.3 The Derivative as a Rate of Change 79 b. Fin the bo s spee an acceleration at the enpoints of the interval. c. When, if ever, uring the interval oes the bo change irection?. s = t 2-3t + 2, t 2 2. s = 6t - t 2, t 6

38 8 Chapter 3: Differentiation 3. s = -t 3 + 3t 2-3t, t 3 4. s = st 4 >4 - t 3 + t 2, t 3 5. s = 25 t 2-5 t, t 5 6. s = 25 t + 5, -4 t 7. Particle motion At time t, the position of a bo moving along the s-ais is s = t 3-6t 2 + 9t m. a. Fin the bo s acceleration each time the velocit is zero. b. Fin the bo s spee each time the acceleration is zero. c. Fin the total istance travele b the bo from t = to t = Particle motion At time t Ú, the velocit of a bo moving along the s-ais is = t 2-4t + 3. a. Fin the bo s acceleration each time the velocit is zero. b. When is the bo moving forwar? Backwar? c. When is the bo s velocit increasing? Decreasing? Free-Fall Applications 9. Free fall on Mars an Jupiter The equations for free fall at the surfaces of Mars an Jupiter (s in meters, t in secons) are s =.86t 2 on Mars an s =.44t 2 on Jupiter. How long oes it take a rock falling from rest to reach a velocit of 27.8 m> sec (about km> h) on each planet?. Lunar projectile motion A rock thrown verticall upwar from the surface of the moon at a velocit of 24 m> sec (about 86 km> h) reaches a height of s = 24t -.8t 2 meters in t sec. a. Fin the rock s velocit an acceleration at time t. (The acceleration in this case is the acceleration of gravit on the moon.) b. How long oes it take the rock to reach its highest point? c. How high oes the rock go?. How long oes it take the rock to reach half its maimum height? e. How long is the rock aloft?. Fining g on a small airless planet Eplorers on a small airless planet use a spring gun to launch a ball bearing verticall upwar from the surface at a launch velocit of 5 m> sec. Because the acceleration of gravit at the planet s surface was g s m>sec 2, the eplorers epecte the ball bearing to reach a height of s = 5t - s>2g s t 2 meters t sec later. The ball bearing reache its maimum height 2 sec after being launche. What was the value of g s? 2. Speeing bullet A 45-caliber bullet fire straight up from the surface of the moon woul reach a height of s = 832t - 2.6t 2 feet after t sec. On Earth, in the absence of air, its height woul be s = 832t - 6t 2 ft after t sec. How long will the bullet be aloft in each case? How high will the bullet go? 3. Free fall from the Tower of Pisa Ha Galileo roppe a cannonball from the Tower of Pisa, 79 ft above the groun, the ball s height above groun t sec into the fall woul have been s = 79-6t 2. a. What woul have been the ball s velocit, spee, an acceleration at time t? b. About how long woul it have taken the ball to hit the groun? c. What woul have been the ball s velocit at the moment of impact? 4. Galileo s free-fall formula Galileo evelope a formula for a bo s velocit uring free fall b rolling balls from rest own increasingl steep incline planks an looking for a limiting formula that woul preict a ball s behavior when the plank was vertical an the ball fell freel; see part (a) of the accompaning figure. He foun that, for an given angle of the plank, the ball s velocit t sec into motion was a constant multiple of t. That is, the velocit was given b a formula of the form = kt. The value of the constant k epene on the inclination of the plank. In moern notation part (b) of the figure with istance in meters an time in secons, what Galileo etermine b eperiment was that, for an given angle u, the ball s velocit t sec into the roll was (a) a. What is the equation for the ball s velocit uring free fall? b. Builing on our work in part (a), what constant acceleration oes a freel falling bo eperience near the surface of Earth? Conclusions About Motion from Graphs 5. The accompaning figure shows the velocit = s>t = ƒst (m> sec) of a bo moving along a coorinate line. 3 3 = 9.8ssin ut m>sec. (m/sec) f(t) 2 4 Free-fall position 6 8 t (sec) a. When oes the bo reverse irection? b. When (approimatel) is the bo moving at a constant spee? c. Graph the bo s spee for t.. Graph the acceleration, where efine.? θ (b)

39 3.3 The Derivative as a Rate of Change 8 6. A particle P moves on the number line shown in part (a) of the accompaning figure. Part (b) shows the position of P as a function of time t. (a) P s (cm) f. When was the rocket s acceleration greatest? g. When was the acceleration constant? What was its value then (to the nearest integer)? 8. The accompaning figure shows the velocit = ƒst of a particle moving on a coorinate line. 2 s (cm) s f(t) f(t) t (sec) (b) (6, 4) t (sec) a. When is P moving to the left? Moving to the right? Staning still? b. Graph the particle s velocit an spee (where efine). 7. Launching a rocket When a moel rocket is launche, the propellant burns for a few secons, accelerating the rocket upwar. After burnout, the rocket coasts upwar for a while an then begins to fall. A small eplosive charge pops out a parachute shortl after the rocket starts own. The parachute slows the rocket to keep it from breaking when it lans. The figure here shows velocit ata from the flight of the moel rocket. Use the ata to answer the following. a. How fast was the rocket climbing when the engine stoppe? b. For how man secons i the engine burn? a. When oes the particle move forwar? Move backwar? Spee up? Slow own? b. When is the particle s acceleration positive? Negative? Zero? c. When oes the particle move at its greatest spee?. When oes the particle stan still for more than an instant? 9. Two falling balls The multiflash photograph in the accompaning figure shows two balls falling from rest. The vertical rulers are marke in centimeters. Use the equation s = 49t 2 (the freefall equation for s in centimeters an t in secons) to answer the following questions. 2 5 Velocit (ft/sec) Time after launch (sec) c. When i the rocket reach its highest point? What was its velocit then?. When i the parachute pop out? How fast was the rocket falling then? e. How long i the rocket fall before the parachute opene?