Section 2.6 Derivatives of products and quotients

Transcription

1 Secion 2.6 Derivaives of proucs an quoiens (3/19/08) Overview: In his secion, we erive formulas for erivaives of funcions ha are consruce by aking proucs an quoiens of oher funcions, an we use hese formulas o suy raes of change in a variey of applicaions. Topics: The Prouc Rule Relae-rae problems The Quoien Rule The erivaive of y = x n for inegers n 2 an Mahemaical Inucion The rae of change of he area of a recangle Imagine ha he sies of he recangle in Figure 1 are changing, so ha he wih w = w(), heigh h = h(), an area A() = w()h() of he recangle are funcions of he ime. We wan a formula for he rae of change of he area in erms of w, h, an heir raes of change. h (I) (III) h h (II) w w w FIGURE 1 FIGURE 2 We consier a nonzero change in he ime from o + an le w an h be he corresponing changes in he wih an heigh. If w an h are posiive, hen a ime +, he wih is w + w an he heigh is h + h, as in Figure 2, an he change A in he area from o + is he area of he hree recangles labele (I), (II), an (III) in Figure 2. Example 1 Soluion Express he areas of recangles (I),(II), an (III) in Figure 2 in erms of w,h, w, an h. Recangle (I) is w unis wie an h unis high an has area w h. Recangle (II) is w unis wie an h unis high an has area h w. Recangle (III) is w unis wie an h unis high an has area w h. The resuls of Example 1 show ha if w an h are posiive, hen A = w h + h w + w h. (1) To verify (1) in general, we erive i algebraically. The area of a recangle of wih w an heigh h is wh, an he area of a recangle of wih w + w an heigh h + h is (w + w)(h + h), so he change in he area is A = (w + w)(h + h) wh = (wh + w h + h w + w h) wh = w h + h w + w h. 137

2 p. 138 (3/19/08) Secion 2.6, Derivaives of proucs an quoiens This esablishes equaion (1). We ivie boh sies of i by he change in he ime o obain A = w h + h w [ ] h + w. (2) We suppose ha w = w() an h = h() have erivaives a. Then w w () = lim an 0 h h () = lim. Also w 0 as 0 since, by Theorem 1 of Secion 2.5, w = w() is coninuous 0 a. Hence (2) gives A A () = lim 0 = lim 0 (w h + h w + w h ) = w()h () + h()w () + (0)h () = w()h () + h()w (). (3) Formula (3) is he Prouc Rule. I is resae in he following heorem wih,w, an h replace by x, f, an g, an A replace by fg: Theorem 1 (The Prouc Rule) prouc, y = f(x)g(x), an If y = f(x) an y = g(x) have erivaives a x, hen so oes heir (fg) = fg + gf (4a) or wih Leibniz noaion, g (fg) = f x x + g f x. (4b) Remember he Prouc Rule as he following saemen: he erivaive of a prouc of wo funcions equals he firs funcion muliplie by he erivaive of he secon, plus he secon funcion muliplie by he erivaive of he firs. Example 2 Fin he erivaive of y = (x 5 + x 2 )(x 1/3 + 1) a x = 1. Soluion By he Prouc Rule, y (x) = x [(x5 + x 2 )(x 1/3 + 1)] = (x 5 + x 2 ) x (x1/3 + 1) + (x 1/3 + 1) x (x5 + x 2 ) = (x 5 + x 2 )( 1 3 x 2/3 ) + (x 1/3 + 1)(5x 4 + 2x). Therefore, y (1) = (1 + 1)( 1 3 ) + (1 + 1)(5 + 2) =

3 Secion 2.6, Derivaives of proucs an quoiens p. 139 (3/19/08) Example 3 Fin he rae of change of he area of a recangle a a momen when he wih is 4 meers, he heigh is 2 meers, he wih is increasing 3 meers per hour, an he heigh is increasing 5 meers per hour. Soluion Equaion (3) saes ha A = wh + hw. We se w = 4, h = 2, w = 3, an h = 5 o obain [ A = [4 meers] 5 meers hour = 4(5) + 2(3) = 26 ] [ + [2 meers] square meers. hour 3 meers hour ] Relae-rae problems In solving Example 3 we sare wih an equaion, A = wh, relaing hree funcions of ime, w,h, an A. We iffereniae he equaion wih respec o o obain an equaion, A = wh + hw, which we use o fin he rae of change of A from w,h, an heir raes of change. This ype of problem, which involves raes of change of relae funcons, is calle a relae-rae problem. Here is anoher example. Example 4 Soluion A he beginning of 1990 he oal populaion of he U.S was million, of whom 51.3% were women, he oal populaion was increasing a he rae of 3.5 million per year, an he percenage of women was ecreasing 0.04% per year. (1) A wha rae was he populaion of women increasing a he beginning of 1990? We le p = p() be he oal U.S. populaion (measure in millions) in year an le F = F() be he fracion ha were women (he percen ivie by 100). Then he populaion of women a ime is W() = F()p(), so ha by he Prouc Rule, he rae of change of he populaion of women is W () = F()p () + F ()p(). Since p(1990) = 248.7, F(1990) = 0.513, p (1990) = 3.5, an F (1990) = , we obain W (1990) = F(1990)p (1990) + F (1990)p(1990) = (0.513)(3.5) + (248.7)( ). = 1.7. A he beginning of 1990 he populaion of women was increasing a he rae of approximaely 1.7 million per year. 1993, p. 15. (1) Daa aape from 1993 Saisical Absrac of he Unie Saes, Washingon, DC: U. S. Deparmen of Commerce,

4 p. 140 (3/19/08) Secion 2.6, Derivaives of proucs an quoiens The Quoien Rule Derivaives of quoiens of funcions are foun by using he following heorem. Theorem 2 (The Quoien Rule) A any value of x where y = f(x) an y = g(x) have erivaives an g(x) is no zero, y = f(x)/g(x) also has a erivaive an f = gf fg g g 2 (5a) or wih Leibniz noaion for he erivaives x g f f x f g x = g g 2. (5b) Remember his resul as he following saemen: he erivaive of a quoien equals he enominaor muliplie by he erivaive of he numeraor, minus he numeraor muliplie by he erivaive of he enominaor, all ivie by he square of he enominaor. Proof of Theorem 2: We consier firs he case where f is he consan funcion 1. We wrie g for g(x) wih a fixe x where g is no zero an has a erivaive. We le x enoe a small, nonzero change in he variable, an we le g = g(x+ x) g(x) be he resuling change g. Since g is coninuous a x, g(x+ x) is no zero for small nonzero x an ens o g(x) as z 0. By he efiniion of he erivaive, x 1 = lim g x 0 x 1 g(x + x) 1 g(x) x = lim x 0 1 g + g 1 g. (6) x We replace ivision by x wih muliplicaion by 1/( x) an ake a common enominaor: 1 g + g 1 g x = 1 ( 1 x = 1 x g + g 1 ) = 1 g (g + g) g x (g + g)g g 1 g = (g + g)g (g + g)g x. Since g = g(x + x) g(x) ens o 0 an g/ x ens o he erivaive g = g (x) as x 0, he las equaion wih (6) gives x ( 1 = lim g x 0 x 1 (g + g)g ) g = g x g 2. (7) This is (5a) wih f = 1 since f = 0. We can now erive (5a) for a general f ha has a erivaive a x by using (7) an he Prouc Rule. We obain f = [ ] 1 1 f = f + f 1 x g x g g x g = f g g + f g 2 = gf fg g 2. QED

5 Secion 2.6, Derivaives of proucs an quoiens p. 141 (3/19/08) Example 5 Wha is R (0) if R(x) = P(x)/Q(x),P(0) = 3, P (0) = 10, Q(0) = 5, an Q (0) = 50? Soluion By (5a) wih P an Q in place of f an g, Example 6 R (0) = Q(0)P (0) P(0)Q (0) 5(10) 3(50) [Q(0)] 2 = 5 2 = = Wha is y x if y = x2 x 4 + a Soluion For all x such ha x 4 + a 0, y x = [ x 2 ] x x 4 + a = wih consan a? (x 4 + a) x (x2 ) x 2 x (x4 + a) (x 4 + a) 2 = (x4 + a)(2x) x 2 (4x 3 ) (x 4 + a) 2 = 2x 5 + 2ax 4x 5 5 2ax 2x (x 4 + a) 2 = (x 4 + a) 2. Example 7 Figures 3 an 4 show graphs of he U.S. naional eb D = D() an he U. S. populaion P = P() as funcions of ime from 1950 hrough (2) (a) Fin approximae values of D(1985) an P(1985) from he graphs. Then raw approximae angen lines an esimae heir slopes o fin approximae values of D (1985) an P (1985). (b) Use your esimaes from par (a) o give he approximae eb per person an he approximae rae of increase wih respec o ime of he eb per person a he beginning of D (billions of ollars) D = D() P (millions of people) P = P() FIGURE 3 FIGURE 4 (2) Daa aape from 1993 Saisical Absrac of he Unie Saes, Washingon, DC: U. S. Deparmen of Commerce, 1993, pp. 15 an 328.

6 p. 142 (3/19/08) Secion 2.6, Derivaives of proucs an quoiens Soluion (a) The graphs show ha D(1985) 1850 billion ollars an P(1985) 240 million people. The poins (, 600) an (1985, 1850) are on he approximae angen line in Figure 5. Therefore, D (1985) = 250 billion ollars per year. Also he 1985 poins (1970, 180) an (1985, 240) are on he approximae angen line in Figure 6 so ha P (1985) = 4 million people per year (b) The eb per person a he begining of 1985 was D 1850 billion ollars P 240 million people housan ollars per person. The rae of change of he eb per person was. = 7.7 [ ] D = PD DP P (240)(250) (1850)(4) P = housan ollars per person per year. (Differen esimaes yiel ifferen answers.) D (billions of ollars) D = D() P (millions of people) P = P() FIGURE 5 FIGURE 6 The erivaive of y = x n for inegers n 2 an Mahemaical Inucion In Secion 2.4 we use he binomial heorem o erive he formula x (xn ) = nx n 1 (8) for posiive inegers n. Here we will prove ha (8) hols for all such n by using is valiiy for n = 1 wih he Prouc Rule an he principle of Mahemaical Inucion, which can be sae as follows: Mahemaical inucion Suppose P n is a saemen involving he parameer n for each ineger n n 0, where n 0 is a fixe ineger. If (i) P n0 is vali an (ii) P n implies P n+1 for every ineger n n 0, hen P n is vali for all inegers n n 0. This saemen of Mahemaical Inucion looks complicae bu acually expresses some very simple logic: if P n0 is vali an P n implies P n+1 for every ineger n n 0, hen P n0 +1 is vali. This implies ha P n0 +2 is vali, which implies ha P n0 +3 is vali, which implies ha P n0 +4 is vali, an so forh. Coninuing his reasoning inefiniely shows ha P n is vali for all n n 0. To esablish (8) for all posiive inegers n, we se n 0 = 1 an le P n be saemen (8). We nee o show ha (i) P 1 is vali, an ha (ii) P n implies P n+1 for all n 1. P 1 is vali because he graph of y = x 1 is he line y = x wih slope 1.

7 Secion 2.6, Derivaives of proucs an quoiens p. 143 (3/19/08) If P n is vali for any n 1, hen by he Prouc Rule, saemen P 1, an saemen P n, we have x (xn+1 ) = x (x xn ) = x x (xn ) + x n x (x) = x(nx n 1 ) + x n (1) = (n + 1)x n. Since his is saemen P n+1, we have shown ha P n implies P n+1 for any n 1. This esablishes (8) for all inegers n 1 by mahemaical inucion. Ineracive Examples 2.6 Ineracive soluions are on he web page hp// ashenk/. 1. Fin he erivaive x [(4x2 + 5)(6x 2 3)]. Do no simplify your answer. 2. Wha is y (2) for y = 3 + 4/x 1 + 8/x? 3. Fin R (1) where R(s) = P(s) Q(s),P(1) = 13, Q(1) = 2,P (1) = 7, an Q (1) = (a) Give an equaion of he angen line o he curve y = (1+x+x 2 )(1 x 1 x 2 ) a x = 1 an C (b) Generae he curve wih he angen line in a suiable winow on z calculaor or compuer. 5. Figures 7 an 8 show he graphs of he wih w = w() (meers) an heigh h = h() (meers) of a recangle as funcions of he ime. Wha is he approximae rae of change of he area of he recangle wih respec o a = 30? w (meers) w = w() 80 h (meers) h = h() (minues) (minues) FIGURE 7 FIGURE 8 In he publishe ex he ineracive soluions of hese examples will be on an accompanying CD isk which can be run by any compuer browser wihou using an inerne connecion.

8 p. 144 (3/19/08) Secion 2.6, Derivaives of proucs an quoiens Exercises 2.6 A Answer provie. CONCEPTS: O Ouline of soluion provie. C Graphing calculaor or compuer require. 1. (a) Suppose ha he wih of a recangle is hel consan a w = 8 an he heigh h = h() increases, as in Figure 9. Express he rae of change A () of he area in erms of h (). (b) Suppose, insea, ha he heigh is hel a h = 6 an he wih w = w() is increasing, as in Figure 10. Express A () in erms of w (). (c) How are he resuls of pars (a) an (b) examples of he Prouc Rule? h h = 6 w = 8 w FIGURE 9 FIGURE (a) Use he Prouc Rule o fin a formula for he erivaive of y = f(x)g(x), where f(x) = x 2 an g(x) = x (b) Fin he erivaive of y = f(x)g(x) by firs simplifying is formula. 3. (a) Use he Quoien Qule o fin a formula for he erivaive of y = F(x)/G(x), where F(x) = x an G(x) = x 2. (b) Fin he erivaive of y = F(x)/G(x) by firs simplifying is formula. BASICS: Fin he erivaives in Exercises 4 hrough 7. Do no simplify your answers in Problems 4 or O x [(5x3 + 2x 2 4)(x 7 2x 5 )] 5. O y x x for y = x O P (5) where P(x) = R(x)S(x),R(5) = 3, S(5) = 4, R (5) = 3, an S (5) = O W (4) where W(x) = Y (x) Z(x),Y (4) = 2, Z(4) = 5, Y (4) = 3, an Z (4) = 6 8. O Give an equaion of he angen line o y = (x 2 + x 3 + x 4 )(x 5 + x 6 + x 7 ) a x = O Figures 11 an 12 give he graphs of iffereniable funcions y = A(x) an y = B(x). Give approximae values of AB, A/B, an of heir firs erivaives a x = 2. 3 y y = A(x) 3 y y = B(x) x x FIGURE 11 FIGURE 12

9 Secion 2.6, Derivaives of proucs an quoiens p. 145 (3/19/08) 10. O Fin y x for y = ax2 + 2 bx wih consans a an b. (Do no simplify he answer.) Fin he erivaives in Exercises 11 hrough 29. C In Exercises 23 hrough 25 check he answers wih a proceure of approximae iffereniaion on a graphing calculaor or compuer. 11. O x [(1 + 3x x2 )(x 2 5)] 12. A x [(x2 + x + 1)(x 3 x 2 x)] 13. x [(x2 6)(x 2 + 4)] 14. O x [(x x2 )(1 + x 1 + x 2 )] 15. A x [(x2 + 5x 1)(x 3 + 2)] 16. x [(x2 + x 2 )(2 + 3x)] 17. O x 2 + x x x 2 2 C 24. A y (0) for y = (5 + 3x + 7x 2 )(x 3 2x + 5) C y 25. for y = 1 + x + x2 x 3 x 3 x=1 18. A x x 1 x 2 2x x 4x O u u u A z z 11 2z x 1 + x + x 2 C 23. O y x (2) for y = 1 x 26. O f (10) where f(x) = g(x)h(x), g(10) = 4, h(10) = 560, g (10) = 0, an h (10) = O Z (1) where Z(s) = X(s) Y (s),x(1) = 4, Y (1) = 10, X (1) = 3, an Y (1) = A y (5) where y(x) = (x 2 + 4)z(x),z(5) = 4, an z (5) = y ( 3) where y(x) = z(x) 1 + x 2, z( 3) = 6, an z ( 3) = 15 In Exercises 30 hrough 34 (a) give equaions of he angen lines o he graphs a he given values of x. C (b) Generae he curves wih he angen lines in suiable winows an copy he rawings on your paper. 30. O y = (1 + x x 4 )(1 x + x 3 ) a x = A y = (1 + x + x 2 )(1 x 1 x 2 ) a x = y = x3 1 x 2 a x = y = (x 2 5x + 1)(x 3 5x 2 + 2) a x = O y = x2 1 x a x = O A he beginning of 1991 here were 2.1 million farms in he Unie Saes wih an average size of 467 acres per farm; he number of farms was ecreasing million farms per year; an he average size was increasing 7 acres per farm per year. (3) Wha was he oal acreage of farms an a wha rae was i increasing or ecreasing a he beginning of 1991? 1993, p (3) Daa aape from 1993 Saisical Absrac of he Unie Saes, Washingon, DC: U. S. Bureau of he Census,

10 p. 146 (3/19/08) Secion 2.6, Derivaives of proucs an quoiens 36. A A wha rae is he area of a recangle increasing or ecreasing a a momen when i is 5 meers wie an 7 meers high, is wih is increasing 3 meers per secon, an is heigh is ecreasing 6 meers per secon? 37. Figure 13 gives he number of inmaes N = N() in U. S. feeral prisons an Figure 14 gives he he percen P = P() who were incarcerae for rug offences in year. (4) (a) Approximaely how many more feeral prisoners were hel for rug offences a he beginning of 1993 han a he beginning of 1979? (b) Wha was he approximae rae of change of he number of feeral prisoners being hel for rug offences a he beginning of 1988? N (housan) N = N() P% P = P() FIGURE 13 FIGURE A he beginning of 1990, annual healh care coss in he U. S. were $2600 per capia an were rising $260 per capia per year. A ha ime he populaion of he U. S. was 250 million an was increasing a he rae of 2.6 million per year. (5) A wha rae were he annual healh coss for he enire counry increasing a he beginning of 1990? 39. A Wha is P (3) if P(x) = x 2 Q(x),Q(3) = 5, an Q (3) = 6? 40. Wha is Z (9) if Z(y) = R(y) S(y), R(9) = 2, R (9) = 4, S(9) = 6, an S (9) = 8? 41. Wha is V (2) where V (x) = x + U(x) x U(x), U(2) = 3, an U (2) = 2? 42. A Wha is an objec s acceleraion in he posiive s-irecion a = 1 (secons) if i is a s = (cenimeers) on an s-axis a ime? 43. An objec is a s = ( )( ) (inches) on an s-axis a ime (hours). Give a formula for is acceleraion in he posiive s-irecion as a funcion of. 44. Wha is 2 y x 2 for y = (x2 1)(x 3 + x)? , p (4) Daa aape from Newsweek, Ocober 17, 1994, New York, NY: Newsweek, Inc., p. 87. (5) Daa aape from 1993 Saisical Absrac of he Unie Saes, Washingon, DC: U. S. Bureau of he Census,

11 Secion 2.6, Derivaives of proucs an quoiens p. 147 (3/19/08) EXPLORATION: Fin he erivaives in Exercises 45 hrough 52. Simplify your answers. 45. A P/z for P = (k z)(k + z) wih consan k 46. y/x for y = (1 + ax n )(1 ax n ) wih consans a, an n ( ) 47. A wih consan K K 48. a 3 wih consans a an b + b 49. A y (1) for y = x2 wih consans A an B A + Bx 50. Figures 15 an 16 give he number N = N() (millions) of MaserCar an Visa accouns an he oal ousaning eb D = D() (million ollars) in he U. S. as funcions of he year. (6) Wha were (a) he approximae average eb per crei car an (b) he rae of change wih respec o ime of he average eb per crei car a he beginning of 1988? N (millions) N = N() D (million ollars) D = D() , , FIGURE 15 FIGURE Figures 17 an 18 give he rae S = S() (pairs per year) a which shoes were purchase in he U. S. an he U. S. populaion P = P() (millions) as funcions of he year. (7) Wha o (a) S() P() an (b) [ ] S() represen, an wha are heir approximae values a = 1982? P() S (million pairs per year) S = S() P (million people) P = P() FIGURE 17 FIGURE 18 (6) Daa aape from Los Angeles Times, November 26, 1993, Los Angeles, CA: The Times Mirror Company. (7) Daa aape from Technology an Environmen by J. Ausbel an H. Slaovich, Washingon, DC: Naional Acaemy of Sciences, 1989, p. 53.

12 p. 148 (3/19/08) Secion 2.6, Derivaives of proucs an quoiens 52. Imagine ha your invesmens are in he sock marke, in real esae, an in livesock. Imagine ha on April 15 he value of your socks is 1.2 million ollars an is rising a he rae of 0.05 million ollars per year; he value of your real esae is 2.1 million ollars an is falling 0.1 million ollars per year; an he value of your livesock is 0.5 million ollars an is rising 0.05 million ollars per year. (a) Wha is he oal value of your invesmens on April 15? (b) Wha percen of your invesmens is in he sock marke on April 15? (c) A wha rae is he oal value of your invesmens increasing or ecreasing on April 15? () A wha rae is he percen of your invesmens in real esae increasing or ecreasing on April 15? 53. Wha is he erivaive of y = G(x) x 54. Wha is W (2) if W(2) = 6 an he erivaive of W(x) x a x = 2 if he angen line o y = G(x) a x = 2 is y = 7+2x? (En of Secion 2.6) is 4 a x = 2?