ma1012_hw15_201012

ma1012_hw15_201012 - ax b 1b (x c 1) 1) c c x x b1 w Ê dy...

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Unformatted text preview: ax b 1b (x c 1) 1) c c x x b1 w Ê dy œ È xb 3 (x ) ’ 2() " x b " Ê y) œ Èa) # +1b(sinc )1)# ˆ x 3) 1b cotc)1“‰ œ Èax# b 1b (x c 1)# ’ ax c x1b(x c 1) “ œ a2x x x b 1b kx 1) 1k bb d È b 1 (x c x bb dx " th dx du ; let u œ 1 b Èx Ê du œ #Èx dx; 2Èx ˆ1 b Èx‰ œ u œ ln kuk b C "x " " Ê 2Èdyˆ1 b Èx‰ ˆ " b t b 1 ‰ y dt œ c # t È ‰ œ (tan x¸ b C œ ln ˆ1 b60. xy b C t)) È2) b 1"Î# (tan ))(2) b 1)"Î# Ê ln y œ ln (tan )dyb " "ln (2) b "1) Ê " dy œ sec ) b ˆ " ‰ ˆ 2 ‰ t " ") y d) tan ) # #) b 1 57. y œ É t b 1 œ ˆ t b 1 ‰ œ Ê ln y œ # [ln t c ln (t b 1)] Ê y dt œ# # ˆ " c t b 1 ‰ t dy " sec ) tan ) Ê d) œ (tan )) È2" b 1 " ) Š )b ‹ œ "asec# )b È2) b 1 b " Ê dy x " É t 1 dx œ t b 1 tanœ # É bt sec x) dx Ê sec"x dx È ; x b tan x Ê du œ asec x tan œb sect#bxb ˆ t c (sec‰x)(tan #)tb11 ’ t(t b 1) “ œ 2Èt (t b 1) œ du 2) b 1 xb dt # u (sec ))(tan )) " " dy " s cos ) " dy œ ) "in ) Ê ) cos Èb 3) " x) b ln (sin )) Ê 64. ) yœ #(c"Î# ) b sin ln y œ ln ) b ln (sin )) c # ln (sec )) Ê y d) œ ’ ) b sin ) c du "Î# 2 sec ) “ y œ œ 'd (ln u) ) sec † u du )œ 2(ln u) b C œ 2Èln (sec x b tan x) b C " dy " tan x) uÈln u 61. y œ t(t b 1)(t b 2) Ê ln y œ " t b ln (t b 1) b ln (t b 2) Ê ln œ " t "1 dy ) sin ) ˆ " dy t b "bˆ "b tb# ‰ dt " Ê d) t(t 1 1) œ ) cot "Î# # tan ‰ " 58. y œ É œbÈsec ) [t(t b 1)]c) c Ê ln)y œ # [ln t b ln (t b 1)] Êy " dt œ c # t b t b 1 y b b t(t dy " " ˆ " b t b 1 b t b # ‰ œ t(t b 1)(t b 2) ’ (t 2y1)(t b 2)b 1)(tb 2) b t(t b 1) “ œ 3t# b 6t b 2 Ê dy œ t(t b 1)(t+2) t b " " 1 1) œ (x(x b 1))"Î# Ê Ê y œ "c "(x(xt(t b 1)) t(t b 1) “ln yc ln (x) b ln (x b 1) Ê y œ t(t b xb12) ln dt œ# ln# É b 1) ’ 2t b "2 œ œ 2t b 1 Ê x dt 2 at b tb b 2 dyxÈx sec1 " "œ Èx(x b 1) ) Ê ˆln‰y œ 2ln ‰ b " ln ax# b 1b c 2 ln (x b 1) Ê y œ " b x c "ˆ )) b # n 1) " 1) Ê y œ b 1)tan )(2x b " 3 y x x b1 3(x b 1) ‰ Èx(x lb (2)ˆb b65.1 ‰ y d)(xœ2x(x b 1) b 1)# œ #È 1 x # ) 2x b " b " " dy " " " x xb x(x b 1 62. y œ t(t b 1)(tÈx2)b Ê ln 2y œx ln1) c ln t c ln (t b 1) c ln (t b 2) Ê y dt œ c t c t b 1 c t b # b " ) ) Ê È) x b 1) 1 ’ x b x b c 3(x 2 ) 59. y œ yw œ b 3 (sin ))"œ () b 13)"Î# sin1) “ Ê ln y œ " ln () b 3) b ln (sin )) Ê y dy œ #() " 3) b cos ) 1 b Ètan b 1 b (x b sin 2) dy (t b 1)(t b 2) b t(t b 2) b t(t b 1) d) " " " "‘ c"# Ê dt œ t(t b 1)(t b 2) <c t c t b 1 c t b # œ t(t b 1)(t b #2) ’ “ t(t b 1)(t b 2) " dy È) b ln ) ’ 1)" Ê y œ # 1b (x c 1)# Ê ln y œÊ cln)ax# b 1bb 32(sin(x)c 2() db 3) bycot )" ˆ x 2x 1 b x c 1 ‰ “b #dœ 3t b 6t b 2 " dy " " "œ c (x b 1) y " 5 5 c x x b1 y ax# y dt (x c 1)# tˆ 166.1 tb#œc 1a‰(2x 3tÈax# blnby(x c 1)# ’ln (xx1b(x c 1) “5œ a(2xc x b 1b kxÊ1k y œ x b 1 c 2x b 1 +1b œ t b bx b b x "Ét b b 1) 2tb œ b Ê 1 œ # [10 ax c bx1) c ln 2x xb 1)] 1) È b 1 (x c b bb t b 1)(t b 2) b t(t b 2) b t(t b 1) " " 60. “Ê 3t#œ )6tÈ221)b 1 œ (tan ))(2)‰b 1)"Î# Ê ln y œ ln (tan )) b " ln (2) b 1) Ê y dy œ sec )) b ˆ # ‰ ˆ #) 2 1 ‰ y œ yw bÉ (x b) ˆ 5 c 5 œ (tan ) b d) tan b t(t b 1)(t b 2) " "# sin (2x ln 2x b c b 63. y œ ))cb 5) Ê b 1)y œx ln1() b 5) 1 ln ) c ln (cos )) Ê y dy œ )b5 c " b cos ) d ) dy os " "‰ ) dy t ‰"Î# " È2Ê " Š sec )"b " c ‹ œ asec# )b È2) )b 1 b tan ) ˆtb1 ()5 1œ Ê ln y œ Ê dy c ln (tb))1)] ) b y " tan ) # ˆ t#) b 1b 1 È 2) b 1 d t œ ˆtanb ‰ ˆ " # [ln ) c dt b tan )‰ t " Ê d) œ " ) Êt " dy œ c " c t by œ tÉ# c) cosÊ ln b 5" " )cln x b ln (x c 2) c ln ax# b 1bd Ê y œ " ˆ " b " c 2x ‰ c t b x(x " 2) ) ) y œ " " y 3x xc# x b1 œ "1 É t b 1 y ˆ dt c t b 1t‰ 67. # É t b 1 ’x b 1 “ œ 2Èt (t b 1)3 t t(t b 1) " " )(t b 2) b t(t b 2) b t(t b 1) 61. y œ t(t b " b 2) 2) "ln y œ ln t b2x (t b 1) b ln (t b 2) Ê " dy œ " b tb1 b tb# 1)(t x(x c Ê ln “ " y dt t t(t b 1)(t b 2) Ê yw œ 3 É x b 1 ˆ x b x c # c x b 1 ‰ dy (t b 1)(t b 2) b t(t b 2) b t(t b 1) " " "‰ ˆ b “ œ 3t# b 6t b 2 " t(t b 1)(t b 2) œ [t(t b 1)]c"Î# Ê lnÊ œ #œ t(t b ln (t b 1)] b t b"1 dy œ c #œ " b t b 1 ‰ b 2) ’ y dt " [ln t b 1)(t+2) t Ê y dt t b # " ˆ t(t b 1)(t t x(x b 1)(x c 2) " ln a # " 1 " b ’ 2t b 1) )68. y œ2tÉ1ax b 1b (2x b 3) Ê ln y œ 3 cln x b ln (x b 1) b ln (x c 2) c " dyx b 1b c ln"(2x b 3)d # É " b 1) " t(t bsin “ œ c 2 at b tb " t(t " " Ê ln c 2) 1 œ )b5 c ) b cos )62. y œ t(t b 1)(t b 2) x(x b 1)(xy œ ln " c ln t c ln (t b 1) c ln (t b22) Ê y dt œ c t c t b 1 c t b # " " " w ˆ x b x b 1 b x c # c x 2x 1 c 2x b 3 ‰ Ê y œ 3 É ax b 1b (2x b 3) b b t(t " " " c" Ê dy œ t(t b 1)(t b 2) <c " c t b 1 c t b # ‘ œ "t(tdb 1)(t b #) "’ (t b 1)(t b 2)b 1)(tb 2) b t(t b 1) “ y ) "Î# dt ln y œ " ln () b 3) b ln (sin )) Ê t t(t b 2) (sin )) œ () b 3) sin ) Ê œ #() b 3) b cos ) # y d) sin 3t b 6t b 2 sin x 1 w w w œa (a) c “t œ b 2tb ) b 3 (sin )) ’ 2() " 69.b cot f(x) b 3t ln (cos x) Ê f (x) œ c cos x œ c tan x œ 0 Ê x œ 0; f (x) € 0 for c 4 Ÿ x  0 and f (x)  0 for ) b 3) 0  x Ÿ 1 Ê there is a relative maximum at x œ 0 with f(0) œ ln (cos 0) œ ln 1 œ 0; f ˆc 1 ‰ œ ln ˆcos ˆc 1 ‰‰ 3 4 4 ) " sin " ln y œ ln ( 1 " 63. y œ "Î#bl5) ŠÊ ‹ œ c " ln)2b 5)" cˆln‰)œ ln ˆcos ˆ") ‰Ê ylndy )œ c " c " Therefore, the absolute minimum occurs at c ln (cos )1 dy œ "sec) œ )b5 2.) 2b cos ) ‰ œ n ln ln d # 2) b 1 œ (tan ))(2) b 1) ) cosÊ È2 y œ ln (tan )) and fln 3 ) b 1) Ê y3 d) œ tan#) b ˆ # ‰ ˆ #) b 1 ‰ ) b # (2 dy b5 " ˆ ) )cos ) ‰ ˆ ) b 5 c " b tan )‰ Ê x )œ 1 with f ˆ 1 ‰ œ c ln 2 and the absolute maximum occurs at x œ 0 with f(0) œ 0. dœ ) ) È 3 n )) È2) b 1 Š sec )) b #) " 1 ‹ œ asec# )b 3 2) b 1 b Ètan b 1 b tan 2) sin (ln x) c w (b) f(x) œ cos (ln x) Ê f (x) œ œ 0 Ê x œ 1; f w (x) € 0 for " Ÿ x  1 and f w (x)  0 for 1  x Ÿ 2 x # Ê there is a relative maximum at x " 1 with f(1) œ cos (ln 1) œ cos 0 œ 1; f ˆ " ‰ œ cos ˆln ˆ " ‰‰ œ " dy " " # # (t b 2) Ê ln y œ ln t b ln (t b 1) b ln (t b 2) Ê y dt œ t b tb1 b tb# œ cos (c ln 2) œ cos (ln 2) and f(2) œ cos (ln 2). Therefore, the absolute minimum occurs at x œ " and # t(t " "‰ b 1)(t+2) ˆ " b t b 1 b t b # x œ 2 with f ˆ b 2) ’f(2)1)(t b 2)b (lnb 2) b t(t bthe absolute maximum occurs at x œ 1 with f(1) œ 1. œ t(t b 1)(t " ‰ œ (t b œ cosb1)(t b 2) and 1) “ œ 3t# b 6t b 2 t t(t 2), # " 3x Ê du œ c3 sin œ È2 ; dy " ˆ" "‰ c 1 2 dt œ # t 12 t b 1 sin 3x " x dx œ '0 6cos 3x dx œ cÈ 1 # duMathematics I: Hc2 ln È2 c ln5. œ ogarithmic differentiation " 2' œ c2 cln kukd 1 2 œ 1 2 ln È2 œ ln 2 59. y œ È)xbb (sin )) œ (#) b 3)"Î# sinomework 1b ln ln (x c 1)dln (sin )œ ʈ y2xdy b #(2 b‰ b œ a 3 u b (x c 1) Ê 1ln y œ " cln axlnb 1b " 2 () b 3) b Ê y ) " " d) œ ) 3) ) Ê # yœ # L 56. y 1 # y # x b1 xc1 Èx b 1 (x c 1) # dy Èx(x " 1 2t b " 2t b 1 b 1) (2x b 1) " “œc "‰ 2x w œˆ " ‰ È Ê dt 1 b" x(x 1) ˆ b 3xÊ y œc2 du œ 6 b 1)t(t bdx; xb1 0œat b tbu œb 1) xœ 2Èx(xÊ1)u dx Ê c # É t(t bsin’3x x 1) x œ 2 Ê 2x(x 1 and œ 1# b # y x xb1 cos ) sin ) ' ' ' Thomas’ Calculus 11 Edition, Chapter 7 Section 2: 55, 59, 61, 63, 67 436 Chapter 7 Transcendental Functions Total: 5 problems ' Answers 55. 59. 61. 63. 67. 2) " " Ê ln y œ ln 170. ln t cf(x) œ x c ln x(tÊ 2)w (x) œ" 1dy œ ;c "xc t 1,1thent bw# € 0 which means that f(x) is increasing c (a) ln (t b 1) c ln b f Ê y dt " if t € b c f (x) cx " t b 1)(t b 2) tb2 b 2tb <c " c t " tb1 b t(t ( " f(1) œ b Ê f(x) œ x c b x € c cb)b # ‘ œœ 1 c"ln #1 ’ (t 1 1)(t b 2)b 1)(tb 2) b t(tln 1) “ 0, if x € 1 by part (a) Ê x € ln x if x € 1 t t(t b 1)(t b ) t(t b 2) 71. '15 (ln 2x c ln x) dx œ '15 (c ln x b ln 2 b ln x) dx œ (ln 2)'15 dx œ (ln 2)(5 c 1) œ ln 2% œ ln 16 " " sin ln y œ ln () b 5) c ln ) c ln (cos )) Ê y dy œ3 )b5 c " b cos ) 0 0 3 d) ) ) 1Î$ 72. A œ ' 4 c tan x dx b '0 tan x dx œ ' 4 c sinxx dx c '0 c sinxx dx œ cln kcos xkd ! 1Î% c cln kcos xkd ! c cos cos " ) b5 ‰ ˆ " c ) b tan )‰ cos ) )b5 œ Šln 1 c ln " ‹ c ˆln " c ln 1‰ œ ln È2 b ln 2 œ 3 ln 2 È2 # # 73. V œ 1'0 Š Èy2b 1 ‹ dy œ 41 '0 3 # 3 " y b1 dy œ 41 cln ky b 1kd $ œ 41(ln 4 c ln 1) œ 41 ln 4 ! 74. V œ 1 ' 2 6 cot x dx œ 1 ' 2 6 cos x sin x 2 1Î# dx œ 1 cln (sin x)d 1Î' œ 1 ˆln 1 c ln " ‰ œ 1 ln 2 # " " " 75. V œ 21'1 2 x ˆ x ‰ dx œ 21 '1 2 x dx œ 21 cln kxkd # œ 21 ˆln 2 c ln # ‰ œ 21(2 ln 2) œ 1 ln 2% œ 1 ln 16 "Î# 2 fpm@itesm.mx ...
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