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# 8 - SECTION 3.8 DERIVATIVES OF LOGARITHMIC FUNCTIONS I 211...

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Unformatted text preview: SECTION 3.8 DERIVATIVES OF LOGARITHMIC FUNCTIONS I: 211 3.8 Derivatives of Logarithmic Functions a 1. The differentiation formula for logarithmic functions. 1 (log 27) : 1 d3: “ mlna lne : 1. 1 d 2:): 2 = 1 2 1 I : _ 2 : f(m) n(:c + 0) :> f (at) 2:2 + 10 do: (at +10) :62 + 10 3. f(6) : 1n(cos0) ;> f’(0) = 1 i (c039) 2 ‘Sing — 1 tang c056 d6 c050 4- f(m) = cos<lnx> 2 m) : —sin(lnac) . é : L0”) 23 1 _ 5. f(z) :10g2(1—3\$) ;» f'(m) d 320) — 3 3 _ (1 —3x)ln2 da: (1 , is simplest when a = 6 because 5-f(\$)=10g10(\$f1)=10glom—10gm(w-1) :» f’(z) = \$1310 — (1— 11m 10 or~ 7. : \5/ = 1/5 , :1 74/51 _;.l_ 1 f(:c) lnm (lnz) => f (cc) 5(lnw) da: (Inz) — 5(Inw)4/5 a; — 5x W 8. f(\$):ln\5/E:ln\$1/5:%lnx :> f’(z)=%%:% _ , _ l ‘L_ 1 lnm _2+ln\$ 9. f(x)—ﬁln\$ => f(:c)—\/;E(x>+(lnx) 2 \$——\$+2—;— 2ﬁ _1+lnt 10. f(t)_ l—lnt => f’(t) _ (1 —lnt)(1/t) — (1+lnt)(—1/t) ¥ (1/t)[(1~lnt)+(1+lnit)] _ 2 (14111)2 (1—1111)2 ‘ t(1—1nt)2 3 11. F(t) : 111% :ln(2t+1)3 — In(3t — 1)4 = 3ln(2t + 1) — 4ln(3t — 1) => , - . 1 1 6 12 . —6(t+3) F . . . _ _ (t) 3 2t+1 2 4 3t—1 _2t+1 3t—1‘Orcombmed’ (2t+1)(3t—1)' 12. h(x) : ln(.z+ x/xz — 1) 2 _‘ __ m+¢xz__1' m ‘ \$2-1 h’(:c)- 1 (1+ 9” ) 1 x/zz—1+x 1 {E 13.g(m)=ln:;:=ln(a—\$)—ln(a+x) => , 2 1 _ _ 1 _—(a+x)—(aex) ~2a 9(32) a—2:( 1) a+z- (a—m)(a+x) :aZ—xz’ 14.F<y)=y1n<1+ey) 2 mow-Hey~ey+ln<1+ey)-1=1T:y+ln<1+ey> _ lnu 15-f(u)—m 3 f'(u)~[1+ln(2u)]-%‘lnu.ﬁ‘2 %[1+ln(2u)—Inu] [1+1n(2u)]2 _ [1+ln(2u)]2 ~1+(In2+lnu)—lnu _ 1+ln2 u[1 +1n(2u)]2 _ u [1 + ln(2u)]2 212 3 CHAPTER3 DIFFERENTIATION RULES 16.y:ln(;c4sin2:c):lnx4+ln(sinw)2=4lnzc+21nsinsc => y’=4-l+2. .1 .cosm2é—I—2cotm x smx as 1 —102:—1 10304—1 17.y=ln‘2 x 5502‘ —> y'— (1 10x)-—————0 2'm—5m2 r 5\$2+x72 , I3u+2_l _ ﬂ , _1 3 _ 3 i —6 18.G(u)—1n 3u_2—2[ln(3u+2) ln(3u 2)] :> G<u>i2<3u+2 3u—2>#9u2—4 19. y : ln(e‘x + see”) : ln(e_z(1+ 55)) =1n(e”’) +ln(1 + x) : —x +ln(1+ 1:) :> 2—x—5x2 1 —1—w+1 m ’2 1 = z y +1+m 1+ac 1+:c 1 2emIn(1+ex) _ Z 2 /# :c ' . w:_____ 20.y—[ln(1+e)] => yﬁ2[ln(1+e)] 1+em (2 1+6“: lnx , ’ m2(1/w)—(1nw)(2x) _ gnu—21:13:) _ 1—2lna: 22' y : 32— :> y _ (332)? ' \$4 — \$3 ,, \$3(—2/a:) — (1 — 21n:1:)(3x2)_ x2(—2 — 3 +6lnw) _ (Elna: 7 5 y 7 (\$3)2 m6 9:4 _ ,_ 1 _ 1 l ,,_ 1 _i :_ 1 23.y#10g1033 => y —:cln10‘lnlO :3 :> y —1n10 :62 93211110 sec 2: tan :6 + sec2 cc 24. y :1n(sec:c + tanac) —> y' — - seem —> y” — SECS!) tangy secac + tan x 23 —1 (m—1)[1—ln(m—1)]+x “5“): [1414224)]? _ [1—ln(ac—1)I2 (ac—1)I1—1n(ac—1)I2 # 2m—1—(m—1)1n(m—1) _ (av - 1)[1 —1n(ga - 1)]2 Dom(f):{m|m—1>0 and 1—ln(:c—1)7é0}={:c|9:>1 and In(:c—1);£1} :{w|x>1 and :c—laéel}={w|:c>1 and w9é1+e}=(1,1+e)U(1+e,oo) l/m _ 1 26. f(x) : 1+1n\$ => f’(:c) = #(1 +1nm)2 [Reciprocal Rule] — —m. Dom(f) : {x | a: > 0 and Ina: 75 —1}={a:|a: > 0 and w ¢ 1/6}: (0,1/e)U(1/e,oo). 2 2 l 2 m2(—2\$) 2 2:133 27. f(9:)=:1: ln(1—3c) : f(;v):2:cln(1—x )+—1—_7=2:1:ln(1—;c )— 1_\$2. Dom(f) : {50] 1 —w2 > 0} = {:L’ | Incl < 1} = (#1,1). 1 28. f(:£)=lnlnlnz :> f'()=m%;H—1;1;; D0m(f)={:c|ln1nw > 0}={a:|1n:c > 1} = {w | as > e} = (6,00). 1: , lnx!x(1/a:)71nsc—1 , _1—1:0 ”“3055; => f(m):—-(1_n_w)’2___(lnm)2 => f(€)- 12 1 30. f(x):m21n:c :> f'(:c):2wln:c+m2(;>:2zlnw—I—x => f’(1):21n1+1:1 SECTION 3.8 DERIVATIVES OF LOGARITHMIC FUNCTIONS 3 213 31. y=f(\$ 9:): lnlnm : f’ac( )— i<£> => f’(e) = l.soanequationofthetangentlineat (6,0) is lnx e 0‘l(m— )or —1;1:—10rm+e —e y — e 5 9 y — e 7 y — - 3 l 1 2 / 12 ~ - ~ 32. y : ln(m — 7) :> y = 3 7 ~3m 2)» y (2) = W = 12, so an equation ofatangent line at (2,0) is x _ 1 —0:12(z—2) or 31:12:13—24. 33. ﬂat) 2 sina: + lnm :> f’(;c) : coszz: + l/w. This is reasonable, 4 because the graph shows that f increases when f ' is positive. and ’ z = 0 when has a horizontal tan ent. ‘ f < > f g 0 AVA A A 61, i V V —2 Ina; \$(1/x) — 111x 1 — Inn: 34. — — '— 1 y m _> y 3:2 2:2 . 1— O 1 —1 y’(1) = 12 = 1 and y’(e) = e2 = 0 :> equations of tangent lines arey — O =1(m — 1) org 2 m — 1 and y— l/e = 0(1) —e) org 2 1/6. 35.y=(2:c+1)5(9:4—3)6 : lny=ln((2\$+1)5(m4—3)6) 2 1 1 ln 51n2\$+1+6lnm4—3 => — ’25. -2 - 3 y: ( ) ( ) yy 2\$+1 +6 2:4—3 4a: 2 ,_ 10 24953 _ 5 4 6 10 2413 y—y(2x+1+x4—3 _(2\$+1) (an —3) <2m+1+\$4—3>i [The answer could be simpliﬁed to y’ = 2(2m + 1)4 (x4 — 3)5(29zc4 + 12c3 — 15), but this is unnecessary] \$2 10 z 35.y=\/Ee (2324—1) => lnyzlru/i-i—lmz2+ln(a:2+1)10 => lny=%lnm+m2+101n(a¢2+1) 1 , 1 1 1 => — — —+2x+10 .2 ’2 \$2 2 1° i 209” yy: 2 ac \$2+1 30 => y ﬁe (ac +1) 2m+2\$+m2+1 sin2 artan4 m 37. y (\$2+1)2 => lny: ln(sin2 xtan4 x)— ln(ac +1) => lnyzln(sinac)2 +ln(tanm)4—ln(m2+1)2 => lny=2lnlsinxl+4lnltanxl—2ln(m2+1) :> 1 , 1 1 1 — 22.. . 4. . 2 _ . . yy smx cosx+ tanx see a: 2 1324—1 29: => y,:sin2xtan:m 2cotm+4SGC2x— 4:1: (\$2+1)2 tanx 932+1 +1 38 y~4\$ lny:llnx2+1 —llna:2 1 => 1' 1 1 1 1 _ _ :_ 2 __._. ( ) 4 ( ) yy 4 \$2+1 m 4 722—1230 :> w :4::+1 1:}:21: :17 cc __l4:c2+1 +226 4x2+1 — 2 \$24 ‘2 x2—1m4—1 _1—\$4 zZ—l n=y 39 y‘x => lnyzmlna: => y/y=\$(1/m)+(lnx)-1 => y' :y(1+ln::) => C:j=a:’”(1+ln:c) 214 I: CHAPTEH3 DIFFERENTIATION RULES 1 ’ _ 4o.y:x1/\$ :> lny=—lna: : l;l(l>+(ln\$) (ﬁl) :> ylzml/xl 111.7) 35 y as x m2 1.2 41.y:ms‘” :> 1ny:1n:vs‘” => lny:sinxln\$ => %=(sinx)-l+(lnx)(cosx) => :6 y’:y(51:m+lnxcosm> => y/:\$Sinm(¥+lnwcosx) I 42'31:(Sin\$)m => 1ny:371n(31n\$) => y—ZI‘ .1 ~c0sx+Iln(Sinm)I‘1 :‘> y sum 3/ 2 (sin any” [x cotz + ln(sin x)] I 43-92(1115CV 2 lny=1n(lnm)x => lnylenlnm => y—:x~—1—.l+(1n1n\$).1 2;, y Ina: as I (I? , a: 1 y :y( +1nlnx> => y =(1nsc) ——+ln1na: :clnzz: 11136 I 44.31239” :5 lnyzlnmlnw:(lnw)2 => y—=21n5€(l> => y’zwlnmclnx) y £13 {I} I 1 x 45.312538 : lnyzemlnzt => y—:e\$~;+(lnm)-em => y'=:ce ez<lnm+l> :c y y’ 1 1 46.y:(lna:)wsac => lnyzcosxlnﬂnsc) :> —=cosx~——lnm.;+(1n1nm)(—sinx) => 3/ y’ : (1n;1:)“°sat ( 008\$ — sinxlnln as) xlnm 1 d 2m+2yy' _ 2 2 I_ 2 2 I__ 2 I 2 /_ I 47.y—ln(\$ +y) :> yﬂx2+y2ﬁ( +31) : yﬂ m2+y2 Z} \$y+yyg2m+2yy 2x 2/ 2/_ I: 2 2_ I: I: => 93y+yy 2yy 2:1: => (cc +y 2y)y 2:1: => y —————x2+y2_2y x 1 I V 1 I I I y 48.29421; => ylnmzsclny :> y~;+(lnx)~y=x-g—j~y+lny :> ylnm#;y=lny~; I lny—y/m y:l—————_ nm~x/y 49. f(.’E) = ln(\$ # 1) :> f'(;c) = 1/(Il — 1) : (ac # 1)”1 => f”(m) : —(:c — 1)"2 :> f”’(m)=2(m—1)_3 ;> fI“)(av)=e2-3(vc—1)‘4 é : f‘")(w)=(-1)"‘1-2-3-4- -(n—1><x#1>‘"=<~1>”"H 50. y = \$8111.13. so Dgy : D8 ’ = D8 (8m7 lna: + 237). But the eighth derivative of x7 is 0, so we now have D8 (8.1671111?) = D7 (8 - 7\$61nm + 8x6) : D7 (8 - 7x61nx) =06 (8~7~6w51nm) =~~=D(8Ixolnx) =8I/z. 51. If f(:c) = 1n(1+sc), then f’(\$) : 141Fm’50flm) z 1. Thus,1im M 2 lim 1122—) = im M = f'(0) : 1. 2—“) LC m—rO :L‘ 22—40 LCﬂO 52. Letm = n/zc. Then n : mm, and as n __, 00, m —+ 00. n 1 mm ' 1 m I m . Therefore, 1im (1 + g) = lim (1+ —> = [ 11m (1 + E) ] = e by Equation6. n—aoo m—voo m m—>oo ...
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8 - SECTION 3.8 DERIVATIVES OF LOGARITHMIC FUNCTIONS I 211...

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