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# All Chapter 5 Notes MTH 162

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This 20 page Class Notes was uploaded by Jack Magann on Thursday February 12, 2015. The Class Notes belongs to MTH 162 at University of Miami taught by Dr. Bibby in Spring2015. Since its upload, it has received 395 views. For similar materials see Calculus 2 in Mathematics (M) at University of Miami.

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Date Created: 02/12/15

Calculus 2 Chapter 52 notes Logarithms Letagt0at1 Then ylogax means ayx In other words log a N equals the exponent that is needed to write N as a power of a Graphs ylnxxgt0 f m ylnx JCqu EX 39l 1 log28y can be written as 23 8 soy 4 2 log5 y through the same process y 2 1 3 log42y soy since4 Z Special Bases Base 10 also called common logarithms loglo x which is commonly written without the base to be log x Base 6 also called natural logarithms loge x which is commonly written as In x 1 De nition ofe e limx01 x x 2718 Properties of logs 110ga101n10 310gaax2xlnexx 210gaa 11ne 1 4 alogax xelnx x solongasxgt 0 Let P and Q be positive 510gaPQ logaPlogaQ lnPQ lnPan 610ga5210gaP logaQ lnPQ lnP an 7 logaPN NlogaPlnPN N lnP ChangeofBase formula lnN loga N E Proof y loga N gt ayzN gtlnaylnN gt y In a In N In a Derivative of Natural log Recall for function fX Difference quotient is f xh f x h fxh fx I h i 0 limh0 if it exists is the derivative of fX 1 e limx01 9c Iffx lnx x gt 0 then f x 2 lnhx lnx 1 xh h 1 Proof f x 11mh0 f 11mh0 E lnT 11mh01n1 h h 1 1 Ifm xmh so x h xm i 1 1 1 i 1 lllrr1ln1 mxm rhino 1n1 mmx Elnglrirnm mm Elne E If f x lnlxl then f x i The proof would be the same Now if the chain rule were applied with this I d l alnw udu u Ex 1 i1nsecx2 2x 1nsecx 2 secxtcmx Ztanx dx dx secx d d d 5 2 1n 2 ln3x 2 1n2x 3 3 2 dx 2x3 dx dx 3x2 2x3 6x213x6 3 j xsinamxz 1 coslnx2 1 2quot x21 4 Find 3 1nx2 y2 2 xy 4 xziyz 2x 23 3 x3quot 3 gt 296 230quot x3y yzy xzy 3393 gt2 X39 xzy y3 3633 xyzyr gt 2x xzy y3 ylx3 xyz I 2xx2yy3 y x3xy2 2y 5 Find the equation of the tangent to the line y x2 lnx2 3 at point 20 yr xz 2quot 2x1nx2 3f 2 164ln1 16 x2 3 y 016x 32 Logarithmic Differentiation Step 1 Take the natural log of both sides of the equation Step 2 Simplify expression using the properties of logarithms Step 3 Take the derivative of both sides and isolate y on one side Ex 2 1 Find y y W Step 1 lny lnquot23 44 Step 2 lny 1nx2 3 41n3x 4 1n xi 1nx2 3 4ln3x 4 1 Step 3gt if 393 3 y39 M3319 3 g lt Squot gtlt 31 5 2 Find y39 y x2 1lnx Step l1ny lnx2 1 quot Step 2 lny lnxlnx2 1 Step 31131 lnx 2x 1nx2 1 x21 y x2 11nxlnx x22 1 1nx2 1 Logarithmic Integration lfu ux gt du u x and thenfdju lnIuI C This general example makes use of u substitution The Ln of the absolute value of u is used if u isn t greater than 0 for all values Ex 1 96161 u x2 1 du Zxdx then rewrite the integral to E f i d 1 2 1 2 51nlx 1 C or Elnx 1 C since x2 1 gt 0 for all real numbers 2 u 1 cosx du Sinxdx 61 lnu C ln1 cosx C 3 f35 u3x5 du3dx 2 df ln3x5 C Have to keep absolute value signs since 3x 5 can be negative 2 4 fx de f1 dx x1nx 3C 6x2 5x 11 5 f 2x3 dx When the polynomial of the nominator is equal to or greater than that of the bottom improper fraction one has to divide the 2 functions Here is how long division between two functions is applied in this problem 3x 2 6x2 5x 11 3x2x 3 4x 11 22x 3 remainder 5 The result will be in the format of a polynomial proper fraction So 6x2 5x 11 5 dex f3x2 3dx 2x2x 21n2x 3C 2x Trigonometric Integrals These are important and should be memorized sin x 1 ftanxdx 2f dx throughu substitution 2 f1ncosx C flnlsecxl C C0596 C0596 2 fcotxdx 2f dx lnlsinxl C 1ncscx C Slnx SCC 9639 SEC xtan 9639 SEC2 xseC x tan x 3 f 1 x CSCXtanx f secxtanx u secxtanx du secxtanx seczx dx sec2 xsec x tan x f dx 1nsecx tanxl C secxtanx 4 fcscxdx 1ncscx cotxC Calculus 2 Chapter 53 quot Natural Exponential Function ex i cxp 1 exylnyx Therefore 1 1n ex x For the derivative for ex we use logarithmic differentiation yex lnylnex gt lnyx Take the derivative ofboth sides gt iy 1 gt y y gt y 2 ex d gt quotquot Therefore ex 2 ex This is a little different when the chain rule is applied at du e e dx dx For the integral f ex 2 ex C EX 1 x 1 x 1 x x 1yEe Slnx Ee cosxge Slnx e cosx 1 1 y 5ex cosx ex s1nx ex Slnx ex cosx 5Zex51nx exsmx 2 xey yex xeyy ey yex y ex gt xexy exy yex ey gt I x x x y Izyexey 31068 e ye e gt y xeLex 1 x2xex 2x2xex 3gtyx2exx lnyx1 x2ex gt EV39ZW gt y ym I 2 x x 2x2xex y x e x2ex 4 fexxex 1dx u substitution u 2 996 1 du exdx 3 fexVex1dx fxZdu u C ex15C Calculus 2 Chapter 54 Derivative of regular log functions If you recall the change of base formula loga N 2 11111 1 Then 10 x lnx i 1nx 1 dx ga dx lna lna dx xlna Derivative and integral of number to the power of X 1 101quot Use logarithmic differentiation to solve this dx 1nylnax gt 1nyxlna gt 3 11y 21na This is true since In a is a constant y aquot In a or with the chain rule 1a aud ulna dx dx 2 faxlnadxaxC faxiaxC lna Ex 1 y 10g2 1 1 1 x 5 1 1 x 1 1 1 1 n ln ln ln 2 y ln2 x2 21n2 x2 ln 22 x x ln4 x x2 2 gx 5603 g x 5603 ln5 sinx 3fx22xdxfx2dxf2xdx x32xC 4f33dxf1dxf131 xdx u1 x du 1dx x f3udu2x i31 xC ln3 Calculus 2 Chapter 56 Inverse Trig Functions These functions can be thought of as doing normal trig equations but backwards There will be proofs for the derivatives of sin 1 x tan 1 x and sec 1 sin 1 x Def1nition Derivative Proof Ex 2 tan 1 x Definition Derivative Proof coszy sin2y 1 so 1x y sin 1 x means siny x The range of answers is restricted to be Only need one positive 1 and one negative 4 quadrant 7T V5 77 Ex51n11 and 51n1 2 2 4 d 1 1 d s1n u dx ll u2 du Recall that coszy sin2 y 1 so cos2 y 1 sin2 y y sin 1x gt siny x Therefore cos2 y 1 x2 gt cosy iVl x2 But in quadrants l and 4 range of sinquot1 x cos y is positive I I 1 I 1 Slnyzx gt y Cosy 1 gt y cosy y 1 x2 1 x I ex ex y gt y 1ex2 1182x y tan 1x gt tany x The range of answers is restricted to be Only need one positive 1 and one negative 4 quadrant Ex tan391 1 E 1 d 1u2 du i 1 dx tan u Recall that coszy sinzy 1 gt 1 tan2x seczx cos2 y cos2 y cos2 y y tan 1x gt tany x Therefore sec2 y 1 x2 gt sec2 y 1 x2 In quadrants 1 and 4 range of tan391 x sec x is positive I I 1 1 tanyZx gt y seczy 1 gt y seczx gt y m 3 sec 1 x Def1nition y 2 sec 1 x means secy x The range of is restricted to 0 S y S and TC S y S 3 On1y need one positive 1 and one negative 3 quadrant Derivative i 1 1 dx sec u u2 1du Proof Recall that 1 tan2 y 2 sec2 y gt tany i sec2 y 1 But in quadrants 1 and 3 range of secquot1 x tan y is positive y sec 1x gt secy x 1 1 Therefore secytany y 1 gt 3 secytany xwxz Ex 1 39 L 1 2x 1 1 fx tan 2x f x x Zx21 2 tan 2x 4962 2 tan 2x 1 1 2 y sec lobe2 4 u x2 4 du x2 4 2x xx2 4 1 2 m m21xx2 4y x 3 DemoW 3 y xtan 1x 1nx2 1 xtan 1x 1nx2 1 x 1 2x x x y tan 1x tan1x x21 2 x21 x21 x21 4 x2 x sin 1 y 2 yequot Use implicit differentiation 2x y sin 1 y 2 yequot y ex W sin 1 y xy yex1 yz y ex1 yz y x ex1 yz 1 yzyex sin 1 y 2x W1 y2yexsin1y2x y x ewayz 5 y 2 sec xsec1 x Use logarithmic differentiation y 2 sec xsec1x gt lny sec 1x lnsec x 1 1 1 yy sec x secx secxtanx1nsecx xm y ysec 1xtan x 125 2 sec xsec1 x sec 1x tan x 125 Integrals of Inverse Trig Functions If u ux and a is a constant Where a gt 0 then 2 sin 1 C j du Vazu2 du 1 u 1 jx2a2 atan aC jal J 1 C um asec a Ex dx exdx du 1 MW 26quot 2996 5 szcm x f l z sin 1 C sin 1 C 2 f2x5dx2f 2x f 5 x2 49 x2 49 x249 2 E 1 E gt1nx 49 7tan 7 C 2 3 f2x x11 dx 2 Need to divide the two functions x 4 2 2x2 x 11 2x2 8 Remainder x 3 x24 3 f2x2x11d 2x32 x x24 x x24 x24 x24 X39 2C gt 2x l1nx2 4 Etan 1 2 2 3dx dx 4fx 9962 25 f3x 3x2 52 u3x du3 a5 gt 1 13xC 5sec 5 Calculus 2 Chapter 57 Hyperbolic Functions exe39x ex e39x nhx coshx Slnh tanh 2 2 coshx 1 coshx sechx cschx coth coshx smhx smhx Graphs rush I K 0 39 sinlu f Ill tunh 39 39 1 FIGURE 1 FIGURE 2 lt l f V 39 lnhl U 6 L h 6 Jlt A Hyperbolic Identities cosh2 x sinh2 x 1 since a hyperbolic function is x2 y2 1 2 2 exe x ex e x ezx2e 2x ezx 2e 2x 4 Proo 1 2 2 4 4 4 Derivatives of Hyperbolic Functions d du d du d du 51nhu coshu coshu Slnhu tanhu sech2 u dx dx dx dx dx dx d du d du cschu cschu cothu sechu sechutanhu dx dx dx dx d du cothu csch2 u dx dx Proofs exe x l isinhx l i x x l x x exex 2 2dxe e 2e e 2 coshx d d sinh x cosh x cosh x sinh x sinh x cosh2 x sinh2 x 1 2 2 tanhx sech x dx dx cosh x cosh2 x cosh2 x cosh2 x d coshx sinh2 x cosh2 x 1 3 cothx csch2 x dx sinh x sinh2 x sinh2 x Integrals of Hyperbolic Functions du du fcoshu dusmhuaC fsmhu ducoshuEC 2 du 2 du fsech u dutanhuEC fcsch u cothuEC du du fsechutanhu du sechua fcschucothu du cschua Ex sinhx l y cosh x sinhx sinhx cosh xsinhx cosh x sinh x sinh2 x cosh x smhx Multiply by inverse y coshxsinhx y cosh2 x sinh2 x 2 sinhx coshx coshx sinh x2 e39x x u u x l 2fe Slnhxdx Usetheidentity fe smhxdx fe 2 dx Zfe 1dx gt llexxlexlxC 2 2 4 2 sinhx dx u coshx du Slnhx 3ftanhxdx 2f cosh x 62 lnlul C lnlcoshxl C Calculus 2 Chapter 58 In previous chapters we were unable to solve such equations as this 62quot 63quot 2 lim x gt0 3 sm2x 2 COSCX 2 Since using the substitution method shows the bottom of the equation to be 0 it s limit would be DNE Does Not Exist However using L Hopital s rule this limit can be solved L Hopital s rule Let f x and gx be differentiable functions If 1 limxa f x limxa gx 0 or 2 limx m f x limx m 906 too Then f x 9 06 limxa imxa if both limits exist gx We can now solve the limits we were not able to in earlier chapters Ex 2x 3x e e 2 o 1 11mx0 smce the 11m1t of both equatlons equals zero we can 3 Sln2x 2 cosx2 0 use L Hopital s rule f x 51 96er 63quot 2 262x 363x g x 2 3 sin2x 2 cosx 2 6 cos2x 2 sinx 2e2x3e3x 23 5 11mx0 Now the subst1tut10n techmque prov1des a real 6 cos2x2 Slnx 60 6 answer Certain problems will require that you make use of L Hopital s rule multiple times EX 1 lim w 3 Now we can use L Hopital s rule x 3x32x 1 oo 6x2 10x oo llmx mo m Slnce 1t stlll equals 00 we use L Hopltal s rule aga1n 12x 10 oo 11mxoo 18x 2 Slnce 1t stlll equals 00 we use L Hop1tal s rule aga1n um 12 2 Xquot 18 18 00 and g are called indeterminate forms however they are not the only ones 0 OIO m Indetermmate Forms 5 0X00 oo oo 00 1 00 What s important to know is that L Hopital s rule only works for 00 and so you have to change the other indeterminate forms into them Ex exe x 2 o 1 11mx0 m 0 Use L Hopltal s rule lim ex e x 0 Use L Ho ital s rulea ain lim exe x 1 9 60 25in2x 0 p g 9 60 4cos2x 2 lntanx oo oo 2 11m x60 1ntan 2x oo 00 SEC2 x lirr1 limx0 cotx sec2 x tan 2x cos2 2x 95 tan 2x Use sin2x 2 sin x cos x so when you cancel everything out you get cos2 x 1 lim 1 9 60 cosx 1 3lim ln2x35x21 oo lim 6x210x 3x47x2 xquot 1n3x47x2 xquot 2x35x21 12x37 Rather then multiplying these expressions out we can just nd the values of the highest degrees of X The limit is then just the fraction that the numbers in front of the highest degrees of X make so 6x210x 3x47x2 18x6 18 lim 11m xquot 2x35x21 12x37 xquot 24x6 24 4 lim 1 1 lim x 0 x gt0 ex1 x x 0 xex x 0 1 ex 0 ex 1 11m gt11m 9 60 xexex 1 0 9 60 xexexex 2 sin 2x 0 5 11m sm 2x csc 3x x60 sin 3x 0 2cos2x 2 lim x 3 cos 3x 3 For the indeterminate forms 1 00 000 let L lim fxgx lnL lnlim Uxgx lim 1anxgx 1 lnL 2 11m gx lnfx 2 11m n11 m 1nfx 1 lim lnL 61 e 906 Ex 1 1 limx0 6quot 3x 10 L 1 lnex3x 4 x so lnL 11mx01nex 3xx 11mx0 9 1393quot 2 E 4 This however is not the answer according to the equation above The full answer is e4 Techniques of Integration Chater 60 1 Division For when there are two polynomials in the fraction and the highest degree of X in the numerator is equal to or greater then that of the bottom Also the equation can not be broken into parts 2 f2x 11 x24 2x2x11 2x2 8 x 3 2x2x11 x 3 f x24 f2 x24 x 3 1 2 3 1 x f2fxz8 fx24 2x lnx 8 tan C 2 Rationalizing Substitution fijr J lett Vx 1 nowputxanddxinterms ofy xt2 1 dx2tdt f f tht t212dtf2t22dt t3 21 vx 13 2 mc 3 Completing the Square f x2x613 dx If the bottom polynomial was in the form of x2 bx or x2 bx the integral could easily be done To do this Beak the last number up for bottom to be perfect square plus or minus removed number x1 x1 dx fx2 6x13 fx 324 Obtain the form x2 bx b2 or x2 bx b2 Now use substitution using the completed square wx 3 dwdx and xw3 f x1 x fW31 x 324 w24 W 4 l 2 1K fwz4fw24 21nw 42tan 2 1nx 6x 13 2tan 1 C All these notes are from a Calculus 2 class run by Dr Patrick Dibby at the University of Miami A majority of the problems used by the teacher and copied in these notes are taken from the textbook Essential Calculus Second edition By James Stewart Citation Stewart James Essential Calculus 2nd ed Np Cangage Learning 2013 Print All these notes are from a Calculus 2 class run by Dr Patrick Dibby at the University of Miami A majority of the problems used by the teacher and copied in these notes are taken from the textbook Essential Calculus Second edition By James Stewart Citation Stewart James Essential Calculus 2nd ed Np Cangage Learning 2013 Print

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