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calc 2 chapter 5.2

by: Jack Magann

calc 2 chapter 5.2 MTH 162

Jack Magann
GPA 3.865
Calculus 2
Dr. Bibby

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About this Document

These notes contain the professors proofs and class examples from chapter 5.2 on the derivatives and integrals of logarithms.
Calculus 2
Dr. Bibby
Class Notes
Calculus 2 chapter 5.2
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This 6 page Class Notes was uploaded by Jack Magann on Monday February 2, 2015. The Class Notes belongs to MTH 162 at University of Miami taught by Dr. Bibby in Spring2015. Since its upload, it has received 256 views. For similar materials see Calculus 2 in Mathematics (M) at University of Miami.

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Date Created: 02/02/15
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 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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