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by: Jack Magann

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# Calc 2 Chapters 5.6 and 5.7 MTH 162

Marketplace > University of Miami > Mathematics (M) > MTH 162 > Calc 2 Chapters 5 6 and 5 7
Jack Magann
UM
GPA 3.865
Calculus 2
Dr. Bibby

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Includes the class notes and examples from chapters 5.6 and 5.7. These include the concept of inverse trig functions and hyperbolic trig functions. Enjoy
COURSE
Calculus 2
PROF.
Dr. Bibby
TYPE
Class Notes
PAGES
7
WORDS
CONCEPTS
Calculus 2 chapter 5.6 and chapter 5.7
KARMA
25 ?

## Popular in Mathematics (M)

This 7 page Class Notes was uploaded by Jack Magann on Friday February 6, 2015. The Class Notes belongs to MTH 162 at University of Miami taught by Dr. Bibby in Spring2015. Since its upload, it has received 93 views. For similar materials see Calculus 2 in Mathematics (M) at University of Miami.

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## Reviews for Calc 2 Chapters 5.6 and 5.7

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Date Created: 02/06/15
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 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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