Class Note for MATH 223 at UA
Popular in Course
Popular in Department
This 8 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 15 views.
Reviews for Class Note for MATH 223 at UA
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 02/06/15
Mathematics 223 Spring 2008 Differentiation and Integration Rules Differentiation Rules De nition The derivative of a function f at a number 017 denoted by f a is M h i M 7 i f a 13 h if this limit exists OR 1 hm M e 1 man 35 7 a Constant Multiple dimcfm cfm Power Rule x nmn l Sum Rule 1 d 1 9M He 9 Product Rule 10 gt gltzgtgtfltzgtg ltxgtgltzgtwltzgt 11 m m m 7 m m Quotient Rule gt g f 915 9 Chain Rule If gm7 then F z f gz g m Trigonometric Functions d7 sinm cosz cosz isinm dtltgti M d we 2ltgt dm anzisecz dm 00x7 cscz d d a secm secm tanz a cscz 7 cscz cotz Exponentials ekm k5 am am lna Logarithms d 1 E 10ga xlna dm In g Inverse Trigonometric Functions d sin 1 1 cos 1 71 i m 7 i m 7 dm 17 0h xlimz d 1 1 1 71 7 t 7 t dm an 1 m2 m CO 1 m2 d sec 1 1 d csc 1 71 7 m i m 01 xv m2 7 1 01 xv m2 7 1 Integration Rules cfz dz cfzdz fzgzdzfz dzgz dz cdzczC nl m dmn10 ny il ldmlnlml0 z Exponentials Trigonometric Functions cosz dz 7 sinz C sinz dz 7 7 cosz 0 sec2zdz tanz C csc2z dz 7 cotz C secz tanz dz 7 secz C cscz cotz dz 7 7 cscz o tanz dz 7 ln secml C cotzdz 7 ln sinml o secz dz 7 ln secz tanzl c cscz dz 7 ln cscz 7 cotml c Inverse Trigonometric Functions 1 idzsin lm 0 Vlim2 1 7 71 Wdz7tan mC dz sec 1z C The Fundamental Theorem of Calculus Suppose f is continuous on 11 l lfgx alt7 then gm 2 f x dz Fb 7 Fm7 where F is any antiderivative of 1 that is7 F f Techniques of Integration Substitution If u gz is a differentiable function Whose range is an interval I and f is continuous on I7 then mmmhwz mw Example Find f xlfwdz Solution Let u 1 7 4z2 Then du 78zdz7 so 7 du zdz Therefore z d lld iz77 iu m 8 1 7gui du Integration by Parts mfmmm7fmmmw udvuU7Udu Example Find fzsinz dz Solution Let u z and dv sinz dz Then du 1dz and v 7 cosz Therefore zsinz dz z7cosz77cosz1dz 7z cosz cosz dz 7z cosz sinz C Trigonometric Integrals Form sinmz cos zdz a If the power of cosine is odd n 2k 17 save one cosine and use cos2z 1 7 sin2z to express the remaining factors in terms of sine sinm cos2k1z dz s1nmzcos2zk cosz dz won 7 sin2zkcosz dz SID Then do substitution with u sinz b If the power of sine is odd m 2k 1 17 save one sine and use sin2z 1 7 cos2 to express the remaining factors in terms of cosine sin2k1z cos zdz sin2zk cos z sinz dz 17 cos2zk cos z sinz dz Then do substitution with u cosz c If the powers of both sine and cosine are even7 use the half angle identities sin2z17 cos2z cos2z 1 cos2z Form tanmz sec z dz a If the power of secant is even 71 2k7 save a factor of sec2z and use sec2z 1 tan2z to express the remaining factors in terms of tangent tanmz sec2kz dz tanmzsec2zk 1sec2z dz tanmz1 tan2zk 1 sec2z dz Then do substitution with u tanz b If the power of tangent is odd m 2k 17 save a factor of secz tanz and use tan2z sec2z 7 1 to express the remaining factors in terms of secant tan2k1z secnzdz tan2zksec 1z secztanz dz sec2z 7 1k sec 1z secz tanz dz Then do substitution with u secz Example ftan3 z sec3 z dz Solution Since the power of tangent is odd we save a factor of secz tanz and use tan2z sec2z 71 So we have tan3z sec3z dz tan2ztanzseczsec2z dz sec2z 7 1sec2ztanzsecz dz Letting u secz we get du secztanz dz So 7 145 7 us C 7 1 5lt gt 1 3lt gt C 7 5 sec x 3 sec x Trigonometric Substitution Expression Substitution Identity a2 7 z2 x asin0 17 sin20 cos20 xaz z2 z atan0 1 tan20 se020 2 7 a2 z asec0 se020 7 1 tan20 Example Find f 7 dm Solution Let x 3sin0 Then dz 3cos0 10 Therefore V97de 7 x979sin20 2 9sin20 917sin20 9sin20 L993cos0d0 3 cos0 10 3 cos0 d0 9 sin2 3 cos0 79 8111209 3cos0 d0 cs02071d0 7cot0 70C Drawing triangles and using that sin0 we nd that cot0 V 97M Therefore 1 g 2 g 2 23 1357 3 7sin71lt gt0 m m 3 Partial Fractions You are inspecting f dz where Pm and Qm are polynomials STEP 1 If the degree of Pm is greater than or equal to the degree of c2z7 do long division to get SQ g g where the degree of Rm is less than the degree of Case I Qm is a product of distinct linear factors 011 b1a2m 2 akm l bk Decompose into P A A A 90 i 1 2 H V k t 011 1 012 2 akz bk Case ll Qm is a product of linear factors7 some of which are repeated Suppose a linear factor akz bk is repeated r times In the decomposition you need A1 A2 AT mm bk mm My m akz my Case 111 Qm contains irreducible quadratic factors7 none of which are repeated Suppose a quadratic factor azz bx 1 appears In the decomposition you need Case lV Qm contains a repeated irreducible quadratic factors Suppose a quadratic factor azz bx c is repeated r times In the decomposition you need 141 B1 142 B2 147 By am2bzc am2bzc2 H39am2bzc739 Example 1 dz Solution First doing long diVision we get that 322232 z 29712 mlg3 Then Az73Bz4 1 Lettingz3weget7B 1 so B Letting z 74 we get 77A 17 so A Therefore 2x3z2712x1 2 1 1 1 1 d if 7 d 0 z2x712 3 0 7m47m73 3 m2 1 1 2 iiiilnlz4llnlzi3l0 z m fwis Consider 2 7 4 1 1 1 1 5 7 F1116 ln1 lt0 y 1114 1M 2 4146 1114 7 1113 1 4 2 1 E 2mlt gt Example dz Solution Consider A BmC DmEi m41 g x2 1 m2 12 7 zm2 1 Then Ax2 12 Bx Czx2 1Dm m4 1 Letting z 07 we get that A 1 Then Am4 2x21Bx4Bm2Cm3CmDm2Emm41 Looking at the coefficient of 3 we nd that C 0 Looking at the coefficient of 4 we get AB 17 so B 0 Then looking at the coef cient of 2 we get 2A B D 2 D 07 so D 72 Lastly7 looking at the coefficient of x we have C E 07 so E 0 Then m41 d 1 2m d z 7 7 7 z mm2 12 m x2 12 Letting u 2 1 we have du 2zdm Thus 2m 1 de du Therefore Rules of Exponents a a 7a 1 a 77 all m a 7 Liam y ay Rules of Logarithms
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'