Calculus I

Warmup 1. Solve the initial value problem.   = 4  a. Condition: When  = 0,  = 4 b. Condition: When  = 0,  = −6 dy x4ydx ft ydy fx.dk a h 4 te h 4 c butyl to µ yl 5th4 x5 5 1C x th4 lyl e lyl e y1 e e ly e eh4 y e ec Let K Iec 1yl e 4 g K e y e 75.4 5 5 a 4 ke0 y 4e 4y ye e or b le k e0 only ybIk6ei y t4e We will know: Eight definite integral theorems We will understand: The definite integral calculates the area under a curve We will be able to: Use definite integral theorems to evaluate integrals Agenda: x Warmup x Go over previous HW x 7.1.1 – Integration by Substitution (not on Friday ’ s test) x HW: 7.1.1 x Tomorrow (Thursday) = Review Day x Chapter 6.1-6.4, 11.2, 11.4 Test Friday, April 12 Chapter 7 – Integration 7.1.1 – Integration by Substitution (reverse chain rule) Evaluate using “guess and check” 1. ∫ 2 cos( 2 )  2. ∫  2 sin( 3 )  check si J Csi hail c as Ef ki t O cos xD 2x as c E_t sin xD xD t O Siu xD 3 2 Ets asks t c t f sink 3 xD t O t si a Ix Point of guess and check examples: Evaluate using substitution (start with same two examples) 1. ∫ 2 cos( 2 )  you can check an anti derivative by doing a derivative most deriv in this section would involve chain rule y f2xa0L dxfdtdx 2xEfcosF2xd0xedu 2x.dxf is sin 2. ∫  2 sin( 3 )  Three important hints for picking : 1.  is usually the “inside function” 2. Derivative of : Whatever you pick for  , the  (derivative) also has to be in the function (only different by a constant multiple) 3. Simple integral : After substituting, you should have an “easy integral” or one directly off of the green sheet. u f sin I 3 qdd aa cos x 3. ∫ 20 3 5 4 +3  4. ∫  2  2 +4  u 5g dfy 20q Jdf du _20g dy ftp.du h u c can 5g4x3 h 5q4 is cbeuhegfmqyquahe eEIssis.ieiitiiivd U e da e 2 0 fe i2 µ 2e Iff du thlulic Ihle 4 tc zh e 4 5. ∫ cos √  √   6. ∫ tan   p x dat Trx 2fcosu d df L sinute 2Siaf tf I faut h ulte h co Boardwork 7. ∫ 7 5  6 +3  8. ∫ sin 7  cos   U d4 6t f f ut du 16h ful t e E h I t 6 3 1 C Zh t6_ f i d dY f ul du YI c C sin8g 7.1.2 – More Integration by Substitution and Definite Integrals Evaluating definite integrals – be careful with notation on limits! Two options: 1. Use original variable and original limits Ex 9: ∫ 1 6+  −8 −11 2. Use substituted variable and substituted limits Ex 9: ∫ 1 6+  −8 −11 u 6tX 1 L du du du DX x 8 fetal x a hl6txDI a Gtx Ufs 6 8 2 d f Ufa 61 11 5 Aptest L.ES tududu dx better method HII hf at ht 51 h2 h5 h z 10. ∫  − 2  4 1 11. ∫ √ln    8 1 I u I u 4 42 16 u 16 dd 2T ucl 12 1 tfue.ge duTdu 2tdtf tIeTi u ihxucs hg gohsr.du.TT xtuH hl o foh8ut.du da t u oh8 u f8 3 g o zce

MAT | 103 | Calculus I | homework | Warmup 1. Solve the initial va

Warmup 1. Given  = arctan( 2   ) determine   . 2. If ℎ() = ( () ) 4 determine ℎ ′ (3) given (3) = 2 and  ′ (3) = 5. 4 h X 4 f f x h 3 4G GD f 3 CT 4 2 3 5 20 8 We will know: Implicit functions We will understand: Shortcuts, when they are possible, make it faster to find the derivative than using the definition We will be able to: Find the derivative and second derivative of implicit functions Agenda: x Warmup x Go over previous HW x 3.7.1 – Implicit Functions x HW: 3.7.1, Gateway Exam Review x First Gateway Exam attempt: Tuesday, Feb 12 x Chapter 3 Test Monday, Feb 18 3.7 – Implicit Functions Explicit vs. Implicit Function Explicit:  =    Example:  = 3 2 − 4 + 5 Implicit: cannot write a single function  = () Example:  2 + 2 = 25 Try solving for : Graph: y 25 y 12 y FEI YE 25 Uy ars x 1. The point (3, −4) is on the circle  2 + 2 = 25. Find the equation of the tangent line to the circle at this point. Why do we have both an  and ? In the past we only had an . Now write it as  2 + (  ) 2 = 25 and take the derivative with respect to : two points with x 3 need to know which point we are using Key idea in implicit differentiation x'that 25 t ter fi ief f eu ft fan d 25 solve for f x slope 2x t 2 fan f x O d 2 f x f x 2x f x 2 x f 4 1 I f m x x ty f x or y x 3 It dx y 13 4 IT I 2. Find   and  2   2 for  2 + sin  = 1. Cx't sin g adf.ci o7 y Fx Itf y dax C 2x cosy 2x cosy y t.cat xyfsinytyOdfI y 2cosy o2gysiagcoF Boardwork: 3. Find   for  −  5 = 9 + 3 . g fan 1 fCx y y Hx f'M q fc D8 f G x y ix y 5y4 y 9y8 y 13 2 It Tf5y 9 yt3 H y on one side all no y on other y 9ys.gl yy yxy 5y4y 9y8y 3x2 yy x 5y4 9y8 3x y y tjIa 4. Find   for cos  +  5 = tan  + ln . sing y 15 4 sedxtly y singyi six se f singy 5 4 seix tyy sing y 5 4 seix y sing fg.gs Isisejx yDor5YtIyYsjeiny yD 5. Find the equation of the tangent line to   = 2 at (1, ln 2). e'T Xy 2 f e xYy CyjJ 2 point ext.G.ytx.gl 2 fGyY f x y xy e2xy y yEdayxy eIy y x dd y e y I Y'd hD ht Y 4h21 I h2 Y l h2 l b 220.307 Y x l th2 Elope y O307 x 693f Warmup – Day 2 1. If 3 = sin  then   = A. 1 √9− 2 B. 3 √9− 2 C. 1 √1−9 2 D. 3 √1−9 2 E. 3 √9 2 −1 2. If  = tan  ,  =  − 1  , and  = ln , what is the value of   at  = ? A. 0 B. 1  C. 1 D. 2  E. sec 2  jrsiiisxiig.es i3 ady o O qw hx i y tan ex tax f sei hx thx And Y seE hx tax taxi Chx date _seichx fxktxtaf.gs faith see.ch fettafyepet oso tF sec4o fE l.E E sees II We will know: Where implicit functions have horizontal and vertical tangents We will understand: Shortcuts, when they are possible, make it faster to find the derivative than using the definition We will be able to: Find tangent lines, including horizontal and vertical ones, of implicit functions Agenda: x Warmup x Go over previous HW x 3.7.2 – Implicit Functions x Gateway Exam first attempt x HW: 3.7.2 x Chapter 3 Test Monday, Feb 18

MAT | 103 | Calculus I | homework | Warmup 1. Given  = arctan( 2

Warmup 1. Find the equation of the tangent line to () at =2 for () = 5  + 4 3 2. For a, b, c as constants determine   for  =  3 +  √ 2 3 −   2 Slope fi X h5 5 12 2 Point f 2 b5 52 12.23 f 21 52 4.23 2515 1962 251 32 57 We will know: Product and Quotient Rules We will understand: Short-cuts, when they are possible, make it faster to find the derivative than using the definition We will be able to: Find derivatives using the product and quotient rules given algebraic expressions, numerical tables, and graphs Agenda: x Warmup x Go over previous HW x 3.3 – Product and Quotient Rules x HW: 3.3 x Chapter 3 Test Monday, Feb 18 3.3 – Product and Quotient Rules Sum and difference rules: ( ± ) ′ = ′ ± ′ Does it work with multiplication and division? Is ( ⋅ ) ′ = ′ ⋅ ′ and (   ) ′ =  ′  ′ ? How could we test these conjectures? Try conjectures on  5 ⋅ 2 and  5  2 x 5 xD I E xD 2x 3 2 2 3 Proof of actual product rule from definition (or PowerPoint): Example: Find the derivative of () =  3 ⋅5  fancy dqfcxj.gcxD hligotlxthtgkthf fl x.gl zero thing flxthl.glxthlftffgx fkd.GG xth fling 4dY H glxthln glH ghj.lcjmoflxt.tn fCxto g x g x f x e fkxt.gfd fcxl.gl I II 9 as a x f g f g akxt 3x2.5 x 5 hn a'Cxl x2.5C3tx.h5 W Alternate product rule proof from PowerPoint:   ( () ⋅ () ) = lim ℎ→0 ( + ℎ) ⋅ ( + ℎ) − () ⋅ () ℎ (add a fancy 0 ): = lim ℎ→0 ( + ℎ)( + ℎ) − ( + ℎ)() + ( + ℎ)() − ()() ℎ = lim ℎ→0 ( + ℎ)( + ℎ) − ( + ℎ)() ℎ + lim ℎ→0 ( + ℎ)() − ()() ℎ = lim ℎ→0 ( + ℎ)[( + ℎ) − ()] ℎ + lim ℎ→0 ()[( + ℎ) − ()] ℎ = lim ℎ→0 ( + ℎ) ⋅ lim ℎ→0 ( + ℎ) − () ℎ + () ⋅ lim ℎ→0 ( + ℎ) − () ℎ = ( + 0) ⋅  ′ () + () ⋅  ′ () = ′ () ⋅ () + () ⋅  ′ () Proof of quotient rule using the product rule: oµ tg Q x g x f 1 1 da Q CH g Cfc xD Q x g k t Q g G f G Q H glad f x Q G g CH Q x f'CH Q x gm da Hxtg g g So our product rule and quotient rule are:   ( () ⋅ () )= ′ () ⋅ () + () ⋅  ′ () or simplified:   ( ⋅ ) =  ′ ⋅  +  ⋅ ′ and   ( () () )=  ′ () ⋅ () − () ⋅  ′ () [()] 2 or simplified:   (   )=  ′ ⋅−⋅ ′  2 Find the derivative function. Ex 1. () = 3 2   Ex 2.  = (5 3 −2 4 )3  f x 3 2 e t 3 2 ex f x 6xe t3x2 f H 3xe 2 t df 5 3 24 3 1 5 224 3D d f5E o 5 5 x't6 5l Ex 3. () = 5  3 5 Ex 4. () = −4 +4 g x 5J 3 5 5 x Gx gA s I gC 5 n5 p Cri cr 4krtfr.fr r4 p'Cr l.lrt4 Ij 4T rtfIrII6 p4r r gr Ex 5. () = 5  5 2 + 3 Ex 6. () = √  3 3 + 2 f z 5eZ 5z2t3z 5e z 5 2 2x 3 z 2 fy 5eZGz fo K x x 3 3 5 Tx 3 x te D Tie KH tx 43x.gg erx.CaxEo Use the chart to find each value.  () ()  ′ ()  ′ () 2 4 3 5 -1 3 2 2 0 7 4 3 4 6 -8 Ex 7. If ℎ() = () ⋅ () evaluate ℎ ′ (2) Ex 8. If () = () () evaluate  ′ (4) h Gd f CH g Cx I t HH g Kd h 2 f 2 g G t f 2 g G 5 3 4 I hY p x ficxl.GG fCx g g Cx D p 4 fkH.GG fCHg fgC4D2 6.4 3f 8 42 24 24 1 I 6 4 8 p e3

MAT | 103 | Calculus I | homework | Warmup 1. Find the equation of

Warmup Find the derivative function. Note: number 1 is easy enough to simplify after you have taken the derivative, but do not worry about trying to simplify number 2 1.  = 2  (3 5 − 4 2 ) 2. () =  2 + 5   + 3 d dd 2 3 5 4 2 12 1,13 5.4 2 i 3.4 – Chain Rule Algebra review of composite functions Ex 1. If () =  2 + 4 + 3 and () =  3 − 2 + 1 determine ( (1) ) Ex 2. If () =  5 and () =  2 − 3 − 5 determine ( () ) f 13 21 1 f o 02 4 O t 3 f gC f gal 9 2 3 50 f glx xZ3x 5µ Ex 3. If () = 4  and () =  5 + 5 3 +1 determine ( () ) Find a function () and () given the following Ex 4. ( () ) = (3 2 + 5) 3 K f g xD f x5t5x f gad 4 5 inside g x 3 2 5 X outside f x 3 Ex 5. ( () ) = √  + 4 Ex 6. ( () )=  3 +2 inside gH e t4x outside fad D inside gal x3t2x outside fly ex Ex 7. Find functions (), (), and ℎ() such that  ( (ℎ () ) )=2 ( 2 +1) 3 furthest inside had 72 1 medium g x 3 Outside f x 2 Derivative of composite functions = chain rule Simple version:   =   ⋅   More precise version: ( () )  = ( () ) () ⋅ ()  Other notation version: ( ( () ) ) ′ = ′ ( () )⋅ ′ () Words version: Derivative composite = derivative outside times derivative inside Examples – find the derivative function. 1. ℎ() = ( 5 + 3 2 ) 23 Warmup h'CH 2365 3 52 6,5 3 4 2 PG 2 2 hkxt 23cxkk.DZ C5x4 6 f X g 5 3 2 f 61 23 2 g x 5x4x6x f gCxD 23 gc D ffg 2365 3 2 d axfcgcxD fkglxD.GG hYxl 23 xE3xYK5x4 T 2. ℎ() = 2  2 +1 3. ℎ() = (  + 1) 5 fcgcxil ax.tl hkxt ffglxD.gl x1fcxl 2xgCxl x41 ln2 2 2x f x ln2 2 g x 2x f glad f x4 2x h2 2 ln2 2 h x 56 174.6 1 5 e tlY 4.  = 5  4 5. () = √  2 − 3 + 2 IF h 5 5 t t 4 hs 5 t 4 t 4t3 h5 f x x 2 3 2 f x t x 2 3 25 5 3 2 6.  =  5 Use the chart to find each.  () ()  ′ ()  ′ () 2 4 3 5 -1 3 2 2 0 7 4 3 4 6 -8 7. Let ℎ() =  ( () ). Evaluate ℎ(2) and ℎ ′ (2) 8. Let () =  ( () ). Evaluate (3) and  ′ (3) dd e5 d_d 5x Same chart as in 3.3 Ons Cts h 2 f god h hg 3 n'GET Y g 2 If donut p 3 g fC3 p x _g fad f x g 2 p 3 g f D f 3 pC3 g 2 O l O donut Use the picture to find each. 9. Let () =  ( () ). Evaluate (1) and  ′ (1) 10. Let () =  ( () ). Evaluate (4) and  ′ (4) 11. Let () =  ( () ). Evaluate (3) and  ′ (3) g 1 fCgCN q fH If'Gallig'd f 2 2 l 2 20 4 fCgl4D r 4 f'Cgc4H g 4 f 5 f 5 I donut E I s gCfC3H Sfx gKfCx fi x gC2 s43 g'CfC3H.fY3 g 2 I g g 2 I 12. Find equation of tangent line to () = ( 2 + 2) 3 at =1 Find the derivative function. 13. () = 2 ( 2 +1) 3 slope point f x 5 25 1 72 f i 14273 362 25 2x 27 f 4 1 6 5 2 1,27 f'd 6l 12 25 y 54 x l t f 11 54 3.42g outside z f x 42.2 CC i5Y outside x3 h2.2cxHP 3CxIi5 Cx2tI fkxt hz.sn P 3Cx4D2.2 14.  = (  + ( 3 + 1) 8 ) 14 15. () =  3  7 outside 146 67118 4 63 118 outs de 8 l4 extCx'tD8Y ex g x3 dzxCx3tl adf.TK xkD8Y ex 8Cx3 iY 3x product f l g t f g g x e em outside e 3 2 e 1 3 e

MAT | 103 | Calculus I | homework | Warmup Find the derivative fun

Warmup: Find   ( 2  5 ) f g f g t f g Gi e t x 65 D 2x e t x e5 da 5 x 2 e5 x e5 7.2.1 – Integration by Parts (reverse product rule) Proof of formula by manipulating derivative product rule:   ( () ⋅ () )= ′ () ⋅ () + () ⋅  ′ ()   ( ⋅ ) =   ⋅+⋅   Add ∫  ⋅  =  ⋅  − ∫  ⋅  to green sheet Note: traditionally the letters are actually ∫  ⋅  =  ⋅  − ∫  ⋅  dx it i i ft.dED zwtc Z w fwdztfz.dw fwdz fw.dz z.w fwdz fz.dwfz.dw zw fwdzf Example: Evaluate. 1. ∫    Three guidelines for choosing  and  in order of importance: 1.  ⋅  = entire original integrand 2. You must be able to integrate  to find  3. It helps if  is simpler than  and if  is simpler than  (or at least not more complicated) 2 X f1dw feXdx Hardpart is often w et choosing what is 2 dat l and what is Dw For first problem just dz DX choose z and dw e d J z dw Z W fw.dz for them After first e d X ex Jex.d problem there are guidelines fxe dx xe e check X ex exec ad Cxle tx ex ex 10 x ex x ex O O v usually simpler means power goes down Examples: Evaluate. 2. ∫  sin(2)  3. ∫  13 ln   choosing we can integrate both and sin 2x so guideline two does not decide wweetitieeIE.f.ibe.it rsiiTneYxax Ez Xftdw n 2 1.2d u 2x de 2 DZ w If Sina du di du 2 dx Xx w Ecosu dz dX w zoos 2x fzidw z.w fw.dz I 2x f Icoslax dx I x cosC2xltItfcosC2xl2dx Ix oskxlxtyfcosu.du txosmtz.fgn.singaI.ge txaskHHy.sina w t sing dz f dt 13 we don't know fluxdx 14 141 so we must choose but't ftp.t.dt z h by guideline 2 Temphasil t ht e that fu ft dt many students struggle 14 t ht 4 c to recognize that 4 t ht can be simplified 4. ∫ ln   Add ∫ ln   =  ln  −  +  to green sheet 5. ∫  2    E hex fdw f1dx choosing once again t W X we don't know dz dx fhxdx so we must do 2 hx h x f DX by guideline 2 x.hx ftdx x.hx X fd.io nineteen does not decide We want 2 X dw eXd x'to be simpler so we like dZ 2 dX Jw da 2x better than fxidx X x.ex fex 2x.dxzz 2xdwi dxdzz 2 de I let F wz ex e 2xe fex 2 dxj xex 2xex 2fex.dx xex 2xex 2ex cffx5 edu int by parts 5 times Boardwork – Some may be substitution, some may be parts 1. ∫( + 2) 7  2. ∫(3 2 + 8) cos( 3 + 4 2 )  z x 12 fdw tfe 7dx u 7x daff w f fe du d 7 dz dX w te du 7 dx xt2 te flye d w f e Xy e f fe dx xf e 4 Lye c x Jcosu du Tx du 3x48 3. ∫  2 ln(5)  2 h 5x fdw.fi dx hC5xljx3 fgE dxdaE 5w xIdz tx.dx x hg5 III dx x hfs I c x hg5 y Warmup 1. ∫ sin(4) cos 9 (4)  2. ∫ 5 ⋅  2  Z 5x flute d u 2x Gosha 9 4sin 40 dO dZ 5d If u cos 40 fzidw z.w fw.dz si iii fix 4 If c 5 Fife tc cos 4O 5x e2 7.2.2 – Integration by Parts Day 2 1. ∫ cos 2   Option one: use half angle formula and double angle formula Formulas from precal that you might use: Half angle: sin 2 = 1 − cos(2) 2 ; cos 2 = 1 + cos(2) 2 Double angle: sin(2) = 2 sin  cos  ; cos(2) = { cos 2  − sin 2  1 − 2 sin 2  2 cos 2 −1 sina.IQ f I d xt f o dx text 4 sxtc lzn tzsinx.us

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