Differentiation. The Product Rule you must rhyme, E I E I O, It s u-prime v plus u v-prime, E I E I O. ANSWER:

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1 Differentition Eple: y = * 3 4 Answer: y' = Constnts, ll by theselves, differentite to zero. Eple: y = 7e 2 {notice: NO VARIABLE} y' = 0 1. Constnt ultiple rule: [ ] [ ], k constnt Eple: [ ] [ ] 2. Su rule: [ ] [ ] [ ] Eple: [ ] [ ] [ ] 3. Generl power rule: [ ] [ ] [ ] "OI!" (outer inner) 4. Product rule: [ ] u'v + uv'. Quotient rule: [ ] 6. Chin rule: ( ) [ ] 7. Eple: y = e y = e [ ] 11. (differentite only the eponent, then just copy the 'e' with the old eponent, ectly s it is) Eple: y e 3+ y e 3+ () Logrithic Differentition: [ ] Prie Nottion f y d d Nottion d f d dy d f d 2 y f d2 d 2 y d 2 Differentite the rguent of the log OVER the rguent of the log copied ectly. Eple: y ln y 'Mrginl' ens differentite! -- To get fro rginl bck to originl (cost, profit, revenue), integrte! () If C() is the cost, then C'() is the rginl cost (b) If R() is the revenue, then R'() is the rginl revenue (c) If P() is the profit, then P'() is the rginl profit. 1 Note: 1st split cople log eqn. into seprte, individul logs using log properties, THEN differentite ech log piece seprtely ccording to rule #10. Eple: y ln [ e ] y ( ) 3 Eple: y The Product Rule you ust rhye, E I E I O, It s u-prie v plus u v-prie, E I E I O. y ln [ ( ) 3 ] Tke ln of both sides Recll: y Use properties of logs to siplify eqn. y + y y [ ANSWER: Differentite (iplicitly) both sides ] Isolte y by ultiplying both sides by y y 3 [ + 6 Lo dee Hi inus Hi dee Lo, Squre the botto nd wy we go! Old Mcdonld knows Clculus e y + ] Recll: y = 2 (+1) 4 ( 2 +2) 3

2 13). Differentite position to get velocity (RATE) 13b). Differentite velocity to get ccelertion 13c). Integrte ccelertion to get velocity 13d). Integrte velocity to get position Differentite Iplicit Differentition 1. Differentite ech ter of the eqution with respect to.. Whenever you differentite the y vrible, tck on y to its derivtive. 2. Get ll the y' ters on one side, ll the non-y' ters on the other side of the equl sign. 3. Fctor out the coon y'. 4. Divide both sides by the y' fctor to get y' ll by itself on one side of the equl sign.. Eple: y 2 y = yy 1y = 0 6yy 1y = y (6y 1) = y = ANSWER 6y 1 Chin Rule Probles 1. Differentite ech ter of the eqution with respect to t(tie).. Tht is, differentite every vrible, then tck on tht vrible s letter with prie. 2. Plug in ll given vlues nd solve for the reining vrible (or its prie) 3. Eple: 3 2 y y 3 = 0 (Note: First ter will require product rule). 6 y y y 2 y = 0 OR Line Infortion 1. To find the eqution of line: ) Find point on the line: ( 1,y 1 ) b) Find the slope, c) Write the eqution of the line: y y 1 = ( 1 ) 2. To find the slope: y2 y1 ) 2 1 b) First derivtive (y') = slope Not finding the slope would be crie. 3. y = + b slope - intercept for of line siply find, y-prie! 4. y y 1 = ( 1 ) point-slope for of line. prllel = Prllel lines hve the se slope. DON T CHANGE THE SLOPE! 1 6. perpendiculr Flip old slope over AND CHANGE THE SIGN 7. X-intercept: (,0) plug in 'zero' for y, nd solve for To get out of this 8. Y-intercept: (0,y) plug in 'zero' for, nd solve for y joint, find the point! To Find the Eqution of Tngent Line 1. Find the slope:. Tke the first derivtive of f() b. Plug -vlue into first derivtive nd solve. c. Tht is your slope. 2. Find the point:. Plug -vlue into the ORIGINAL EQUATION, f(), nd solve for f() (this is the y-vlue of the point). b. Your point is the vlue (,f()) 3. Write the eqution of the line: y y 1 = ( 1 ). Solve for y in ters of, to put it into slope-intercept for, y = + b. 2 s(t) Position s(t) s'(t) Velocity s'(t) s''(t) Accelertion s''(t) Integrte Clculus DEVIL

3 (divide by whtever nuber is in front of the then just copy the 'e' with the old eponent, ectly s it is) {Eple: Integrtion Eple: y = Answer: y = 3 +C 0. = k + C, k is constnt Constnts pick up the vrible when integrted [ ] e d e C When the power of the denointor is positive one, The nturl log is your best chu! To find the vlue in BETWEEN, siply use the CURVY THING! Clculus lincoln 7. = F(b) F() Fundentl Theore of Clculus (i.e. the re under the curve between two points, & b) Initil Vlue Probles ; then use given f() = b, tht is, the point (,b) to solve for C Eple: f () = ; f(1) = 9 ( ) = C, tht is, y = C We re given f(1) = 9, or the point, (1,9), so plugging in for & y gives: 9 = (1) 3 2(1) 2 + 8(1) + C Solve for C nd rewrite the eqution: 9 = C 9 = 7 + C C = 2 ANSWER: f() = Averge Vlue of Function [ ] The verge vlue integrl Eple: [ ] [ 3 ] [ 3 ] [ ] [ ] [ ] [ ] [ 3 3 ] [ 3 3 ] [ 3 3 ] [ 6 3 ] ANSWER Integrtion by Substitution A Buddhist or Confucin would integrte by SUBSTITUTION! 1. Define u = g(), where g() is chosen so tht g'()d will get rid of nsty ter in the integrl. 2. Replce g() with u nd d with the djusted g'()d with du. 4d Eple: 3. Integrte the resulting function of u. ( 2 +3) ; let u = 2 +3; du = 2d, so 2du = 4d 2du 4. Rewrite the nswer in ters of by replcing u by g(). Rewrite integrl: 2 u du 2 u4 u 4 u u 4 Substituting bck in for u: ( 2 +3) 4 ANSWER c (hypotenuse) Hndy Forule b 1. Pythgoren Theore: 2 + b 2 = c 2 2. Qudrtic Forul: Given 0 = 2 + b +c, then 3 b ± b c Oop, Loop, Doop Dee Dorul, To Solve This Eqution, Use the Qudrtic forul!

4 Choose only ONE of these four situtions: OR OR OR Guidelines for Integrtion by Substitution: 1. Let u equl the highest power polynoil epression. 2. If n e polynoil is present, let u equl JUST the polynoil prt of the eponent, NOT the whole eter. 3. If there is n (e polynoil epression) Power, let u equl the e polynoil epression, i.e. ONLY wht is in the prentheses. 4. If nturl log is present, let u equl the ln(rguent). Uber Iportnt!!! The substituted epressions MUST tch the epressions of the integrl EXACTLY. If they don t, they MUST be djusted (by ultiplying or dividing by constnt). Eples (corresponding to guideline nubers bove): Let: u = 3 2 This is the highest power polynoil epression. du = (3 2 2)d Notice how this tches EXACTLY the epression in RED in the originl proble. Now, substitute into the originl proble, u (the green) nd du (the purple) to get uch esier proble to integrte. Let: u = 2 This is just the polynoil prt of the eponent. du = 2d Notice this does NOT tch EXACTLY the epression in RED in the originl proble; it will hve to be ADJUSTED until it does tch EXACTLY. 1 du = d 2 By dividing both sides by 2, n EXACT tch is chieved. Now, substitute into the originl proble, u (the green) nd du (the purple) to get uch esier proble to integrte. Notice: Pull the constnt, 1 outside the integrl to ke things esier ( + ) 3 Let: u = e 3 3 This is just wht is in the prentheses, NOTHING ELSE!. du = 3e Notice this does NOT tch EXACTLY the epression in RED in the originl proble; it will hve to be du = 3 e 3 2 ADJUSTED until it does tch EXACTLY. 1 du = e3 2 d By dividing both sides by 3, n EXACT tch is 3 chieved. Now, substitute into the originl proble, u (the green) nd du (the purple) to get uch esier proble to integrte. Notice: Pull the constnt, 1 outside the integrl to ke things esier Let: u = ln() This is the ln(rguent).. du = 1 d Notice how this tches EXACTLY the epression in RED in the originl proble. Now, substitute into the originl proble, u (the green) nd du (the purple) to get uch esier proble to integrte. 4

5 Properties of Eponents Properties of Logriths 1. 0 = 1 1. ln(1) = 0 For ll rel nubers, b, N, M>0, 1: b b 2. ln(e) = 1 Eples: 1 3. ln(e ) = b b 1 4. e ln() =, >0 Eple:. ln( 2 +1)( 3 3) = ln( 2 +1) + ln( 3 3) 6. ln( 2 +1) 3 ( 3 3) 2 = 3ln( 2 +1) + 2ln( 3 3) 7. ln (2 +1) 2 3 2ln ln n n.. ln(m N) = ln(m) + ln(n) 6. n n 6. ln(n ) = ln(n) M N n n 7. ln ln( M) ln( N) 3 8. e = ln() = n n n 9. (negtive) Odd = negtive 3 9. (negtive) Even = positive Eple: Copound Interest & Eponentil Growth/Decy Note: since Siple Interest: 1. I = Prt I = $ interest erned; P = principl (strting ount); r = interest rte; t = tie (in yrs.) 2. A = P(1+ rt) A = $ finl ount; P = principl (strting ount); r = interest rte; t = tie (in yrs.) Copound 'n' ties per yer: nt r 3. A P 1 n A = $ finl ount; P = principl (strting ount); r = interest rte; n = # ties copounded per yer; t = tie (in yrs.) Copound Continuously, Popultion Growth, Rdioctive Decy: 4. A(t) = Pe rt A = finl ount; P = strting ount or present vlue; r = growth/decy constnt or interest rte, t = tie () Rte: A'(t) = r A(t). Doubling tie: Tripling tie: r Eples: ln 1 Hlf-life: 2 ln(2) doubling tie r hlf life ln(3) r Qudrupling tie: tripling tie log y b y, b 0, b 1 b ln(4) r, etc. qudrupling tie,etc.

6 How to solve M/Min Probles: 1. Revenue: R() = ites price. Price bbrevited with the letter, p. Ites represented by the letter,. The Cpitl bove ens ties. 2. Profit: P() = Revenue Cost. 3. To find the nuber of ites needed to get iu or iniu: () Differentite either the revenue or profit eqution. (1) If trying to iize revenue, differentite revenue eqution. (2) If trying to iize profit, differentite profit eqution. (b) Set the first derivtive equl to zero (c) Solve for. 4. To find the ctul iu or iniu vlue for revenue or profit: () Tke the -vlue fro 3.(c) bove (b) Plug tht -vlue into the originl revenue or profit eqution.. To find the idel price to chrge: () Tke the -vlue fro 3.(c) bove (b) Plug tht -vlue into the originl price eqution. 6. Eple: Clculus Prophet () We re told the price of ite, s function of the nuber of ites,, is: p() = (b) Then the revenue eqution, s function of the nuber of ites,, will be: R() = ites price = [ ] = (c) If we re told the cost to produce ite, s function of the nuber of ites, is: C() = (d) Then the profit eqution, s function of the nuber of ites,, will be: P () = R() - C() = [ ] [ ] P () = (e) To iize the profit, tke the derivtive of the eqution in (d) bove, set it equl to zero, nd solve for '.' This tells you the perfect nuber of ites to produce to iize your profit: P ()= = 0 8 = 24 = 3 ites (1) This ens when you ke 3 ites, you will hve iized your profit. (f) The ctul iu profit would be: P(3) = 4(3) 2 +24(3) 22 = $14 (g) You should price your ites t: p(3) = 3(3) + 14 = $ (h) If you chrge $/ite, you cn epect to sell 3 ites, thus iizing your profits t $14. 6

7 Eple: tht The price of selling n ite,, is given by: p = The totl cost (in dollrs) for copny to produce nd sell ites per week is C() = How ny ites ust be sold to iize the profit? Wht is tht iu profit? How uch should be chrged per ite to rech tht iu profit? Grph the profit function lbeling the verte, points of syetry nd ll intercepts. Revenue is ite ties price: R() = ites price ( ) R() = Profit is Revenue inus Cost: P () = R() - C() = [ ] [ ] P() = X-vlue of the verte (which represents the nuber of ites you need to sell in order to iize your profit) is found by setting the first derivtive of the profit equl to zero nd solving for. P () = = 0 = 1 ite (eciting, isn t it?) Y-vlue of the verte (which represents the ctul iu profit) is found by plugging in the -vlue fro the step bove into the ORIGINAL Profit function. P() = P (1) = (1) 2 This ens if you sell (1) + 11 = 16 Verte: (1, 16) ite, you will iize your profit t $16. Plug the -vlue of the verte into the price eqution to find out how uch to chrge per ite in order to iize your profit. p = p = p = 4(1) + 30 p = $26/ite Y-intercept: P(0) = (0) (0) + 11 = 11 Y.I.: (0, 11) X-Intercepts: 0 = b ± b c ± ± ± ± ± ( ± ) ( ± ) ± ± X. I.: (2.8, 0) ( 0.8, 0) Ites to Sell to Miize Profit: 1 ite Price to Chrge per ite: $26/ite Miu Profit: $16 Clculus Prophet 7

8 How to Grph Polynoil 1. Tke FIRST & SECOND derivtives () Set first derivtive = 0 nd solve for. (b) Plug in -vlue(s) into originl function nd get y-coordinte(s). (c) Plug in -vlue(s) for ech coordinte into second derivtive. (i) If second derivtive is +, then grph siles (in) t tht coordinte. Note: Ignore the vlue of the second derivtive. We just cre if it is '+' or '-'. Nothing else! (ii) If second derivtive is, then grph frowns () t tht coordinte. Note: Ignore the vlue of the second derivtive. We just cre if it is '+' or '-'. Nothing else! (d) Set second derivtive = 0 nd solve for. (e) Plug in -vlue(s) into originl function nd get y-coordinte(s). (i) These re the inflection points. 2. Find y-intercept(s) {by setting = 0 in the originl function & solving for y} 3. Find -intercept(s), if esy (generlly only with even functions i.e. 2, 4, etc.) {by setting y = 0 in the originl function & solving for usully by fctoring or the qudrtic forul} 4. Deterine end behvior, syptotes, etc. (i.e. As, y?: As, y?) () End Behvior - Polynoil Grph: + even As As even As As + odd As As odd As As. Plot ll points, end behvior, syptotes, then grph. Intervl Nottion f() f() f() f() f() f() f() f() 1. Intervl Nottion: left to right (or down to up). {sllest, lrgest } OR {sllest y, lrgest y} ) > or < or or round prenthesis: ( ) b) or squre brckets: [ ] c) Join two or ore sets with the union sybol: U 4 3 d) Eples: 1) > 4 ( 4, ) 2) 3 (, 3] 3) 2 & 3 (, 2) U ( 2, 3) U (3, ) 4) (, ) U (, ) 2 3 8

9 How to Drw Grph fro its Derivtive (How to get n igun fro drgon shdow) 1. The zeros (-intercepts) of the grph of f'() re the iu or iniu vlues of f(). () If f'() goes fro negtive to positive, f() hs iniu. (b) If f'() goes fro positive to negtive, f()hs iu. 2. Wherever f'() is negtive (i.e. BELOW the -is), f()is DECREASING (hs negtive slope). i.e. the ldybug on f() is going "downhill" 3. Wherever f'() is positive (i.e. ABOVE the -is), f()is INCREASING (hs positive slope). i.e. the ldybug on f() is going "uphill" 4. Wherever f'() hs positive SLOPE i.e. the ldybug on f'() is going "uphill" (tht is, the second derivtive f''() is positive), f()smiles (concve up).. Wherever f'() hs negtive SLOPE ). i.e. the ldybug on f'() is going "downhill" (tht is, the second derivtive f''() is negtive), f() FROWNS (concve down). 6. Wherever f'() (the first derivtive) hs iu or iniu (i.e. the zeros of the second derivtive, f''()), the grph f() hs n INFLECTION POINT. 7. The degree of the FUNCTION itself is TWO ore thn the # of turning points ("bups") on the first DERIVATIVE grph, f'(), tht is, one ore thn the DEGREE of f'(). 8. The Y-vlues of the FUNCTION re rbitrry nd cn NOT be deterined fro the grph of the first derivtive, f'(). 9. Eple: f'() f'(,1.) f f() incresing (., 1.) U (3, ) f() decresing (,.) U (1., 3) f() reltive iu = 1. {no info on 'y' vlue} f() reltive iniu =. & = 3 {no info on 'y' vlue} f() concve up (siles) (, 1) U (2, ) f() concve down (frowns) (1, 2) f() inflection points = 1 & = 2 {no info on 'y' vlue} slope of f() t = 1. Note: Y-vlues re rbitrry. X-vlues re fied. Note: There re 2 turning points ("bups") on f'(), therefore the degree of the FUNCTION is two ore thn the bups on f'(), tht is, 4. P.S. Just dding 1 to the nuber of bups on f(), s you did in Mth 121, is not lwys relible ethod for finding the degree of f(). You re better served dding 2 to f'(), which ALWAYS works. degree of f() 4 {two ore thn the nuber of "bups" on f'()} 9

10 ALWAYS MOVE LEFT TO RIGHT IGNORE Y-VALUES STEP 1 Using the zeros (-intercepts) of f () Drgon, Identify the & in -vlues for F() Igun. 1. When f () Drgon goes UP through the -intercept, drw CUP.. Hving n UP dy? SMILE! 2. When f () Drgon goes DOWN through the -intercept, drw FROWN.. Hving DOWN dy? FROWN! In other words: Quickie 3-Step Sury: 1. Plot es nd ins. 2. Plot I.P.s. 3. Connect the dots. Moving fro left to right, Does f () Drgon go UP s it crosses the -is? NO Moving fro left to right, Does f () Drgon go DOWN s it crosses the -is? YES YES X-intercept on f () Drgon is the -vlue of the MINIMUM on F() Igun. Drw SMILE t tht -vlue on your grph of F() Igun The y-vlue is rbitrry your choice. X-intercept on f () Drgon is the -vlue of the MAXIMUM on F() Igun. Drw FROWN t tht -vlue on your grph of F() Igun The y-vlue is rbitrry your choice. f () Drgon F() Igun EXAMPLE B D A B C D C A Refer to the -intercepts on f () Drgon A. On f () Drgon, the grph goes UP through = 4, so we SMILE t = 4 on F() Igun. B. On f () Drgon, the grph goes DOWN through = 2, so we FROWN t = 2 on F() Igun. C. On f () Drgon, the grph goes UP through = ½, so we SMILE t = ½ on F() Igun. D. On f () Drgon, the grph goes DOWN through = 2, so we FROWN t = 2 on F() Igun. Reeber: y-vlues re rbitrry, so you cn put the wherever you wnt so long s your es re bove your ins. 10

11 STEP 2 IDENTIFY THE INFLECTION POINT X-VALUES FOR F() Igun 1. The ius nd inius on f () Drgon, re the inflection points on F() Igun. EXAMPLE f () Drgon F() Igun The purple rrows on f () Drgon indicte the -vlues of the inflection points (purple strs) on F() Igun. Agin, s with the & ins, the y-vlues for the inflection points re your choice so long s they re between the es & ins. STEP 3 Connect the dots to coplete your F() Igun 1. Answer ll relevnt questions regrding F() Igun. EXAMPLE F() Igun Reeber: Intervls re -vlues only! Moving fro left ( ) to right (+ ) 1. F() decresing Wherever ldybug goes downhill. Alwys increse nd decrese between MAX & MIN 2. F() incresing Wherever ldybug goes uphill. 3. F() reltive iniu (- vlue only) = 4, = ½ 4. F() reltive iu (- vlue only) = 2, = 2. F() concve down (frowns) 6. F() concve up (siles) 7. F() inflection points = 3, = 1, = 1 8. degree of F() "degree" is TWO ore thn # turning pts. ( bups ) on f () Drgon, not F() Igun 9. End Behvior: Alwys sile & frown between INFLECTION POINTS 11

12 To Find the Rte, DIFFERENTIATE! How To Do Height/Velocit y/tep/interest Probles Eqn A: QUANTITY: How ny, How uch, How fr, How high, How hot, etc. Eqn B: 1st Derivtive of Eqn. A: RATE, velocity, speed: How Fst, t wht rte, etc. Note: If no eplicit vlue is given for Quntity or Rte, then Tie is probbly zero (t=0) Tips: 1. "On the ground" ens Quntity = 0 2. Ape or verte ens Rte = 0 3. Before n event strts, t = 0 Given: Quntity: "How Much" (i.e. feet, $, etc.) Given: Rte i.e. "How Fst" (i.e. $/yr., fps, etc.) Plug Quntity into Eqn A (Quntity eqn). Solve for Tie, t Plug Rte nuber into Eqn B (1st Derivtive). Solve for Tie, t Were you sked to solve for tie? YES DONE! Were you sked to solve for tie? YES DONE! NO NO Plug t vlue into Eqn B (1st Derivtive) Solve for Rte. Plug t vlue into Eqn A (Quntity eqn). Solve for Quntity. Given: vlue for Tie (t) Do I need "How Fst?" ( rte) YES Plug t vlue into Eqn B (1st Derivtive) Solve for rte. Eple: You invest $00 t 3%. At wht rte is the blnce growing when the ount in the ccount is $200? A t e. 3t A t e. 3t e. 3t A. e e. 3t A. $ yr. ln [ ]. t. yrs. Fster ( Trevor ) Method: (Aount in the ccount t tie, t) * (interest rte) $200 *.03 = $7/yr. Trevor ethod ONLY works with A=Pe rt probles! NO Plug t vlue into Eqn A (Quntity eqn). Solve for Quntity. DONE! 12 Eple: The height of eteor in erth s tosphere is given by: s(t) = 12t t How fst is it going when it strikes the erth? s(t) = 12t t s (t) = 24t + 24 when hits erth height is zero 0 = 12(t 2 2t 120) s (12) = 24(12) = 12(t 12)(t + 10) s (12) = 264 fps 12 seconds = t negtive becuse eteor is heding DOWN 10 seconds = t Silly Answer (neg. tie)