Let's start with the formulas. In fact for our purposes we need only the following three: d e e dx = (1.3)

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Jeffrey Francis
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1 . Diffrntiation Ms. M: So y = r /. And if yo dtrmin th rat of chang in this crv corrctly, I think yo'll b plasantly srprisd. Class: [chckls] Ms. M: Don't yo gt it, Bart? dy = r dr, or rdr, or rdrr. Har-d-har-har, gt it? From: Th Simpsons In this chaptr w rviw som basic comptational tchniqs from calcls. Of cors, it taks mor than mchanical skills to proprly s ths tools. On nds som ndrstanding of th idas bhind th maniplations. W will rviw thos in latr chaptrs as th nd dvlops. Lt's bgin by rcalling som fndamntal tchniqs of diffrntiation. W mak no attmpt hr to jstify ths procdrs. Th comptation of drivativs rlis on two componnts: formlas for th drivativs of lmntary fnctions rls plaining how to diffrntiat fnctions that ar pt togthr from th lmntary ons Lt's start with th formlas. In fact for or prposs w nd only th following thr: (.) d k ( ) k d d d k = (.) ( ) = (.) d (ln ) = d Hr w hav sd th diffrntial notation to indicat th drivativ. Somtims it is mor convnint to s th prim notation ( ) = as shorthand for th drivativ. In (.) it is important to raliz that th ponnt k can b any ral nmbr, not jst a positiv intgr. Eampl.: Find th drivativ of. Soltion: / Th prssion is first convrtd to pr ponntial form =. W now s (.) to carry / 4/ ot th diffrntiation: d ( ) = =. Th last convrsion is optional. Gnrally 4/ d it is mor convnint to work algbraically with a fractional form having no ngativ ponnts, bt this dpnds on th final objctivs.!

2 To procd frthr rqirs th basic rls of diffrntiation. Rl.: Th drivativ of a sm is th sm of drivativs. In symbols: ( f ( ) + g( )) = f ( ) + g ( ). A similar rslt applis to sms of thr or mor fnctions.! Rl.: If c is a constant thn ( cf( )) = cf ( ). In words, th drivativ of a constant tims a fnction is th sam constant tims th drivativ of th fnction.! Ths rls allow on to diffrntiat polynomials as wll as othr combinations of th lmntary fnctions listd abov. Eampl.: Find th drivativ of ach of th following prssions: a) 4 + b) ln + t t + c) t Soltion: a) Th drivativ comptation is a dirct application of Rls. and.. This is somtims rfrrd to as trm-by-trm diffrntiation. d d d d ( 4 + ) = ( ) 4 ( ) + (). d d d d Eqation (.) (sing k = 0 ) and Rl. imply that th drivativ of any constant is zro. Th drivativs of th othr trms ar obtaind dirctly from (.). Ths, d d d d ( 4 + ) = ( ) 4 ( ) + () = 8. d d d d b) Convrt th fractional trm to ponnt form as w did in Eampl.; thn diffrntiat trmby-trm. d d d + ln ( ) (ln ) 4 = + = +. d d d Th last answr may b combind into a singl fraction with common dnominator of = + =. c) Hr th indpndnt variabl is t so w nd to compt d / dt. Althogh th prssion is a fraction, sinc th dnominator is a pr powr w can convrt th fraction into a sm by dividing ach trm in th nmrator by th dnominator t. Not that this is actly th rvrs of th simplification procss w carrid ot at th conclsion of b).. 4

3 t t + t t / = t t t + = +. t t t t Th last prssion can now b diffrntiatd trm-by-trm. d / 9 5/ 9 (t t+ t ) = t t =. 5/ dt t t If dsird, th answr may b prssd as a singl fraction. Th radr shold show that th final rslt is 9t + 4t + 4.! t Rls. and. ar qit straightforward. To handl prodcts and qotints mor laborat rls ar ncssary, sinc nfortnatly th drivativ of a prodct is mphatically not th prodct of th drivativs. Rl.: Prodct Rl ( f( g ) ( )) = f ( g ) ( ) + f( g ) ( )! Rl.4: Qotint Rl d f( ) f ( g ) ( ) f( g ) ( ) = d g( ) g ( )! Th Prodct Rl has a form that is asy to rcall and allows an immdiat gnralization. Each trm on th right sid contains th drivativ of on of th factors on th lft, with othr factors lft ntochd. Ths ar thn addd. Th sam principl works for thr or mor factors. For ampl, ( f( gh ) ( ) ( )) = f ( gh ) ( ) ( ) + f( g ) ( h ) ( ) + f( gh ) ( ) ( ). Th Qotint Rl is not so simpl to rmmbr. Th nmrator rsmbls th right sid of th Prodct Rl. Howvr, bcas of th sbtraction w mst b carfl not to switch th trms. Doing so will giv an incorrct sign for th answr. Eampl.: Compt th drivativ of ach of th following prssions: a) c) 5 ( ln )( ) + + b) d) Soltion: + 5

4 a) W s th Prodct Rl. (Altrnativly, w cold pand th prssion into a sm and diffrntiat trm-by-trm. This is sally a bad ida as it dstroys any bnfit that might accr from th factord strctr of th original prssion.) b) W can writ d (( + ln + )( )) = + ( ) + ( + ln + )5. d =. Now w s th Prodct Rl. d d d ( ) = ( ) + ( ) = + = d d d Th sam rslt will b obtaind blow sing th Chain Rl. (S Eampl.5) c) W can writ =. Now w s th Qotint Rl. d d () ( ) d ( ) d d = = = = d Th rslt may also b obtaind mor asily sing th Chain Rl as in Eampl.5. d) d + ( + )( ) ( + )( ) = d ( ) (.4) W s th rslt in c) (togthr with Rl.) to diffrntiat th trms in th nmrator. W hav ( + ) = and ( ) = +. Sbstitting ths in th right sid of (.4) givs d + ( ) ( + ) = d ( ) This answr can b considrably simplifid. Sqaring th trms in th nmrator givs ( ) = + = +,. sinc 0 = =. Similarly, ( + ) = + + = + +. On sbtracting w gt ( ) ( ) 4 + =. This givs as or final answr 6

5 d + 4 = d ( ) Or final rl, th Chain Rl, is th most powrfl and also th most difficlt to apply and stat. 5 Sppos w want to find th drivativ of y = ( ). Considr first how this prssion is 5 valatd. Th valation procss involvs two stps: ( 5) = = y. In ffct, th variabl srvs as a link btwn th indpndnt variabl and th dpndnt variabl y. Th linking variabl is a dpndnt variabl in th dfining qation = and an 5 indpndnt variabl in th qation y =. If w know how to diffrntiat ach of ths links, th Chain Rl shows s how to find dy / d. Rl.5: Chain Rl Sppos y is a fnction of a variabl, and is in trn a fnction of th indpndnt variabl. Thn dy dy d =.! d d d Obsrv th powrfl symbolism. Th drivativs that ar mltiplid on th right sid appar to symbolically yild dy / d throgh cancllation of th prssion d. Althogh not mathmatically accrat, this is a sfl fiction for rmmbring and gnralizing th rslt (S Eampl.4d blow). Eampl.4: Find th drivativs of th following prssions. a) y 5 = ( 5) b) y =.! c) y = ln( + ) d) y = ln Soltion: a) Introdc th intrmdiary variabl = 5. Th Chain Rl givs = 5, as discssd abov. W can writ y 5 =, and dy dy d = = 5 (6 ) = 5( 5) 6= 0( 5). d d d Not that in th prssion dy / d it is ncssary to sbstitt for its dpndnc on. Th final answr shold only involv th variabl. 7

6 b) Introdc th intrmdiary variabl givs =. W thn hav y = and =. Th Chain Rl dy dy d = = =. d d d c) W hav y = ln, whr dy = dy d = = d d d +. = +. From th Chain Rl d) W hav y =, whr = w, and w= ln, a chain with two links. Th Chain Rl tnds to mltipl links. As indicatd abov, th notation gids yo to th corrct rslt. ln dy dy d dw / = = w = = d d dw d w ln.! In Eampl.4 w followd fairly closly th pattrn stiplatd in th statmnt of th Chain Rl. In practic, introdcing intrmdiary variabls plicitly oftn maks th work nwildy. To mak it asir to prform ths calclations withot having rcors to th tra variabls, it is sfl to kp in mind som spcial cass of th Chain Rl. Onc on mastrs ths spcial pattrns th gnral sitation will sm qit transparnt. Rl.6: Spcial Chain Rl a) If is a fnction of thn d k k ( ) k d =. d b) If is a fnction of thn ( ) =. d d c) If is a fnction of thn (ln ) =.! d Ths of cors all follow from th Chain Rl. For ampl, to dmonstrat c) writ Thn y = ln. d (ln ) dy dy d = = = =. d d d d In ffct, to find th drivativ of ln with rspct to w first diffrntiat ln as if wr an indpndnt variabl, obtaining /. To tak accont that in fact dpnds on w mltiply th rslt by, so in this contt plays th rol of a dpndnt variabl. A similar intrprtation applis to th othr statmnts. 8

7 Eampl.5: Compt drivativs of th following prssions: a) b) c) (ln( + ) ) 5 Soltion: a) Hr = so w find sing Rl.6b) that.b) d d ( ) = ( ) = (S also Eampl d d d d b) ( ) = ( ) = (S Eampl.c) d d c) Sinc th prssion is a 5 th powr w first w s Rl.6: d d d d 5 4 ((ln(+ ) ) ) = 5(ln(+ ) ) (ln(+ ) ) Thn w mst compt th drivativ of th prssion that is insid th powr. Hnc, or final answr: d (+ ) 6 (ln( + ) ) = = d + +. d 5 0(ln(+ ) ) ((ln( + ) ) ) = d + 4.!. Antidiffrntiation or Indfinit Intgration Givn a fnction f( ), can w find a fnction F( ) with th proprty that F ( ) = f( )? Sch a fnction F( ) is calld an antidrivativ of f( ). In gnral, finding an plicit formla for an antidrivativ of f( ) is mch mor difficlt than diffrntiating th sam fnction. In fact, an plicit antidrivativ in trms of lmntary fnctions may not ist. In this sction w will considr som tchniqs for finding antidrivativs. Eampl.6: Find all antidrivativs of th fnction f( ) =. Soltion: W can asily gss an answr sing what w know abot diffrntiation. If w diffrntiat w gt, which is not qit what w want. Howvr, dividing or initial gss by will do th trick, as th nwantd will b cancld. Namly, 9

8 d =. d Having fond on antidrivativ w asily constrct infinitly many in th form C +, whr C is an arbitrary constant. In fact, according to a basic thorm in calcls, all antidrivativs can b obtaind from any on in this way.! It is convnint to hav a notation for th antidrivativs of a fnction, whthr or not w can find an plicit formla. Bcas antidiffrntiation has a fndamntal connction with th valation of dfinit intgrals, th intgral sign, withot limits of intgration is sd to rprsnt th antidrivativs of a fnction. In this notation, th comptation abov wold b indicatd by writing d= + C. Th symbol is an intgral sign and d is calld an indfinit intgral. Ths indfinit intgration is simply anothr trm for antidiffrntiation. Th prssion d that is part of th intgral notation hlps kp track of th variabl of intgration, as w will considr blow whn w discss changing that variabl. It also ariss bcas of th connction with dfinit intgrals. Althogh not absoltly ssntial, it is bst not to omit it. Evry rl of diffrntiation whn viwd in rvrs givs ris to a rl of antidiffrntiation. For instanc, corrsponding to th basic formlas (.), (.) and (.) w hav (.5) k + k d= + C, if k (.6) k + d C = + d = + C (.7) ln As with any prportd indfinit intgration, th answrs can b chckd by diffrntiation. Hr, th right sids diffrntiat to giv th prssion following th intgral sign. In (.5) obsrv th rstriction on k, sinc clarly th right sid is not vn dfind whn k =. Th cas k = is covrd by (.7). Not that to b tchnically corrct, whn is ngativ th antidrivativ of th fnction f( ) = / is ln, not ln. Rl. and Rl. giv ris to corrsponding principls of intgration..! Rl.7: ( f ( ) + g( )) d = f ( ) d + g( ) d This statmnt mans that an arbitrary antidrivativ of th sm f( ) + g( ) can b obtaind by adding antidrivativs of th smmands f( ) and g. ( ) From Rl., if F ( ) = f( ) thn 0

9 ( cf( )) = cf ( ) = cf ( ) so cf( ) is an antidrivativ of cf ( ). Writing this statmnt in th intgral notation yilds.! Rl.8: cf( d ) = c f( d ) Oprationally, this rl is sally invokd by saying that a mltiplicativ constant can b passd from ndr th intgral sign. Eampl.7: Compt ach of th following antidrivativs: a) ( + ) d b) Soltion: a) Using Rl.7 w hav + d From Rl.8 and (.5) w hav ( + ) d = d + d d. (.8) / / d = d = + C = + C *. Similarly th last two intgrals in (.8) valat as d= + C and d = + C!. Adding ths w can combin th thr arbitrary constants into a singl constant D. Ths or final answr is 4 / ( + ) d= + + D. b) W split th fractional intgrand into a sm of trms with dnominator and thn intgrat ach trm as w did in a). + d = d ln C + = + +.! Th Prodct Rl sd in rvrs givs ris to an intgration tchniq calld intgration by parts, which w will not prs. Rvrsing th Qotint Rl is too rstrictiv to b of practical s. Th Chain Rl lads to a fndamntal tchniq of intgration, known as chang of variabls. W will dscrib how this works only in th contt of th spcial forms dscribd in Rl.6. For instanc, Rl.6b) implis that 4 d d =. (.9)

10 Thrfor, th antidrivativ of th right sid of (.9) is th fnction, or d = + C. Th factor in th intgrand disappars in th answr for th sam rason that th sam factor appars in th diffrntiation (.9). Both ar consqncs of th Chain Rl. W can thn rintrprt Rl.6 in th langag of intgration. Rl.9: Sppos is a fnction of, thn a) b) k+ k d = + C, providd k. k + d = + C. c) d = ln + C.! Applying ths rls rqirs that th intgrand hav a spcial form. In any spcific intgration th difficlty lis in matching th intgrand to on of ths tmplats, prhaps aftr making som prliminary maniplations. To do this with assranc and fficincy rqirs som practic. Th basic stratgy is to idntify th part of th intgrand that plays th rol of and to vrify that th factor appars in th appropriat plac. For typs a) and b) this will oftn b clar bcas prssion appars in a distinctiv part of th intgrand. For typ c) th slction of may b mor difficlt (s Eampl.8c). Eampl.8: Evalat ach of th following indfinit intgrals: a) + d b) Soltion: d c) d ln / + d = ( + ) d. This sms to fit th pattrn of Rl.9a) with = + and =. With this slction for w hav a) W hav " / / / + d = d = + C = ( + ) + C. #$%$& b) Th ky obsrvation is to s that =, so th nmrator is actly th drivativ of ( ) th dnominator. Hnc w can apply Rl.9c) with =. This givs d = d = ln + C = ln + C.

11 c) This is mor difficlt. Yo mst first notic that (ln ) = /, so with = ln th intgrand may b writtn = = =. ln ln d Thrfor, = d = ln + C = ln ln + C ln.! Th ampls w jst solvd appar to rqir a spcial strctr, which is indd corrct. Howvr, w can dviat from that strctr a bit and still carry ot th intgration. Eampl.9: Evalat + d. Soltion: Procding as in Eampl.8a), w lt = +. Sinc = th intgrand is missing th factor from th drivativ. Howvr, w can insrt th missing factor and, so as not to chang th val of th intgrand, compnsat by mltiplying th intgrand by th rciprocal /. Rl.8 allows s to pass th tra constant factor of / otsid th intgral. Ths w hav, Factor / cancls th insrtd factor of. Factor of maks this. + d = + d = + d. Th last intgral was valatd in Eampl.8a). Using th antidrivativ obtaind in that rslt w hav 9 / / + d= ( + ) + C = ( + ) + C.! Th valation tchniq dscribd in th prvios ampls can b pt into a mor standardizd form by making s of th diffrntial trm d, which ntil now appars as an annoying ncmbranc. Th tchniq for doing this is calld th Mthod of Sbstittion. Jst as in th Chain Rl of diffrntiation, th nw sbstittion variabl will fnction both as an indpndnt and dpndnt variabl. Eampl.0: Us th Mthod of Sbstittion to valat + d. Soltion: As abov w lt =. = +. W nt compt d, th diffrntial of, dfind as d d

12 = + d = d Th ida of th mthod is to prss th ntir original intgrand (inclding th diffrntial trm d ) in trms of and d. Clarly + d = / d /, sinc d = d/. Ths, sing Rl.8 w hav / / + d = d/= d. So far has bn thoght of as a dpndnt variabl. Now in th last intgral w trat as an indpndnt variabl, sing (.5) to do th valation. This givs 9 / / / + d = d = + C = ( + ) + C.! To smmariz, th Mthod of Sbstittion mploys for stps: i) Slction of th nw variabl and th qation that rlats it to th original intgration variabl. ii) Transformation of th ntir original intgrand, inclding th diffrntial trm, to an prssion involving th nw variabl and its diffrntial d. iii) Evalation of th nw intgral, with as th intgration variabl. iv) Back-sbstittion for in trms of to prss th soltion of th original problm. Eampl.: Try sing th Mthod of Sbstittion to valat + d. Soltion:! If w lt = + thn d = d. Eprssing d in th lattr qation in trms of d, w hav d / d d =. Th intgral wold thn bcom. Unfortnatly, th in th dnominator is not a constant and so w cannot rmov it from ndr th intgral sign. Using th dfining qation = +, th tranos trm cold b liminatd. Indd solving th lattr qation for givs = which can b sbstittd for th in th dnominator. Ths w wold / arriv at th corrct statmnt that with = +, + d= d. Unfortnatly, th lattr intgral dos not appar any asir to valat than th original and w mst concld that 4

13 th mthod has faild. It trns ot that a diffrnt sbstittion, involving trigonomtric fnctions, sccds in this cas. Howvr, w will not prs th dtails.! Rl.9 is oftn prssd in diffrntial form, which maks it asir to rmmbr. In fact th statmnts appar idntical to (.5), (.6), and (.7), cpt th qantity is in rality a dpndnt variabl, notationally disgisd to rsmbl an indpndnt on. Rl.0: If is a fnction of thn (.0) k+ k d= + C, if k (.) k + Of cors sinc d. Smmary d C = +! d = + C (.) ln = d ths formlas mrly rpat Rl.9 with a diffrnt notation. Th following tabl smmarizs th diffrntiation and intgration rls discssd in this rviw. Th tabl is arrangd so that paralll rls for th two procsss appar adjacnt to ach othr. Rl # Diffrntiation Intgration Rl. & Rl.7 ( + v) = + v ( + v) d = d + v d Rl. & Rl.8 ( c) = c, c constant c d = c d Rl. ( v) = v + v Intgration by parts (not discssd) Rl.4 Rl.6 & Rl.0 v v = v v d k k k d = Not sd k+ k d = + C, k, k + whr d = d Rl.6 & Rl.0 Rl.6 & Rl.0 d d = = + d C d d ln = ln C d = +.4 Erciss In rciss to 8 compt th drivativ of th givn prssion. Try to simplify yor answr, if it sms appropriat. 5

14 ln 4. ln 5. (ln ) ln 0. ( ) ( + ) (+ ) + t 5. ln 4. ln(ln ) / 7. k, k a constant 8. In rciss 9 to 7 compt th indfinit intgral. ln( + ) t 9. ( + ) d 0. ( + ) d. d. 5. d. ( + ) d 6. 5 k d, k a constant 4. ( + ) d + d d 7. ( + ) ( + ) 8. + d ( ) 9. + d (*Hint blow) 0. + ln d. d. / d. + d d d d 6. + ( + ) *(Hint: Using long division writ th intgrand in th form B A +, for som constants A and B.) + 4 d 6

Higher order derivatives

Higher order derivatives Robrto s Nots on Diffrntial Calculus Chaptr 4: Basic diffrntiation ruls Sction 7 Highr ordr drivativs What you nd to know alrady: Basic diffrntiation ruls. What you can larn hr: How to rpat th procss of