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 diffrntiation to obtain drivativs of drivativs, that is, highr drivativs. Onc w hav computd th drivativ of a function f ( ), w nd up with anothr function f ( ). Why not tak th drivativ of this nw function, and thn its drivativ and so on? Bcaus w hav bttr things to do with our tim! But what if that rptition provids somthing usful? And it dos, so lt us st th notation and trminology for this simpl concpt. Dfinition Th drivativ of th drivativ of a function f ( ) is calld th scond drivativ of that function and is dnotd by on of th symbols: f ( ) y d y d Thr is a rason for that, but it is linkd to crtain oprations that ar don in advancd calculus. For now, think of it just as a strang quirk with which you nd to liv! And bing ambitious, w do not stop hr. Dfinition Th drivativ of th scond drivativ of a function f is calld th third drivativ of that function ( ) and is dnotd by on of th symbols: f ( ) y d y Now th gam can b continud, so as to dfin th fourth, fifth,, n-th drivativ of a function, for any intgr numbr n. d Why is th ponnt of placd diffrntly on th numrator and dnominator? Diffrntial Calculus Chaptr 4: Basic diffrntiation ruls Sction 7: Highr ordr drivativs Pag 1
Dfinition Rpating th procss of diffrntiation n tims gnrats th n-th drivativ of a function, which is dnotd by on of th symbols: ( n f ) ( ) ( n) y n d y d If n, thn th n-th drivativ is also said to b a highr drivativ of ordr n. Now you hav changd th notation again: numbrs in brackts? I d lov to tak crdit for this notation, but it is on that has bn usd univrsally for a long tim. Aftr th third drivativ, w would nd to insrt many prim symbols and thy would b difficult to count. Adopting th notation of ponnts in brackts kps things asir. OK, but what is all this stuff usful for, bsids bing a prvrsly fun gam? It turns out that ths highr drivativs hav many concrt manings in applications. Rmmbr that a drivativ indicats a rat of chang, so ach highr drivativ rprsnts th rat of chang of th prvious on. Th most common and famous applications of this ida ar for graphs and motion functions. Tchnical fact Th scond drivativ of a function f ( ) rprsnts th rat at which th slop changs. This is calld th concavity and w shall analyz it latr in mor dtail. n Proof Tchnical fact If () t is a function of tim that dscribs th position of a moving objct, thn: 1. Th first drivativ () t vlocity of th objct.. Th scond drivativ () t acclration of th objct.. Th third drivativ () t of th objct. rprsnts th rprsnts th rprsnts th jrk Just think of th maning: th scond drivativ tlls us how fast th vlocity changs and that is acclration. Th third drivativ tlls us how fast th acclration changs and that tlls us how jrky th motion is. How to us vn highr drivativs will bcom apparnt in th study of infinit sris. I can wait! For now, can w s how th gam is playd? Eampl: f ( ) + 1 To comput th first drivativ of this function, w bgin with th quotint rul, followd by addition, powr and constant ruls for th numrator, and linar for th dnominator: Diffrntial Calculus Chaptr 4: Basic diffrntiation ruls Sction 7: Highr ordr drivativs Pag
( ) f / / ( + ) ( 1) ( 1) ( + ) ( 1) ( ) ( + ) 1 ( 1) Bfor computing th scond drivativ, w bttr rarrang th numrator to mak our job asir: f '( ) 1 ( ) Now w us th sam procdur (quotint rul followd by othrs for ach brackt) to gt th scond drivativ: ( ) f / / ( ) ( 1) ( 1) ( ) ( 1) ( )( 1) ( 1)( ) ( 1) Rarrangmnts of this fraction ar possibl and asy, but not rquird, unlss w nd to comput th third drivativ, somthing that you may want to do as a furthr rcis for you. f + Eampl: ( ) ( ) 5 By using th chain, powr and linar ruls w can comput th first fw drivativs of this function asily: ( ) ( ) 4 f 5 + You may b tmptd to combin th cofficints, but princ tlls us not to do that yt, and hr is why. Lt us look at th nt fw drivativs: f ( ) ( + ) ( + ) ( ) ( ) ( ) 4 4 5 4 5 4 f 5 4 + 5 4 + Thrfor, by kping th cofficints sparat w can idntify a pattrn for ach group of factors that may simplify th notation. W kp going: Rmmbr that 4 and 5 rspctivly. f ( ) ( ) ( ) 4 4 f 5 4 + ( 4 ) ( ) ( ) ( ) 5 5 f 5 4 and ( n ) ( ) 0 5 f if n f ( 5 ) ( ) ar calld highr drivativs of ordr Th product of succssiv intgrs that you s in this ampl is somthing that oftn appars whn computing highr drivativs. Mathmaticians hav agrd on a spcial symbol for that, on that you may hav sn bfor. Dfinition For any positiv intgr n, th product of th first n intgrs is dnotd by n! : n! 1 n and it is calld th factorial of n. For a numbr of rasons, both thortical and practical, w assum that, countr to intuition: 0! 1 I can s th us of th notation, but to say that 0!1 maks no sns! I may agr, cpt for th fact that it dos! It turns out and it is not difficult to chck that th numbr of ways of placing n objcts in ordr is n! Wll, how many ways do w hav of ordring NO objcts? Just on, right? But if this sms lik a twist of logic (to s that not having anything to ordr is on way to ordr), hr is anothr rason rlatd to highr drivativs. Diffrntial Calculus Chaptr 4: Basic diffrntiation ruls Sction 7: Highr ordr drivativs Pag
Eampl: f ( ) 1 1 In ordr to comput th drivativs of this function, it is convnint to writ it as a ngativ powr, so as to avoid th quotint rul. ( ) ( 1) 1 f Now lt us comput its highr drivativs and kp th cofficints sparat, as w did bfor: f ( ) 1( 1) ( ) ( ) ( )( ) 1 1 ( ) ( )( )( )( ) 4 1 1 f f ( ) ( ) ( )( )( )( )( ) 5 4 4 f 1 4 1 Notic that if w lt 0 ( n ) n ( ) ( ) ( )( ) n 1 f 1 n! 1 n and assum 0!1, this formula bcoms: ( 0 ) 0 ( ) ( 1) ( 0! )( 1) ( 1) 0 0 1 1 f So th 0-th drivativ is th original function, as it should b! You will s mor ampls of this mthod in th futur. For now, it is tim to practic on th basics you hav sn hr. n Notic that w hav a pattrn dvloping, sinc thr ar: As many ngativs as th ordr of th drivativ A factorial qual to th ordr of th drivativ An ponnt givn by th ngativ of th ordr minus 1 A powr of qual to th ordr of th drivativ. This pattrn will continu, so that: Summary Highr drivativs ar obtaind by succssivly computing th drivativ of a lowr ordr drivativ. Th ordr of a drivativ rfrs to how many tims diffrntiation has bn prformd, starting from th original function. For simpl functions, highr ordr drivativs may dvlop a pattrn that can b summarizd in a singl formula, oftn including a factorial numbr. Common rrors to avoid Whn looking for a pattrn for th highr drivativs of a function, don t stop too soon: you may nd at last 5-6 drivativs bfor it bcoms clar. Diffrntial Calculus Chaptr 4: Basic diffrntiation ruls Sction 7: Highr ordr drivativs Pag 4
Larning qustions for Sction D 4-7 Rviw qustions: 1. Dscrib what a highr drivativ is.. Dscrib th notation usd for highr drivativs. Mmory qustions: 1. Prsnt two corrct notations for th scond drivativ of a function f ( ). Prsnt two corrct notations for th n-th drivativ of a function f ( ). What is th physical maning of th scond drivativ of a position function? 4. What information dos th scond drivativ contain about th graph of a function? 5. Which formula rprsnts n! as a product? 6. What is th valu of 0!? Computation qustions: In qustions 1-, comput th scond and, if not too mind-boggling, third drivativ of th givn function. 1.. 1 + 7 7 1 +. 7 4 4. 7 4 + 5. + 5 + 6. 7. 8. 9. + 1 4 Diffrntial Calculus Chaptr 4: Basic diffrntiation ruls Sction 7: Highr ordr drivativs Pag 5
1/ 10. ( ) 5 11. ( ) 4 1. y 1. 5 4 + 15. 16. 17. 18. 6 + + ( 1) 5 + 5 5 + + 5 5 0. 1.. 5 4 + 1 1 5 5 14. 6 19. In qustions -9, dtrmin th gnral formula for ( n f ) ( ).. f ( ) ln 4. f ( ) 1 + 5. f ( ) 5 6. f ( ) 7. ( ) ( ) f 5 8. f ( ) f 9. ( ) Thory qustions: 1. What is th 100 th drivativ of y 50?. Which functions hav a scond drivativ qual to 0?. What dos th 4 th drivativ of a function tll us about its rd drivativ? 4. If th product rul is ndd to comput th drivativ of a function, which othr rul will b ndd to comput th scond drivativ? 5. If th first drivativ of a function rquirs th chain rul, which rul will crtainly b appropriat for th scond drivativ? Diffrntial Calculus Chaptr 4: Basic diffrntiation ruls Sction 7: Highr ordr drivativs Pag 6
6. What is th rlation btwn factorials and highr drivativs? 7. ( n + ) What quadratic prssion rprsnts th valu of ( n ) 1! 1!? 1. Assum that ( ) final answr. Proof qustions: f is a function with both first and scond drivativ. What is th scond drivativ of th function y f ( ). Whn computing th gnral formula for? Collct lik trms and factors in th ( n f ) ( ), you may run into a product of th form 5 7 ( n 1), that is, th product of th first n odd numbrs. For our purposs, it is nough to dnot such product as I hav just don, but you may wondr if thr is a factorial-basd way to rprsnt this numbr. Of cours it is NOT ( n 1 )!, sinc such formula includs all vn numbrs, and som authors dnot it by ( n 1 )!! (doubl factorial). But thr is a nic formula for it: ( n 1 )! 1 5 7 ( n 1) n 1 ( n 1 )! For som wknd fun, you task is to show that this formula is corrct by using th mthod of induction, that is, by showing that: a) Th formula works for n1. b) If th formula works for any n, it also works for n + 1. Application qustions: 1. What is th acclration of an objct moving on th ais so that its position at tim t is t 5?. An objct falls through a forc fild so that its hight is givn by th function ( ) ht 0 ( + t) in mtrs and sconds. Comput th formula that provids th acclration of this objct in trms of tim. What qustions do you hav for your instructor? Diffrntial Calculus Chaptr 4: Basic diffrntiation ruls Sction 7: Highr ordr drivativs Pag 7
Diffrntial Calculus Chaptr 4: Basic diffrntiation ruls Sction 7: Highr ordr drivativs Pag 8