Calculus concepts derivatives

Size: px

Start display at page:

Download "Calculus concepts derivatives"

Branden Boone
6 years ago
Views:

1 All rasonabl fforts hav bn mad to mak sur th nots ar accurat. Th author cannot b hld rsponsibl for any damags arising from th us of ths nots in any fashion. Calculus concpts drivativs Concpts involving diffrntiations luvis /9/0

2 Calculus Concpts-Diffrntiation Calculus is usd in mathmatics to solv problms such as dtrmining th gradint of crtain functions, dtrmin th rat of chang and solv maimum and minimum problms. It is an trmly important topic that has many applications. THESE NOTES DO NOT COVER ALL THE CONCEPTS OF CALCULUS BUT JUST A BASIC INTROUCTION. THEY DO NOT SHOW HOW TO USE GRAPHICS CALCULATOR Concpt- Using first Principls Calculus hlps us dtrmin th gradint (slop) of a curv at a particular point. Tak th graph y y Say Point A has th following coordinats (, ) (,9) And say Point B has th following coordinats (( h),( h) ) ( h,( h) ) Now if w would lik to dtrmin th gradint btwn ths two points w us th formula gradint Now follow th logic and s what happns: Gradint AB AB AB ( h) ( ) ( h) () ( h) (9) ( h) () (9 6 h h ) (9) 6h h AB h h(6 h) AB h AB 6 h ris run ( h) () ris run Pag of 6

3 Now w hav th gradint of th lin joining ths points togthr bing AB 6 h If now w mad h 0 thn AB 6 This mthod is th bginning of calculus. It is calld finding th drivativ from first principals, in othr words finding th quation of th slop for a particular function, in th abov th quation of th parabola. f ( h) f ( ) f( ) lim h0 h This is th main quation w us whn w want to find from first principals th gradint of a particular function. Eampl Find from first principals th drivativ of f ( ) Solution: Stp First w find f ( h) ( h) h Stp Nt stp w follow th quation for first principals f ( h) f ( ) f( ) lim h0 h ( h ) ( ) f( ) lim h0 h h f( ) lim h0 h f( ) lim h0 Now w lt h 0,but sinc thr is no h f( ) Eampl Find th drivativ of f ( ) from st principals. Stp : Find f ( h) f ( h) ( h) ( h)( h)( h) ( h h )( h) ( h h h ) Notic how I lft th brackts in to hlp m with th nt stp Stp : Put it into th gnral quation and work out slowly Pag of 6

4 f ( h) f ( ) f( ) lim h0 h f( ) lim h0 ( h h h ) ( ) ( h h h ) f( ) lim h0 h h h h f( ) lim h0 h f h h ( ) ( ) lim( ) h0 Now w lt h 0 f ( ) h Skill Buildr Qustion Using First Principls find th drivat of th following functions Answrs f ( ) f( ) f ( ) 5 f( ) f ( ) f ( ) 4 f ( ) 5 f ( ) 5 f ( ) f ( ) Concpt - Finding th drivat of functions fastr Thr is an asir way to find th drivativ of th function abov and that is by using th following rul: n If f ( ), whr n is a positiv or ngativ numbr thn f ( ) n ( n ) If th function starts with f( ) thn th drivat can b shown as f ( ) whras if th function starts with y thn th drivat is shown as Eampl Find th drivat for th following: Pag of 6

5 f ( ) 5 4 f ( ) (4) (5) f ( ) 0 (4) () Skill buildr Qustion Using th shortcut mthod find th drivativ Answrs f ( ) f( ) f ( ) 5 f( ) 5 f ( ) f ( ) 4 4 f ( ) 5 f ( ) 5 5 f ( ) f ( ) 6 f ( ) 4 4 f ( ) 6 7 f ( ) f ( ) 8 4 f ( ) 7 f ( ) f ( ) f ( ) 7 0 f( ) f( ) 0 6 Concpt-- Finding th gradint at a particular point of a function So what if w ar askd to find th gradint at a particular point, lt s say = of th function do w go about it? f ( ), how Answr Stp Equation Writ down th quation f ( ) Tak th drivativ using th short cut f ( ) 4 Substitut = into th drivat to find th valu of f () 4() th gradint at that particular point So th gradint at = is Skill buildr Qustion Using th shortcut mthod find th drivativ Find th gradint at = f ( ) f( ) f ( ) 5 f () 0 f ( ) f( ) f ( ) f ( ) 5 f ( ) 8 f ( ) f ( ) 4 4 f( ) 6 Pag 4 of 6

6 Concpt-4- Rcognizing how to rwrit crtain quations into an asir form so you can tak th drivativ. Normal form Easir to us form Drivativ y y y y y y Eampl 4 Problm: Find th gradint at (, ) for th graph Solution: f ( ) W must find th drivat of th function firstly and thn substitut = into th drivat to find th gradint. But hr w nd to b carful! Stp : Put th function into an asir form f ( ) f ( ) f ( ) f ( ) ( ) Now substitut = f () () () f () f () Skill Buildr Find th gradint at = for th graph f ( ) Find th gradint at for th graph f ( ) Pag 5 of 6

7 Concpt 5- Using th chain rul y u n du du Usd to diffrntiat composit functions in othr words whr on function is insid as it wr of anothr function 4 Eampl of composit functions: y ( ) How do w us th chain rul? Lt s look at a stp by stp ampl find of 4 y ( ) y ( ) 4 du du 4u 6 4( ) 6 Always call th insid function u 4 u y u du 6 4u du Whn doing ths typs of qustions you will nd to b trmly carful and do not forgt th brackts and st it out as abov. Find of 5 4 y ( ) du du 4 4 u (6 0 ) 5 4 4( ) (6 0 ) Always call th insid function u 5 4 u ( ) y u du u du Pag 6 of 6

8 Skill Buildr Qustion Using th chain rul find th drivativ Answrs f ( ) 4 f ( ) 5 f ( ) 5 f ( ) 5 4 f ( ) f ( ) Concpt-6- Using th Product Rul Usd to diffrntiat two functions that ar multiplid togthr y u v whr u and v ar functions Watch out somtims you will hav to us th chain rul along with th product rul so things can gt a littl long and confusing y uv dv du u v Workd out ampl Find of y ( )( ) Always call th insid function u Call th othr function v u v du dv ( ) dv du u v ( )( ) ( )( ) ( )( ) ( )( ) W can obviously simplify th abov answr and I will lav it up to you to obtain a simplr answr. Pag 7 of 6

9 Concpt 7- Combination qustions involving chain rul and product rul Find of 4 y ( 5) Stp - Us Product rul y uv dv du u v dv du u v 4 (( 5) ) ( 5) ( ) 4 ( 5) ) ( )( 5) ( 5) ( ( )( 5)) ( 5) ( 5 ) u v( 5) du dv? Nd to us th chain rul dv ( 5) 4 Us on th chain rul on a sparat part and you will find that th Concpt-8- Quotint Rul Usd whn on function is dividd by anothr function Hr if you ar not carful of ngativs you will dfinitly mak a mistak u y, v 0 v du dv v u v Workd out ampl Find of Always call th top function u Call bottom function v Lt u v du dv Pag 8 of 6

10 du dv v u v ( ) (( ) Watch out with Ngativs and do not forgt th brackts Concpt-9- Drivativs of Eponntial functions Find of y k Now w us th chain rul du du Putting vrything in w gt th following u k k k Always call th top function u Lt u k so th function thn bcoms du k Function bcoms y u Now th drivativ of th ponntial function is th ponntial function u This rul abov is only valid for ponntial functions of bas If you ar askd to diffrntiat ponntials othr than bas such as to work out th valu at a crtain point. Eampl If f( ) find f () Solution: Us graphics calculator: ndriv (^,, ) =.958 y you thn must us your calculator Pag 9 of 6

11 For composit ponntial functions i.. y f ( ) us chain rul. Eampls to s how th rul is usd for ponntial functions Problm Find of y ( ) Workd out solution Us th chain rul Lt u du 6 Lt y u du u Now using th chain rul du du u (6) (6) Skill Buildr Qustion Using th chain rul find th drivativ Answrs f ( ) f ( ) 5 5 f ( ) 4 5 f ( ) 5 f ( ) 7 ( ) f ( ) Pag 0 of 6

12 Concpt-0- Drivativs of logarithmic functions Find of y log( k ) Follow th stps on th right Normally with practic you will b abl to do this far quickr by rmmbring th dfinition. Always call th top function u y k Now w divid it by k y k k k y k Now w tak th drivativ y k But k y so k k So flipping th drivativ ovr ( not supposd to do this) If askd to find th drivativ of a logarithmic function othr than bas must us th graphics calculator (Lik th ponntial cas) Find of log ( ) Chain rul application now du du (6 ) u 6 Us th chain rul u du 6 y log u du u Pag of 6

13 Skill Buildr Qustion Using th chain rul find th drivativ Answrs f ( ) log ( ) f ( ) log (4 ) f ( ) log (5 ) 4 f ( ) 4 log ( ) 5 f ( ) log ( ) 7 f ( ) log ( ) Concpt-- Drivativs of trigonomtric functions y sin k k cos k y cos k k sin k y tan k k sc k Somtims w can onc again nd to us th chain rul whn tackling qustions in trigonomtric functions y sin( f ( )) f ( )cos( f ( )) y cos( f ( )) f ( )sin( f ( )) y tan( f ( )) f f ( )sc ( ( )) Trigonomtric functions can only b diffrntiatd if th angl is in RADIANS! If is in dgrs thn w should chang it to radians. Find if y sin Solution: Convrt dgrs to radians Pag of 6

14 80 y sin( ) 80 cos( ) Now w can convrt back to dgr form cos 80 Anothr Eampl Solution Find of y sin( ) Rcogniz w must us th chain rul hr Lt u du 6 Now w hav y sin u cosu du Chain rul du du cos u(6) (6 ) cos( ) WATCH OUT IF YOU ARE ASKED TO TAKE THE DERIVATIVE OF SOMETHING THAT LOOKS LIKE THIS y sin y (sin ) Find of y sin First rcogniz that y sin y (sin ) Now w apply th chain rul Using th chain rul lt u=sin du cos y u u du Pag of 6

15 du du u(cos ) sin cos Concpt--Combinations of th various ruls W can hav qustions that will forc us to us combinations of th various ruls such as th chain rul along with th product rul. It is important in ths instancs to st out th qustion natly and procd mthodically in solving ths qustions Eampl Find of y sin 4 First rcogniz that this involvs th product rul y sin y (sin ) Now w apply th chain rul y uv dv du u v Substituting all th information vry carfully w gt th following 4cos 4 sin 4 Skill buildr Using th product rul lt u= du v sin 4 dv 4cos 4 Qustion Using th chain rul find th drivativ Answrs f ( ) 4 5 f ( ) ( ) 5 5 f( ) f ( ) ( )log ( ) 5 f ( ) Pag 4 of 6

16 Concpt - Application of Calculus-Equations of straight lins To find th quation of a tangnt to a curv y f ( ) at point (, y ) us th quation y y m( ), whr m f ( ) To find th quation of a normal (th lin prpndicular to th tangnt) to th curv y f ( ) at point (, y ) us th quation y y ( ), whr m f ( ) m Find th quation of th tangnt to th curv y at point = Stp - Find th coordinats of y at = Stp - Tak th drivat of th curv Find th gradint at = Th quation of th tangnt is y () () y y () 4 y y m( ) y 4( ) Skill Buildr Qustion Find th quation of th tangnt to th curv y at point = Answrs Find th quation of th normal to th curv log y at 0. 4 Find th quation of th normal to th curv Find th quation of th normal to th curv y at point = y 6 at point = 5 Th lin y is a tangnt to th parabola y p q. Find th valus of p and q. Pag 5 of 6

17 Concpt 4- Application of Calculus-Stationary Points A Stationary point is any point on a graph y f ( ) whr f( ) 0 Thr ar basically typs of stationary points. Local Minimum Point. Local Maimum Point. Stationary point of inflion Stationary Points Graphs Local Minimum Point Notic at = w hav a stationary point which is a minimum < a f ( a) <0 = a f( a) 0 > a f ( a) >0 Notic hr th following Maimum occurs at =- < b f () b >0 = b f () b =0 > b f () b <0 Pag 6 of 6

18 Stationary Point of Inflion Notic that somthing uniqu occurs at (-, ) this is a point of inflion <c f () c >0 = c f () c =0 > c f () c >0 Anothr typ of inflction point <c f () c <0 = c f () c =0 > c f () c <0 POINTS OF INFLEXION WILL HAVE A DERIVATIVE WITH A SQUARED FACTOR IN THEM Eampl: ( ) thus = point of inflion If askd for th coordinats and th typ of stationary points thn you will nd to do th following:. Find f( ) 0. Tst ithr sid of th stationary point for th valus of th gradint. It probably is bst to s how th graph looks and it is asy to s what typ of stationary point it is. Find all stationary points of th following graph f ( ) 5 Stp - Always a good ida to graph it and s what is actually happning Pag 7 of 6

19 Stp - You can s immdiatly whr th stationary points ar: =0 (ma) and =.5 (min) Stp - Tak th drivativ of th graph And st th drivativ to zro to find th stationary points Now w nd to show that for ampl = 0 is a ma point < 0 f ( a) >0 f ( ) 5 f ( ) 6 0 Lt f( ) ( 5) or Chos a valu on th lft of =0 say = - and put that into th quation for th drivativ and s th valu it givs you. f ( ) 6 0 6( ) 0( ) > 0 f ( a) >0 f ( ) 6 0 So do s that th gradint gos from a positiv to a ngativ thus this mans that = 0 is a maimum. (W always considr from lft to right) 6() 0() X = 0 is a maimum point thrfor as w can clarly s from th graph abov Rpat abov procdur for othr point to show it is a minimum If w try th sam procdur for th othr stationary point w will discovr that it is a minimum. Is thr a quickr way of working out th natur of th stationary points? Graph thm! Pag 8 of 6

20 Skill Buildr Find th stationary points of f ( ) Find th stationary points of f ( ) ( 4)( )( ) Answrs Concpt 5-Application of Calculus-Maimum and Minimum Problms Ths problms can b quit challnging, so hr blow ar th stps you will nd to considr if you ar to solv thm succssfully.. Draw diagram if rlvant. Dfin variabls. Dtrmin variabl to b maimizd( or minimizd) and st up quation for this variabl 4. Calculat th drivativ, st it to zro and solv 5. Dtrmin th natur of th stationary point 6. Dtrmin th valu ( lik finding th y valu) 7. Sktch th graph, watch th nd point valus Eampl Som psychologists bliv that a numrical masur of a child s larning ability during th arly yars of lif is approimatly dscribd by th function L 0.6 log for 0 4, whr is th ag in yars. Qustions Qustion: Us calculus to dtrmin th ag whn a child s larning ability is bst. Lav answr in act form. Solution- Find th drivativ of L() and st it to zro Qustion:. Dtrmin th rang of a child s larning ability btwn th ag of 0 and 4. Lav answr in act form Solution- Using th prvious answrs w nd to sub into original quation and to dtrmin L() dl 0.6 dl log 0.6log 0.6 log 0.6log log Lt 0, hnc 0.6log or Whn, ma L Whn 4, L ,. As 0, L. Pag 9 of 6

21 th rang is 5, 0. A cubic function of th form child s larning ability. Show that th maimum valu of b 4a whn 6 4b. 6 M a b 6.5 for 0 4 can also b usd to approimat a M is dm b 6.5 a 6.5 a dm Lt 0 Ma M = a 6.5 b 6.5 a 6.5 b 6.5 for 0 4, b 6.5 0,. 4b. 6 4b 4b a b b a 6 6 b 4a. 6 b Find th valus of a and b to 4 dcimal placs such that L and M hav th sam maimum valu at th sam ag. Find th othr two ags ( dcimal placs) such that L and M giv th sam larning ability of a child in ach cas. b 6. Lt Lt, b b 5 4a 6 0, a M Us graphics calculator to sktch both L and M. Dtrmin th -coordinats of th two intrsctions, and. 4 Pag 0 of 6

22 Concpt-6- Linar approimations LINEAR APPROXIMATIONS Using th following f ( h) hf ( ) f ( ) w can find th approimation of crtain function quickly, if has bn changd by a small amount h f ( h) hf ( ) f ( ) Valu of function with original valu Approimat chang in f() whn has bn changd by h Nw valu Eampl If 4 4 f ( ) us calculus to find appro valu for 0.9 Solution Notic that th qustion is asking for th valu of a function whn th original valu changd from = to =0.9 X= h=-0. (small chang in ) f ( h) hf ( ) f ( ) f ( 0.) 0. f () f () 4 f ( 0.) 0. f () Somtims w ar sking an approimat chang to y whn has bn changd by a small amount thn w nd to us th following formula: y Rmmbr whn answring qustions involving linar approimations dtrmin whthr th qustion is asking for a chang in y ( y, which quals hf ( ) or or whthr it is asking for a nw valu for th function f ( h) Pag of 6

23 Eampl Th radius of a circl incrass from 4cm to 4.0 cm. Find using calculus th approimat incras in ara Solution: Obsrv that th qustion is asking for A r da r and r 0.0 dr da A r dr A (4) 0.0 A 0.08 A Etra Eplanations on th rat typ of qustions Essntially rats tnd to us th chain rul for solving thm. Lt us hav a look at a fw ampls to s how this works. Eampl-A Th radius of a circular disc is incrasing at a constant rat of 0.0 cm/s. Find th rat at which th ara of th disc is incrasing whn th radius is 0 cm. Working out List down what thy hav givn us and pay particularly attntion to th units of th information givn Radius of disc incrasing at a rat of 0.0 cm/s dr Notic that it is giving us th radius pr scond.0 dt Find th rat at which th ara is incrasing whn da radius is 0 cm? dr Now what is th rlationship btwn ara and radius of a circl? Th formula is A r so that would man that if w took th drivativ w would gt th following: da r dr Now this is whr w us th chain rul da da dr dt dr dt da r r dt Pag of 6

24 da Now substitut th radius of 0 and w gt th answr namly: r r.04* *0 4 dt So what is th ky to doing ths typs of rat qustions? -Simpl look at th units thy giv you - ths will suggst how to structur th chain rul -B carful to work out th ara or volum of th initial shap Eampl-B An oil slick in th shap of a circl is incrasing at a rat of 0 th slick whn th ara is 00? Solution: cm / s. Find th rat of incras of th radius of Stps Rad th qustion carfully. Working out Look at th qustion carfully, what hav thy givn you and what do you nd to find out? It is circl so ara of a circl is A r It is incrasing at a rat of 0 cm / s Ar r r 00 What dos that man? Look at th units That should suggst that da dt 0 cm / s Now w ar askd to find th rat of incras of th radius? Maning what? dr Find? dt So writ down th chain rul and you will hav th following: da da dr dt dr dt Now w must substitut rlvant information da r dr da da dr dt dr dt dr 0 r dt dr 0 0 dt r 00 Pag of 6

25 Concpt-7- Whthr a function is diffrntiabl at a point Limits and continuity f at a Somtims th valu of a function f may gt clos to crtain valu L as gts clos to a from both sids of a. W say L is th limit of f as approachs a. This ida is prssd by th notation, lim f L or lim f a h L, whr h. a If approachs a from th lft, If approachs a from th right, h0 a a may not b dfind, but, h is a ngativ valu., h is a positiv valu. Eampl f is undfind at, i.., f. 5. lim f. 5 or f h. 5, Th abov function lim h 0 i.. th limit of f ists as approachs and it is.5. Th function is discontinuous at. f dos not ist. Howvr, as from ithr sid of Eampl Pag 4 of 6

26 Th function lim f h0 discontinuous at. or lim f h f is dfind at, and, i.. th limit of Diffrntiability of a function at a point on an intrval f. Howvr, as from ithr sid of, f. f ists as approachs and it is. Th function is f is diffrntiabl at a if it is continuous and thr is no abrupt chang in its gradint at a A function, i.. th sction on th lft of a is smoothly joind to th sction on th right, and th curv appars to b a straight sction (local linarity) in th immdiat nighbourhood of a. A function is not diffrntiabl at an nd point. A function f is diffrntiabl at ach point of th opn intrval p, q. Eampl Discuss th diffrntiability of th following functions at f is diffrntiabl on a closd intrval q a p, if (a) (b) a a (c) (d) a a () a (a) Th function has an abrupt chang in its gradint at a, it is not diffrntiabl at a. (b) Th function is not continuous at a, it is not diffrntiabl at a. (c) Th function is not continuous at a, it is not diffrntiabl at a. (d) Th function is undfind at a, it is not diffrntiabl at a. () It is an nd point of th function at a, it is not diffrntiabl at a. Eampl Discuss th diffrntiability of th following absolut function,. f on th unboundd intrval Pag 5 of 6

27 f is diffrntiabl at vry point cpt th nd point and th vrt. f, or,. is diffrntiabl on th intrval THESE NOTES DO NOT COVER ALL THE CONCEPTS STUDENTS WILL NEED TO KNOW IN CALCULUS- STAY TUNE FOR THE INTEGRATION SECTION Pag 6 of 6

Differentiation of Exponential Functions

Differentiation of Exponential Functions Calculus Modul C Diffrntiation of Eponntial Functions Copyright This publication Th Northrn Albrta Institut of Tchnology 007. All Rights Rsrvd. LAST REVISED March, 009 Introduction to Diffrntiation of