M2 Ch6 PDF - Differentiation

 V App/JattJon, ofOlffnrntlauon...

 V App/JattJon, ofOlffnrntlauon ( Quick Review 1 1. Derivative of a Function The dcnva tl\e of a funcll._ dyl on J - f( t) at r = u 1s denoted by f'(o) or - , which 1s the slo J'IC of the tangent to the curve y = /( r) at x = u dt , (or the slope of the curve J' = /( t) at r = u). 0 2 Basic Rules of Differentiation If 11 =f(x) and 1· -- g( r) arc d'" bl 111crcn1111 c functions and k 1s a con~1an1, then we ha,e: d (:1) -(k) = 0 dt J (b) clr x" =11x - 1, where II is a non-tcro real number d du (c) -(k 11 ) = k - ,Lr Jr Ju Jr J Ju J.- (d) -(II J + 1 ) = -clr + -clr , -(11 - 1·) = - - - (Addition rule) dr dr dr dt ,1 ,1, du (e) - (111 ) = u - + 1 - (Product rule) ,x 1 "- ti, tlu ti, man The max imum spee d of the faste st 11 race is often called the fastest man in the world. t How faS can (f) :!...(!!.] = i·~ - ~ (Quollcnt rule) The world record holde r of the I00 m men's dt I' ,,.! this fastes t man run? that 100 m divided by 3. Chain Rule 100 m race in 9.58 second s. Docs 11 mean Suppo se the world record holde r runs the 1 his averag e JU.S 9.58 s (::: 10.44 mis) b the maxim um speed that the man can run? In fact, 10.44 ml d sis hich 1s the rate of. 0 u:1 )' = /(11) and 11 = g(.r) w d1ffcrcntiablc function of r. then introd uce the conce pt of msw111011ea11.s spec If)' is a differentiable function of II and 111s a speed. To find his top speed, we need 10 distan ce travell ed by the runner is instan t dunng the race. If the chang e of dbtan cc travel led at a partic ular Jy ,ly du st a certain instant is the in antaneous ~=;;;;·;z; lim As over an i11fi11i1e.wnal time interval at plotte d agains t time. the limit o.,-o 6.t speed at that instan L 4. Differentiation of Trigonometric Functions J (DuWlO C ur,-dJcd J cl ,I. (c) d\(lan.r) = sec·., ,I. (b) -(cos t) = -sm r (a) -(Sin.\)= COSX cix rq,rcscnu I he "' Logarithmic Functions A, and 5. Differentiation of Exponential Functions lnl'UDl,U>c'OW. speed -+.. ::::: :....- ----- - 1(T1mcJ ',, (11) d,:e = e ' (b) ~ln.r =.!.,w here.r > 0 clr.r 0 insta,,;; jneous sooed lil~l!:I ! ,ntw,,1es1mal ii!A J ll'J  ~ -~ ~ -- -~ -~ -~ -- -- ,,.,- -6-.-,m -p_ ll_c l_t_ D_ l_ff _er _e_ n_t l_a _tl_ o_n -~ -~ -~ V Ap plk...- otD II~ ---- ---- -~- --~ ---- n~n d_'...._ ,.----·\ 1 Tangents to Curves If)' 1< nol cxrrc:ssc,d In 1cm u or, cxrh runcuon of x. We can cilly , we 111y 1ha1 J' is an 1mphcil r/.< of :\n 1mrhc,1 func uon by follo In Chu plcr S. we huvc kam 1 1h111 wing lhc 11cr below. lhc slope or 1hc 1angcn1101hc y - /(x) nl J)Olnl l'(.t, , yJ u llJVCn cur~c "" by: Strp I: O111. c~1m1c bo1h mies or 1hc llJ''Cn cquu uon w11h rc:spc cl 10 ,t, rco...- nhng y ns n function of x. /'(. ) - - 1s called the seco nd dcnv d, J,. nllvc or y wi1h respect 10.t nnd d noted by J'~ f'(., ) or,... ,s c dt (Re vie w Exercise p 0 Fin d~ for each or1h c following Jt runcuon (I - 6) I. 2. ,.~--,'-3 x+2 Find the ~qua llon of "1ni;aco sc-2 mn< 1hc cqu:u1on or 1hc 1nngcn110 lhc poi111 I'. cuo c at ye-- I 5. y.' ;j; a. - a On the other hand, f(x) is decreasing on [a, oo). Within this interval, [a, oo) x in [a, oo) means that x a. - a the slopes f'(x) of the curve are negative or zero. lnfinjte < a. (-oo, a) x in (-oo, a) means that x - a Interval (-oo, aJ x in (-oo, aJ means that x Sa. - - a " (-oo, oo) x in (-oo, oo) means that x can be any values. ·11~w~ strictly increasing function Er.t~taw~ -····· ··················································:············:······iiii~·~·········.~·;;~~s1ng.lunction strictly increas,~g..,..,P.!u~ creasing function ii&il{w~ strictly decreasing function M.l§~R.wl'A increasing ~ltl strictly decreasing... e decreasing 111 Interval rs:fri·- 6.10  7 i In general, we have the following result: Quick Prac tice~ Theorem 6.1 Let f(x) = -2x3 + 1Sx2- 36x + 12. Find the range of values Let /(x) be a differentiable function on the interval (a, b) of x for and which f(x) is continu ous on the interval {a, b). (a) decreasing, (b) increasing. (a) If/'(-") =::: 0 for all x in (a, b), then /(x) is increasing on [a, b]. If f'(x) > 0 for all x in (a, b), then /{x) is strictly lncreaslna on (a, bi (b) If /'(x) $ 0 for all x in (a, b), then /(x) is decreasing on [a, b]. If f'(x) < 0 for all x in (a. b), then /{x) is strictly d«ttaslna on (a, bi ID Local Extre ma The converse of the above stateme nts also hold. In the graph of y = x3 - 2.x2 + x + l in Example 6.4, it is obvious that The proof of Theore m 6.1 is out of the scope of this textboo k. i there is a 'peak' at x = and a 'trough' at x = l. In fact, a 'peak' and a i,ou.J 'trough' on the graph of a function is known as a local maximum and a local minimum of the function respectively. They are collectiv ?, ( ely )( ? Find the ranges of values of x such that f(x) Is Increasing called local extrema. b Examp le !fl! and decreasfng 'Extn:mn' 15 the plural form or 'c11.tn:mum'. 0 Local maximum and local minimum arc defined as follows: Let /(x) = x 3 - 2.x2 + x + l. Find the range of values of x for which /(x) is y (a) decreasing, Afunction /(x) is said to have a local maximum at x =.:r0 (b) increasing. ir/(~)~ /(x~ for all x in an interval (a, b) containing x 0. local muimum/{:co) Solution The point (x0, /(xJ) is called a maximum point of the graph of y=/(x). /(x) = x 3 -2.x2+ x+ l f'(x) = 3x2 -4x+ 1 Note: Local maximum is also known as relative maximum. 0 a Xo b = (3x- l)(x-1 ) Factorize lhe expression. A function /(x) is said to have a local minimum at x = x y I I I x o slightly? (b) (I) X < c slightly? (II) X > C slightly'! , slopc= O -----t-----+ X -----1------.x b d Graph off(x ) near thnna ximwn Graph off( x) near the mlnfmwn point poillt B D 6.14 ·stationary poi~t It~ 6.15  Th e results on the pre AppllcatJons ,!Dljftrrntlat vious page arc useful lon atta ins a local ma xim for determining wheth um or a local minimu er a point following theorem. m. They are summarize d in the Quick Pr ac tic e~ Theorem 6.2 First Der Find the local maximum ivative Test and local minimum of /(x ) = 4x3 - 9,xl + 6x. Let /(x ) be a different 5, lv ~ 1f( icJ.::-o iable fun ctio n on an inte rval containing -'"o· / (a) If f'( xJ = 0 and the sig n of f'(x ) cha Find the stationary poin ts of a curve X:: ,. I.o-,r¾_, negative as x increases nge s fro m pos itiv e to thr oug h x , then /(x J It is given a curve y = 0 is a local maximum. (x- 1) 3~. (b) If f'(x r) = 0 and the sig n of f'(x ) cha (a) Fin d all the station positive as x increases nge s fro m neg ativ e to ary points of the curve. thr oug h x , then /(x J 0 0 is a local minimum. (b) Detennine whethe r each of them is a maximu m point or a min z.I L'I C L \ - imum point (..._ Ex am ple Solutfon _ _. __ , __ _ _ Find_the _loca _ l extrema of ____ a function ____ ____ _ _,oo -- - - (a) y = (x- 1) 3c I - Fin d the loc al ma xim dy um and local min imu d m of /(x ) =x3 - 3x2 - dx = (x- 1) 3 dx (~ d 3 24x + 3. +c dx (x- 1) Solution = (x- 1)3e,. + 3(x -1) 2c Tips for Stu den ts = (x- 1) 2c(x +2 ) Step s for finding loca., l extr ema of a differentiable func.dy tion f(x)-. Solving dx = 0, we hav I. Solv e f'(x ) = 0. e x = -2 or l. r is always positive. 2. Not e tbe cha nge in tbe sign of dy near tbe valu Wb enx =- 2,y = (-2 -:::-ft(1' 4 -q dz - 0 ' + 0 + 0 + Lo cal ma xim um (-2 , -27 e-2) is a minimu. dy Fro m the table. the sign m point; The 51gn of d:t changes from of f'(x ) changes =/(- 2) from positive to negative as x increases negative to positive as increases tbro ugb -2. x = (-2 ) 3 - 3(- 2) 2 -24 (-2 ) + 3 thro ugh -2. By first derivative test, (1, 0) is neither a maxim um point nor a minimu dy docs DOI change sign /(x) attains a local = 31 maximum at x = -2. m point. d.t as X increases through I. Local min imu m Fro m the table, the sign Quick Practice GD of f'(x ) changes. It is give =/( 4) from oegalive to positive as·'" increases n a curve y = x 2e3x thro ugh 4. = 4 3 - 3(4) 2 - 24(4) + 3 By first derivative test, (a) Find all the station ary points of the curve. /(x) attains a local =- 77 minimum at x = 4. (b) State whether eac. = h of them is a maximum. pomt or a mmunum po int 6.17  ' Appllcotlons of Dlffaffltlatlon Exercise 6B ) 29, It is given tho t /(x) =x3 + kx2- 6x + J, where k is a constant. The curve JI =f(x) is increasing on -3 S x S-1 and decreasing on-1 s x s 2. Level 1/ Find the value of k. (b) Find the maximum and minimum points of the curve. For each of the following functions (I - 6), find the range of values of x for which the function is (a) increasing, (b) decreasing. 30. The curve JI =x3 + ax2 +bx+ I has a stationary point at P(-4, 49). (a) Find the values of a and b. 1. f(x) =3x 2 + 4x (:wmplc 6,D 2. f(x) =-4x2 + I Ox - 5 3. /(x) =x3-3x (b) Determine whether Pis a maximum point or a minimum point. 4. f(x) = 31. s. f(x) =xe-2x 6. f(x)=- X 2 It is given a function y =(3x - 7k)(x - k) 3, where k is a positive constant. x 2 +4 dy {a) Show that - dx =12(x- 2k)(x-k) 2 t he stationary Fmd points of the graph of the following functions. State whether each f them is a maximum point, a minimum point or neither. r, - 16) 0 (b) (i) Find the turning point(s) of the graph of the function. State whether it is a maximum or a minimum point. 1. y = 3x 2 - 6x + 7 (:_h.imple 6.~ (Ii) If the local minimum of the function is -4, find the value of k. 8. y=1+12x-2x2 9. y = x3 - 6x2 + 10 10. y = 2x3 - x2- 4x - 2 11. y = + 4x3 + 3 12. y =2- - x +I X (EJ Second Derivative Test and Point of Inflexion 13. y=(x-1) 3 15. 14. rJ Concavity of a Function y = (lnx) 2 for x >0 16. y = X + COS X for O S X S 1Z' < Suppose we are given a steel wire AB. If we bend the two ends of the steel wire upward, the resulting curve A'B' is said to be concave upward. Level 2/ B' Find the local extrema of the following functions. (17- 24) )s 17. y =x 2 (x-2) 2 (-Example 6.S) 18. y= (x-1)(2x-1) 3 OJ A' __) , x+I ',,' 19. y= In X ,C 2 20. X +2x+2 y = - forx>O X On the other hand, if we bend the two ends of the steel wire downward, the 21. y=e2x-2ex-3 resulting curve A' B' is said to be concave downward. 22. y= 3x 23. y=- e2x 24. y = e..., Find the maximum points and minimum points of the graph of the following functions. (25 - 28) 25. y =x + 2 sin x for O 5 x 5 1I 26. y=xe2.1-2-sx x'-2x 27. y= e2x 28. y =2- - - - l 3x +2x+ I - 6.18 ~;;~~;~;·~p~-~d-·'§~'.tlt.J concave downward l!!lflill-lt.J ·-------·-- ··-·------- ------ 6.19 O,lculu, 1,r.cJal/On : AppllcatJons of D1pmn In fact, the concavity of the graph of a di1Tcrcntiable function is related to The signs of /"(.t) arc summarized below: , "-' + lr + 5 1 x _..,. the second derivative of the function. Consider the graphs of y = g(x) and y = h(x) in the following figures. x 0 for all x in the interval. then the graph of y = f(x) is concave upward on the interval. ffi At the maximum point A, the slope of the curve is zero (i.e. f'(a) = 0) and the graph is concave downward on an interval containing a (i.e. /"(a) < 0). {b) If f"(x) < 0 for all x in the interval, then the graph of y =/(x) is concave downward on the interval. On the other hand, at the minimum point B, the slope of the curve is.z~ro = (i.e. /'(b) 0) and the curve is concave upward on an interval contammg b (i.e. /"(b) > 0). - - Find the ranges of values of x such that y = f(x) Is concave In fact, the converse of the above facts are also true, as stated below: upward and concave downward Second Derivadve Test = It is given that /(x) x3- 6x2 + 2x + 5. Find the ranges of values of x for = which the curve y f(x) is concave upward and concave downward. cfmed on the interval (a, b) containing xo. and /"(xo) < 0, then /(x) has a local maximum Solution /(x) = x -6x2+ 2.x+ 5 3 (b) :%o = 0 and /"(xo) > 0, then /(x) has a local minimum 2 = f'(x) 3x - 12.x + 2 atx= x0 = /"(x) 6x - 12 6.20 = 6(x-2) The proof of the theorem is out of the scope of this textbook. - 6.21 J Calculus Appl1catJons of Dljfmntlatlon d Find the extrem by the secon Find the ~rem f derivative test e Points of a curve derivative test e po nts a curve by I O the second. 9.,_ h h is given that the equa tion of 8 c urve 1s y == x + _. ll is given that the curve y = f3x +cos ~. w ere OS.'< s ,r, X curv. dy and - d2y (a) Find the stationary poin u of the c. (1) Fmd - dx dx2· (b) State whether each of the.. point. r m ,s mruumum pomt or a minimum 8 (b) Find the stati onar y · e and state whether each 0 I ts a maxi. mum pPoin them 1s. curv of the Solution omt or a mini mum point. (a) ,, =13x + COS2.'( Solution dy -dx = 13 -2sin 2.x ti d f (~ ·. y == x+! th (cos 2.T) -sin 2.T th (lx) )\l\e\.Y. I v\-\1 V\, (a) it~~"- X. dy - Solving - =O' we have· dy >O 9 -== 1- - Think Further,.!,.Q01 dx dx x2 18 r f' d the maximum ond minimum the curve in Examp le 6.8 t9 13 - 2sin2.x = o dly I~ points o. -;;;= by using the first dcrivnuvc test. 2.x =.[j. - XJ dy sm - 2.x =3 Tl 2 or 2.x =7r - tr 3 (b) S0 Ivmg - · dx = 0 we have. Tl tr x=- or x=- 9 6 3 1-- =o x2 1l /"3,r I x 2 9 = Wh enx6=-, y=- +-2· 6 x=- 3 or x= 3 Whe n x = - 3, y = -6. When x = !!., y = fi ,r _ ! 3 3 2· Whe nx =3, y =6. The statio - -l"J,r + - and (,r s arc (,r nary pomt - -l"J- I) ,r - -. 1) and (3, 6). 3 3 2 The stati onar y poin ts are (-3, -6) 6' 6 2 d2y 18 y dly dx2 x.,-3 =(-3) < 0 y x+ i 9 (b) - dx 2 =-4c os2. x ,td (sin 2.\') = cos 2.'1'~(2 t d.t x) (-3, -6) is a maxi mum poin t. Dy second derivative test... ,r =-4 cos3-=-2 < 0 )' dx 2 x - y = ,/f;c + cos 2.x 6... [6 6 2. of the curve. tr,./3,r + ·]·1s a maxi.mum pomt (3, 6) is a mini mum poin t. Dy second derivative test Quick Practice ( 6.8 )... 2,r =-4 cos3- = 2 > 0 dx2 I I It is given that the equation of a curve is y =2x3- Sx2- 4x + 3. x -J I ' X 0 ,r ,r... 1 of thecurve. 2 dy d y fi ] t umpom 6 ) 1t m1m [ -3' -3- -2 1Sam. 1t (a) Find - and - dx2 dx the curve and state whether each of them is (b) Find the stationary points of 6.22 a maximum point or a minim um point. - 6.23  V AppllcalJoffl of Dljfnff'tlarlon FM Quick Practice (j:i) fdlnd the local extrema of a functlon when the second erlvatlve test docs not work It Is given thnt y sin 2x-.\', where oS x S Tt. Find the local extrema of the function /(x) =.~e-,., (a) Find the stntionary points of the curve. (b) Stnte whether each of them is 8 maximum point or a minimum point. Solution I J(x) ::::::i x3e_,, Allllough the second derivative test is useful to distinguish extreme points, the test docs have some limitations. Suppose a function f(x) is given. From Theorem 6.4, we know that when = f'(xo) O and /"(xo) :I- 0, then /(x) has a local extremum at X Xo = I f'(:'C) r:=::k:tt= (1)±6J::-;-91-t,s-0::, _t(naltaim-itrlocmnpjnumpn Z # J ,:: :1.btmifii rin~ Whenx =0, - - - - - + - - - - + :c 0 ·: f'(x) =3x2 and f"(x) =6x f"(O) = Oe [0 - 6(0) + 6] = 0 0 2 f'(x) =4x3 and J"(x) = 12x2 ·: f'(x) =-4:x3 and r 0 (Ill) Fin d the point(s) or inflexion or the cu.. rvc. (2, -16 ) is a minimum poi (c) Hen ce, ske tch the curv nt. e y = x(x - 4)2. (iii) Solving f"( x) = 0, we have 12x2 - 16 =0. 2 x = - - or x =2- Sketch the curve of a ratio nal function Sklllsto L-l Tackle Problems lfffl 13 ,/3 2 Consider the functio Th e signs of /"( x) are sum marized below: n f(x ) = Xx _+ 5 and the graph of y =J(x). 2 - (a) Fin d the domain 2 2 of /(x ). x< -- x = -2- --2< x - (b) Find the x- and y-in 13 13 13 13 x= rf./3_ f°(x ) cbongcs sign o5ix terc epts of the graph. 2. -~. -.: through - '- nnd (c) Find the range of valu f ( x) + 0 - 0 + mi;,.....,......-3 c...-3 es of x for which the gra ph is ~r (i) increasing, (ii) decreasing. Wh en x= - ~, y= [- ~) 8 Hence, find the turning '-s [- point(s) of the graph. =- :- (d) Fin d the asymptote (s) of the graph. Wbenx= ~, y= [,M -s (e) Using the results [~ ]' of (a) to (d), sketch the gra =- ~- ph. 2 -so] Solutfon -- - and [- 2 - so] - are p01 nts o r·m ne,ao [ 13· 9 rt 9 n. 2 X +5 (c) y y=x '-8. x' Reminder I (a) f(x ) = - - is x- 2 undefined when x = 2. The curve is symmetrical Check all the features foun Th e domain of /(x ) is the abo ut the )'-axis. d: set of all real numbers (0, 0) symmetric about the y-ax except 2. ---4----- is -....- - - - - + - - + JC (b) Wh en x= O, y= 0 +s 2 s 0 2 =- 2. y-interceJ)t = 0, _ Always mark the turning points, points of x-interc:eJ)ts = - /i, 0, /i maximum point: (0, 0).. s inflexion, x-intereePts The y-mtercept is-. and y-intercepl of the JDinjmum points: (-2. -16 ), 2 graph. (2,- 16) (-2, -16 ) (2, -16 ) Wh eny =O , points of inflexion: (-i,-~),(J,,-~) x1 +s= O x- 2 x2 +5 =0 x has no real roo ts. Th e gra ph has no x-interc epts.  (c) 1l"lx)=- ~-' + s Appllcatloru ofDtjf,rrntlotlon - ' x-2 (x- 2)(2x)- (x2 + 5)(1) (e) 1 f'(x) = - - - -2 - - Remfndl,r (x-2) Check all the features (ound: x 1 -4x-5 = (x - 2) 2 1 (S, 10) , y-intcrccp1 - i, I ,,'y x+2 no x-intcrcrplJ (x + l)(x- 5) I =----- 2 ;-'.,., I min. point (S, 10). (x- 2) mu. point (-1, -2) lncn:a.\inf on x S-1 and Solving f'(x) =0, we have x = -1 or 5. ·" s. dccrcnsina on -~- - -- -Is x < 2 and 2 < " S S The signs of f'(x) are summarized below· Whcnx-2 ,y-+a:>. I Remember lo 1 2 When x-r. JI - -. S X x--1 -1. x 2 -6 =0 When X - -r. y - +a>. Whcn x--r., --. x=± /6 The curve should be close The x-intercepts are 16 and -16. to the line y = I when.r tends (ii) Solving f'(x) =0, we have: I Quick Practice (6.1 ij) to infinity. 4x =0 Consider the function /(x) = 1- x2. 12 (x2-4) 2