Differential calculus

Transcription

1 A The derivtive of tn (k) B Second derivtives C Anlsing the behviour of functions using the second derivtive D Derivtives of inverse circulr functions E Antidifferentition involving inverse circulr functions F Implicit differentition Differentil clculus res of STuD Differentil nd integrl clculus, including: Derivtives of inverse circulr functions Second derivtives, use of nottions ƒ () nd d, nd their ppliction to the nlsis of grphs of functions, including points of inflection (tretment of concvit is not required) Applictions of chin rule to relted rtes of chnge nd implicit differentition; for emple, implicit differentition of the reltions + 9 nd +. ebookplus The derivtive of tn (k) introction Digitl doc 0 Quick Questions In this chpter we etend the functions tht cn be differentited to include f () tn (k) nd the inverse circulr functions sin (), cos () nd tn (). It is ssumed tht the stndrd results for the differentil clculus re fmilir to the student s re the proct rule, quotient rule nd chin rule. These results re listed in the tble below. f () n n n log e (k), > 0 e k sin (k) cos (k) tn (k) u () v () u ( ) v ( ) g[h()] ke k k cos (k) k sin (k) k sec (k) f () u () v () + u () v () (proct rule) u' ( ) v ( ) u ( ) v' ( ) ( quotient rule ) v h () g [h()] (chin rule) 96 mths Quest Specilist mthemtics for the Csio Clsspd

2 sin ( k) Since tn ( k), its derivtive cn be found using the quotient rule: cos ( k) If f () tn (k), sin ( k) cos ( k) d v u v v u dv where k is constnt, then using the quotient rule with u sin(k) nd v cos(k): k cos ( k)[cos ( k)] sin ( k)[ ksin ( k)] f ( ) cos k k cos( k) + k sin( k) cos k k[cos k + k ( ) sin ( )] cos ( k) (b fctorising the numertor) k( ) sincecos k + ( ( ) sin ( k) ) cos ( k) k sec (k) If f () tn (k), then f () k sec (k). Worked Emple Differentite the following epressions with respect to. tn (6 ) b tn c Find the nswers to prts nd b using CAS clcultor. Think Differentite b rule: tn (k), then ksec ( k) where k 6. Write tn (6) 6 sec ( 6) b Differentite tn where k. b rule Multipl the derivtive b. b tn sec 8 sec Chpter Differentil clculus 97

3 c On the Min screen, using the soft kebord, tp: ) - Complete the entr line s: d (tn (6)) d tn Press E fter ech entr. Note: The ClssPd operting sstem does not hve reciprocl functions; therefore it uses the identit for secnt squred ( + tn ()). Write the solutions. For, d For b, d 6 [tn ( 6)] [cos ( )] 6 8 tn cos Worked Emple If f () tn ( + ), find f (). Think Write Let u + to ppl the chin rule. f () tn ( + ) Let u +. Find. + Epress f () in terms of u. So f () tn (u). Find. Find using the chin rule. sec () u ( + ) sec (u) 6 Replce u with the epression +. f () ( + ) sec ( + ) if tn [ f ()] then f () sec [f ()]. 98 Mths Quest Specilist Mthemtics for the Csio ClssPd

4 WorkeD emple Differentite the following with respect to. log e [tn ( )] b ( + ) tn () ebookplus Tutoril int-079 Worked emple Think Let u tn ( ) nd consequentl ppl the chin rule. Find. WriTe log e [tn ( )] Let u tn ( ). sec ( ) Epress in terms of u. log e (u) Find. Find using the chin rule. u So u sec ( ) 6 Replce u with the epression tn ( ). 7 Epress in terms of sin ( ) nd cos ( ) onl in order to simplif the rtionl epression. 8 Epress the division of rtionl epressions s multipliction. sec ( ) tn ( ) sin ( ) cos ( ) cos ( ) cos ( ) cos sin ( ) ( ) 9 Cncel out the fctor of cos (). sin ( ) cos ( ) 0 Use the double ngle formul to simplif sin ( ) cos ( ). sin ( 6 ) Stte the nswer in simplest form. nd hence 6 sin( 6) or 6 cosec ( 6). b ( + ) tn (). Let u + nd v tn (), nd ppl the proct rule. If uv, then u () v () + u ()v () b ( + ) tn (), with u + nd v tn (). The proct rule sttes u () v () + u ()v () Chpter Differentil clculus 99

5 Substitute for u (), v (), u () nd v (). u () + nd v () sec () ( + ) tn () + ( + ) sec () Simplif if possible. [( + ) tn () + ( + ) sec ()] Worked Emple Find the eqution of the tngent to the curve + cos () + tn () where π, mnull b using clcultor. Think To find the eqution of tngent line to curve t point, the coordinte of the point is needed s is the grdient of the curve t tht point. Find when π to estblish the coordinte. Write + cos () + tn () If π + +, π cos π tn π π π The coordinte is π π, + Find to estblish the grdient function. Evlute when π to find the grdient of the tngent t π. sin ( ) + sec ( ) If π +, sin π sec π + cos π + so the eqution of the tngent is: Substitute π π +, nd m into the eqution for stright line: m ( ) where m is the grdient nd (, ) is point on the line. Simplif the eqution nd mke the subject. + π π π + + π + 00 Mths Quest Specilist Mthemtics for the Csio ClssPd

6 b On the Min screen, tp: b Action Clcultion tnline Complete the entr line s: tnline( + cos() + tn(),, π ) Then press E. Write the solution. Eqution of the tngent is + REMEMBER. If tn (k) then. If tn [ f ()] then ksec ( k). f () sec [ f ()]. Eercise A The derivtive of tn (k) WE Mnull differentite ech of the following with respect to. tn () b tn () c tn (7) d tn () e tn ( ) f tn ( ) g tn ( ) h tn () i tn ( ) j tn k tn l tn m tn 7 n 8 tn o 6 tn 9 p tn 6 8 q tn r 7 tn WE For ech of the following find f () if f () equls: tn ( + ) b tn ( + ) c tn ( ) d tn ( ) e tn ( + ) f tn( 8) g tn (7 ) h tn ( ). Chpter Differentil clculus 0

7 MC If e tn (), then: cn be found b using: A direct rule B the proct rule C the chin rule D the quotient rule E grphicl methods b is equl to: A sec (e tn () ) B sec (e tn () ) C sec (e tn () ) D tn (e sec () ) E e sec () WE Differentite the following with respect to if equls: log e [tn (6)] b etn () c tn (e ) d sin [tn ()] e cos [tn ()] g sin () tn () i tn () For ech of the following find f () if f () equls: tn () b cos () tn f [tn ()] h log e () 6 + tn 7 j tn (). d e tn ( ) e e tn ( 8 ) f g tn( ) h sin ( ) tn ( ). c ( 6) tn () tn( ) 6 If f: [0, p] R, f () tn (), find the coordintes of the points on the grph where the grdient is equl to: b c d 0. 7 If f: [ - p, p] R, f () tn, find the coordintes of the points on the grph where the grdient is equl to: b c 0. 8 Find the eqution of the tngent to the curve with eqution tn () t π 6. 9 WE If f () tn () find the eqution of i the tngent nd ii the norml t the point on the curve where π. B 0 Show tht there re no sttionr points for the grph of tn () for ll vlues of. Eplin wh the grdient of tn () is lws positive. Second derivtives Suppose tht f (). Then the derivtive, lso known s the first derivtive, is written s f () or. B differentiting f (), the second derivtive, f () or d is obtined. 0 Mths Quest Specilist Mthemtics for the Csio ClssPd

8 The process of obtining f () or d from f () is lso clled double differentition. The second derivtive is commonl referred to s f double dsh or d squred, squred. WorkeD emple Find d if +. Think WriTe Epress s. + Find the first derivtive. Differentite to obtin. d + or + WorkeD emple 6 Find, using clculus, f ( ) if f ( ) is equl to: e cos () + log e () b sin( ). c Repet prts nd b using clcultor. ebookplus Tutoril int-080 Worked emple 6 Think Differentite e cos () b the chin rule short cut nd log e () b rule to obtin f (). Epress in inde nottion so it cn be differentited. Differentite f () to obtin f (). Use the proct rule to differentite sin ()e cos (). WriTe f () e cos () + log e () f () sin () e cos () + f () sin () e cos () + f () cos () e cos () + sin () e cos () or e cos () [sin () cos ()] b Epress in inde nottion. b f( ) sin ( ) ( ) f( ) sin Chpter Differentil clculus 0

9 Epress f () s proct. Differentite f () using the proct rule to obtin f (). Differentite f () using the proct rule to obtin f (). sin( ) f ( ) sin ( ) + cos ( ) f ( ) sin ( ) cos ( ) cos( ) sin ( ) Simplif b collecting like terms. sin ( ) cos ( ) sin( ) 6 Simplif f () b tking out the fctor. or [ sin ( ) cos( ) sin ()] sin( ) cos ( ) sin ( ) c On the Min screen, tp: Action Commnd Define Complete the entr line s: Define f () e cos() + ln() Then press E. Complete the entr line s: d f ( ( )) Then press E. Write the solution. The second derivtive, cos ( ) f''( ) [ sin ( ) cos( )] e 0 Mths Quest Specilist Mthemtics for the Csio ClssPd

10 On the Min screen, complete the entr line s: sin( ) Define f( ) d f ( ( )) Press E fter ech entr. Write the solution. The second derivtive, cos ( ) f ( ) sin ( ) 6 You m rerrnge the nswer to form similr to tht given in the solution tht ws obtined mnull. Tp: Action Trnsformtion fctor Then highlight nd cop the previous nswer nd pste it, s shown on the right. Then press E. 7 Write the solution. The second derivtive: ( ) sin ( ) cos ( ) f ( ) Worked Emple 7 k d If e nd + 0, find the vlue of k. Think Write Find k e k from. e k ke Chpter Differentil clculus 0

11 Find d. ke Substitute nd d into the eqution + 0. k k k k so ke ke + e 0 Tke k e out s fctor. e ( k 6k + 8) 0 Fctorise the qudrtic function of k. (Note: e k cnnot equl zero.) k k e ( k)( k ) 0 6 Stte the solutions. Therefore k or k. REMEMBER. is the second derivtive of with respect to. It is found b differentiting.. If f () then d f (). Eercise B Second derivtives WE Find d if: + 7 b + c 6 7 d + + e 6 + f + g + h 6 + i + 9 j + k +. MC If sin ( ), then: is equl to: A cos( ) B cos( ) C cos ( ) D sin E cos 06 Mths Quest Specilist Mthemtics for the Csio ClssPd

12 b is equl to: A D sin cos( ) B WE 6 Find f () if f () is equl to: E sin ( ) cos ( ) cos ( ) sin ( ) C sin ( ) cos ( ) log e () b log e () c sin () + cos () d sin + e e e + f e + g tn () h tn ( ) + i j sin ( ) m sin( ) n [cos ( )] cos( ) k e sin () l cos (e ) o log e [cos ()]. For ech of the following functions f (), mtch the grph which could represent f (). f () + b f () c f () d f () sin () e f () log e () f f () e g f () A B C f"() 0 f"() 0 f"() 0 D E F f"() f"() 0 f"() 0 0 G f"() H I f"() f"() Chpter Differentil clculus 07

13 C If cos (), show tht d. 6 If sin (), show tht: d If log e (), show tht: ( + ). 8 If e, show tht: d +. 9 We 7 Find the vlue of k if e k nd d. 0 If e k nd d 0, find the vlue of k. The position of prticle trvelling in stright line is given b the eqution: (t) t t t + 7, where hs units in cm nd t is in seconds. Find: dt, tht is, the velocit t n time t b, tht is, the ccelertion t n time t dt c when nd where the prticle momentril stops; tht is, when dt 0 d the minimum velocit. nlsing the behviour of functions using the second derivtive ebookplus The second derivtive of function cn be used for testing the nture of sttionr points. The rte t which the grdient of function chnges indictes the tpe of sttionr point we re deling with. first derivtive function We hve seen how the first derivtive of function, f () or Interctivit int-06 Second derivtive cn tell us where function hs positive grdient, negtive grdient or zero grdient (sttionr point). For emple, let us look t the functions f () nd f (). Function f () Emine the grph t right. Since f (), f () nd f () 0 t 0. If f () 0 At 0, f () is sttionr point.. If f () < 0 This occurs when < 0. So if < 0, then f () is decresing function, one with negtive grdient for ll < 0.. If f () > 0 < 0 f() is decresing > 0 f() is incresing 0 0 f() is neither incresing nor decresing This occurs when > 0. So if > 0, then f () is n incresing function, one with positive grdient for ll > 0. Consequentl, t 0, minimum sttionr point occurs. f() 08 mths Quest Specilist mthemtics for the Csio Clsspd

14 Function f () Emine the grph t right. Since f (), f () nd f () 0 t 0. If f () 0 At 0, f () is sttionr point.. Here, f () > 0 for ll vlues of ecept zero. Consequentl, t 0, sttionr point of (horizontl) inflection occurs. Second derivtive function Similrl the second derivtive f ( ) or cn tell us where the grdient function f ( ) or is incresing or decresing or neither (tht is, chnging from incresing to decresing or vice-vers). Let us look t the sitution when f () is greter thn, less thn nd equl to zero. f() Function f () > 0 Emine the grph of f () t right. f '( ) 6 f '() 6 f () f () f '( ) f '() f () f '( ) f '() When f () > 0, the grdient function f () is incresing. Tht 0 is, s increses f () increses. Grdient is lws incresing from left to right Function f () < 0 Emine the grph of f () - t right. f () f '( f '(0) 0 ) f ' () f () 0 f () When f () < 0, the grdient of f () is decresing. Tht is, s increses f () decreses. f '( ) f '( ) 6 f '() f '() 6 f() Function f () 0 Emine the grph of f () t right. f () f () f () 6 when 0, f () 0 < 0, f () < 0 > 0, f () > 0 < 0 f() is incresing f() 0 0 f() is neither incresing nor decresing Therefore the grdient is decresing left of 0 nd incresing right of 0. The point (0, 0) is clled point of inflection. As f (0) 0, the point (0, 0) is lso sttionr point, so it is clled sttionr point of inflection. > 0 f() is incresing Grdient is lws decresing from left to right f '( ) < 0 f '() is decresing f() f '() f '( ) f '() 0 > 0 f '() is incresing Chpter Differentil clculus 09

15 Points of inflection Points of inflection occur when the second derivtive chnges sign. A tngent drwn t point of inflection crosses the grph t tht point (see Figure ). Sometimes points of inflection re lso sttionr points (see Figure ). f"() 0 < f "() 0 < Tngent Point of inflection [ f"() 0] f"() < 0 Figure Tngent Tngent Tngent Sttionr points of inflection [ f "() 0 nd f '() 0] f"() < 0 Figure The second derivtive cn tell us the nture of n sttionr points or where points of inflection occur on grph. When f () 0 there will be sttionr point t. In ddition:. the point will be locl minimum sttionr point t if f () > 0 s well. the point will be locl mimum sttionr point t if f () < 0 s well. the point will be locl sttionr point of inflection t if f () 0 nd f () chnges sign t s well. It is lso possible to hve point of inflection t point other thn sttionr point. Tht is, when f () 0 nd f () chnges sign t, there will be point of inflection t, but the point is not necessril sttionr point since there is no requirement tht f () 0. An emple is f () tn t 0. Notes. Tpes of sttionr points (,, bove) cn lso be determined using the first derivtive test either side of.. The second derivtive test is usull more efficient thn the first derivtive test in determining mimum or minimum sttionr points, but not for sttionr points of inflection.. Displing curve on grphics clcultor will ssist in determining the nture of n sttionr points. Worked Emple 8 Find n sttionr points, nd their nture, if f () ( )( + ). b Sketch the grph of f (), clerl indicting ll sttionr points nd es intercepts. Think Write Epnd f () so it cn be differentited esil. f () ( )( + ) f () ( ) Find f (). f () Find f (). f () Solve for where f () 0. For sttionr points, f () 0 ( ) 0 0 or 0, or 0 Mths Quest Specilist Mthemtics for the Csio ClssPd

16 Find f (0) nd f (0) to determine one sttionr point. 6 Find f nd f sttionr point. for the second At 0, f (0) 0, f (0) 0 nd f (0). Therefore (0, 0) is locl mimum sttionr point. At, f Therefore nd f 6, is locl minimum sttionr point. 7 Find f nd f sttionr point. for the third At, f nd f 6 Therefore, is locl minimum sttionr point. b Evlute f (0) for the -intercept. b f (0) 0 so the -intercept is 0. Solve the fctorised form of f () 0 for the -intercepts. f () ( )( + ) 0 hs solutions 0,, so the -intercepts re, 0 nd. Sketch the grph of f (). Check the grph using clcultor. f() ( )( + ) (, ) 0 (, ) Worked Emple 9 If + find: n sttionr points of inflection b n other points of inflection. Think Write Find. + Chpter Differentil clculus

17 Solve for where 0 (tht is, find the sttionr points). For sttionr points, 0. 0 ( ) 0 0 or Find d. 6 Evlute d where 0. At 0, (0) 6(0) 0 Find where 0. nd (0) (0) + 6 Evlute d where. At, () 6() (therefore, minimum sttionr point). 7 Stte n sttionr points of inflection. The point (0, ) is sttionr point of inflection. b Solve d 0 b For points of inflection, d ( ) 0 0 or Find t onl, since sttionr point hs lre been determined t the point (0, ). Check the sign of d either side of. At, () If, d () 6() If, d 9 (from prt ) Stte the other point of inflection. Therefore the point (, ) is point of inflection (not sttionr). For Worked emple 9, the second derivtive sign digrm verifies tht t 0 nd there re indeed points of inflection s f () chnges sign t these points. This verifiction is not usull required but in rre cses it will show tht wht seems to be point of inflection is in fct not. For emple, if f () f () 0 when 0. Hence sttionr point occurs t 0. For f () 0, solution occurs when 0. Thus, it ppers, there is sttionr point of inflection t 0. But the sign digrm of the second derivtive (see figure t right) shows tht f () does not chnge sign t sign of d + + f"() 0 Mths Quest Specilist Mthemtics for the Csio ClssPd

18 Therefore, there is not sttionr point of inflection t 0. The first derivtive test will verif tht t 0 there is locl minimum sttionr point s shown in the figure below (left). The sketch grph of f () is shown in the figure below (right). + + f '() 0 0 f() WorkeD emple 0 For the function f () log e (), find: ll sttionr points b n points of inflection c nd sketch the grph of f (). Think WriTe ebookplus Tutoril int-08 Worked emple 0 Find f () nd set it equl to zero. f ( ) For sttionr points, f () 0 0 Multipl the eqution b. or 0, 0 Solve f () 0. Thus. Evlute f (). f () (0) 6 Find f (). f () 6 6 Evlute f (). f () Stte n sttionr point nd its tpe. Therefore (, 6) is locl mimum. b Set f () 0 nd multipl the eqution b. b f () 6 For points of inflection, f () or 6 0, 0 Solve f () 0. 6 Thus or pproimtel Chpter Differentil clculus

19 Since 0.79 is the onl solution there re no points of inflection becuse the implied domin is > 0. c There is verticl smptote t 0. c 0 smptote Sketch the grph of f (). Check the grph using CAS clcultor. Therefore there re no points of inflection s 0.79 is outside the implied domin, > (, 6) f() log e () Worked Emple Find the minimum vlue of the grdient to the curve: f() + 7. Think Write Find the eqution of the grdient function f (). f () Find the rte of chnge of the grdient function, f (). f () Set f () 0. Let f () Solve f () Evlute f ( ) which is the minimum vlue of the grdient s f () is positive qudrtic. f 6 ( ) ( ) + ( ) which is minimum s f () is positive qudrtic eqution. 6 Stte the minimum vlue of the grdient. Therefore the minimum grdient of f () is. 7 Verif using CAS clcultor. REMEMBER. If f () 0 nd f () > 0, then locl minimum sttionr point occurs t.. If f () 0 nd f () < 0, then locl mimum sttionr point occurs t.. If f () 0 nd f () 0, nd f () chnges sign t, then sttionr point of inflection occurs t.. If f () 0 nd f () chnges sign t, then point of inflection occurs t without the point being sttionr point. Mths Quest Specilist Mthemtics for the Csio ClssPd

20 eercise C ebookplus Digitl doc WorkSHEET. nlsing the behviour of functions using the second derivtive You m use grphics or CAS clcultor to ssist with drwing grphs. We 8 For ech of the following functions find the sttionr points nd the nture of the sttionr points using the second derivtive test. f () b g() c ( + ) d ( )( + ) e h() ( )( + )( + ) f f () + 6 g g() h f () + i h() j + We 8b Sketch the grph of ech function in question, clerl indicting ll sttionr points nd es intercepts. mc The function f () + + : hs point or points of inflection when is equl to: A 0 nd B 0 onl C onl D 0 nd E 0 nd b hs sttionr point of inflection: A (, ) B (0, ) C (, ) D (, ) E (, 0) mc The minimum grdient of the curve with eqution 6 8 is: A 0 B C D 0 E mc If f () 0 when nd, nd f () nd f ( ), then f () hs: A sttionr points of inflection when nd B sttionr point of inflection when nd locl mimum when C locl mimum when nd locl minimum when D locl minimum when nd locl mimum when E sttionr point of inflection when nd locl minimum when 6 We9 Give i n sttionr points of inflection nd ii n other points of inflection for ech of the following functions. 6 b f( ) c d g() + + e e f g f () e h g() 8 log e () 7 Show tht log e () does not hve n points of inflection. 8 Show tht the grph of the function g() e hs no points of inflection. Find its sttionr point. 9 We 0 Find i sttionr points, nd ii points of inflection, nd iii sketch the grphs with the following rules: f () b + 8 c f( ) d g ( ). 0 We Show tht the mimum vlue of the grdient to the curve f () + + is. Find the mimum vlue of the grdient to the curve 0 +. Sketch the grph of the function, showing ll intercepts with the es, the sttionr points nd n points of inflection. (Verif with clcultor.) The downwrd displcement of meteor t seconds fter hitting the surfce of the ocen is given b: t t d 0t+, where 0 t 0 nd d is in metres. Find the depth of the meteor fter 0 seconds. b Find the velocit, v, t n time t. Chpter Differentil clculus

21 c Find the mimum velocit of the meteor. d If the depth of the ocen where the meteor strikes is 600 metres nd the meteor disintegrtes 0 seconds fter hitting the ocen, does the meteor rech the ocen floor? d Derivtives of inverse circulr functions The derivtive of sin, If then So or sin, < < nd > 0 π π < < π π sin ( ), < < sin () cos( ) 0 < < cos( ),cos ( ) nd π π Since sin ( ) cos( ) (from digrm t right) or cos( ) so, < <. Therefore, if f( ) sin then f ( ), < <. π π 0 0 sin ( ) 6 Mths Quest Specilist Mthemtics for the Csio ClssPd

22 WorkeD emple Find the derivtive of: sin Think b sin (6 ). WriTe Differentite b rule where. If sin then 6 b Epress 6 s. b If sin (6) 6 then sin 6 ebookplus Tutoril int-08 Worked emple Differentite b rule where 6. 6 Tke 6 out s fctor of the 6 denomintor to remove the frction from the squre root. Simplif the derivtive. 6 6 ( 6) Note: Emple (b) could lso be done b the chin rule, using the substitution u 6. The derivtive of cos, If then Thus, or > 0 cos, nd 0 p cos( ), 0 p cos ( ) sin ( ), sin 0 nd 0 < < p. sin ( ) Since cos sin( ) or sin( ) (from digrm t right) π π 0 cos ( ) Chpter Differentil clculus 7

23 So, < <. Therefore, if f( ) cos then f (), < <. Worked Emple Find f () if f () is equl to: cos b cos. Think Write Differentite b rule where. f( ) cos f ( ) b Epress s. b f( ) cos cos Differentite b rule where. f ( ) Tke out s fctor of the denomintor. Simplif f (). ( ) The derivtive of tn If then tn, R nd π π tn ( ), < < tn () sec ( ) ( + tn ()) π π < < π π 0 tn ( ) 8 Mths Quest Specilist Mthemtics for the Csio ClssPd

24 or + (from the digrm t right) ( + ) + + Therefore, if f ( ) tn then f (), R Worked Emple Find f () if f () is equl to: tn Think b tn 8. Write Differentite b rule where. f( ) tn Thus, f ( ) + b Epress 8 s. b f( ) tn 8 8 tn 8 Differentite b rule where. f ( ) 8 + Tke out s fctor of the denomintor. 6 ( 9+ 6) Simplif the derivtive b dividing 8 b Worked Emple Find if is equl to: sin ( + 7) b cos ( ) c sin tn. Think Write Let u + 7 so the chin rule cn be pplied. sin ( + 7) Let u + 7. Chpter Differentil clculus 9

25 Find. Epress in terms of u. sin (u) Find. Appl the chin rule. So 6 Replce u with + 7. u u ( + 7) b Let u so the chin rule cn be pplied. b cos ( ) Let u. Find. Epress in terms of u. cos (u) Find. Appl the chin rule. So 6 Replce u with. u u ( ) c Let u tn pplied. so the chin rule cn be c sin tn Let u tn. Find. + Epress in terms of u. sin (u) Find. Appl the chin rule. So 6 Replce u with tn. cos ( u) cos( u) + cos tn + 0 Mths Quest Specilist Mthemtics for the Csio ClssPd

26 Worked Emple 6 Find the eqution of the norml to the curve with eqution: cos t the point where, mnull b using clcultor. Think Write Find when. cos When cos π 6 π Find. Substitute into to find the grdient of the tngent t. Find the grdient of the norml grdient of norml grdient of tngent. When, the grdient of the tngent is The grdient of the norml π Substitute, nd m into the eqution of stright line rule: m( ) where m is the grdient nd (, ) is point on the line. Therefore, the eqution of the norml is: π ( ) 6 Simplif the eqution. π + (or 6 + π ) Chpter Differentil clculus

27 b On the Min screen, tp: Action Clcultion norml Complete the entr line s: Norml cos,, Then press E. b Write our solution in n pproprite form. π The eqution of the norml is +. REMEMBER. If f ( ) sin f ( ), then < <. If f ( ) cos f ( ), then < <. If f ( ) tn f ( ), then R + Eercise d Derivtives of inverse circulr functions WE,, Find the derivtive of ech of the following epressions with respect to. sin d cos g tn j tn m cos. p tn 08. b sin e cos h tn k sin n cos 7 q sin 6 c sin 8 f cos 6 i tn 7 l sin 0. o tn r tn 0 Mths Quest Specilist Mthemtics for the Csio ClssPd

28 WE b Find if is equl to: sin () b sin () c sin () d sin (8) e cos () f cos (6) g cos (7) h cos (0) i tn () j tn (9) k tn () l tn (). Using the results of question, or otherwise, stte the derivtive of: sin (b) b cos (b) c tn (b) where b is rel, positive constnt. MC Consider the function with the rule f( ) sin. The miml domin of f () is: A π π, B D [, ] E b The miml domin of f () is: π π, C [, ], π π A, B [, ] C (, ) D [, ] E (, ) MC Let f( ) cos 7. When epressed in the form f( ) cos, the vlue of is: A 7 b f () is equl to: 7 A 9 9 D 9 B 7 B E C MC The derivtive of tn 8 is equl to: A D B E WE b, b Find f () if f () is equl to: cos e tn b cos 7 f tn 8 D 7 c cos 9 g tn 7 C C E d cos 8 h tn 9 i sin j sin k sin 6 7 l sin 8. Chpter Differentil clculus

29 e 8 Using the results of question 7, or otherwise, stte the derivtive of with respect to : b sin b b cos b c tn where nd b re rel, positive constnts. 9 WE Find mnull, then check with clcultor. If is equl to: sin ( + ) b sin ( ) c cos ( ) d cos ( + 8) e tn ( + ) f tn (6 7) g sin + h cos + i tn j sin ( ) k cos (7 ) l tn (8 ) m sin n cos Find the derivtive of ech of the following epressions with respect to. + cos () b sin c sin( ) + tn d log e ( 6) + cos e e 7+ + tn () + f sin ( ) g tn i sin () + cos () k tn sin h cos () j sin [cos ()] l cos tn o tn. 9 Find the grdient of the grph of sin t the origin. b Hence, find the eqution of the tngent to this curve t the origin. WE 6 Find the eqution of the norml to the curve cos () t the point where it crosses the -is. Find the eqution of the tngent to the curve sin when. Find the coordintes of the point where the mimum grdient of f( ) tn occurs ( is constnt). b Find the mimum grdient of f (). Antidifferentition involving inverse circulr functions We now know tht: d d sin, cos, < < < < Mths Quest Specilist Mthemtics for the Csio ClssPd

30 d It therefore follows tht: tn, R + sin + c, >0 cos + c, >0 tn + c When finding ntiderivtives of inverse circulr functions, the integrnd should be epressed in one of the stndrd forms bove nd then integrted. WorkeD emple 7 Differentite tn Think nd hence find. + WriTe Write the eqution. tn Differentite tn b rule where. + Epress the result using integrl nottion. Therefore tn + + c. WorkeD emple 8 Find the ntiderivtive for ech of the following epressions: 0 b c Think The ntiderivtive is n inverse sine function of the form sin where. b Tke out s fctor. b The ntiderivtive of is 9 n inverse cos function of the form cos where 7. WriTe sin + c 9 9 cos 7 + c (or sin 7 + c) ebookplus Tutoril int-08 Worked emple 8 Chpter Differentil clculus

31 c Tke out s fctor. c The ntiderivtive of is n 6 + inverse tn function of the form tn where. Alterntivel, prts, b nd c cn be nswered using CAS clcultor. On the Min screen, using the soft kebord, tp: ) - P Complete the entr line s: Press E fter ech entr. Note: The clcultor finds the second form of the ntiderivtive in prt b. Also, it does not include the constnt. You will hve to do tht ourself. Write our solutions, remembering to include the constnt of integrtion tn + c sin + c tn sin 7 + c + c Worked Emple 9 Find ech of the following indefinite integrls: Think b 96 c Use the substitution u. Write Let u. 6 Mths Quest Specilist Mthemtics for the Csio ClssPd

32 Find. Mke the subject. Rewrite the integrl in terms of u. u u Antidifferentite b rule. sin ( u) + c 6 Replce u with. sin ( ) + c b Use the substitution u. b 96 Let u. Find. Mke the subject. or Rewrite the integrl in terms of u. 96 Antidifferentite b rule. 6 Replce u with. c Use the substitution u. c Let u. 9 u 9 u u cos + c cos + c Find. Mke the subject. Rewrite the integrl in terms of u Epress the integrl in stndrd form u 9 + u + u 9 Chpter Differentil clculus 7

33 8 6 Antidifferentite b rule. u tn + c 7 Replce u with nd simplif the surd fctor. 8 tn + c Worked Emple 0 Find the ntiderivtive of Think Divide + into + 6 to mke the rtionl epression ntidifferentible. + 6 b first simplifing the rtionl epression. + Write Simplif + 6 b long division: so ) Rewrite s two seprte integrls Epress in stndrd form Antidifferentite both integrls. 8 tn + c REMEMBER... + sin c 0 +, > + c > 0 cos, tn c + Eercise e Antidifferentition involving inverse circulr functions WE 7 Differentite tn 6 6 nd hence find 6 +. b Differentite cos nd hence find. 9 8 Mths Quest Specilist Mthemtics for the Csio ClssPd

34 c Differentite sin nd hence find. WE 8 Find mnull the ntiderivtive for ech of the following epressions. Check our nswers using CAS clcultor. b c 9 d e f g h i j 6 + Differentite sin. Hence find. Em tip Tke cre tht ou distinguish between nd. The nture of the problem chnges considerbl if the wrong term is used. VCAA Assessment report 00 WE 9 Find ech of the following indefinite integrls. d g 9 b e h i j k 9 6 MC The equivlent epression to is: A D b ( ) The ntiderivtive of B ( ) E is: A cos () + c B cos + c C D cos + c E sin + c C c f l ( ) cos ( ) + [ VCAA 00] c Chpter Differentil clculus 9

35 6 mc The ntiderivtive of 9 + is: A tn 7 + c B 7 tn + c C tn + c D tn 7 + c E tn c 7 We 9b Find f () if f () is equl to: b c d e + 9 f + 6 g h i j k + l m n o ( + ) ( ) ( ) p q r 9 ( + ) + ( ) 6 + ( + ) s t 8 u v. + 8 We 0 Find the ntiderivtive of ech of the following epressions b first simplifing the rtionl epression b c d e f g Differentite sin ( ) nd hence find. 0 Differentite cos ( ) nd hence find ( ). Differentite tn ( ) nd hence find ( + ). A curve hs grdient given b + nd its grph psses through the origin. Find the eqution of the curve. A curve hs grdient given b + +, π 8. Find the eqution of the curve. ebookplus Histor of mthemtics Sofi Kovlevsk 0 mths Quest Specilist mthemtics for the Csio Clsspd

36 f implicit differentition Your techer writes the eqution + on the bord nd sks ou to find. You crefull trnspose the eqution to give ± nd differentite using the chin rule to correctl obtin. There is different w the technique of implicit differentition where ech term in the eqution + is differentited with respect to. Thus + d( ) d( ) d() + d( ) + 0 d 0 d +. ± thus Remember lws to be wre of s n independent vrible nd s dependent vrible, nd to use the chin rule to llow differentition of functions of such s with respect to. WorkeD emple Differentite the eqution + to find in terms of, using clculus b using clcultor. ebookplus Tutoril int-08 Worked emple Think Differentite ech term in the eqution with respect to. Appl the chin rule to eplicitl differentite with respect to. Mke the subject. Substitute for nd write in terms of. WriTe + ( ) d( ) d( ) + ( ) ±. ; < Chpter Differentil clculus

37 b On the Min screen, tp: Action Clcultion impdiff Complete the entr line s: impdiff( +,, ) Then press E. b Substitute for s in prt (which is preferble in this strightforwrd eqution) or continue to use the clcultor to mke the subject in the eqution. Tp: Action Advnced solve Complete the entr line s: Solve( +, ) Then press E. Epress the domin, - 0 shown in the screen in more pproprite form. Tke cre to chnge to < s is in the denomintor in the solution. Write our solution, remembering to include the domin. ± < < ; < < Worked Emple Differentite the following equtions to find in terms of nd. b Think Differentite ech term in the eqution with respect to. Appl the proct rule to the left-hnd side. Write d( ) ( ) ( ) d( ) ( ) ( ) + + ( ) ( ) Mths Quest Specilist Mthemtics for the Csio ClssPd

38 Appl the chin rule to ( ) to eplicitl differentite with respect to. Collect the like terms nd trnspose to mke the subject. b Differentite ech term in the eqution with respect to. Appl the proct rule to the left-hnd side. Appl the chin rule to ( ) to eplicitl differentite with respect to. Collect the like terms nd trnspose to mke the subject. b ( ) ( ) + + ( ) ( + ) + ( ) d () d ( ) ( ) ( ) + 0 ( ) + + Worked Emple For +, find the grdient of the tngent,, t the point (, - ). Hence, determine the eqution of the tngent t this point. Think Differentite ech term in the eqution with respect to. Write + ( ) d + ( ) ( ) Appl the proct rule to the first term. +. ( ). ( ) + 0 Simplif nd mke the subject. + ( + ) 0 + Clculte the vlue of t (, - ) b substitution. At (, - ), ( ) + 8 Chpter Differentil clculus

39 On the Min screen, tp: Action Clcultion impdiff Complete the entr line s: impdiff( +,, ) Then press E. Cop nd pste the nswer without the nd complete the entr line s shown t right. Then press E. 6 The grdient of the tngent t (, - ) is 8. Use this informtion nd the generl eqution of stright line to determine the eqution of the tngent. m ( ) 8 ( ) REMEMBER. d ( ") d ( ") dn. ( ) nn. Eercise f Implicit differentition MC The epression ( ) is equl to: A B C D E WE Differentite the following equtions mnull to find in terms of. Check our nswers with clcultor. - 7 b c + d + e - f - g h + i - Em tip The chin rule will usull be required in implicit differentition problems. Mke sure tht ou understnd how it is pplied. For emple, prt of the problem m require d d ( ) ( ). VCAA Assessment report 007 [ VCAA 007] Mths Quest Specilist Mthemtics for the Csio ClssPd

40 ebookplus Digitl doc WorkSHEET. ebookplus Digitl doc Investigtion Rte of chnge of ngle Using the technique of implicit differentition, find for ech eqution. + b + c + d 0 e + f ( + ) 6 mc ( ) is equl to: A B D. +. E. +. C We Differentite the following equtions mnull to find in terms of nd. Check our nswers with clcultor. + b + c 7 d + e + f + g + h i 6 Use implicit differentition to find mnull in the following. Check our nswers with clcultor b + 0 c d ( ) 0 7 The eqution for circle of rdius units centred on the origin is + 6. Find the grdient of the tngent to the circle t point (, b) on the circle in terms of onl. 8 Consider the eqution. Use implicit differentition techniques to find. 9 We For + 6, find the grdient of the tngent,, t the point (, ). Hence determine the eqution of the tngent t this point. 0 For + + 0, find the grdient of the tngent,, t the point (0, ). Hence determine the eqution t the tngent t this point. Chpter Differentil clculus

41 Summr Derivtives of the tngent nd inverse circulr functions d [tn ( )] sec ( ) d [tn ( f( )] f ( )sec [ f( )] d sin, > 0 d cos, > 0 d tn + Using the second derivtive to nlse the behviour of functions If f (), then the first derivtive is denoted b or f (). The second derivtive is denoted b or f (). Sttionr points nd points of inflection cn be estblished from the second derivtive:. If f () 0 nd f () > 0, then locl minimum sttionr point occurs t.. If f () 0 nd f () < 0, then locl mimum sttionr point occurs t.. If f () 0 nd f () 0, nd f () chnges sign t, then sttionr point of inflection occurs t.. If f () 0 nd f () chnges sign t, then point of inflection occurs t without the point being sttionr point. Antiderivtives involving inverse circulr functions c 0 sin +, > cos + c, > 0 tn c + + Implicit differentition d ( n) d ( n) dn. ( ) nn. 6 Mths Quest Specilist Mthemtics for the Csio ClssPd

42 chpter review Short nswer Find the derivtive of: tn (e ) b + tn ( ). Given tht f () log e (), find f (). If g() ( )( )( + ), find: g () b g (). For the function h() + 7 find: ll of its sttionr points nd their nture b n points of inflection. Sketch the grph of the function +, clerl lbelling ll sttionr points nd -intercepts. 6 If sin, find nd b d. 7 If tn (), find nd b d. 8 Differentite ech of the following: sin ( + ) b cos ( ) c tn ( + ). 9 Find the ntiderivtive of 0 If f () nd f (), find f (). Find f () if f () nd f (0.). 9 For the eqution ( ), use the technique of implicit differentition to find. The eqution for prticulr ellipse is +. Use implicit differentition techniques to determine n epression for nd hence find the coordintes of the point on the ellipse where the tngent to the curve is equl to. Determine f () if f () is equl to: cos () b sin ( ) c sin [cos ()]. Determine f () if f () is equl to: cot () b sin [log e ()] c log e [cos ()]. 6 Sketch the grph of: f :(0, ) R, f () + log e (), clerl indicting n es intercepts, smptotes nd sttionr points. 7 On the sme es s the grph of f () shown below, sketch the grph of: f () 8 If f () (, ) 0 9 (, ) f() b f( ). nd f π, determine f (). 9 Find for the following, using implicit differentition. + + c ( - ) + ( + ) 9 b 0 For + 8, find the grdient of the tngent,, t the point (, ). Hence, determine the eqution of the tngent t this point. Find the eqution of the tngent to the curve - + t the point P (, ). [ VCAA 007] Consider the reltion Find n epression for in terms of nd. b Hence find the ect vlue of when. [ VCAA 006] Given the reltion + +, find the grdient of the norml to the grph of the reltion t the point in the first qudrnt where. [ VCAA 008] A verticl post three metres high csts its shdow on the horizontl ground. Find the length of the shdow s function of the cute Chpter Differentil clculus 7

43 ngle between the sun s rs nd the ground. If the ngle is incresing t rte of 0. per minute, find the rte t which the length of the shdow is incresing when the ngle is. b A serchlight is 8 km from stright costline nd mkes one revolution ever 0 seconds. How fst is the spot of light moving long wll on the cost, if the wll is 0 km from the serchlight? Multiple choice The derivtive of tn () with respect to is: A sec () B sec () C sec ( ) D sec ( ) E The derivtive of tn A 9 sec C sec E sec ( ) is equl to: B sec D 9sec If f () log e (tn ()), then f () equls: A C sec ( ) tn ( ) E sin ( )cos ( ) B tn ( ) If e tn (), then is equl to: A e [sec () tn ()] B e [ sec () tn ()] C e [ sec () tn ()] D 0e sec () E e sec () D sec () tn () The second derivtive of 6 + is: B 0 C 0 D 0 0 E If g() e, then g () is equl to: A 0 e ( + ) C e 0 E 0 e B 00 8 e D 0 e 7 If h() log e ( ), then h () is equl to: A 0 B C D E 8 The second derivtive of 9 cos A C 6 sin sin B D is: cos 6 cos E cos For questions 9 to consider the function f () f () hs locl mimum sttionr point t: A, 7 B (., 6.78) C D (., 6.06) E (, 6) 0 f () hs locl minimum sttionr point t: A (., 7.8) B, 0 C (.79, 6.) E (, 7) D, 6 f () hs point of inflection t: A (, 6) B (, 6) C, 6 D (, 7 ) E, 7 Use the grph below to nswer questions nd. The grph of g() hs: 0, 0 A sttionr points B point of inflection C sttionr points nd points of inflection D sttionr points of inflection E no sttionr points of inflection 8 Mths Quest Specilist Mthemtics for the Csio ClssPd

44 g () 0 when is equl to: A nd B nd C nd D, nd E nd If e k nd d + + 0, then k is equl to: A 0 or B or C or D 0 or E 0 or The grdient of tn () when π 6 is: A B C 8 E D 8 6 The grdient of f () tn () is equl to zero when is equl to: A π 6 C π E p B π D π 7 If f( ) sin 0 then f () is equl to: A B C D E If g() cos () then g () is equl to: A B C D + E 9 The derivtive of tn is equl to: 6 6 A + 9 B 9+ C 9 E + 9 D 9 ( + ) 0 The ntiderivtive of + is: A tn () + c B 0 tn () + c C tn + c D tn ( ) + c E tn + c The ntiderivtive of is equl to: 9 A sin () + c B sin + c C cos () + c D cos + c E 6 sin () + c The ntiderivtive of A sin (6) + c C cos ( 6) + c E cos (6) + c is equl to: 6 B cos (6) + c If + 8 then is equl to: A + 8 B 8 C 8 E + 8 D 6 cos 6 + c D The epression d ( ) is equl to: + b + c + d + e + The rte t which tpe of bird flu spreds throughout popultion of 000 birds in certin re is proportionl to the proct of the number Chpter Differentil clculus 9

45 N of infected birds nd the number of birds still not infected fter t ds. Initill, two birds in the popultion re found to be infected. A differentil eqution, the solution of which models the number of infected birds fter t ds, is: dn ( 000 N) k dt 000 b dn dt c dn dt kn ( )( 000 N) kn( 000 N) d dn dt kn( 000 ( N + )) e dn kn ( + )( 000 N) dt [ VCAA 007] 6 The slope of the curve - 7 t the point where - is: - b - c d e 7 [ VCAA 006] Etended response The cost, $C, of procing litres of prticulr romtic oil on n d is C Find the set up cost, i.e. the cost incurred before n oil is proced. b Find the cost of procing 0 litres of oil. The verge cost per litre of oil proced, A, is A C. c Epress A s function of, A(). d Find the verge cost of procing 00 litres. e Find the minimum verge cost nd the number of litres proced for the minimum cost. f Show tht the rte of chnge of cost is the sme s the verge cost for the number of litres procing the minimum cost. The proportion of the originl popultion of kngroos remining, P(t), t ers fter culling ws introced t onto n islnd cn be modelled b the function: P() t tn, t 0. π Sketch the grph of P(t). b Wht percentge remins er fter culling begn? c B wht percentge hs the popultion dropped fter ers? d When is the popultion hlf of its originl number? e Find the rte of chnge of the popultion proportion fter ers. f When is the rte of decline of the popultion proportion gretest? g When is the rte of decline of the popultion proportion per er? 0π B strnge coincidence, the shpe of hillside for prt of frm follows the eqution f :[0, 6] R, f () Using the first nd second derivtive of f (), show tht is locl mimum nd is locl minimum. b Find the coordintes of ech of the sttionr points. Let A be the locl mimum, B be the locl minimum nd C be point to the right of B such tht t tht point, c, f (c) f (). c Find the vlue of c to deciml plces. The owner of the frm suggests tht becuse there is gull between the points A nd C, he could use tht to crete dm to store wter. d Sketch grph of the function f :[0, 6] R, f () - + +, lbelling the points A, B nd C. Shde in the region bounded b the horizontl line AC nd the curve f () nd hence use integrtion to determine the size of the shded re. 0 Mths Quest Specilist Mthemtics for the Csio ClssPd

46 In Chemistr, functions of the tpe f () e b (where nd b re positive rel numbers) re used to nlse the behviour of electrons inside toms. Show, b using f () nd f (), tht f () hs onl one sttionr point nd tht this point is locl mimum. b Give the coordintes of the sttionr point in terms of nd b. Consider the function f( ) tn where is positive rel number. Show tht the mimum grdient of f () is nd tht this occurs t 0. b Show tht there is point of inflection t 0, using the propert tht point of inflection occurs t when f () 0 nd f () chnges sign t. 6 The eqution k represents n hperbol when plotted on Crtesin grph. Find n epression for in terms of nd when k; k > 0. b Find n epression for in terms of nd when () k; k > 0. c Find n epression for in terms of nd when ()n k; k > 0 nd n J +. d Comment on the results. 6 7 The function f :[0, ) R where f( ) hs first nd second derivtives with rules given b: + 9 ( ) 99 ( 6 + ) f ( ) nd f ( ). ( + ) ( + ) Find the coordintes of the mimum turning point of the grph of f nd use n pproprite test to verif its nture. b i Write down polnomil eqution which, when solved, will give the -coordintes of the points of inflection of the grph of f. ii Find the coordintes of the two points of inflection of the grph of f. Give our nswers correct to one deciml plce. c Sketch the grph of f, clerl indicting the loction of n intercepts with the es, the mimum point nd the two points of inflection. [ VCAA 008] ebookplus Digitl doc Test Yourself Chpter Chpter Differentil clculus

47 ebookplus CTiviTieS Chpter opener Digitl doc 0 Quick Questions: Wrm up with ten quick questions on differentil clculus. (pge 96) A The derivtive of tn (k) Tutoril We int-079: Wtch how to differentite composite nd proct functions. (pge 99) We6 int-080: Wtch tutoril on how to use clculus to find second derivtives of f(). (pge 0) C Anlsing the behviour of functions using the second derivtive Interctivit Second derivtive int-06: Consolidte our understnding of the use of second derivtives. (pge 08) Tutoril We 0 int-08: Lern how to use clculus to find sttionr points of function including points of inflection, nd use this informtion to sketch the grph of the function. (pge ) Digitl doc WorkSHEET.: Find first nd second derivtives, sttionr points nd their nture, nd sketch grphs of polnomils. (pge ) D Derivtives of inverse circulr functions Tutoril We int-08: Wtch how to differentite inverse trigonometric sine functions. (pge 7) E Antidifferentition involving inverse circulr functions Tutoril We 8 int-08: Wtch how to ntidifferentite function where the ntiderivtives re inverse trigonometric functions. (pge ) Digitl doc Histor of mthemtics: Lern bout Sofi Kovlevsk, nineteenth centur mthemticin. (pge 0) F Implicit differentition Tutoril We int-08: Wtch how to differentite reltion using implicit differentition. (pge ) Digitl doc WorkSHEET.: Find first nd second derivtives, sttionr points nd their nture, nd find tngent nd norml equtions. (pge ) Rte of chnge of ngle: Investigte the rte of chnge of n ngle. (pge ) Chpter review Digitl doc Test Yourself: Tke the end-of-chpter test to test our progress. (pge ) To ccess ebookplus ctivities, log on to mths Quest Specilist mthemtics for the Csio Clsspd