Introution to Differentition Introution to Differentition Curriulum Rey www.mtletis.om
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Introution to Differentition Differentition is proess tt works on most funtions. Te wole point of ifferentiting funtion, is to fin te grient of tngent to tt funtion (t speifi point). One use of tis is fining te eqution of te tngent for speifi funtion. Answer tese questions, efore working troug te pter. I use to tink: How mny tngents n funtion ve? Wt is te ifferene etween tngent n sent to funtion? Look t te imge elow, wt soul te vlue of e to turn te sent into tngent? y Sent Tngent + Answer tese questions, fter working troug te pter. But now I tink: How mny tngents n funtion ve? Wt is te ifferene etween tngent n sent to funtion? Look t te imge elow, wt soul te vlue of e to turn te sent into tngent? y Sent Tngent + Wt o I know now tt I in t know efore? 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Bsis A Tngent to Curve A tngent t point is strigt line tt toues urve only one t tt speifi point. A sent uts troug te urve: Tis is tngent y Point of ontt y Tis is sent Tis is sent Tis is tngent E urve n ve n infinite numer of tngents euse tere n e tngent t every point. y y Tis is euse tese grps n ve tngent t every -vlue. Wt oes tis men? Look t tis emple: Look t tis grp of f () elow Tngent t - y Tree tngents ve een rwn to f () A tngent t - - - - - 0 - - - - A tngent t - A tngent t Tngent t - Tngent t L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Bsis Grient of Tngent It turns out tt nlysing tese tngents is very useful in mtemtis. A prtiulrly useful prt is te tngent s grient. Sine te tngent is strigt line (wit eqution y- y m ^ - ) its grient is m. Here is ow to fin te grient. To fin te grient of te tngent to f () t, strt y rwing sent troug f () utting te urve t te point (, f ()). y ( +, f ( + )) (, f ()) f () Sent Tngent If is me smller, ten te sent is BETTER pproimtion of te tngent (euse it s loser). y ( +, f ( + )) (, f ()) f () Sent Tngent + Te grient of te sent is: + Te grient of te sent is: m y - y - m y - y - m f^+ - f^ ^+ - m f^+ - f^ ^+ - ` m f^+ - f^ ` m f^+ - f^ From tis, it s esy to see tt te smller te vlue of, te loser te sent moves towr te tngent. Tis mens s moves towr zero, te sent moves towr te tngent. n never e zero sine it is in te enomintor. So, te grient of te tngent t n e foun using tis it: " 0 f^+ - f^ Te numer foun in tis it is te grient of te tngent t. It is lso lle te slope of f () t. Te grient of te sent is lle te verge rte of nge of y wit respet to. Te grient of te tngent is lle te instntneous rte of nge of y wit respet to. 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Bsis Here re some emples of fining te grients of some tngents to grps. Answer tese questions for f^ + Fin te grient of te tngent to f () t Fin te grient of te tngent to f () t - Te grient of te tngent t is given y: " 0 f^+ - f^ ` Te grient of te tngent t is given y: " 0 " 0 " 0 " 0 f^+ - f^ 6^+ + ^+ @ - 6 + @ + ^ + ^ + " 0 Te grient of te tngent t is given y: " 0 f^+ - f^ ` Te grient of te tngent t - is given y: " 0 6^- + + ^- + @ -6^- + ^- @ " 0 " 0 " 0 f^- + - f^- - ^ - ^ - " 0 - ` Te grient of te tngent t is m. ` Te grient of te tngent t - is m -. Here is grpil representtion of wt ws just lulte. y 8 7 6 Tis tngent (t -) s grient of m - Tis tngent (t ) s grient of m - - - - 0 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Bsis Te Derivtive Te erivtive (written s fl^) is te funtion for te grient of te tngent in terms of, not. fl^ " 0 f^+ -f^ Tis is lle te Derivtive Funtion Ten, fl^ is te grient of te tngent to f () t te point. Consier te funtion f^ + Fin te erivtive, fl^. fl^ " 0 f^+ - f^ [( ) ( )] [ ] + + + - + " 0 + + " 0 ^ + + " 0 ^+ + " 0 + Fin f l^. Fin f l^-. From, fl^ + ` f l^ ^+ Notie, tis is te sme s in te previous emple on te previous pge. From, fl^ + ` f l^- ^- + - Notie, tis is te sme s in te previous emple on te previous pge. Te proess of fining te erivtive is lle ifferentition. Te nottion f l() is not te only wy to write te erivtive. Tese re ll te ifferent wys to write te erivtive of f (): fl^ y [ f^ ] D[ f^] yl Te ove five nottions ll men te sme ting, te erivtive. 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Bsis Here re some more emples of fining te erivtive. Fin te following If f^ - + ten fin fl^. fl^ " 0 f^+ - f^ 6^+ - ^+ + @ -6 - + @ " 0 + - " 0 ^+ - " 0 - Fin yl if y - +. y Fin if y. yl " 0 f^+ - f^ yl " 0 f^+ - f^ 6- ^+ + @ -6- + @ " 0 ^+ - " 0 - - 6 " 0 + + " 0 ^-- 6 " 0 ^ + + " 0-6 Fin D 6 - @. e If g ^ + ten fin gl^. D 6 - @ " 0 f^+ - f^ gl^ " 0 f^+ - f^ 6^ + - @ -6 - @ " 0 6^+ + ^+ @ - 6 + @ " 0-6 " 0 + 8+ " 0 ^+ 6 " 0 ^+ 8+ " 0 6 8 + ` gl^ 8^+ 7 6 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Questions Bsis. Use te grps elow to nswer te questions. Drw ny two tngents to f () elow. y Drw tngent t -, n. y - - - - 0 - - - - -6 - - - - - 0 - - - - - -6-7. Answer te questions out tese grps.. y. - - - - 0 f^ - - - - - -6-7 f^ 7 6 y - - - - - 0 6 - - - Wi of te lines in te grp is sent? Wi of te lines in te grp is sent? Fin te grient of tis sent. Fin te grient of tis sent. Wi of te lines in te grp is tngent? Wi of te lines in te grp is tngent? Fin te grient of tis tngent Fin te grient of tis tngent 00% Introution to Differentition Mtletis 00% P Lerning L 7 C SERIES TOPIC NUMBER
Introution to Differentition Questions Bsis. Use te grp of f () elow to nswer tese questions: Wt re te oorintes of point A? y f^ B Wt re te oorintes of point B? A + Wt is te grient of te sent pssing troug points A n B? Wt nees to ppen to so tt te sent moves loser to te tngent t? e Write te formul for te grient of te tngent t. 8 L C 00% Introution to Differentition SERIES TOPIC NUMBER Mtletis 00% P Lerning
Introution to Differentition Questions Bsis. Fin te grient of te tngent to g^ - t: -. Fin te grient of te tngent to g^ - + t: - 00% Introution to Differentition Mtletis 00% P Lerning L 9 C SERIES TOPIC NUMBER
Introution to Differentition Questions Bsis 6. f^. Fin fl^. 7. q^. Fin D6 q^@. y 8. y + +. Fin. 9. Differentite p^ 7 -. 0. Fin 6- @. 0 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Knowing More Te erivtive of f ^ is given y: fl^ " 0 f^+ - f^ Wen tis formul is use it is esrie s ifferentiting y first priniples. Tis long formul oesn t ve to e use ll te time euse tere re sortuts. A Sortut for Eponent Derivtives Look t tese erivtives wi were foun using first priniples: 6 @ 6 @ 6 @ 8 6 @ 9 6 @ Cn you see pttern in tese? Here is te sortut, using te seon emple to emonstrte: 6 @ # Te new eponent is less tn te originl Te eponent multiplies te term In generl tis sortut formul n e use for tese type of prolems: [ ] n # n n - You n see te proof for tis t te en of te ook Here re some emples: Fin tese erivtives: [ ] [- ] [ ] # - # - -0 - # # - 0 [ 0] 0 [ 0 ] - 0# 0 0 0 Py prtiulr ttention to emples n ove. From emple, it n e seen tt te erivtive of liner term is just te oeffiient n no powers of. From emple, te erivtive of ny onstnt (no vriles) is 0. So te sortut n tully e written s sortuts, if is onstnt n n is wole numer ten: [ ] n # n n - [ ] [ ] 0 Liner term Constnt 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Knowing More Negtive n Frtionl Eponents Te sortut lso works wen te eponents re negtive or frtions. Here re some emples of negtive eponents 6 - @ 6 - @ - ; E -# - - - -- -# -6-6 - -- Weter to leve your nswer in negtive eponents or s frtions (positive eponents) epens on te question - [- ] -# - 6 6 - -- Te sme rule pplies wen te eponents re frtions. Emples of erivtives wit frtionl eponents 6 @ 6 - @ 6 @ 6 @ # - # - - # - 6 @ - - - # - Rememer te ifferent nottions for te erivtive. Fin te erivtives of tese funtions D 6 @ Fin yl if y 6 pl^if p ^ 6 # 6 - y' # 6 - - pl^ 0 # 6 - After some prtie you my fin tt you n skip te mile step n move strigt to te nswer. L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Questions Knowing More. Differentite te following y first priniples, n ten use te sortut. Sow ot erivtives re equl. f^ g ^. Fin tese erivtives: D 6 @ yl if y 6 fl^if f^ 7 6 @ e ml^ if m ^ 000000 f 60 @ g D 0 6 - @ 60 0 @ 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Questions Knowing More. Fin tese erivtives. Leve your nswers wit negtive eponents if neessry. 6 @ 6 - @ 6 @ 8 B e 6 - @ f 6 - @ g 9-8 B 0-8- B i 68 @ j 6 @ k 6-7 @ l 6 8 @ - - L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Knowing More Aing n Sutrting Up until now only te erivtive of single term s een foun. Tt is, someting like or 6 @ 6- - @. However, wt woul ppen if tere is more tn one term like 6 + - @? In te se of ition (+) n sutrtion (-), to fin te totl erivtive, just fin te erivtive of e term. Fin tese erivtives ; + - E 6 @ + 6 - @- 6 @ - - - ^# + ^- # + 0 Tese steps n e left out if you oose. - 0 - - 6 67 + 8 - + + 0@ 7 8 6 6 - @ + 6 @- 6 @ + 6 @ + 60@ 6 - - - - - ^- # 7 + ^# 8 - ^# + ^6# + 0 Tese steps n e left out if you oose. - - 7 + - 6 + Te forml wy to write tis is s te sum rule (f () + g()) n te ifferene rule (f () - g()) for ifferentition. Te Sum Rule: 6 f^+ g ^ @ fl^+ gl^ Te Differene Rule: 6 f^- g ^ @ fl^- gl^ [ f ^ + g ^ ] Proof 6 f^+ + g^+ @ - [ f^+ g ^ ] " 0 f^+ - f^ g ^ + - g^ + " 0 " 0 fl^+ gl^ [ f ^ - g ^ ] Proof 6 f^+ - g^+ @ - [ f^- g ^ ] " 0 f^+ - f^ g ^ + - g^ - " 0 " 0 fl^- gl^ So, oring to te sum rule, te erivtive of sum is foun from te sum of te erivtives. Similrly, for te ifferene rule, te erivtive of ifferene is te ifferene of te erivtives. Tis is NOT te se for prout ^# or quotient ^'. 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Questions Knowing More. Fin tese erivtives: 6 + - @ 6 @ + [ ] + [-] Are te erivtives in n equl? Is tis wt you were epeting?. Fin tese erivtives: 9 ; 6 + - E 9 ; 6 E+ + - 8 6 B @ Are te erivtives in n equl? Is tis wt you were epeting? y 6. Fin for tese funtions: y + y + y + y ^ - + e y + - f y 6 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Knowing More Te Prout Rule After working wit te sum rule n ifferene rule your instint my e tt prouts work te sme wy ut tey o not: 6 f^: g ^ @! fl^: gl^ Here is n emple to prove tis: Let f^ - n g^- - 7 Fin 6 f ^ : g ^ @ Fin fl^: gl^ f g 6 ^ : ^ @ 6^ - ^- - 7@ 6- - + + @ f g l^ : l^ 6 - @ : 6- - 7@ ^6 - ^- -8 - + 6 + - 6 Are te erivtives in n equl? No, tey re not equl. Tere is speifi rule use wen fining tese erivtives lle te prout rule. Prout rule 6 f ^ : g ^ @ f l^ : g ^ + g l^ : f ^ You n see te proof for tis t te en of te ook Here re some emples using te prout rule: Fin tese erivtives 6 ^ - ^ - - 7@ Let f^ - n g ^ -- 7 6 ^ + + ^ + @ f () g() f l() g() gl() f () ` 6 ^ - ^-- 7@ ^6 - ^-- 7+ ^-^ - -6 - + + - + Let f ^ + + n g^ + -8 - + 6 + f () g() f l() g() gl() f () ` 6 ^ + + ^+ @ ^+ ^+ + ^^ + + 6 + 7+ + + 9+ 6 9 + 6+ 8 00% Introution to Differentition Mtletis 00% P Lerning L 7 C SERIES TOPIC NUMBER
Introution to Differentition Questions Knowing More 7. Fin tese erivtives: 6 ^+ ^- @ 6 + @ : 6 - @ Are te erivtives in n equl? Is tis wt you were epeting? 8. Fin tese erivtives using te prout rule: 6 ^ + ^ - @ 6 ^ - + ^6-6@ 6 ^7- ^ - + 8@ 6 ^ - @ e 8 ` + jb f ; - + m + ^ E 8 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Knowing More Te Quotient Rule Te quotient rule is use to ifferentite funtions wit vrile enomintors. It is esy to mke te mistke tinking te erivtive of quotient is te quotient of te erivtives. Tis is not true: Here is n emple to prove tis: Let f^ + n g^ f^ fl^ ; E! g ^ g l ^ Fin ; f^ g ^ E Fin fl^ g l^ f^ ; E ; + E g ^ 6 + @ f l() + n gl() fl^ ` + gl^ + It is ler tt tese re not equl. Tere is speifi rule use wen fining tese erivtives lle te quotient rule. Quotient rule f^ ; E g^ fl^: g^- gl^: f^ 6g^@ You n see te proof for tis t te en of te ook Here re some emples using te quotient rule: Fin tese erivtives f () fl^ g() gl^ f () f () fl^ g() gl^ f () ; + E ^+ ^ - ^^ + ^ - ^ ^ -^ ^ - ; E ^ g() 6 g ^ @ g() 6 g ^ @ + - - 7 7-9 + 6 6 7 6 + 6 ^6 + 6 6 6 6 + 6 00% Introution to Differentition Mtletis 00% P Lerning L 9 C SERIES TOPIC NUMBER
Introution to Differentition Questions Knowing More 9. Fin tese erivtives: 6 + @ 68- @ ; + E 8- Are te erivtives in n equl? Is tis wt you were epeting? 0. Answer tese questions using te quotient rule: Fin f l() if f () - + Fin D ; E - Fin y if y + Fin ; - + - E - e Fin ; + E - f Fin l^if ^ - g Fin - ; E + Fin - + ; E + 0 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Knowing More Wt out someting like ^ + 7? Of ourse it is possile to multiply tis rket wit itself 7 times to fin 00 polynomil n ten ifferentite. However, tis oul tke long time n wt out ifferentiting ^ +? Te Cin Rule Also lle te funtion of funtion rule Look t te epression ^ +. Tere is n insie funtion + n n outsie funtion 7. Let s ll te insie funtion g^ + n te outsie funtion f^ 7. Ten te originl epression n e written s 7 f^g^ ^ +. 7 A funtion of funtion is lle omposite funtion n tey ve teir own rule for ifferentition, lle te in rule. Cin rule 6 f ^ g ^ @ f l^ g ^ : g l^ You n see te proof for tis t te en of te ook In oter wors tis n e si s: Te erivtive of omposite funtion is te erivtive of te outsie funtion, witout nging te insie funtion, times te erivtive of te insie funtion. Here re some emples: Fin tese erivtives 6 ^ + @ 7 Te insie funtion is g^ + Te insie funtion is g^ - - Te outsie funtion is f () 7 Te outsie funtion is f^ 8 6 ^ - - @ 8 ` 6 ^ + 7 @ 7-7^ + # ^ + 6 7^ + ^ + ` 6 ^ - - 8 @ 8 8 ^ - - # ^6 - - 0 7 8 ^ - - ^6 - - You n leve your nswers in tis form witout epning te rkets. Te sme rule n e use for ny insie n outsie funtion, even roots. Here re some emples: Fin tese erivtives 6 @ - Write s frtionl eponent 8 ^- + B Write s frtionl eponent 8^ - B 8^- + B Insie funtion Outsie funtion Insie funtion Outsie funtion - 6 ^ - ^ - ^- + ^- + Derivtive of outsie funtion Derivtive of insie funtion Derivtive of outsie funtion Derivtive of insie funtion 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Questions Knowing More. Fin tese erivtives using te in rule: ^ + 6 ^ + 6 ^- - ` + j e 8 ^ + - f + g 6 ^ - + ^- + - + i + j + m L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Questions Knowing More. Let s sy m^ -. Rewrite tis wit frtionl eponent. m() is omposite funtion. Ientify te outsie funtion n te insie funtion. Use te in rule to fin ml(). Hene, use te prout rule to fin l() if ^ ^ - ^ -.. Let q^ ^ + - +. 8 q() is omposite funtion. Ientify te outsie funtion n te insie funtion. Use te in rule to fin ql(). Hene, use te quotient rule to fin sl^if s ^ ^ + - +. 8 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Using Our Knowlege Comining Rules for Differentition Rememer, tere re severl rules to keep in min wen fining erivtives, epening on te emple. Wi rule is require to ifferentite tese? 6 ^ - + ^ - @ f () g() ; + E f () g() ^ - 6+ 0^- + ^- 0^ - + 8 - + 6 + 0 6 ^ - + @ Prout rule ^+ 0^ - ^^ + - Quotient rule ^ - + ^ - 8 Cin rule Up until now only one of tese rules s een neee to fin erivtive. Sometimes more tn one of tese rules is neee to fin erivtive. Here re some emples: Fin tese erivtives Tis is omintion of te prout rule n te in rule 6 ^ + ^- @ ^ + ^^- + ^^ + f () () fl^ g ^ gl^ ^ + ^ - + ^ + f^ Te in rule is require to fin just fl^ sine f () is omposite funtion In tis emple, te in rule n te prout rule were use. Tis is euse to fin prt of te prout rule, fl^, te in rule ws neee. () fl^ g ^ gl^ f^ ^ - ^ + 6 ^ + + ^ - @ - ^ - ^ + ; E g() 6 g ^ @ 8 - + 0 + 0 In tis emple, te prout rule n te quotient rule were use. Tis is euse to fin prt of te quotient rule, fl^, te prout rule ws neee. L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Using Our Knowlege Sometimes te omintion of rules neee is te sme rule, more tn one. Hve look t tese emples: y Fin for e of te following y 6 ^ + + @ Outsie funtion Insie funtion y ` 6^ + + @ # 6^ + : + @ Derivtive of outsie funtion Derivtive of insie funtion 6^ + + @ 6 + 7@ Tis is te in rule of te in rule (te in rule ws use twie). In ses like tese, lwys strt wit te outsie funtion n move in, until te inner-most funtion s een ifferentite. Of ourse, tese n get very omplite. It my e require tt e rule is use. y Fin for e of te following f () y ^- ^ - ^ + g() fl^ g ^ gl^ f () y ` 6^^ - + ^ - ^- @ ^ + - 6^ + ^@ ^- ^ - 6^ + @ In tis emple, more tn one rule for ifferentition ws use: Te prout rule ws use to fin f l(). 6 g ^ @ Te in rule ws use to fin gl(). y Te quotient rule ws use to fin. So ll rules were use one. y Usully, simplifying n epression similr to in tis emple wouln t e neessry. However, it n e simplifie to: y ^8 - - + ^ + - 6 ^ + ^ - - 6 + 6 ^ + 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Questions Using Our Knowlege. Consier te funtion ^ ^+ ^-. Wi ifferentition rules woul you nee to fin l()? Fin l(). ^. Consier te funtion f () +. Wi ifferentition rules woul you nee to fin f l()? Fin f l(). ^ + + ^ - 6. Consier te funtion y. y Wi ifferentition rules woul you nee to fin? y Fin. 6 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Questions Using Our Knowlege 7. Fin te erivtives of tese funtions: 8 f^ ^ - + - g ^ ^ + ^ - y - ^ - m ^ ^+ ^- - 7 e p ^ 6 ^7 - ^- + 6 - f t ^ 6 ^^+ + - 8@ 00% Introution to Differentition Mtletis 00% P Lerning L 7 C SERIES TOPIC NUMBER
Introution to Differentition Using Our Knowlege Te Seon Derivtive Sometimes it is useful to fin te erivtive of erivtive, in oter wors fining te erivtive twie. Up until now funtions ve only een ifferentite one. Tis is lle te first erivtive wi oul ve tese for nottion: fl^ Te seon erivtive s tis nottion: 6 @ y fll^ y 6 @ Nottion like tis, mens ifferentition must e one twie to fin te erivtive of te erivtive. Fin te following: fll^ if f^ ^ - fl^ # ^ - ` fll^ # 6 6 + + @ 8 6 + + @B 6 + + 0@ 6 + y if y ^ - gll^if g ^ - + y y ; E Quotient rule gl^ ^ - ^ -^^ - + ^ 6^- + ^ @ Prout rule - - 6 69 - @ 8 - Quotient rule ` gll^ ^ - ^ -^8^ - - 6 ^ 6 8-6 - + 6 + 8 6 e y if y ^- + - + y y ; E Cin rule - 6 # ^- + # ^- @ 0 6- ^- + @ 0 9-66^- + + 60^- + # ^- ^- @ -66^- + - 660 ^- + 0 9 Prout rule 8 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Using Our Knowlege Higer Orer Derivtives It is possile to fin te seon erivtive (te erivtive of te erivtive), so it must e possile to fin tir erivtive n fourt erivtive n so on. Tis is wt ppens wen funtion is ifferentite multiple times. Of ourse tis lso s its own nottion. Te tir erivtive oul e written s one of tese: flll^ y 6 @ Te first nottion (wit te ses) n eome quite long, te iger te orer of te erivtive. For emple, it n e time onsuming to write fllllllllll^ for te 0t erivtive. A simpler wy to write tis is f (0) (). Usully for n $ tis nottion is use for te n t erivtive: f n^ n y n n 6 @ Here re some emples: Let f^ 6 n g^ - n y ^- 8 Fin flll^ Fin f ^ ^ fl^ 6 from, f () () 0 ` ` fll^ 0 flll^ 0 ` f () () 60 Fin f ^ ^ Fin g ^ ^ ^ from, f ^ 60 gl^ 8-6 ^ ` f ^ 70 ` gll^ - 6 ` glll^ 8-0 8 ` g ^ ^ 8 e y Fin y 8 ^ - # ^- 7 7 y ` 7 8 # ^ - # ^ - 6 6 y ` 6 76 # ^ - # ^ - ^ ` g ^ 0 Cin rule 00% Introution to Differentition Mtletis 00% P Lerning L 9 C SERIES TOPIC NUMBER
Introution to Differentition Questions Using Our Knowlege 8. Fin te following: 6 - + - + @ 6 @ + - y - were y + ; + E 9. Use te prout n in rules to fin te following: 6 fll^if f^ ^ - - gl^if g ^ ^ + 0 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Questions Using Our Knowlege 0. Use te quotient n in rules to fin te following: fll^if f^ + ll^if ^ - + + mll^if m ^ ^ + 7. Let q^ 7. (i) Wt oes qlll^ men? (i) Wt oes q ^7 ^ men? (ii) Fin qlll^. (ii) Fin q ^7 ^. 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Questions Using Our Knowlege ^. Let p^ ^ + ^ - + - n q^. ^ - Fin pll^. Fin qll^. L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Tinking More Te Grient of Tngent A tngent is just strigt line. Tis mens tt tngents will ve te eqution y m +. Tese sttements re ll equivlent: Te erivtive of f () n e use to fin te grient of te tngent t ny point. f l() is te vlue of te grient of te tngent to f () t te point. Just s reminer, ere is n emple: Fin te grient of te tngent to ^ - + t te following points - First, fin l() l^ - + First, finl^ l^ - + ` l^ -^- + ` l^- - ^ + - ` Te grient of te tngent t - is. ` Te eqution of te tngent is y +. ` Te grient of te tngent t is -. ` Te eqution of te tngent t y - +. y Tis tngent (t ) s grient of - -7-6 - - - - - 0 6 7 8 9 - - Tis tngent (t - ) s grient of - - - -6 Te ove meto fins te grient (te vlue for m) for te eqution of te tngent y m +. Te only vlue missing to know te entire eqution for te tngent is te -vlue. 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Tinking More Te Eqution of Tngent Fining te eqution of tngent y- y m ^ - involves two steps: Fin te vlue of te grient m. Sustitute point ^, y into y- y m ^ -. Fin te eqution of te tngent to f^ - t fl^ - ` f l^ ^ - ^ ` Te grient of te tngent t is. ` Te eqution of te grient is y- y m ^ -. Fin te vlue of te grient of f () t f^ ^ - ^ 9 ` Te point (, 9) is on te tngent n on f (). Sustitute (, 9) into y- 9 ^- ` y- 9 - ` y - 6 Fin te point (, y) t te point Sustitute te point into y- y m ^ - ` Te eqution of te tngent to f () t te point s te eqution y - 6. Fin te eqution of te tngent to g^ os^ t te point gl^-sin^ r ` gl r sin r ` j- ` j- ` Te eqution of te tngent is y- y -^-. g r os r ` j ` j 0 ` Te point (, 0) is on te tngent n on g(). First, fin te vlue of te grient of g() t r Fin te point (, y) t te point r Sustitute (, 0) into y- y -^ - ` y- 0 -^ - ` y - + ` Te eqution to te tngent t r is y - +. Sustitute te point into y- y m ^ - L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Tinking More Te Eqution of Norml Wt is norml? A norml is line perpeniulr to te tngent t te point of ontt (tngent toues te funtion). Te igrm elow sows te ifferene etween tngent n norml: y Tis is funtion f () - - 0 6 - - Tis is tngent to f () t - - A norml is lso just strigt line, wi mens it lso s te eqution y- y m ^ -. Sine te norml n te tngent re perpeniulr, it mens tt te prout of teir grients is -. Rememer, two lines (wit grients mn m) re perpeniulr if n only if m# m -. For tngents n normls te rule oul e written s: Tis point is ommon to te funtion, te tngent n te norml Tis is norml to f () t. Te norml n te tngent interset t rigt ngle t m # m - T N T is for Tngent N is for Norml Here is n emple wi fins te norml of funtion t speifi -vlue: Let f^ + -. Fin te norml to f^ t given tt te tngent t s te eqution y 9 - Te grient of te tngent is mt 9 ` mn n e foun using mt # mn - ` 9# mn - ` m N - 9 ` Te eqution of te norml is y- y - _ 9 - i. Fin te point (, y) on f () t : f^ ^ + ^ -. ` Te point (, ) lies on f (), te tngent to f () n te norml to f () (it is te ommon point). Sustitute (, ) into te eqution of te norml. y- y - _ - i 9 y- - _ 9 - i ` y - - 9 + 9 ` y - 9 + 9 + ` Te eqution of te norml is y - 9 9 + 9 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Tinking More Here is n emple requiring tngent n norml: Let t() - 6 + Fin tl^. tl^ - 6 Use te prout rule Use t'() to fin te grient of te tngent to t() t -. tl^- ^- - 6 Tis is te grient ` Te grient of te tngent to t () t - is ` te eqution of te tngent is y- y ^-. Fin te eqution of te tngent to t () t -. t^- ^- - 6^- + -8 ` Te point (-, -8) is ommon to te tngent n t (). Fin te point ^y, t te point - Sustitute tis point into te eqution of te tngent. y-^- 8 ^-^- ` y + 8 ` Te eqution of te tngent is y + 8. Sustitute te point into y- y m ^ - Fin te eqution of te norml to t () t -. mt (from ove) ` mn# m - ` mn # - ` m N - T So te eqution of te norml is y- y - ^-. Sustitute te point (-, -8) into tis eqution. y-^- 8- ^-^- y+ 8 - - ` y - - ` Te eqution of te norml t - is y - -. Fin te grient of te norml using perpeniulr lines Sustitute te point into y- y m ^ - 6 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Questions Tinking More. Lel te following on tis igrm: y Te tngent to f () t -. Te tngent to f () t. - - - - - 0 - Te norml to f () t -. - - - - Te norml to f () t.. Fin te grients of te tngents to tese funtions: g ^ - + - 7 t te point -. ^ ^ - t te pont -. 8 f^ ^+ ^ - + t te point 0. k ^ - t te point -. - 00% Introution to Differentition Mtletis 00% P Lerning L 7 C SERIES TOPIC NUMBER
Introution to Differentition Questions Tinking More. Fin te eqution of te tngents to tese funtions: f^ - t te point -. ^ ^ + - t te point -. p + ^ t te point. ^ ^+ ^ + t te point 0. - 8 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition 6. Consier te funtion f^ - +. + Fin fl^. Questions Tinking More Fin te grient of te tngent to fl^ t te point. Fin te eqution of te tngent t. Fin te grient of te norml t te point. e Fin te eqution of te norml t te point 00% Introution to Differentition Mtletis 00% P Lerning L 9 C SERIES TOPIC NUMBER
Introution to Differentition Questions Tinking More 7. Fin te eqution of te tngent n te norml for te funtion t^ ^ - + ^- + - t te point. 0 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Tinking More Proofs for Some Differentition Rules Te Sum Rule Te Differene Rule 6 f^+ g ^ @ 6 f ^ @ + 6g ^ @ 6 f^- g ^ @ 6 f ^ @ - 6g ^ @ Proof 6 f ^ + g ^ @ Proof 6 f ^ - g ^ @ 6 f^+ + g^+ @ - 6 f^+ g ^ @ " 0 6 f^+ - g^+ @ -6 f^- g ^ @ " 0 f^+ - f^ g ^ + - g^ + " 0 " 0 f^+ - f^ g ^ + - g^ - " 0 " 0 fl^+ gl^ fl^- gl^ Te Prout Rule 6 f ^ : g ^ @ f l^ : g ^ + g l^ : f ^ Proof f^+ : g^+ - f^: g ^ 6 f ^ : g ^ @ " 0 A n sutrt f^ : g ^ +. Sutrting n ing te sme term s no effet on te originl epression. " 0 " 0 f^+ : g^+ - f^: g ^ + 6 f^ : g ^ + - f^ : g^+ @ f^+ : g^+ - f^ : g ^ + + f^ : g^+ - f^: g ^ g ^ + 6f^+ - f^@ + f^6g ^ + - g^@ " 0 g ^ + 6f^+ - f^@ + f^6g ^ + - g^@ " 0 f^+ - f^ g ^ + - g^ g ^ + + f^ " 0 " 0 g ^ : fl^+ f^: gl^ 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Tinking More Te Quotient Rule f^ ; E g ^ fl^: g ^ - gl^: f^ 6g ^ @ Proof f^ ; E g ^ f^+ f^ - g ^ + g ^ e " 0 " 0 f^+ : g^- f^: g^+ o g ^ : g ^ + " 0 f^+ : g^- f^: g^+ g ^ : g ^ + : A n sutrt f^: g ^. Sutrting n ing te sme term s no effet on te originl epression. " 0 " 0 f^+ : g^- f^: g^+ + 6 f^: g ^ - f^: g ^ @ g ^ : g ^ + : f^+ : g^- f^: g^+ f^: g^- f^: g^+ g ^ : g ^ + : g ^ 6f^+ - f^@ - f^6g^+ - g ^ @ # " 0 g ^ : g ^ + f^+ - f^ g ^ + - g^ # g ^ - f^ m " 0 g ^ : g ^ + # ^g ^ : fl^- f^: gl^ 6g ^ : g ^ @ fl^: g ^ - gl^: f^ 6g ^ @ Anoter wy to write first priniples y fl^ " f^- f^ - ^, f^ Sent ^, f^ Tngent L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Tinking More Te Cin Rule (lso lle te funtion-of--funtion rule) 6 f^g^@ fl^g ^ : gl^ Proof Use te lterntive first priniples for ifferentition. Let ^ f^g^ ` l^ " " " f^g^ - f^g ^ - f^g^ - f^g ^ g ^ - g ^ # - g ^ - g ^ f^g^ - f^g ^ g ^ - g ^ # g ^ - g ^ - fl^g^ # gl^ ` l^ fl^g^: gl^ ` l^ fl^g^: gl^ Te Power Rule 6 @ n # n n- Proof Using first priniples, let f^ ` fl^ " 0 f^+ - f^ n ^ + - " 0 n ^+ - : " 0 n n n - - - - ^ $$$ : + + + + + + - " 0 n n n n n n n : ^ + + $$$ + + " 0 n- n- n- n- Tis is lle te Binomil teorem : n - 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition More Questions Bsis:. Wi its epression is given te erivtive of te funtions t. ^+ - " 0 ^ -^- " 0 " 0 - ^- - " 0. Wt is te erivtive of y? 0 6 9. Wt is te vlue of ^8? 8 8 8^+ - 8. Prtie using funtion nottion: If f ^ -, fin f ^ n f ^. Given tt fl^ ^ +, fin f l^0 n f l^.. Use te grp elow to nswer te questions: y Wt re te oorintes of te point A? + Wt re te oorintes of te point B? A f^ B e Wt is te grient of te sent pssing troug points A n B? Wt nees to ppen to so tt te sent moves loser to te tngent t? Write te formul for te grient of te tngent t. 6. Wi of tese lines re tngents to te urve f^? Eplin. f g i f^ e 7. Fin e of te following erivtives using first priniples: If f ^, fin fl^. 6 + + @ If g^ -, fin D 6 g^@. e If q^, fin ql^. f For y - 7 y, fin. 6@ were is some onstnt. 8. Fin te grient of te tngent to ^ - t: ^, - 9. 0. If y + fin te vlue of if te erivtive equls 9 wen. For wi vlue of o te funtions y 6 + n y + 8 ve tngents wit te sme grient? L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition More Questions Knowing More:. Wi of tese is te erivtive of? y y y 0 y 0... Wt is te rule for ifferentiting te funtion f^? f n n - l^ f n- n - l^ ^ fl^ ^n- n fl^ n Wi of tese oes not give you te erivtive funtion of te funtion f ^? 6 8 @ 6 f ^ @ D 6 7 ^ fl^ Differentite y first priniples, n ten verify y using te sortut rules. g ^ 7 - y n 8. Fin tese erivtives: D 8 67 @ 60-90 @ If y -, fin yl. 6. Derive te following: 6 0 @ 6 - @ ;- 6 E 6 - @ e 8 B f 66 @ g 6 @ - - ; - E 8 9 7. Fin yl for e of tese funtions: y + 7 y + 7 y + 7 y - 9 + e 6 y ^ - 8 f y + 7 8-8 g y y e y ^+ p^ were ^, p ^ re funtions of. 8. Fin tese erivtives using te prout rule: 0 0 6 ^ - @ 6 ^ - ^ + @ 8` - j^ - B 6 ^ - ^- @ 9. Fin tese erivtives using te quotient rule: 8 + B ; - E + - ; E - G + - 0. Fin tese erivtives using te in rule: 6 ^ - 0 @ 6 ^ 7-9 00 @ 6 6 - @ 7 ; + me 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition More Questions Using Our Knowlege:.... Wi rules of ifferentition oul e use to fin te erivtive of y +? + Wt is te erivtive of ^ + 000? One pplition of te in rule. 000 One pplition of te quotient rule. 000 999 + Using ot te prout n in rule fin tese erivtives: 00 6 ^ - @ 00 6 ^+ ^ + @ 6 ^ - ^ + @ Te in rule n quotient rule. 000^ + 999 y Looking t te vlue of te eponents wi funtion s seon erivtive of te from is onstnt? y 0 y 6 8 y y 6 Te prout rule n te in rule. 000^ + 000 6 6 were ; ^ - m + E. Using ot te quotient rule n in rule fin tese erivtives: 8 + B ; ^ - 0 E ^ - G - 00 ; - m E + 6. Using ot te prout rule n quotient rule fin tese erivtives: 6 8 ^+ G + ; ^ + + me 7. 8. Fin e of tese erivtives y simplifying te lger first: Use te quotient rule iretly to fin te erivtive. 9 - G 9-0 ; E Sow tt 6 + ^+ ^- G ^ + +, n ten ifferentite. 7 G 9 9. Fin te following seon erivtives: 6 6 - @ 6 6 ^ - @ D ; + E ; - E + 0. Fin te following iger erivtives: 0 6 @. 6 ^ ^ If f ^ + + +, fin f. If y +, fin yll. 6 y 7 Fin, of te funtion y 6 ^ -. e Fin D 6 ^+ @. 8 f ^ D 6 + + + + @ 6 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition More Questions Tinking More:..... 6. 7. 8. 9. 0. If te grient of tngent line to urve is -, wt is te grient of te norml line t te sme point? - - Te tngent to te urve y + t ^, rosses te -is t te point: 0, ` 7 j `, 0 7 j ` 7, 0 j 0, 7 ` j Te funtions f ^ + n g^ + ve tngents wit te sme grients for prtiulr vlue of. Wt is tt vlue? 0 - - - Fin te grient of te tngents to tese funtions: ^ ^ + 6 t te point were 0. k ^ - t te point ^0,. + Sow tt te eqution of te tngent to te funtion y - t te point were is 68- y- 0. Fin te grient of te normls to tese funtions: ^ t te point were. k - ^ ^ + t te point, + ` 6 j. Sow tt te eqution of te norml to te funtion ^ - t te point were is - + 7y- 0. Te line y + is te tngent line to te funtion f ^ - wen -. Fin te vlue of. Te urve y s two points were tngents ve grient -. Fin te oorintes of tese points. + Sow tt te eqution of te norml to te prol y t ^, is + y- 0. By solving te two equtions y n + y- 0 simultneously fin te oorintes of were te norml line gin meets te prol. 00% Introution to Differentition Mtletis 00% P Lerning L 7 C SERIES TOPIC NUMBER
Introution to Differentition Answers Bsis:... n. n f^ f^ Sent Tngent Tngent Sent m - m m - m -. ^, f^ ^+, f^+ m nees to get smller. More formlly, s moves towr zero, te sent moves towr te tngent. e f^+ - f^ Te grient of te tngent t is given y f^+ - f^. " 0.. Te grient of te tngent t is m 9. Te grient of te tngent t - is m -7. Te grient of te tngent t is m -8. Te grient of te tngent t - is m 0. 6. fl^ y 7. D 6 q^@ 9 8. + 9. pl^ - 0. 6- @ - 8 Knowing More:. fl^ 0 gl^.. e e D 6 @ yl fl^ 7 6@ 0 ml^ 0 f 6@ 0 0 g D 0 0 9 6 - @ - 0 9 60 @ 00 6 @ 8 6 - @ -8 6 @ 0 0 8 B 6 - @ - - f - g - 9 - - 0 8 B - - 0-6 @ - -9 8 8 B 8 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Answers Knowing More:. i 68 @ - j 9 6 @ - k 67 @ - - - l 6-8@ - - 8. 6 6 + - @ + 6 @ + [ ] + [- ] 6 + Tese two erivtives re equl. You woul epet tis to e so oring to te Sum Rule: 6 f ^ + g ^ @ f l^ + g l^. 9 - - 6 7 + - - - - 9 - - ; E 6 7 ; E+ + - - - - 8 6 B @ Tese two erivtives re equl. You woul epet tis to e so oring to te Sum Rule: 6 f ^ + g ^ @ f l^ + g l^ 6. y - y - e y y 6-6 - + - ^ ^ f y y + - ^ ^ 7. 6 ^+ ^- @ 6 + @ : 6 - @ No, tese results re not equl. Tis is s epete sine 6 f ^ : g ^ @! f l^ : g l^. 8. 9. e f g 8 6 ^ : ^ @ - + 6 - f g 8 6 ^ : ^ @ - + 6-6 f g 0 8 6 ^ ^ @ - - + 6 + 6 6 f ^ : g ^ @ - - 9 7 6 f ^ : g ^ @ - f 6 f ^ : g ^ @ - 8 + + ^ 6 + @ f^ ^ - - - ; E- g ^ 68- @ ^8- No, tese results re not equl. Tis is s epete sine f^ fl^ ; E!. g ^ g l 0. f^ ; E g ^ - f^ ; E g ^ ^- ^ - f^ 0 ^+ ; E g ^ ^ + f^ 8 ^ - 8- ; E- g ^ ^ - e f^ ; E g ^ ^ - f f^ ; E g ^ + ^- g f^ ; E g ^ + ^9 + + ^ + f^ ; E g ^ + ^ + 8-9- ^+. 6 ^+ 6 @ ^+ 6 ^ + 6 @ ^ + 6 6 ^- - @ -^- - ^9 + ; 8 7 9 ` + + - - je e 8 8 7 6 ^ + - @ ^ + - ^ - f 6 + @ + + g 8 ^ - + 6 B ^ - : - + 6 8 ^80 7 0 : - + - + - + - ^ ^ B - + - + - - - i ; + E j m ; + + m E + 00% Introution to Differentition Mtletis 00% P Lerning L 9 C SERIES TOPIC NUMBER
Introution to Differentition Answers Knowing More:. - ^ - Te insie funtion is g ^ - Te outsie funtion is f^. 6 - @ - ^ 7 6 f : g - - + l^ 6 ^ ^ @ - -. Te insie funtion is g ^ + - + n te outsie funtion is f^ 8. 6 ^ + - + @ 8^ + - + : ^ + - ql^ 7 f^ : ^ + - + : ^7 + 6 - - ; E g ^ Using Our Knowlege:. To ifferentite e ^ + n ^ - te in rule is require. Wile te prout rule is require to ifferentite te overll funtion, ^. l^ ^+ ^- ^7 -. To ifferentite ^ + te in rule is require. Wile te quotient rule is require to ifferentite te overll funtion, f^. fl^ ^ + ^ - 6. To ifferentite ^ - te in rule is require. To ifferentite te numertor, ^ + + ^ -, te prout rule is neee. Wile te quotient rule is require to ifferentite te overll funtion. y f^ 6^ + ^ - + 6 ^ - @ ^ + + @ ^ - 6^ + + ^ - @ ; E g ^ 7. 6 ^ - + - 8 @ 8^ - + - 7 ^6 - + gl^ ^+ ^ + ^ - + 00 ^ - ^ + ml^ ml^ - + 6 + ^ - 6^- ^+ @ ^ - 7 - ^66^+ ^- @ ^ - 7 e f " 67 ^ - ^ - 8@ ^- + 6 + 66^- + 6 ^- 0+ 6@ ^7 -,: ^ - -^6" ^7 - ^- + 6, pl^ ^ - 6 6 tl^ 6^^+ + - 8@ 6^^+ + : : ^+ : - 8@ 8. 6 0 6 6 - + - + @ - + - - 6 6 + @ + y 6 6 + @ - - - - 6 ; E - + 9. fll^ ^ - 60^6 - + 7^ - @ - - - - gll^ ^ + 6 ^- + + ^ + @ 0. fll^ ^ + ll^ ^+ 6^0+ + 0 ^+ @ - ^+ ^0 + mll^ 6 0 L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Answers Using Our Knowlege:. qlll^ mens te tir erivtive of funtion q^ 7 q ^ ^ mens te fift erivtive of funtion q^ 7 qlll^ 0 ^ q ^ 0. 9 7 pll^ ^ + ^7 + 7-8 - 6 6 qll^ 00 + 80-70 + 80 + 78-80+ ^ - Tinking More:.. Norml f ^- Tngent f ^- gl^- 8 l^- -6 f l^0 kl^- - 6 Tngent f ^- Norml f ^-. Te eqution of te tngent to f^ t - is y. Te eqution of te tngent to ^ t - is y +. Te eqution of te tngent to p ^ t is y - +. Te eqution of te tngent to f ^ t 0 is y. 6. ^- ^ + - ^+ ^ - + fl^ ^ + Te eqution of te tngent t is y -. 6 e Te eqution of te norml t is y - +. Te grient of te tngent t is. Te grient of te norml t is -. 7. Te eqution of te tngent t is y -. Te eqution of te tngent t is y - + 7. More Questions. ^+ - " 0. 0. 8. f^ f l^0 8 f^ 76 f l^ 87. A^, f^ B^+, f^+ As point B moves loser to point A, pproes zero. e f^+ - f^ m f^+ - f^ f l^ " 0 6. Line, e, g, re tngents s tey just tou te urve. Line,,, f re not tngents s tese ut te urve. Line i is lso tngent even toug it ppers not to e. It toues te urve t te flt spot. 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Answers More Questions 7. fl^ f y l^ + - D 6 g^@ - e ql^- f 6@ 0 8. l^ 6 l^- - 9. 0. y. 0. f n n - l^. D ^6 7. gl^ - fl^- - 8 7. D 67 @ 6-90 -9 60 @ -900 or - 900 y - l - or - -9 6. 6 @ 0 0 9 6 - @ -8 7 6 or ;- 6 6 E - 7 g - - 6 @ - or - - - 6 - @ or e 8 B -9-0 ; E- or - 8 0 f 66 @ - or or 7. yl 8+ 7 y - yl - 7 - l + 6 9 yl 8-7 e yl 6-60 f yl 9-7 - g yl yl e yl l^+ pl^ 8. 9 9 yl 0-0 yl + - y - - l - + - yl - 0 + + 6-6 9. yl ^ + yl - 6 ^ + yl - 0 + ^ - yl 9 + - - - ^ + - 0. yl 0^- 9 7 99 6 yl 00^ - 9 ^ - 8 yl 6 - - 6 - y 7 6 - l ` + j - + m. Te in rule n quotient rule.. 000^ + 999. y 8 6. yl ^- 99 ^- 0 yl ^ - ^ + ^ + yl 99 yl ^ + ^0 + 00+ - - ^+ y y 9 + l - -. l yl -6^ - ^+ ^ - 00^- yl - 0 ^ + 99 7 ^+ ^6 + + 7 + 6. yl yl ^ + 6 ^ + + ^ + L SERIES C TOPIC NUMBER 00% Introution to Differentition Mtletis 00% P Lerning
Introution to Differentition Answers More Questions 7. yl 6 + @ 8. - 7-0 0 > H ; E- 6 9 ^+ ^- G - ^ + 9-7 G - 9-7 9. 66 - @ D ; + E - 9 - - yl ^- ^ - ; - E 6 + ^ + 0. 70 7 f ^ 70 00^ - e 6880^ + f ^ - ^ + ^ + 8 9..,. ` 0 j 7 -. 0 9 6. Norml grient 6 Norml grient 9 8. 9. ^0, n ^-, -. 0. + y- 0, 9 `- j 00% Introution to Differentition Mtletis 00% P Lerning L C SERIES TOPIC NUMBER
Introution to Differentition Notes L C 00% Introution to Differentition SERIES TOPIC NUMBER Mtletis 00% P Lerning
www.mtletis.om Introution to Differentition