# METHODS OF DIFFERENTIATION

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1 J-Mathematics MTHODS OF DIFFRNTIATION The process of calculatig derivative is called differetiatio.. DRIVATIV OF f() FROM TH FIRST PRINCIPL : f( ) f() Obtaiig the derivative usig the defiitio Lim Lim f '() 0 0 d derivative usig first priciple or ab iitio or delta method. is called calculatig Illustratio : Differetiate each of followig fuctios b first priciple : (i) f() ta (ii) f() e si Solutio : (i) f'() ta( h) ta lim h0 h h0 ta( h ) ta ta( h) lim h tah lim. ( + ta ) sec. A s. h0 h (ii) f'() si( h) si e e lim h0 h si(h) si si e si( h) si lim e h0 si( h) si h si si( h) si e lim h0 h e si cos A s. Do ourself - : ( i ) Differetiate each of followig fuctios b first priciple: (a) f() (b) f(). DRIVATIV OF STANDARD FUNCTIONS : Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 f() f'() f() f'() (i) (ii) e e (iii) a a a, a > 0 (iv) / (v) log a (/) log a e, a > 0, a (vi) si cos (vii) cos si (viii) ta sec (i) sec sec ta () cosec cosec. cot (i) cot cosec (ii) costat 0 (iii) si, (v) ta, R (vii) cosec, (iv) cos, (vi) sec, (viii) cot, R 77

2 J-Mathematics. FUNDA MNTAL THORMS : If f ad g are derivable fuctios of, the, (a) (c) d df dg (f g) (b) d d d d (cf) d d dg df (fg) f g kow as PRODUCT RUL d d d df c, where c is a costat d (d) df dg g f d f d d d g g where g 0 kow as QUOTINT RUL (e) If f(u) & u g () the Note : I geeral if f(u) the du. kow as CHAIN RUL d du d du f '(u).. d d Illustratio : If e ta + log e, fid d. Solutio : e.ta + log e O differetiatig we get, d e ta + e sec + log + Hece, d e (ta + sec ) + (log + ) A s. Illustratio : Solutio : If log O differetiatig we get, + e si + log 5, fid d. d log d d + d d (e si ) (log5 ) d d log. + e si + e. cos + loge 5 Illustratio 4 : Solutio : Hece, If ep log d ta [ + ta (log ) + sec ] + e (si + cos) +, the d equals - 78 loge 5 [ + ta (log )] + sec [ + ta (log )] + sec + [ + ta(log)] Takig log o both sides, we get log ta + ta (log ) O differetiatig, we get ta (log ) ( ) / d + ta (log ) + sec (log ) [ + ta (log )] + sec (log ) A s. + [ + ta(log)] As. Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

3 J-Mathematics Illustratio 5 : If log e (ta ) Solutio : log e (ta ) O differetiatig we get,, fid d.... ta ( ) Do ourself - : ta ta A s. (i) Fid d if - (a) ( + ) ( + ) ( + ) (b) e 5 ta( + ) 4. LOGARITHMIC DIFFRNTIATION : To fid the derivative of a fuctio : (a) which is the product or quotiet of a umber of fuctios or (b) of the form [f()] g () where f & g are both derivable. It is coveiet to take the logarithm of the fuctio first & the differetiate. Illustratio 6 : If (si ), fid d Solutio :. (si ) O differetiatig we get, d (si) +. cos si d (si ) (si) cot A s. Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 Illustratio 7 : If / / ( ) / 4 4 / 5 ( ) ( 4) fid d Solutio : + ( ) 4 ( ) 4 ( 4) 5 O differetiatig we get, Do ourself - : (i) d 4 9 ( ) 4( ) 6 5( 4) d ( ) 4( ) 5( 4 ) Fid if (ii) d Fid d if 4 e.e.e.e A s. 79

4 J-Mathematics 5. DIFFRNTIATION OF IMPLICIT FUNCTIONS : (, ) 0 (a) (b) To fid /d of implicit fuctios, we differetiate each term w.r.t. regardig as a fuctio of & the collect terms with /d together o oe side. I the case of implicit fuctios, geerall, both & are preset i aswers of /d. Illustratio 8 : If +, the fid d. Solutio : Let u ad v du dv u + v d d Now u ad v u ad v 0 du u d + d ad dv v d + d du d d ad dv d d d + d 0 d.. A s. Illustratio 9 : If si cos si cos..., prove that cos si. d cos si Solutio : Give fuctio is si cos ( ) si cos or + + cos ( + ) si Differetiate both sides with respect to, cos si d d d ( + ) cos + si d ( + + cos si) ( + ) cos + si d or Do ourself -4 : (i) cos si A s. d cos si Fid, if + si( ) d (ii) If + e + 0, fid ', also fid the value of ' at poit (0,0). 80 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

5 J-Mathematics 6. PAR A MTRIC DIFFRNTIATION : If f( ) & g( ) where is a parameter, the d / d d / d. Illustratio 0 : If a cos t ad a(t sit) fid the value of d at t Solutio : a si t d a( cos t) d t t Illustratio : Prove that the fuctio represeted parametricall b the equatios. ; t t t satisfies the relatioship : ( ) + (where d ) t Solutio : Here t t t Differetiatig w.r. to t d 4 dt t t t t Differetiatig w.r. to t dt t t A s. d / dt t ' d / dt Sice t ' or (') + ' As. t ( ') Do ourself -5 : (i) (ii) Fid d at t if cos 4 t & si 4 t. 4 Fid the slope of the taget at a poit P(t) o the curve at, at. Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 7. DRIVATIV OF A FUNCTION W.R.T. ANOTHR FUNCTION : Let f () ; z g () the / d f '() dz dz / d g '() 8. DRIVATIV OF A FUNCTION AND ITS IN VRS FUNCTION : If g is iverse of f, the (a) g{f()} (b) f{g()} g'{f()}f'() f '{g()}g'() Illustratio : Differetiate log e (ta ) with respect to si (e ). Solutio : d (log d(loge ta ) e ta ) d d(si (e )) d si (e ) d cot.sec e. / e e e si cos A s. 8

6 J-Mathematics Illustratio : If g is iverse of f ad f'(), the g'() equals :- + + [f()] + [g()] oe of these Solutio : Sice g is the iverse of f. Therefore f(g()) for all d f(g()) d for all f'(g()) g'() g'() f '(g()) + (g()) As. Do ourself -6 : ( i ) Differetiate with respect to. ( i i ) If g is iverse of ƒ ad ƒ () + si; the g () equals: + si + cos g() cos(g()) 9. HIGHR ORDR DRIVATIVS : Let a fuctio ƒ () be defied o a iterval (a, b). If ƒ () is differetiable fuctio, the its derivative ƒ '() [or (/d) or '] is called the first derivative of w.r.t.. If ƒ '() is agai differetiable fuctio o (a, b), the its derivative ƒ "() [or d /d or "] is called secod derivative of w.r.t.. Similarl, the rd order derivative of d d d w.r.to, if it eists, is defied b d d d ad deoted b ƒ '''() or ''' ad so o. Note : If f() ad g() where '' is a parameter the d / d d / d & d d d d d d d I geeral d d d d d d d d Illustratio 4 : If f() + f'() + f''() + f'''() for all R. The fid f() idepedet of f'(), f''() ad f'''(). Solutio : Here, f() + f'() + f''() + f'''() put f'() a, f''() b, f'''() c...(i) f() + a + b + c f'() + a + b or f'() + a + b...(ii) f''() 6 + a or f''() + a...(iii) f'''() 6 or f'''() 6...(iv) from (i) ad (iv), c 6 from (i), (ii) ad (iii) we have, a 5, b f() A s. 8 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

7 Illustratio 5 : If a (t + si t) ad a( cos t), fid Solutio : Here a (t + si t) ad a ( cos t) Differetiatig both sides w.r.t. t, we get : d dt a( + cos t) ad dt a (si t) d d. t t d a si t si.cos t ta a cos t t cos Agai differetiatig both sides, we get, d d Hece, t dt sec d sec t / a cos t d d t 4 sec 4a t sec a t cos J-Mathematics Illustratio 6 : f() ad g() are iverse fuctios of each other the epress g'() ad g''() i terms of derivative of f(). A s. Solutio : f '() d ad d g '() Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 Do ourself : 7 g '() f '()...(i) Agai differetiatig w.r.t. to d g ''() f '() f ''() g ''() (f '()) d d. d f '()...(ii) Which ca also be remembered as F '() l () m() () + l '() m '() '() + u() v() w() u() v() w() 8 f ''(). (f '()) f '() d d d d d d (i) If e the fid ''. (ii) Fid " at /4, if ta. (iii) Prove that the fuctio e si satisfies the relatioship '' ' DIFFRNTIATION OF DTRMINANTS : If f() g() h() F() l () m() (), where f, g, h. l, m,, u, v, w are differetiable fuctios of the u() v() w() ' ' ' f () g () h () f() g() h() f() g() h() l() m() () u '() v '() w '() A s.

8 J-Mathematics Illustratio 7 : If f() 0 6, fid f'(). Solutio : Here, f() 0 6 O differetiatig, we get, f'() d d d () ( ) d d d 0 6 d d d + d d d d d d 0 6 d d d or f'() As we kow if a two rows or colums are equal, the value of determiat is zero f'() 6 ( ) Therefore, f'() 6 A s. Do ourself : 8 (i) If e ƒ(), the fid ƒ '(). (ii) If si ƒ() 5 the fid ƒ ' ().. L ' HOˆ PITAL ' S RUL : (a) This rule is applicable for the idetermiate forms of the tpe 0 0,. If the fuctio f() ad g() are differetiable i certai eighbourhood of the poit 'a', ecept, ma be, at the poit 'a' itself ad g'() 0, ad if lim f() lim g() 0 or a a the f() f '() lim lim a g() a g '() provided the limit (+ or ). lim f() lim g(), a a f '() lim eists (L' Hôpital's rule). The poit 'a' ma be either fiite or improper a g '() (b) Idetermiate forms of the tpe 0. or are reduced to forms of the tpe 0 0 or (c) trasformatios. 84 b algebraic Idetermiate forms of the tpe, 0 or 0 0 are reduced to forms of the tpe 0 b takig logarithms or b the trasformatio [f()] () e ().f(). Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

9 J-Mathematics Illustratio 8 : valuate Solutio : lim 0 si lim 0 / lim e 0 cos ec cot si lim e 0 si log e log e lim 0 cos ec e (applig L'Hôpital's rule) Illustratio 9 : Solve Solutio : si lim e 0 cos Here 0 si lim 0 cos e lim 0 log si si. lim 0 log si si log si lim log si cos lim si cos si e e A s. form {applig L'Hôpital's rule} Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 Illustratio 0 : valuate cos si lim 0 cos si e lim /. / cos lim cos 0 e Solutio : Here, A lim e log A lim log log e 0 lim Do ourself : 9 loga A e or e lim ( 0 form) log e log lim form {applig L'Hôpital's rule} ta (i) Usig L' Hopital ˆ ' s rule fid (a) lim 0 si ta (ii) Usig L' Hopital ˆ ' s rule verif that : (a) lim 0 INTRSTING FACT : / A s. e A s. (b) lim 0 e (b) ( lim ) 0 I 694 Joh Beroulli agreed to accept a retaier of 00 pouds per ear from his former studet L'Hôpital to solve problems for him ad keep him up to date o calculus. Oe of the problems was the so-called 0/0 problem, which Beroulli solved as agreed. Whe L'Hôpital published his otes o calculus i book form i 696, the 0/0 rule appeared as a theorem. L'Hôpital ackowledged his debt to Beroulli ad, to avoid claimig authorship of the book's etire cotets, had the book published aomousl. Beroulli evertheless accused L'Hôpital of plagiarism, a accusatio iadvertetl supported after L'Hôpital's death i 704 b the publisher's promotio of the book as L'Hôpital's. B 7, Beroulli, a ma so jealous he oce threw his so Daiel out of the house for acceptig a mathematics prize from the Frech Academ of Scieces, claimed to have bee the author of the etire work. As puzzlig ad fickle as ever, histor accepted Beroulli's claim (util recetl), but still amed the rule after L'Hôpital. 85

10 J-Mathematics. ANALYSIS AND GR APHS OF SOM IN VRS TRIGONOMTRIC FUNCTIONS : ( a ) ta f() si ta ( ta ) Importat poits : / (i) Domai is R & rage is, (ii) f is cotiuous for all but ot differetiable at, (iii) for o eistet for d for D / D (iv) Icreasig i (, ) & Decreasig i (, ) (, ) ta if 0 ( b ) Cosider f() cos ta if 0 Importat poits : (i) Domai is R & rage is [0, ) (ii) (iii) Cotiuous for all but ot differetiable at 0 d for 0 o eistet for 0 for 0 / f() 0 (iv) Icreasig i (0, ) & Decreasig i (, 0) ( c ) ta f( ) ta ta ( ta ) Importat poits : (i) Domai is R {, } & rage is, (ii) (iii) It is either cotiuous or differetiable at, d o eistet (iv) Icreasig i its domai (v) It is bouded for all 86 f() / - 0 / Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

11 J-Mathematics ( d ) ( si ) if f() si ( 4 ) si if si if Importat poits : (i) Domai is [, ] & rage is, (ii) Cotiuous everwhere i its domai (iii) Not derivable at (iv) (v), if (, ) d if (, ) (,) Icreasig i, ad Decreasig i,, D 0 I I D (e) cos if f() cos (4 ) cos if cos if Importat poits : (i) Domai is [, ] & rage is [0, ] (ii) Cotiuous everwhere i its domai Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 (iii) Not derivable at (iv) (v), if, d if,, Icreasig i Decreasig i GNRAL NOT :, &,, Cocavit is decided b the sig of d derivative as : d d 0 Cocave upwards ; D d 0 Cocave dowwards d / I O I D 87

12 J-Mathematics Illustratio : d si co t d 0 Solutio : Let si cot. Put cos 0, si cot cos cos si cot (cot ) si cos. As d Illustratio : If f() si the fid (i) f'() (ii) f' (iii) f'() Solutio : ta, where si (si) ( ) f() ta ta ( ta ) f'() (i) f'() 5 Do ourself : 0 (i) If cos (4 ) The fid (a) ƒ ' (ii) f' 8 5, (b) ƒ ' (0), (c) ƒ ' (iii) f'( + ) ad f'( ) + f'() does ot eist A s. 88. Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

13 J-Mathematics Miscellaeous Illustratios : Illustratio : If a( ), the prove that Solutio : Put si si () si si () d - cos + cos a(si si) cos cos a cos si cot a cot (a) si si cot (a) differetiatig w.r.t. to. d 0 d hece proved A s. Illustratio 4 : Fid secod order derivative of si with respect to z e. Solutio : / d cos dz dz / d e Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 d d cos d dz d dz. d e dz e si cos e e. e si cos A s. e d Illustratio 5 : If (ta ) the prove that ( + ) d Solutio : (ta ) Differetiatig w.r.t. ta d d ta () Agai differetiatig w.r.t. d d d + ( + ) d d ( ) d d A s. 89

14 J-Mathematics Illustratio 6 : Obtai differetial coefficiet of ta with respect to cos Solutio : Assume u ta, v cos The fuctio eeds simplificatio before differetiatio Let ta sec u ta ta cos ta si ta ta sec v cos sec cos cos cos cos u v du dv. A s. ANSWRS FOR DO YOURSLF : (i) ( a ) ( b ) : (i) ( a ) + + ( b ) 5e 5 ta ( + ) + e 5 sec ( + ) : (i) ( + ) (ii) ( ) 4 : (i) cos( ) cos( ) (ii) 5. (i) (ii) 6 : (i) ( )() (ii) D 7 : (i) '' 4 + ' (ii) : (i) e( si + cos ) (ii) 9 9. (i) ( a ) (b) e ' t e 90, 0 : (i) ( a ) 6 ( b ) ( c ) 6 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

15 J-Mathematics XRCIS - 0 CHCK YOUR GRASP SLCT TH CORRCT ALTRNATIV (ONLY ON CORRCT ANSWR). If sec ta sec ta the d equals - sec (sec ta ) sec (sec ta ) sec (sec + ta ) sec (sec + ta ) 4. If ad a + b, the values of a & b are - d a, b a, b a, b a, b d. Which of the followig could be the sketch graph of? d ' 0 ' ' 0 ' ' 0 /e 4. Let f() + l( ) & g() + 5 l( - ), the the set of satisfig the iequalit f'() < g'() is - 7, 5. Differetial coefficiet of 7,, (, ) m m m m m m.. ' w.r.t. is - ' 7, 0 lm 0 e ' Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 6. If m p m m p m p p mp e 7. If cos If f() If f() F I HG 4 K J F I HG 4 K J m / p e log a the d ; the / / si F HG F HG F, the f' HG 4 4 log log 4 I K J I KJ I KJ f 0 f ' is - the d at p / m e 000 F I HG 4 K J F I HG 4 K J / / F HG F HG p m e is equal to - oe of these 4 log log 4 I KJ I KJ 000 9

16 J-Mathematics 0. If... the a b a b a b d a b ab a ab b - a ab b b ab a. If the d.. If m. ( + ) m+, the d is -. If ( ) ( ) 0, the ( ) ( ). ( ) d d equals - 4. If e + e + 0, the d equals - e ( ) (e ) 5. If e e...to, > 0 the d is - 6. If e ( ) (e ) ad +, the d ( ). ( ) ( ) e ( ) (e ). (e ) oe of these oe of these 7. The derivative of si w.r.t. cos 8. Let g is the iverse fuctio of f & f '() 5 0 a a 0 0 9, ( > 0) is -. If g () a the g'() is equal to - a 0 a 9. Let f() si ; g() d F & h()log e & F() h[g(f())] the d is equal to - cosec cot ( ) 4 cosec ( ) cot cosec 0. If ƒ (), g() 0 ad h(), the ƒ '(h'(g'()) 5 a a 0 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

17 J-Mathematics. If ƒ & g are the fuctios whose graphs are as show, let u() ƒ (g()); w() g(g()), the the value of u'() + w'() is does ot eist ƒ (,4) g (6,) f'() g() ad g'() - f() for all real ad f(5) f'(5) the f (0) + g (0) is oe of these. If f(), the the value of...( times ) f '() f ''() f '''() ( ) f ''''''''' () f()... -!!!! 0 4. A fuctio f() has secod order derivative f"() 6( ). If its graph passes through the poit (, ) ad at that poit the taget to the graph is 5, the the fuctio is - ( + ) ( + ) ( ) ( ) 5. If ƒ () +..., the ƒ (0) + ƒ '(0) + ƒ ''(0) ƒ''''... times (0) is equal to -!! ( )! ( ) ( ) ( ) ( )( ) 6 cos 6. Let f () si ta. The f '() Limit 0 7. If f is differetiable i (0, 6) & f'(4) 5 the f(4) f( ) Lim 5 5/4 0 0 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 8. If f(4) g(4) ; f (4) 9 ; g (4) 6 the Limit 4 f( ) g 0 is equal to - oe of these SLCT TH CORRCT ALTRNATIVS (ON OR MOR THAN ON CORRCT ANSWRS) 9. The slope(s) of commo taget(s) to the curves e & e si ca be - / e e 0. If + ( + ) 0, which of the followig is true? e ' + ' ( ) ' + e 0 ' e 9

18 J-Mathematics. If, the ' equals - (log ). 6.. If t & t the d d equals- t Noe of these. If g is iverse of ƒ ad ƒ () + ( > 0) the g'() equals- g() 5 ƒ '() (ƒ()) CHCK YOUR GRASP ANSWR KY X R CI S - Que A s. B C C D B D C B A D D C B A C Que A s. A C B D C B C B D A B D A A, B A,B,C Que. A s. B, D C A,C 94 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

19 J-Mathematics XRCIS - 0 BRAIN TASRS SLCT TH CORRCT ALTRNATIVS (ON OR MOR THAN ON CORRCT ANSWRS). If fofof () ad f (0) 0, f '(0), the fid '(0) If p() is a polomial of degree, the d d d d is equal to - p'''(). p'() p''(). p'''() p(). p'''() oe of these. If is a fuctio of the d d 0 4. If ta ta ta the d 5. If d 0. If is a fuctio of the the equatio becomes - d d d d 0 is equal to- d d 0 sec ta ta + sec ta ta + sec ta ta (cosec + cosec 4 + cosec 6) sec sec sec sec + sec + sec e e the equals - d d d 0 e e e e Let... the - d Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 7. If + + the d 95 has the value equal to - 8. The fuctios u e si ; v e cos satisf the equatio - du dv v u u v d d d u v d 4 d v d u 9. Two fuctios f & g have first & secod derivatives at 0 & satisf the relatios, f(0), f'(0) g'(0) 4g (0), g''(0) 5 f''(0) 6 f(0) the - g(0) f if h () g the h'(0) 5 4 g ' Limit 0 f ' oe of these ( ) ( ) oe of these if k() f(). g() si the k'(0)

20 J-Mathematics 0. If + b, the d equals - b b b. If c, the is equal to - d c c. Lim 0 is - equal to 0 equal to equal to o eistet. Select the correct statemets - The fuctio f defied b f() for for is either differetiable or cotiuous at. The fuctio f() is twice differetiable at 0 If f is cotiuous at 5 ad f(5) the Lim f(4 ) eists. If Lim (f()+g()) ad Lim (f() g()) the Lim f(). g() ma ot eist. a a a m 4. Let Lim 5. where m, N the - 0 is idepedet of m ad is idepedet of ad depeds o m log Lim 0 si log cos si cos has the value equal to - is idepedet of m ad depeds o m is depedet o both m ad 4 oe of these BRAIN TASRS ANSWR KY X R CI S - Que A s. C C C A,B,C A, C A,C,D A,B,C,D A,B,C A,B,C B, C Q u e. 4 5 A s. A,B,C C B, C A C 96 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

21 J-Mathematics XRCIS - 0 MISCLLANOUS TYP QUSTIONS TRU / FALS. Let u() ad v() are differetiable fuctios such that u v. If f(), the f'(f()) for > 0 () 7 If u '() v '() p ad u() ' v() q the p q p q. If f(0) a, f'(0) b, g(0) 0 ad (fog)'(0) c, the g'(0) c b 4. The differetial coefficiet of f(log) w.r.t. log where f() log is 5. f'(si) (f(si))' 6. If t + t 8, t t 5, the d at (, ) is 6 7 log MATCH TH COLUMN Followig questios cotais statemets give i two colums, which have to be matched. The statemets i Colum-I are labelled as A, B, C ad D while the statemets i Colum-II are labelled as p, q, r ad s. A give statemet i Colum-I ca have correct matchig with ON statemet i Colum-II.. Colum-I Colum-II Graph of f( ) Graph of f'( ) (p) (q) Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 (r) (s) 97

22 J-Mathematics. Colum-I Colum-II If f() + +, the f'( + ) at (p) 0 is If f() log (log ), the f'(e e ) is equal to (q) 0 For the fuctio ta 4 if d sec + p, the p is equal to 98 (r) 8 If f() + si, the (s) 4 4f'(8f(f())) is equal to ASSRTION & R ASON These questios cotai Statemet I (assertio) ad Statemet II (reaso). Statemet-I is true, Statemet-II is true ; Statemet-II is correct eplaatio for Statemet-I. Statemet-I is true, Statemet-II is true ; Statemet-II is NOT a correct eplaatio for statemet-i. Statemet-I is true, Statemet-II is false. Statemet-I is false, Statemet-II is true.. Statemet-I : Let f() is a cotiuous fuctio defied from R to Q ad f(5) the differetial coefficiet of f() w.r.t. will be 0. B e c a u s e Statemet-II : Differetiatio of costat fuctio is alwas zero. A B C D. Statemet-I : Derivative of B e c a u s e Statemet-II : si si cos with respect to cos for. A B C D. Cosider ƒ () & g() ƒ ''(). Statemet-I : Graph of g() is cocave up for >. B e c a u s e Statemet-II : d ( )! (ƒ ()) d ( ) ( ), N A B C D COMPRHNSION BASD QUSTIONS Comprehesio # Let f( ) f() f() at 0 ad follows the fuctioal rule g k Let g'(0) 0 is for 0 < <. +,, R. f() is differetiable ad f'(0). Let g() be a derivable fuctio g() g() k (k R, k 0, ) Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

24 J-Mathematics. The statemet lim h 0 f is odd f( ) f( h) h f is either odd or eve lim h 0 f() f( h) implies that for all R h f is eve othig ca be cocluded Comprehesio # 4 A operator is defied to operate o differetiable fuctios defied as follows. If ƒ () is a differetiable fuctio the ƒ ƒ h ƒ lim h 0 h g() is a differetiable fuctio such that the slope of the taget to the curve g() at a poit (a, g(a)) is equal to e a (a+) also g(0)0. O t he basis of above i format io, a swer t he fol low i g que st ios :. g at is 4 { + } 4e 96 4e 9(4e). ( ( + )) lim 0 g cos 4 4 Tr ue / False. T. T. T 4. T 5. F 6. T Match the Colum. (q); (s); (p); (r). (s); (q); (q); (s) Assertio & Reaso. A. C. A Comprehesio Based Questios ANSWR KY MISCLLANOUS TYP QUSTION X R CI S - Comprehesi o # :. C. A. A,C Comprehesio # :. B. A. A Comprehesio # :. A. A. B Comprehesio # 4 :. C. D. A 00 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

25 J-Mathematics XRCIS - 04 [A] CONCPTUAL SUBJCTIV XRCIS a. If b 5 / 4 / ad d vaishes at 5 the fid a b.. If 4 4 the fid d. If f'() ad f( ) the fid d at. + t 4. If t ad + t t. Show that d d +. f ( ) 5. If f () e for all N ad f 0 () the show that d f () f ().f ()...f (). d 6. If prove that ' + ', where ' deotes the derivative of w.r.t.. 7. Let f() +... Compute the value of f(00).f'(00). u 8. If ta & sec, u u u 0,, ; prove that 0 d 9. If ta + si ta, the fid d for (, ). 0. If cost cost & sit sit, fid the value of (d / d ) whe t ( / ).. If a b c, Prove that ' a b c ( a)( b)( c) ( b)( c) ( c) a b c [J 98] Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65. If arcsi e. Prove that d ( ), d ( ) > 0.. Let f() 4, ad let g be the iverse of f. Fid the value of g' at f() 9 d 4. If [(a) + a ], prove that ( + ) d + d 5. If sec cos ; sec cos, the show that ( 4) ( 4) 0 d. 6. (a) Differetiate cos (b) If w. r. t. ta -, statig clearl where fuctio is ot differetiable. si ( 4 ) fid /d statig clearl where the fuctio is ot derivable i (,). 0

26 J-Mathematics 7. Suppose f ad g are two fuctios such that f, g : R R, f() ad g() the fid the value of.e g() ' f + g'() at. (a b cos ) c si 8. Determie the values of a, b ad c so that Lim 0 5 Solve usig L' Hoˆ p it a l 's rule or series epasio. (Q.8 Q.) cos ( ) 9. Lim 0 0. Lim 0 si. If a Lim a a a a fid 'a'.. Lim log (ta ) 0 ta. If 4 ( a) ( a) 4 f() ( b) ( b) 4 ( c) ( c) the 4 ( a) ( a) 4 f '(). ( b) ( b) 4 ( c) ( c) Fid the value of. CONCPTUAL SUBJCTIV XRCIS ANSWR KY X R C IS - 4 ( A ) (a) Not differetiable at 0 (b) Not derivable at / 8. a 0; b 60; c zero.. a.. 8 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

27 J-Mathematics XRCIS - 04 [B] BRAIN STORMING SUBJCTIV XRCIS. If z ad f(), show that : d f 4 z z d dz d dz. Prove that if a si + a si a si si for R, the a + a + a a. Show that the substitutio z ta chages the equatio d cot 4 cosec 0 d d to (d /dz ) Fid a polomial fuctio f() such that f() f'() f''(). 5. If Y sx ad Z tx, where all the letters deotes the fuctio of ad suffies deotes the differetiatio w.r.t. the prove that X Y Z X Y Z X s t X Y Z If a.( ), prove that s t d If be a repeated root of a quadratic equatio f() 0 & A(), B(), C() be the polomials of degree., 4 & 5 respectivel, the show that A() B() C() A( ) B( ) C( ) A '( ) B '( ) C '( ) is divisible b f(), where dash deotes the derivative. 8. If ta ta ta ta +... upto terms Fid /d, epressig our aswer i terms. g(), 0 / 9. Let g() be a polomial, of degree oe & f() be defied b f(), 0 Fid the cotiuous fuctio f() satisfig f'() f( ) 0. Let f( ) f() f() a for all real ad. If f() is differetiable ad f'(0) eists for all real permissible values of 'a' ad is equal to 5a a. Prove that f() is positive for all real. Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65. Fid the value f(0) so that the fuctio f(), 0 is cotiuous at 0 & eamie the differetiabilit e. If 4. of f() at 0. a si b c Lim 0. ( ) ( ) eists & is fiite, fid the value of a, b, c & the limit. BRAIN STORMING SUBJCTIV XRCIS ANSWR KY X R C I S - 4 ( B ) if f() / if 0. f(0), differetiable at 0, f'(0 + ) (/); f'(0 ) (/). a 6, b 6, c 0 ; /40 0

28 J-Mathematics XRCIS - 05 [A] J-[MAIN] : PRVIOUS YAR QUSTIONS. If f(), f'(), the. f() lim [AI - 00] () () () (4) 4 log [] lim, N, (where [] deotes greatest iteger less tha or equal to )- [] [AI - 00] () Has value - () Has values 0 () Has value (4) Does ot eist. If log, the d [AI-00] () log () log ( ) () ( log ) (4) log 4. If cos cos ad si si, the d [AI-00] () si () cos () ta (4) cot 5. If the ( + ) + [AI-00] () () () (4) Noe of these f '() f "() f '"() ( ) f () 6. If f(), the the values of f()...!!!! is- [AI- 00] () () () (4) 0 7. Let f() be a polomial fuctio of secod degree. If f() f( ) ad a, b, c are i A.P. the f'(a), f'(b) ad f'(c) are i- [AI- 00] () Arithmetic-Geometric Progressio () Arithmetic progressio (A.P.) () Geometric progressio (G.P.) (4) Harmoic progressio (H.P.) 8. If e () e...to, > 0, the d () 9. If m. ( + ) m+, the d () is - [AI -004] 04 () (4) is- [AI -006] () () (4) 0. Let be a implicit fuctio of defied b cot 0. The '() equals :- () log () log () (4) [AI -009]. Let f : (, ) R be a differetiable fuctio with f(0) ad f'(0). Let g() [f(f() + )]. The g'(0) :- [AI-00] () 4 () 4 () 0 (4) Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65

29 J-Mathematics. d equals :- [AI- 0] () d d d () d d d () d d (4) d d d. If sec(ta ), the d at is equal to : [J-(Mai)-0] () () () (4) Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 PRVIOUS YARS QUSTIONS ANSWR KY XRCIS-5 [A] Que A s Que. A s. 05

31 J-Mathematics (b) Let f ad g be real valued fuctios defied o iterval (, ) such that g"() is cotiuous, g(0) 0, g'(0) 0, g"(0) 0, ad f() g() si. Statemet- : Lim 0 [g() cot g(0)cosec] f "(0) a d Statemet- : f ' (0) g(0) Statemet- is true, statemet- is true ad statemet- is correct eplaatio of statemet-. Statemet- is true, statemet- is true ad statemet- is NOT the correct eplaatio for statemet- Statemet- is true, statemet- is false. Statemet- is false, statemet- is true. [J 008, +] 6. If the fuctio f() e ad g() f (), the the value of g'() is [J 009, 4] 7. Let si ƒ( ) si ta, cos where d. The the value of (ƒ( )) 4 4 d(ta ) is [J 0, 4] Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\04.MOD\MOD.p65 PRVIOUS YARS QUSTIONS ANSWR KY. ( a ) A ; ( b ) a. (a) C ; (b) B ; (c) B, (d) g' (0) 0. C 4. D 5. ( a ) A ; ( b ) A XRCIS-5 [B] 07

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