METHOD OF DIFFERENTIATION
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1 EXERCISE - 0 METHOD OF DIFFERENTIATION CHECK YOUR GRASP 5. m mn n ( m n )(n ) ( n )( m ) ( m )(m n ) 0 7. co y / y / loga m m n n (m n )(n )( m ) d () d 0 tan y y log a tan y tan log a Now diffrntiating bot id, w gt y d 0 d y log a 8. ƒ() ( ) 00. ( ).99. ( ).98...( 00) 00. Tak log & tan diffrntiat w gt Now 0. y ƒ '() ƒ() ƒ '(0) ƒ(0) ƒ(0) ƒ '(0) 5050 a b y y (b y) ab ay aby + ay + y b + y ab + ay b d d d b ab ay. ( + y) y ( + ) y + y( y) 0 ( y) ( + y + y) 0 Now y [do not atify t givn quation] + y + y 0 y ( ) d ( ) ( ) 8. g() ƒ () ƒog() ƒ '(g())g'() g'() g'() a 0 ƒ '(g()) ƒ '(a) a. ƒ '() g() and g'() ƒ() ƒ '(g()) Now d d [ƒ () + g ()] ƒ() ƒ '() + g() g'() ƒ()g() g()ƒ () 0 ƒ () + g () contant ƒ (5) + g (5) ƒ (0) + g (0) 8. ƒ() n 8. ƒ '() n n-, ƒ"() n(n ) n... ƒ ""... n tim () n! Now ƒ() 4 ƒ '() ƒ "()...n tim ƒ "" ()...!! n! n n(n )!! ƒ() 9 6 g()... n! n! 4 ( ) n 0 ƒ '() g '() ƒ() g() 9. y & y in y'...(i) & y' (in co)...(ii) quating (i) & (ii) ( in + co) 0 0 in + co 0 co in co co in co 0 / lop can b &. f () g() f(g()) Diffrntiating bot id, f'(g()) g'() g'() Now f'() + So g'() g() gof() g'(f()) f'() f() at & f'() 5 f '(g()), g'() (g()) g'()f'() g'() /5
2 EXERCISE - 0. y p() y d p'() y d + d y d EXERCISE p '() y p"() p"() 4y d + (p'()) y p"() 4y d d y d p()p"() (p'()) d [p'()p"()+p()p'"() p'()p"()] d y d. d d d p()p'"() d d d d d d d d d. d. d d Now put t valu of d & d d + y d 0 On olving w gt d. y y c y d Tru and Fal :. u( ) v ( ) u '() v '() + 7 u() 7v() y d 7 p & q 0 o in 0 y d p q p q 0 Matc t Column :. (A) f'() + f'( + ) ( + ) + f'( + ) at 0 i 4 d y y y y y y y 4. m (n ) n 0 ( n) 0 ( / ) y y y y (By rationalizing Nr. or Dr.) n m n a 0 c n( n) / 0 m (applying L'Hopital' rul) m n n( n) 0 m / m Again diffrntiating (n ) tim n! 0 n n ( ) m m 0 5. log co. 0 log co log in log in log co 0 0 log co tan 0 tan 0 cot cot log in log in 0 MISCELLANEOUS TYPE QUESTIONS tan 4 tan (B) f() log log() log log (log ) (log ) log f'() log(log()) log.. log (log) log.(log()) (log ) BRAIN TEASERS f'( ) 0
3 (C) (D) y ntan 4 c 4 d co tan in 4 c Hnc p 0 f() + in f() ( + ) ( ) in ( + ) ( ) in wn ( + ) ( ) in wn < Now 8f(f()) 0 At 0 f'() [( ) in + ( + ) in + ( + ) ( ) co ] 4f'(0) 4 Artion and Raon :. Hint : Statmnt I : f() i contant function Statmnt II : It i tru Comprnion # : f( + y) f() f(y) + y f(0 + 0) f(0) f(0) + (0) (0) f(0) and f'(0) (givn) Alo f'() 0 0 f() f(0) f'() + Intgrat it f( ) f() 0 + f'(0) + f() + + c f() + + [f(0) c ]. n( + + ) Domain R. y log /4 ( + + ) Now f(). g(0) nc rang i (, ] g(0) g(0) k g(0) 0 (a k ) g'() g(0) kg(0) g g g() g() g() g g 0 g() + c g() [g(0) 0] g'(0) Now ( ) + 0 For concidnt pt. D 0 ( ) 4 0, Comprnion # :. LHD 0 0 f( a ) f( a) f(a ) f(a) 0 f(a ) f(a). RHD f '(a) f '(a ) f '(a) f '( a) 0 Sinc drivativ of vn function i odd & vic vra.. 0 f( ) f( ) f'( )... (i) 0 f( ) f( ) f() f( ) and f'()... (ii) 0 from (i) and (ii) f'() i odd function and nc f() i vn function. EXERCISE - 04 [A] ƒ 0 ( ) 5. ƒ () ƒ ( ) ƒ () ƒ () imilarly f n () d d Now f n ()...( n ) tim ( ) d d f n ( ) f n. ( ( ) ) On diffrntiating it compltly w gt d f n () d f n ( ) f ( n ) ( ) f ( n ) ( ) f 0 ( )..... CONCEPTUAL SUBJECTIVE EXERCISE 6. y f n (). f (n )...f () + n y + n y' + y' +...(i) + +
4 + + y' Put t valu of + y y' + n y'. y in y y in (i) n ( + y ) in y y y( yy') y y' yy' y y y y y + yy' y + yy' y' y Again on diffrntiation w gt ( y ) y" ( y). f() 4 & f() 9 For 6, 6 ( > ) Now y f() f (y) g(y) d g'(y) 4 8 ( y )y ' y( yy ') y d d d ctan + in tan 4 nc n. c. tan + nco n ( in) n tan [c n + co n ] ntan y 4 y 4 n d 4 (a b co ) c in 0 5 Now uing L opital rul b in (a b co ) c co For it to it a + b c 0...(i) b(in co ) b in c in Again 0 0 b(co co in ) b co c co 60 0 b cco b in 0 60 For it to it b + c 0...(ii) b in l im 0 60 b 60, c 80, a 0 log tan c tan l im l im. 0 log tan 0 c tan EXERCISE - 04 [B] BRAIN STORMING SUBJECTIVE EXERCISE. z d z Now. d d ( z ) d ' d z (y ').. d d d z. d d f d. z n tan d in z + z 4 d y + z 4 d y Now d. d in d d d in. d d in co in d d in d + in co d coc cot d d Now put valu from givn quation coc 4y 0 + 4ycoc 0...(i)
5 4. f() f '()f "() Lt t dgr of f b n. Comparing igt powr on bot id n n + n n Lt f() a 0 + a + a + a f() f'()f''() (8a 0 + 4a + a + a ) (a 0 + a + a )(6a 0 + a ) Comparing cofficint of 7. ƒ() c( ) F() A() B() C() A( ) B( ) C( ) A '( ) B '( ) C '( ) Now F() 0 ( ) i root of F(). A '() B '() C '() F'() A( ) B( ) C( ) A '( ) B '( ) C '( ) Now F'() 0 i a factor of F '(). So ( ) mut b rpatd at lat two tim in F(). F() i diviibl by f(). 8a 0 8a 0 a Rt all ar zro. ƒ() ( ) 0 f() 4 9 (by L opital rul) ( ) 0 X X tx 5. X X X tx tx X X X X tx X t Xt C C C, C C tc X X 0 0 X X Xt X X X Xt t X X Xt X X Xt t X X t X X Xt t X (R R X R ) X X t Xt X t t ƒ(0) ƒ '(0 + ) 0 0 ( ) 0 ( ) 4... ( )! 0 ( ) 8 () 8 (...) (...)!!! 0 (...)! Similarly ƒ '(0 ).
6 EXERCISE - 05 [A] JEE-[MAIN] : PREVIOUS YEAR QUESTIONS. y log y y log log y y logy log d logy + y y d d (logy + ) d ( log y) 4. co co y in in d d d in + 6co in co 6in co d co 6 in co in 6 co in 6. f() n f() f '()! put n co in co cot in co in + f "()! f "'()!... n ( ) f () n! f() for ri f() f '() 0 f() put n f() f'() f"() f() o ri f() f '() + f "() +!! 0 Put f() f'() f"() 6 f() f"'() 6 ri f() f '()! + f"()! f "'()! 7. f() i a polynomial function f() a + b + c f() f( ) a + b + c a b + c b 0 a, b, c in A.P. a c b a c f() a + b + c f'() a + b f'(a) a + b f'(c) ac + b f'(b) ab + b tn f'(a), f'(b), f'(c) f'(b) 0 f'(a) a f'(c) a o tat f '(a), f '(b),f '(c) ar in A.P. 8. y y... > 0 d? y+ y+ d + d d 0. cot y 0 cot y 4 cot y 4 cot y coc y cot y + coc y diff. w.r. to at, cot y coc y y ( + log) [ coc y coc y cot y] coc y [coc y + cot y] d. g() [f(f() + )]. d g'() f (f() + ) f ' (f() + ) f '() Put 0 g'(0) f(f(0) + ) f '(f(0) + ) f '(0) f(( ) + ) f '(( ) + ) f '(0) f(0) f '(0) f '(0) 4( ) () () 4 d d d d.. d d d d. d. y c(tan ) d d d d d
7 EXERCISE - 05 [B] JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS. (b) Lt P() a + b + c P(0) 0 c 0 P() a + b P() ( a) + a P'() ( a) + a > 0 put 0, a > 0, a < S {( a) + a ; 0 < a < }. 5. (a) g( + ) log(f( + )) log + log f() g( + ) log + g() g( + ) g() log g'( + ) g'() g"( + ) g"() g" g" 4 g" 4 g" 9... g" N 4 g" N (N ) By adding Hnc g" N g" (N ) (b) g() co g(0) L im 0 in g '() co g() in L im 0 0 co Now f() g() in f'() g'() in + g() co f'(0) 0 f''() g''() in + g'() co g() in + g'() co f''(0) 0 Givn it f''(0) & alo f'(0) g(0) So S(I) & S(II) bot ar corrct but S(II) i not corrct plaination of S(I) 6. f() + /, g() f () g'(f()).f'() Put f() + / 0 g' ().f'(0), f'() + /. g'() 7. Lt ƒ( ) in wr tan in co in in tan co ƒ( ) tan d(ƒ( )) d(tan ) in tan co, 4 4
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