MA1200 Exercise for Chapter 7 Techniques of Differentiation Solutions. First Principle 1. a) To simplify the calculation, note. Then. lim h.

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1 MA00 Eercise for Chapter 7 Techiques of Differetiatio Solutios First Priciple a) To simplify the calculatio, ote The b) ( h) lim h 0 h lim h 0 h[ ( h) ( h) h ( h) ] f '( ) Product/Quotiet/Chai Rules a) 8 6 b), c), so 6 d) Quotiet rule: ( ) 5 8 e) Quotiet rule: 5 ( ) 999 f) Chai rule: 000( 5 ) ( 0) g) Product rule + Chai rule: ( )()( ) ( ) ( ) (6 ) h) Chai rule: 6si ()cos() 5 si( ) i) Chai rule: l l(l ) j) Product rule: l k) Chai rule: cot l) Chai rule: ta m) ) Chai rule: e si( e ) o) Product rule: = e ( cos cos si )

2 p) Quotiet rule + Product rule: ( 5) e ( si cos ) e ( 5) ( ) q) Quotiet rule: r) Quotiet rule: ( ) ( s) Product rule + Chai rule: cos()cos() si()si() si(l() ) t) Chai rule: cos(l[cos(l() )] ) cos(l() ) a) Differetiate both sides: So y ' / y (We eed to use Chai rule o the term, ad the we get a ) b) Differetiate both sides: y So y' y I the middle term we used Product rule c) Differetiate both sides: a) d ( y ) y ( y ) y d dt dt 9 t 9 t t 9 whe t = Whe Collectig like terms, we get ( si )(6 5) 6 ) So 9 Taget: y 5 ( ), y Normal: y 5 ( ), y 9 9 (Recall: eq of lie is ; slope of taget = ; slope of ormal slope of taget = ) 7 b) We ca do as i part a) to get t whe t = d t Or we ca write ad we sub i Taget: 5 y 5 y 7 5 a) Take log The use Chai rule o LHS ad Product rule o RHS: So cot cot ( ) csc l( ) b) So ( e e ( ) ) e l( )

3 c) The The (si5 ) [ l(si 5) 5( )cot 5] 6 a) So i geeral, y ( ) e b) Use the formula: If, the So y 5(6 )si(6 7 ) c) Write The So y ( )!( ) (There is o, eed to write )!! 6 ( )! ( )!! d) y whe ( )! 0 (First we kow if differetiate more tha times, all terms become 0; we also kow if differetiate more tha times, oly the term remais We differetiate a few times to guess the geeral form above) Leibiz s rule 7 If, the Leibiz rule states that a) The puttig ito the formula above ad otig that ad for, we get b) y e ( ) e ( 6 ) e [ ( ), () ( ) ( )( )] whe So for

4 y ( ) cos( ) ( 7)cos cos ( 7) ( ) cos ( ) ( ) ( ) cos whe whe c) So for, y ( ) ( ) ( )!( ) whe whe Miscellaeous 8 If y si, prove that y'' y' y 0 Proof: y si y' si cos y" cos cos si cos si So y' ' y' ( ) y cos si si cos si si si cos cos si si 0 d a b c 9 If u a b c, prove that ( u) d u Proof: du d d a b a b a b c ( a b c) ( a b c) (a b) d d d u ( a b c) The d du a b a b u a b a b c a b c ( u) u u d d u u u u

5 if c *0 Fid the values of a ad b (i terms of c) such that f '( c) eists, where f ( ) a b if c Solutio: if c if c f ( ) f ( ) a b if c c a b if c if c I order that f '( c) eists, f () must be cotiuous at c That is, lim f ( ) lim lim f ( ) ac b f ( c) c c c c c c ( ac b) f( ) f( c) lim lim lim c lim c lim c lim c c c c c c c c c c c c c f( ) f( c) ab( acb) a( c) lim lim lim a c c c c c c f ( ) f ( c) f ( ) f ( c) That f '( c) eists meas lim lim a c c c c c Therefore, a ad put i ac b c c c a We have b c b c So b c Fially, we have c c c c c b c A fuctio y of is defied by the equatio si y msi y Epress y eplicitly i terms of mcos Hece, or otherwise, show that d m cos m Proof: si y msi y si cos y cos si y msi y si cos y m cos si y si y si si si ta y y ta cos y mcos mcos mcos The si si d m cos cos si d ta m cos m cos d m cos mcos d d si mcos si mcos m m cos m cos 5

6 Let cot y, fid d Solutio: cot l cot l d y y csc l cot y cot cot ycsc l csc l d cot Solutio: d f ta ta f ' 0 d Show that f ' 0, where f ta ta t Cosider the parametric curve where t y t Fid, d y ad the equatio of the taget lie to the curve at the poit, d d Solutio: Method : t y y y t The d y y, d d Method : d d y d d y d d d dt dt dt dt dt dt dt dt dt d d d d d d d dt d d d d dt dt dt dt d d t t dt dt t So t y t t d y d t 6t dt dt d d y d d y dt dt dt dt dt t6t t dt dt d d d t d d dt 6 cot

7 dt dt d d t As a result, t d, d y dt d d t The slope of the taget lie to the curve at the poit, d Therefore, the equatio of the taget lie to the curve at the poit y I additio, t t also t t t 0 7, is: 5 By Leibitz s theorem o repeated differetiatio, fid the th derivatives of the fuctio y = e cos Solutio: By Leibiz s theorem o repeated differetiatio, we have i i i i 0 0 d d f dgd f d g dgd f d f d g fg g Ci f i i C i, where f, g, i i 0 0 d d i d d d i0 d d d d! Cr, r 0,,,,, ad 0! r! r! i d f i Now, f e e, i 0,,, d i i d g i i d g i i gcos cos, si, i 0,,, i i d d So i i d d cos d e e cos Ci i i d d d i0 If m where m is a oegative iteger, the i i d d cos d e e cos Ci i i d i0 d d m i m i cos d e d d cos d e Ci i i C i i i d d d d i i i0 i0 m m i i i i Ci cose Ci sie i0 i0 If m, where m is a oegative iteger, the i i d d cos d e e cos Ci i i d i0 d d m i m i d cos d e d cos d e Ci i i C i i i d d d d i i i0 i0 m m i i i i Ci cos e Ci sie i0 i0

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MATHEMATICS. 61. The differential equation representing the family of curves where c is a positive parameter, is of

MATHEMATICS. 61. The differential equation representing the family of curves where c is a positive parameter, is of MATHEMATICS 6 The differetial equatio represetig the family of curves where c is a positive parameter, is of Order Order Degree (d) Degree (a,c) Give curve is y c ( c) Differetiate wrt, y c c y Hece differetial