Differential Calculus-I

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Clement Mason
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1 UNIT I Differetil Clculus-I. INTRODUCTION Clculus is oe of the more beutiful itellectul chievemets of hum beig. The mthemticl stu of chge motio, growth or dec is clculus. Oe of the most importt ide of differetil clculus is derivtive which mesures the rte of chge of give fuctio. Cocept of derivtive is ver useful i egieerig, sciece, ecoomics, medicie d computer sciece. The first order derivtive of deoted b, secod order derivtive deoted b, third order derivtive b d so o. Thus b differetitig fuctio f (), times, successivel, we get the th order derivtive of deoted b or D () or (). Thus, the process of fidig the differetil coefficiet of fuctio gi d gi is clled Successive Differetitio.. th DERIVATIVES OF SOME STANDARD FUNCTIONS Below we obti formuls for the th order derivtives of some stdrd fuctios. () th derivtive of e Let e. The b differetitig successivel, we obti e, e e e

2 A TEXTBOOK OF ENGINEERING MATHEMATICS I Thus, we hve the formul I prticulr, D ( e ) D ( e ) e...() e...() () th derivtive of log ( b) Let log( b). The we fid, b successive differetitio b ( ) ( b) ( )( ) ( b) ( )( )( ) ( b ) Thus, we hve the formul I prticulr, D [log( b)] ( ) ( ) ( )! ( b) ( )! ( b)...() D [log] () th derivtive of ( b) m Let m ( b) Differetitig successivel, we get ( ) ( )!...() m ( b) m m( m ) ( b) m

3 DIFFERENTIAL CALCULUS-I m( m )( m ) ( b) m m ( )( m ) L ( m ) ( b...(5) m m ) This formul is true for ll m. Followig re some prticulr cses. Cse (i): Suppose m ( ve iteger) I equtio (5) becomes, I prticulr, D [( b) ] ( ) ( ) ( b)!...(6) D ( )!...(7) Cse ( (ii): Suppose m is positive iteger d m >. The formul (5) becomes D [( b) m ] m( m) ( m )( m)( m) ( b) ( m)( m) m m! m ( b ( m )! ) I prticulr, D ( m m! m ) ( m )! Cse ( (iii): Suppose m is positive iteger d > m. From (6) we ote tht...(8)...(9) m m D [( b) ] m! I differetite further, the right-hd side gives zero. Thus, m D [( b) ] 0 if > m...(0) I prticulr, D ( m ) 0 for > m...() Cse ( (iv): Suppose m, i this cse formul (5) becomes, D b ( )( )( ) K( ) ( b) ( )! ( b )...()

4 A TEXTBOOK OF ENGINEERING MATHEMATICS I Cse (v): Suppose m is egtive iteger. Let us get m p, so tht p is positive iteger. The formul (5) becomes, ( ) pp ( ) ( p ) D ( b) P p ( b ) p p( p ) ( p ) ( ) p p ( b ) I prticulr ( ) ( p )! ( p )! ( b ) ( p )! D p ( ) ( p )! () th derivtive of cos ( b) Let cos( b). Differetitig this successivel, we get si( b) cos( bπ/ ) p p...()...() si( bπ/ ) cos( b π / ) si( b π/ ) cos( b π/ ) cos( b π/) Thus, we obti the formul D [cos( b)] cos( b π / )...(5) I prticulr, D (cos) cos( π/ )...(6) (5) th derivtive of si( b) Let si( b)

5 DIFFERENTIAL CALCULUS-I 5 Differetitig successivel, we get cos( b) si( b π / ) d π cos( b π / ) si( b / ) Thus, we hve the formul, cos( b π /) si( b π/) d si( b π / ) D [si( b)] si( b π /)...(7) I prticulr, D (si ) si( π / )...(8) (6) th derivtive of e si (b c) Let e si( b c) e si( b c) be cos( b c) For computtio of higher-order derivtives it is coveiet to epress the costts d b i terms of the costts k d defied b k cosα, b ksiα So tht Thus, Therefore, k b, α t ( b/ ) e [ k(cos α )si( b c) k(si α )cos( b c)] ke si( b c α) [ k e si( b c α ) be cos( b c α)] ke [ k(cos α )si( b cα ) k(si α ) cos( b cα)] k e si( b c α)

6 6 A TEXTBOOK OF ENGINEERING MATHEMATICS I Proceedig like this, we obti Thus, we hve the formul D [ e si( b c)] I prticulr, k e si( b c α ) / ( b ) si{ b c t ( b / )}...(9) D [ e si ] / e si( π / )...(0) (7) th derivtive of e cos(b c) Let e cos( b c) The e cos( b c) be si( b c) e [ k(cos α )cos( b c) k(si α )si( b c)] Therefore, cos( b c α) ke k[ e cos( b c α) be si( b c α)] k e cos( b c α) Proceedig like this, we obti k e cos( b c α) Thus, we hve the formul D [ e cos( b c)] I prticulr, D ( e cos) (8) th derivtive of m ke [ k(cos α ) cos( b cα) k(si α )si( b cα)] / Let m Tkig logrithm o both sides, log Differetitig w.r.t., we get / ( ) b e cos{ b c t ( b / )}...() e cos( π /)...() m log m log

7 DIFFERENTIAL CALCULUS-I 7 ( log ) m m log ( m log)( m log) ( mlog) ( mlog ) ( mlog) ( mlog) m D [ ] ( mlog ) m where WORKED EXAMPLES Emple : Fid the th derivtive of the followig fuctios: (i) si (ii) cos. Solutio: (i) si si (si si) \ D (si ) D (si) D (si) where so tht (ii) cos π π si si cos ( cos) cos ( cos) [ cos cos] [Usig the formul (5)]

8 8 A TEXTBOOK OF ENGINEERING MATHEMATICS I where cos ( cos) \ cos cos ( cos) cos 8 cos cos 8 8 cos cos 8 \ D (cos ) D D (cos) D (cos) 8 8 π π 0 cos cos 8 π π cos cos 8 Emple : Fid the th derivtive of the followig: (i) si h si (ii) e si (iii) e (iv) e si cos (v) e cos si Solutio: (i) sih si ( e e ) si si hsi [ e si e si] D (sihsi) [ D ( e si ) D ( e si )] [0 / { si( t ) si( t e e )}] Usig the formul (9) si h e e

9 DIFFERENTIAL CALCULUS-I 9 (ii) We hve si ( cos) Therefore, D ( e si ) D ( e ) D ( e cos ) / e ( ) [5 e cos( t ( ))] / e [( ) 5 cos(t ())] (iii) We hve cos (cos cos) D ( e cos ) D ( e cos) D ( e cos) [( ) / cos{ t e (/ )}] / e [( ) cos{ t / e [ cos{ t / (5) cos{ t (iv) We hve si cos (si si ) Therefore, D ( e si cos) D ( e (v) We ote tht cos [(( ) si) D ) ( e (/ )} (/ )}] si ) / e [(( ) e si{ t / ) e si{ t / / [0 si( t si ( cos) si ) (/ )}] ( )}] ( )}] si( π / )]

10 0 A TEXTBOOK OF ENGINEERING MATHEMATICS I \ D ( e cos si ) si si cos si (si si ) si si D ( e si) D ( e si ) / e [( ) si{ t (/ )} / ( 9) si{ t (/ )}] Emple : Fid the th derivtive of the followig: (i) Solutio: (i) ( )( ) ( )( ) B prtil frctios ( )( ) A B (ii) The A ( ) B ( ) Tkig A / Equtio (), becomes B / ( )( ) ( ) ( ) ( )( )...() D D D ( )( ) ( )! ( )! ( ) ( ) [Usig the formul ()] ( )! ( ) ( )

Repeated multiplication is represented using exponential notation, for example:

Repeated multiplication is represented using exponential notation, for example: Appedix A: The Lws of Expoets Expoets re short-hd ottio used to represet my fctors multiplied together All of the rules for mipultig expoets my be deduced from the lws of multiplictio d divisio tht you