Save from: 1 st. class Mathematics الرياضيات استاذ الماده: م.م. سرى علي مجيد علي

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1 Sve from: st clss Mthemtics الرياضيات استاذ الماده: م.م. سرى علي مجيد علي

2 Chpter One Consider n rbitrry system of eqution in unknown s: A B.( n mn m m m n n n n B r bm n n m m m b n n b n in i i il ( The coefficient of the vribles nd constnt terms cn be put in the form: mx nx mxn mn m m n bm b b n n..( Let the form ( i n n u A mn m m...(4 Is clled (mxn mtrix nd donted this mtrix by: [ij] i,, m nd j,,..n. We sy tht is n (mxn mtrix or ةلمكت The mtrix of order (mxn it hs m rows nd n columns. For exmple the first row is (,, n And the first column is m

3 (ij denote the element of mtrix. Lying in the i th row nd j th column, nd we cll this element s the (i,j - th element of this mtrix Also n nx b b bm mx is (nx [n rows nd columns] Is (mx [m rows nd column] Sub Mtrix: Let A be mtrix in (4 then the sub-mtrix of A is nother mtrix of A denoted by deleting rows nd (or column of A. Let A Find the sub-mtrix of A with order ( ny sub-mtrix of A denoted by deleting ny row of A Definition.: 4 5 6,, Tow (mxn mtrices A [ij] (mxn nd B [bij] (mxn re sid to be equl if nd only if: ij bij for i,..m nd j,.n

4 Thus two mtrices re equl if nd only if: i. They hve the sme dimension, nd ii. All their corresponding elements re equl for exmple: 4 5 (7 4 Definition. If A [ij] mxn nd B [bij] mxn re mxn mtrix their sum is the mxn mtrix AB [ ij bij]mxn. In other words if two mtrices hve the sme dimension, they my be dded by ddition corresponding elements. For exmple if: A Then AB 7 nd B Additions of mtrices, like equlity of mtrices is defined only of mtrices hve sme dimension. Theorem.: Addition of mtrices is commuttive nd ssocitive, tht is if A, B nd C re mtrices hving the sme dimension then: A B B A (commuttive A (b C (A B C (ssocitive 4

5 Definition. The product of sclr K nd n mxn mtrix A [ij] mxn is the nn,n mtrix KA [kij] mn for exmple: 7 6( 6( 6( (5 6( 6( 6 Appliction of Mtrices Definition.4: If A [ij] mxn is mxn mtrix nd B [bjk] nxp n nxp mtrix, the product AB is the mxp mtrix C [cik] nxp in which n Cik j ij bik Exmple: if A b nd B b b x A B b b b b b b Exmple : Let A nd B 4 A B

6 Note.: in generl if A nd B re two mtrices. Then A B my not be equl of B AB nd B A BA. For exmple A A B B A if A B is defined then its not necessry tht B A must lso be defined. For exmple. If A is of order ( nd B of order ( then clerly A B is define, but B A is not defined.. Different Types of mtrices: Row Mtrix: A mtrix which hs exctly one row is clled row mtrix. For exmple (,,, 4 is row mtrix Column Mtrix: A mtrix which hs exctly one column is clled column mtrix for exmple is column mtrix. Squre Mtrix: A mtrix in which the number of row is equl to the number of columns is clled squre mtrix for exmple 4 is squre mtrix. A mtrix (A (n n A is sid to be order or to be n n-squre mtrix. 6

7 7 4 - Null or Zero Mtrix: A mtrix ech of whose elements is zero is clled null mtrix or zero mtrix, for exmple is ( null mtrix. 5 Digonl Mtrix: the elements ii re clled digonl of squre mtrix ( nn constitute its min digonl A squre mtrix whose every element other thn digonl elements is zero is clled digonl mtrix for Exmple: or 6 Sclr Mtrix: A digonl mtrix, whose digonl elements re equl, is clled sclr mtrix. For exmple,, 5 5 re sclr mtrix 7 Identity Mtrix: A digonl mtrix whose digonl elements re ll equl to (unity is clled identity mtrix or (unit mtrix. And denoted by in for Exmple I Note.: if A is (mxn mtrix, it is esily to define tht AIn A nd lso ImA A Ex: Find AI nd IA when A 7

8 8 Solution: IA And AI Tringulr Mtrix: A squre mtrix (ij whose element ij whenever i j is Clled lower tringulr mtrix.simillry y squre mtrix (ij whose element ij whenever is clled n upper Tringulr Mtrix For exmple:, re lower tringulr mtrix And,, re upper tringulr Definition.4: Trnspose of mtrix The trnspose of n mxn mtrix A is the nxm mtrix denoted by A T, formed by interchnging the rows nd columns of A the ith rows of A is the ith columns in A T. For Exmple: A T A 9 Symmetric Mtrix: A squre mtrix A such tht A A T is clled symmetric mtrix i.e. A is symmetric mtrix if nd only if ij ji for ll element. 4

9 For Exmple:, b Skew symmetric Mtrix: A squre mtrix A such tht A A T is clled tht A is skew symmetric mtrix. i.e A is skew mtrix ji -ij for ll element of A. The following re exmples of symmetric nd skew symmetric mtrices respectively (,( b 4 ( symmetric (b Skew symmetric. Note the fct tht the min digonl element of skew symmetric mtrix must ll be Zero Determintes: To every squre mtrix tht is ssigned specific number clled the determintes of the mtrix. ( Determintes of order one: write det (A or A for detrimentl of the mtrix A. it is number ssigned to squre mtrix only. The determinnt of ( mtrix ( is the number itself det (. (c Determinnts of order two: the determinnt of the. mtrix c b d Is denoted nd defined s follows: c b d d bc Theorem.: determinnt of product of mtrices is the product of the determinnt of the mtrices is the product of the determinnt of the mtrices det (A B det (A. det (B det (A B # det det B 9

10 (C Determintes of order three: (i the determinnt of mtrix is defined s follows: (ii Consider the ( mtrix Show tht the digrm ppering below where the first two columns re rewritten to the right of the mtrix. Theorem.: A mtrix is invertible if nd only if its determinnt is not Zero usully mtrix is sid to be singulr if determinnt is zero nd non singulr it otherwise..5 prosperities of Determinnts ( det A det A T where A T is the trnspose of A. ( if ny two rows (or two columns of determintes re interchnged the vlue of determinnts is multiplied by -. ( if ll elements in row (or column of squre mtrix re zero. Then det (A

11 (4 if two prllel column (rows of squre mtrix A re equl then det (A (5 if ll the elements of one row (or one column of determinnt re multiplied by the sme fctor K. the vlue of the new determinnt is K times the given det. Exmple; Exmple: (6 if to ech element of selected row (or column of squre mtrix k times. The corresponding element of nother selected row (or column is dded. Exmple: row ( row ( (7 if ny row or column contin zero elements nd only one element not zero then the determinnt will reduced by elementry the row nd column if the specified element indeterminte.

12 .6 Rnk of Mtrix: we defined the rnk of ny mtrix tht the order of the lrgest squre sub-mtrix of whose determinnt not zero (det of submtrix Exmple: Let A find the rnk of A Since A of order Rnk Since the rnk Minor of mtrix: Let A n n n n nn Is the squre mtrix of order n then the determinnt of ny squre sub- mtrix of with order (n- obtined by deleting row nd column is clled the minor of A nd denoted by Mij..8 Cofctor of mtrix: Let A be squre mtrix in (4 with mij which is the minors of its. Then the Cofctor of defined by Cij (- ij Mij Exmple: Let A 4 6 Solution: The minor of element 7 is 4 4 M det (4 7 find the minor nd the cofctor of element 7. i.e (denoted by tke the squre sub-mtrix by deleting the second rows nd third column in A. the Cofctor of 7 is

13 C ( M ( Adjoint of mtrix: Let mtrix A in (4 then the trnsposed of mtrix of cofctor of this mtrix is clled djoint of A, djoint A trnsposed mtrix of Cofctor. The inverse of mtrix: Let A be squre mtrix. Then inverse of mtrix {Where A is non-singulr mtrix} denoted by A - nd A - dj ( A det A. method to find the inverse of A: To find the inverse of mtrix we must find the following: (i the mtrix of minor of elements of A. (ii the Cofctor of minor of elements of A (iii the djoint of A. then A - A dja Exmple: let A 4 Find A - ( Minors of A is Mij ( Cofctor of A is (- Mij ( Adj of A 7 (4 det 6 7.

14 4 A Properties of Mtrix Multipliction: (KA B K (AB A (KB K is ny number A (BC (AB C (A B C AC BC 4 C (A B CA CB 5 AB BA (in generl For exmple: Let A nd B A B B A A B B A 6 A B but not necessrily A or B For Exmple: A, B A B, B A But A B 7 - C C C C 8 A I I A A where I is identity mtrix

15 9 (A B T B T A T A - A A.A - I. Crmer s Rule Let the system of liner question s b ( i b The system (i cn put in the form: b ( ii b If D Then the system (ii hs unique solution, nd Crmer s rule stte tht it my be found from the formuls: b b, D b b D Exmple: solve the system 9-4 So, the system cn put in the form, 9, 4 5

16 9 4 4 D 7, D D 7 Let the following system in the unknowns: b b b The system (I cn be put in the form: b b b (II If D The system hs unique solution, given by Crmer s rule: D b b b, D b b b D b b b Exmple: solve the system

17 7 By crmer s rule. The system ( become Since D det Crmer s rule gives the solution:

18 Chpter Two Function Numbers: N set of nturl numbers N {,,, 4 } I set of integers {., -, -, -,,,, } Note tht: NCI A set of rtionl numbers ρ : ρ nd q reint egers q q 4 7 Ex:,,, 5 Note tht: ICA 4 B set of irrtionl numbers { : is not rtionl number} Ex:,, 7 5 R: set of rel numbers set of ll rtionl nd irrtionl numbers Note tht R AUB Note: the set of rel numbers is represented by line clled line of numbers: (ii NCR, ICR, ACR, BCR Intervls The set of vlues tht vrible my tke on is clled the domin of. The domins of the vribles in mny pplictions of clculus re intervls like those shows below. open intervls is the set of ll rel numbers tht lie strictly between two fixed numbers nd b: In symbols In words b or ( q, b The open intervl b 8

19 Closed Intervls contin both endpoints: In symbols b or [, b] In words the closed intervl b Hlf open intervls contin one but not both end points: In symbols: in wrds b or [,b] the intervl less thn or equl cb To less thn b b or [,b] the intervl less thn less thn or equl b Ex: find the domin of Y The domin of is the closed intervl Y The domin for is open intervl becuse is not defined B - y or 9

20 The domin for is the hlf open Ex: the eqution the domin Y Y Y y 4 ( 4 The domin for is U Definition: A function, sy f is reltion between the elements of two sets sy A nd B such tht for every A there exists one nd only one Y B with Y F(. The set A which contin the vlues of is clled the domin of function F. The set B which contins the vlues of Y corresponding to the vlues of is clled the rnge of the function F. is clled the independer vrible of the function F, while Y is clled the dependnt vrible of F. Note: Some times the domin is denoted by DF nd the rnge by RF. Y is clled the imge of.

21 Exmple: Let the domin of be the set {,,,,4}. Assign to ech vlue of the number Y. The function so defined is the set of pirs, { (,, (,, (,4, (,9, (4,6 }. Exmple: Let the domin of be the closed intervl. Assign to ech vlue of the number y. The set of order pirs (, y such tht And y is function. Note: Now cn describe function by two things: the domin of the first vrible. the rule or condition tht the pirs (, y must stistfy to belong to the function. Exmple: The function tht pirs with ech vlue of diffrent from the number y f ( Note : Let f ( nd g ( be two function. - ( f ± g( f ( ± g( - ( f. g( f (. g( - ( f f ( ( if g ( g g( Exmple: Let f (, g( evlute f ± g, f. g nd f g

22 So: ( f ± g( f ( ± g( ± ( ( f. g ( f (. g( ( ( f f ( ( ( { : } g g( Composition of Function: Let f ( nd g( be two functions We define: ( fog ( f ( g( Exmple: Let f (, g( 7 evlute fog nd gof So: ( fog ( f [ g( x] f ( 7 ( 7 ( gof ( g[ f ( ] g( 7 fog gof Inverse Function Given function F with domin A nd the rnge B. The inverse function of f written f, is function with domin B nd rnge A such tht for every y B there exists only A with f ( y. Note tht: f f Polynomils: A polynomil of degree n with independent vrible, written f n (x or simply f ( is n expression of the form: fn ( q... n Where o n...(* q,,..., n re constnt (numbers. The degree of polynomil in eqution ( * is n ( the highest power of eqution Exmple: (i f ( 5 polynomil of degree one.

23 5 (ii f ( 7 polynomil of degree five. (iii F ( 8 polynomil of degree Zero. Notes: The vlue of which mke the polynomil f ( re clled the roots of the eqution ( f ( Exmple: ( is the root of the polynomil F ( Since f ( Exmple: F ( Liner function if F ( b. Even Function: F ( is even if f (-x F (x Exmple: - F ( ( is even since f ( ( ( f ( - F ( cos ( is even becuse f ( cos ( cos ( f ( Odd Function: If f ( f ( the function is clled odd. Exmple: - f ( is odd since f ( f ( - f ( Sin ( Sin f (. Trigonometric Function: Sin ϕ c

24 Cos ϕ c b tn ϕ b 4 Cotn ϕ 5 Sec ϕ b tnϕ c cosϕ b 6- CSC ϕ c Sinϕ Reltion ships between degrees nd rdins ϕ In rdius r s 6 o πr r π rdius o π 8 rdius.74 rdin 8 ο rdin deg ree π 6 ο rdin π 8 o π rdins.459 rdins o π π. 754 rdins 6 8 tn sin cos Cos Cot Sin tn C 4 Sin

25 5 cos Sec sin Csc Sin Cos tn Sec Csc Cot Sin y Cos Cosy Sin y Sin ± µ ( Siny Sin Cosy Cos y Cos ± ± ( y x y x y tn tn tn tn tn ( µ ± ± B A Cos B A Sin Sin B SinA - B A Sin B A Cos B Sin Sin A B A Cos B A Cos b Cos A Cos 4 - B A Sin B A Sin Cos B A Cos x Cos x Sin x Cos x Cos Cos Sin Sin Cos Cos Cos Sin Sin ϕ π ϕ Sin Sin ( ϕ π ϕ Cos Cos ( ϕ π ϕ tn ( tn B

26 Degree O o o 45 o 6 o 9 o 8 o 7 o 6 o θ rdius O π 6 π 4 π π π π π Sin θ O O - O Cos θ O - O tn θ O Cos ( ϕ nπ Cosϕ Sin ( ϕ nπ Sinϕ Cos ( ϕ Cosϕ Sin ( ϕ Sinϕ π Cos ( ϕ Cosϕ tn ( π ϕ tnϕ π tn ( ϕ Cotϕ Grphs: The set of points in the plne whose coordinte pirs re lso the ordered pirs of function is clled the grph of function. Exmple: Grph function we crry out three steps y, x 6

27 Mke tble of pirs from the function s y (, y - 4 (,4 - (-, (, (, 4 (,4 Plot enough of the corresponding points to lern the shpe of the grph. Add more pirs to the tble if necessry. Complete the sketch by connecting the points. Exmple: y 7

28 y (,y O (, O (, Absolute Vlue: We define the bsolute vlue function y, the function ssign every negtive number to non-negtive, which corresponding points. The bsolute vlues of : Then: if if -. b, b - b b - C C C y f ( y (,y - ( -, (, (, 8

29 Exmple : y f ( b y f ( y (,y O (, (, Exmple : y f ( y (,y (, Domin: Rnge: y - - (-, (, (, (, (, Domin: Rnge: y 9

30 y tnx Domin: All rel numbers except odd integer multiplies of π Rnge: y

31 Limits: We sy tht L is right hnd limit for f ( when pproches C for the right, written Lim f ( C L. Similry, L is the left hnd limit for f ( when pproches C for the left, written Lim f ( C. L, Then Lim f ( L, C Lim f ( Lim f ( x If nd. only if C C Exmple: ± Lim Lim Lim Theorem ( ( ( If Lim f ( L, Lim g( L c c Then Lim [ f ( ± g( ] L ± L - c

32 Lim [ f ( g( ] L L f ( L - Lim ( if L g L 4 - Lim c [ K f ( ] KL Theorem - Lim K K, K is constnt Where K is constnt - Lim[ c n... n ] c c... nc n Lim smx Lim Cos Sin Lim Exmple: Evlute - - Lim (4 4 Lim 4( Lim ( x ( ( 4 Lim 4 ( ( 4 Sin Sin 4 Lim Lim ( 4

33 tn x 5 Lim Sin Cos Sin Lim. Cos Sin Lim. Cos Infinity s Limits Evlute: - Lim Sin Lim. Lim Cos ( ( ( Lim Lim x Lim 5 x x Lim x Theorem If f ( g ( h( nd Lim f ( Lim h ( L then L is the limit of g(x Exmple: Evlute Sin Lim -

34 - Lim Sin ( Continuity Definition: A function f is sid to be continuous t following conditions re stisfied: f (C is defined C provided the Lim f ( x C exists Lim f ( x f ( C x C Theorem Any Polynomil 4 P (... n (n Is continuous for ll 4... n R( ( n, bn 4 bo b b... bn Is continuous t every point of its domin of definition tht is t every point where its denomintor isd not zero Ech of the igonometric function Sin, Cos, tn, Cot Sec, nd Csc, is continuous t every point of its domin of definition. Exmple Lim ( Cos Cos π Solution Lim ( Cos ( Cos ( Cos π Cosπ 4

35 Exmple f ( Discontinuous t Lim Lim Lim does not exist f ( discontinuous Exmple : check the continuity of the function f ( SOL f ( Lim ( f ( Lim f ( The function continuous t Problems Q// find Domin, rnge nd sketch ech of the following: - y - y - y 4- y 5 - y 6 - y 5

36 7 - y 8 - y Sin 9 - y Sin - y Cos Q // Evlute ech of the following limits: - t Lim t y - Lim - y 5y b Lim y y 4 - y Lim y 5y 6 y 5-4 Lim 6 - t Lim t t 7 - t Lim t t t 5t tnθ Lim 8 - θ θ Sin θ Lim 9 - θ θ - Sin Lim - Sin5 Lim Sin - Lim Sin - Sin Lim 4 - Sin Lim Lim tn Csc 4 5-6

37 Chpter Three Specil Function - Exponentil function (i y e, e. 7 Domin: R Rnge: R (, (ii y e Domin: R Rnge: R: (, (iii y, Domin: R Rnge: (, Logrithmic Function: (i Common logrithmic function (Log y Log Domin: (, Rnge: R Y 7

38 (ii Nturl Logrithmic function (Ln x y Ln x e, e.7 Domin: (, Rnge: R Theorem: If nd re positive numbers nd n is ny rtionl number, then (i Ln (ii Lne (iii Ln Ln Lnx (iv Ln ( Ln Ln (v Ln n n Ln. Note: Ln e ( ( Ln e ( e Y e y e (4 (5 Lim Ln e Lim (6 Ln Ln 8

39 Hyperbolic Functions: The Hyperbolic Functions re certin combintions of the exponentil e functions e nd they re: (i Hyperbolic Sine (Sinh: y Sinh, Sinh e e Domin: R Rnge: R (ii Hyperbolic Cosin (cosh y Coshx, Cosh e e Domin: R Rnge: (, (iii Hyperbolic tngent (tnh y tnh x, tnh x e e e e Domin: R 9

40 Rnge: ( -, (iv Hyperbolic cotngent (coth y coth, coth Domin: R - { } e e e e Y Coth Rnge: { y : y or y } (v Hyperbolic Secnt (Sech y Sech, Sech e e Domin: R Rnge: (o, (vi Hyperbolic cosecnt (Csch y Csch, Csch Domin: R - { } Rnge: R - { } e e 4

41 Reltionships mong Hyperbolic Function - Cosh Sinh - Sech tnh - Coth Csch Functions of negtive rguments - Sinh( Sinh - Cosh ( Cosh - tnh ( tnh 4 - Coth ( Coth 5 - Sech ( Sech 6 - Csch( Csh Addition Formul: - Sinh ( ± y Sinh Coshy µ Coshy Sinhy - Cosh ( ± y Cosh Coshy ± Sinh Sinhy Double ngle formul: - Sinh Sinh Cosh - Cosh Cosh Sinh Sinh Cosh Inverse Trigonometric Function Inverse Sine (Sin - y Sin rc Sine Siny. 4

42 Y is the ngle whose Sine is Exmple: 45 ο π Sin, Sin Domin: [, ] π π Rnge:, Principle vlues of y. Inverse Cosine ( Cos y Cos rc Cos Cosy y is the ngle whose cosin is Exmple: ο Cos Cos Domin: [, ] Rnge: [,π ] ο Inverse of tngent (tn - y tn rc tn tn y y is the ngle whose cosin is Exmple: 45 ο tn tn 45 Domin: R π π Rnge: (, 4

43 4 Inverse of cotngent (Cot - y Cot rc Cot Coty Domin: R Rnge: (, π - 5 Inverse secnt (Sec - y Sec rc Sec Secy Domin: (, ] Υ [, Rnge: π π, Υ π 6 Inverse Csc (Csc - y Csc rc Csc Cscy Domin: (, ] Υ [, Rnge: π π, Υ, Inverse hyperbolic Functions Inverse hyperbolic sine ( Sinh - y Sinh Sinhy 4

44 Domin: R Rnge:R - Inverse hyperbolic cosin (Cosh - y Cosh Coshy Domin: [, Rnge: [, Inverse hyperbolic tngent (tnh - y tnh tnh y Domin: (-, Rnge: R 4 Inverse hyperbolic cotngent (Coth - y Coth Cothy Domin: { Υ } Rnge: R / { } 44

45 5 Inverse hyperbolic secnt (Sech - y sec h sec hy Domin: (, ] Rnge: y 6 Inverse hyperbolic cosecnt (Csch - y Csch Cschy Logrithmic Form of Inverse hyperbolic Functions Theorem: the following reltionships hold for ll in the domin of the stted inverse hyperbolic. Functions: - Sinh Ln ( - Cosh Ln ( 45

46 - tnh Ln ( ( 4 - Coth Ln ( ( 5 - Sech Ln ( 6 - Csch Ln ( Prove tht: * Sinh Ln ( Sol Let y Sinh * Sinhy Since Sinhy y y e e z y y e e z y e y e y y y y y e z e e e e y e ± 4 4 ± y Since e then y e y Ln ( or Sinh Ln ( Exmple: Sinh Ln ( Ln (.88 46

47 * tnh Ln ( Sol Let y tnh tnh y y y e e y e e y y ( e e y y y e e y y y y ( e e e e y y e e y e ( y e y Ln ( tnh Ln ( Exmple: tnh ( Ln ( Ln Problems: Q: find domin, rnge nd sketch ech of following: π π - y Sin, y - y Cot, y π - π π y tn, y 47

48 4 - y π π csc, y 5 - y Cot, y π 6 - y y Sec, y π 7 - y Ln 8 - y Ln 9 - y e - y e - y Sinh - y Cosh - y Cosh 4 - y tnh 5 - y Coth 6 - y Coth 7 - y Sech 8 - y csc h Q: Prove tht: - Sinh Ln (, - Coch Ln (, - tnh Ln (, 4 - Coth Ln (, 5 - Sech Ln (, 48

49 49 6 -, ( Ln Csch Q// Discuss the Continuity of the following functions t the given points: ( t f - { ( t if if if f - ( t f 4 - ( t for for Sin Cos f 5 - ( t for for f ( t for for f 7 - ( π t Lnx Sin f

50 8 - f ( t 9 - f ( for for t Q4 // Simplify ech of the following: - Ln e - Ln ( e - ( Ln e 4 - Ln ( e 5 - Ln ( e 6 - Ln ( e 7 - Ln( e Ln( e Ln Ln e - Ln e e - Ln ( - Ln (, e 5

51 - e Ln Ln e 4 - Lny 5

52 Chpter Four The derivtive Derivtive of function: Let y f ( nd let P(, y be fixed point on the curve, nd Q (, y y is nother point on the curve s see in the figure y f (, nd y y f ( y f ( y Divided by y f ( f ( The slope of the curve f ( is M tnφ M y f ( f ( We define the limit my exist for some vlue of. At ech point where limit does exist, then f is sid to hve derivtive or to be differentible. 5

53 5 Rules of Derivtions: C constnt n Positive integer dx du LnC C f t Cons C U C f d du e f e f d du U n f U f d du U Where V uv vu f v u f d du V d dv U f UV f d dv dx du d dy f V U f Cn f n f f d dy f y U U u u n n n n n.. ( tn ( ( ( ] [ ( ] [ (, ( ( ( ( ( ( ( ( ( ( ± C f y (

54 Derivtive of trigonometric functions: (sin u ' cos u du (cos u ' - sin u du (tn u ' sec u du 4 (cot u ' - csc u du 5 (sec u ' sec u tn u du 6 (csc u ' - csc u cot u du Derivtive of hyperbolic functions: sinh u cosh u du cosh u sinh u du tnh u sech u du 4 cot u - csch u du 5 sech u - sech u tnh u du 6 csch u - csch u coth u du derivtive of the inverse trigonometric functions: (sin - u ' du/(-u / (cos - u ' - du/(-u / (tn - u ' du/u 4 (cot - u ' - du/u 5 (sec - u ' du/ u(u - / 6 (csc - u ' - du/ u(u - / derivtive of the inverse of hyperbolic functions: (sinh - u ' du/( u / (cosh - u ' du/ ( u - / (coth - u ' du/-u if u > 4 (tnh - u du/-u if u <

55 5 (sech - u ' - du/ u(-u / 6 (csch - u ' - du/ u(u / ex: find y ' of ( y [ln (x ] ( y 4 x Sol: ( y ' [ln (x ] [/(x] 9[ln (x ] / (x ( y ' 4 x ln4 {Applictions of derivtive} Velocity nd ccelertion Ex: find velocity nd ccelertion t time t to moving body s S t 5t 4t -. Sol: V ds/dt 6t t 4 A dv/dt t- Theorem: Prove tht: D(sin - u /(- u / (du/dx Proof Let y sin - sin y u u[-,] y [ Π/, Π/] Cos y dy/dx du/dx dy/dx /cos y du/dx Since cos y sin y this implies tht Cos y ( - sin y / Cos y ± ( - u / Cos y is positive between Π/ nd Π/ Dy/dx /(- u / (du/dx D(sin - u Ex: find dy/dx for the following functions: ( y tn (x

56 ( y x sin - x ( x / ( y cosh - (sec x sol: ( y ' sec (x 6x 6x sec (x ( y ' x/( x / sin - x - x/( x / sin - x ( y ' [/(sec x / ] sec x tn x sec x tn x/(sec x / Implicit reltions: Ex: find dy/dx if x 5 4x y y 5 sol: 5x 4 4x y (dy/dx 4y 5y 4 (dy/dx (x y 5y 4 dy/dx - 5x 4-4y dy/dx ( - 5x 4-4y /(x y 5y 4 Chin Rule - If y f (x, nd x x (t, then y y x t x t - If y f (t, nd x x (t, then y x y t x t dy dt dx dt dy dx Ex: find, nd of x t nd y t Sol: dx dt. dy dt t, dy dx LHopitl's Rule dy dt dx dt t

57 Let f nd g be two functions which re differentible in n open intervl I contining the point c nd let g ' (x. if f ( x lim x c g( x,,.,.,.,,, then lim x c f ( x g( x lim x c ' f ( x. ' g ( x Ex: evlute Sol: ( lim x sin x x x x x x lim sin (, x ( x ln x x lim x sin x x cosx lim lim x x x x ( lim x ln x. x ln x lim x ln x lim x x x lim ln x lim x lim x x x x x x Series (Power series: If { n } is sequence of constnts, the expression: x x. n x n. is clled power series in x. n n n x

58 (Tylor's series: If function f cn be represented by power series in (x-b clled Tylor's series nd hs the form: f ( x f ( b ' f ( b( x b f '' ( b( x b!... f n ( b( x b n! n... Exmple: Find Tylor series expnsion of cos x bout point Sol: f(x cos x f( cos ( f ' ( x sin x, f ' ( sin f '' ( x cosx, f '' ( cos f ''' ( x sin x, f ''' ( sin f iv( x cosx, f iv( cos ( x ( x 4 ( x cosx 6...! 4! 6! (Mclurin series: when b, Tylor series clled Mclurin series. Exmple: Find Mclurin series for the function f(x e x Sol: f(x e x f( e f ' (x e x f ' ( e f '' (x e x f '' ( e f ''' (x e x f ''' ( e e x x (x /! (x /!.

59 Chpter five (INTEGRALS The process of finding the function whose derivtive is given is clled integrtion, it's the inverse of differentition. Definition:(indefinite integrl A function yf(x is clled solution of dy/dxf(x if df(x/dxf(x. We sy tht F(x is n integrl of f(x with respect to x nd F(x c is lso n integrl of f(x with constnt c s.t D(F(x cf(x. Formuls of Integrtion: ʃ dxx c. ʃ dx ʃdx ʃ(du ± dv ʃdu ± ʃdv. 4 ʃx n dx(x n /n c 5 ʃ(u n du(u n /n c 6 ʃe u du e u c 7 ʃ u du( u /ln c 8 ʃ du/u ln u c. Exmple: Solve the differentil eqution: dy/dxx. Sol: dyx dx since d(x x dx, then we hve: ʃ dy ʃ x dx ʃ d(x dx y x c. 9 methods for finding integrls:

60 ''Integrl of trigonometric functions'': ʃ cos u du sin u c ʃ sin u du - cos u c ʃ sec u dutn u c 4 ʃ csc u du-cot u c 5 ʃ sec u tn u du sec u c 6 ʃ csc u cot u du - csc u c ''Integrl of hyperbolic functions'': cosh u du sinh u c ʃ sinh u du cosh u c ʃ sech u du tnh u c 4 ʃ csch u du -cot u c 5 ʃ sech u tnh u du - sech u c 6 ʃ csch u coth u du - csch u c Integrl of the inverse trigonometric functions: ʃ du/(-u / {sin - u c or -cos - u c } ʃ du/u {tn - u c or -cot - u c} ʃ du/ u(u - / sec - u c or -csc - u c} Integrl of the inverse of hyperbolic functions: ʃ du/( u / sinh - u c ʃ du/ ( u - / cosh - u c ʃ du/-u tnh - u c if u < nd ʃ du/-u coth - c if u > 4 ʃ du/ u(-u / - sech - u c 5 ʃ du/ u(u / -csch - u c

61 ex: evlute: ʃ (5x 4-6x / x dx ʃ cos x dx ʃ cos x dx sol: ʃ (5x 4-6x / x dx 5 ʃ x 4 dx-6 ʃ x dx ʃ x - dx 5 -x -/x c ʃ cos x dx sin x/ c ʃ cos x dx / ʃ( cos x dx /[ʃ dx ʃ cos x dx] /[x sin x/] c /x /4 sin x c - ''Integrtion by prts'' Let u nd v be functions of x nd d(uv u dv v du By integrtion both sides of this eqution (w.r.t x ʃ d(uv ʃ u dv ʃ v du this implies (uv ʃ u dv ʃ v du ʃ u dv (uv - ʃ v du ex: find ʃ x e x dx sol: let ux du dx nd let dv e x dx from ʃ u dv (uv - ʃ v du ʃ x e x dx x e x - e x c 4- ''Integrls involving ( - u /, ( u /, (u - /, - u, u, u - '' (A u sin Φ replces - u - sin Φ (- sin Φ cos Φ (B u tn Φ replces u tn Φ sec Φ (C u sec Φ replces u - sec Φ - tn Φ

62 Ex: find ʃ dx/x (4 - x / Sol: Let xsin Φ dx cos Φ dφ ʃ dx/x (4 - x / ʃ cos Φ dφ/4 sin Φ(4-4 sin Φ / ʃ cos Φ dφ/4 sin Φ(cos Φ ʃ dφ/4 sin Φ /4ʃ csc Φ dφ - /4cot Φ c now from x sin Φ sin Φ x/ cos Φ (- x /4 / /(4 - x / -/4cot Φ c (-/4(4 - x / /x 5-'' Integrls involving x bx c '' First, We put the eqution s (x bx c. Second, if, we tke s mutble by the sides of the eqution which hs x, [x (b/x] c. Third, put nd sub to the eqution [(/ the number multiplied by x], [x (b/x (/4(b/ - (/4(b/ ] c. Fourth, rewrite the eqution s [x (b/x (/4(b/ ] c - (/4(b / Lst, the eqution become [x (/ (b/] c - (/4(b / nd suppose u x (/ (b/ to become [u] c - (/4(b / Ex: Find dx/(4x 4x Sol: 4x 4x (4x 4x 4(x x 4[x x (/4 -(/4] 4[x x (/4] - 4[x/]. Let u x/ 4[x/] 4u. Since u x/ x u (/ dx du

63 dx/(4x 4x du/(4u / du/(4u / tn - u / tn - (x/. 6-''method of prtil frctions'' If the integrl of the form f(x/g(x s.t f(x nd g(x re poly. And degree of f(x< degree of g(x we cn crry out two cses: Cse i If ll fctor of g(x re liner, by the following ex: Ex: find ʃ dx/ x x Sol: / x x /(x-(x A/(x- B/(x [A(x B/(x - ] /(x- (x Ax A Bx B (ABx (A-B A-B (AB A A/ put in eq.( B -/ ʃ dx/ x x ʃ /(x-(xdx ʃ [A/(x- B/(x ]dx ʃ [(//(x- (//(x ]dx /ʃ /(x- / ʃ /(x dx / ln x- -/ ln x c Cse ii If some of the fctors of g(x re qudrtic, by the following ex: Ex: find ʃ (x x dx/ (x - x x Sol: (x x / (x - x x (x x / x (x (x (x x / (x (x [A/(x ] [(Bx C/ (x ] [A(x (Bx C (x ]/ (x (x x x A(x (Bx C (x x x (A B x (B C x (A-C A B

64 B C A - C - A -7/5, B 4/5, C /5 (x x / (x (x (-7/5/(x [(4/5x (/5]/ (x And ʃ (x x dx/ (x (x (-7/5 ʃ dx/(x (4/5 ʃ x dx/(x (/5 ʃ dx/ (x -(7/5 ln x- (/5 ln x /5 tn - x 7-''further substitutions'' Some integrls involving frctionl powers of the vrible x my be simplified by substitution x u n where n is the lest common multiple of the denomintors of the exponents. Ex: Iʃ (x / dx/(x / sol: let xu 6 dx 6 u 5 du I ʃ (u 6 / (6 u 5 du/(( u 6 / 6 ʃ u u 5 du/u 6 ʃ u 8 du/u By long division u 8 /u u 6 u 4 u - (/ u I 6ʃ u 8 du/u 6 ʃ [u 6 u 4 u - (/ u ] du (6/7 u 7 (6/5 u 5 u -6u 6 tn - uc (6/7 x 7/6 (6/5 x 5/6 x / -6x /6 6 tn - (x /6 c 8-''rtionl functions of sin x nd cos x'' If the integrl tht is rtionl function of sin x or cos x or both, cn be chnged s following: Let z tn(x/ x/ tn - z x tn - z dx dz/( z cos(x/ /( z /, sin(x/ z/( z /

65 sin x sin(x/ cos(x/ z/( z cos x cos (x/ sin (x/ (- z /( z from [cos(x/ x/] ex: find I ʃ dx/(- sin x cos x sol: I ʃ dz/( z (- [z/( z ] [(- z /( z ] ʃ dz/( z ( z -z- z /( z ʃ dz/(-z ʃ dz/(-z -ln -z c -ln -tn(x/ c 9-''evluting integrls of the following types'' (A sin(mx sin(nx (/ [cos(m-nx- cos(mnx] (B sin(mx cos(nx (/ [sin(m-nxsin(mnx] (C cos(mx cos(nx (/ [cos(m-nx cos(mnx] Ex: ʃ sin(4x sin(x dx ʃ (/ [cos(4-x- cos(4x]dx ʃ (cos x- cos 7xdx sin x (/7sin 7x c {definite integrl} The definite integrl like indefinite integrl but there is limit to the integrl like ʃ b f(x dx F(- F(b. Ex: evlute ʃ x dx Sol: ( 4 /4 8/4 Applictions of definite integrl {re under the curve}

66 Ex: find the re under sin x bdd by x nd x π nd x-xis Sol: A ʃ π sin x dx ʃ π sin x dx- ʃ π π sin x dx (cos π-cos ( cos π-cos π (--(-(-4 {re between two curves} Ex: find the re bdd by y - x nd y -x Sol: Y - x y -x - x -x x x- (x-(x x, x - A ʃ - [(-x -(-x]dx [x (x / (x /] Double integrls When the integrl hve two signls of integrl to two prmeters x nd y clled double integrl, like ʃ ʃ f(x,y dx dy. The benefit of like integrls is to find the volume of things. Ex: find the volume of f(x,y x y limited by x(, nd y (. ʃ ʃ x y dx dy ʃ [(x /y dy] ʃ [( /- ( /]y dy ʃ [(7/- (/]y dy ʃ (6/y dy (6/ʃ y dy [(6/(y /] [(/ y ] (/ 4- (/

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