7. Differentiation of Trigonometric Function

Transcription

1 7. Differentiation of Trigonoetric Fnction RADIAN MEASURE. Let s enote the length of arc AB intercepte y the central angle AOB on a circle of rais r an let S enote the area of the sector AOB. (If s is /60 of the circference, AOB = 0 ; if s = r, AOB = raian). Sppose AOB is easre as a egrees; then (i) csc = cot C an π S = α r 60 Sppose net that AOB is easre as θ raian; then (ii) s = θ r an S = ½ θ r A coparison of (i) an (ii) will ake clear one of the avantages of raian easre. TRIGONOMETRIC FUNCTIONS. Let θ e any real ner. Constrct the angle whose easre is θ raians with verte at the origin of a rectanglar coorinate syste an initial sie along the positive -ais. Take P(, y) on the terinal sie of the angle a nit istance fro O; then sin θ = y an cos θ =. The oain of efinition of oth sin θ an cos θ is the set or real ner; the range of sin θ is y an the range of cos θ is. Fro sinθ tan θ = an sec θ = cosθ cos θ it follows that the range of oth tan θ an sec θ is set of real ners while the oain n of efinition (cos θ 0) is θ ± π, (n =,,, ). It is left as an eercise for the reaer to consier the fnctions cot θ an csc θ. In prole, we prove sin θ li = θ 0 θ (Ha the angle een easre in egrees, the liit wol have een π/80. For this reason, raian easre is always se in the calcls) This instrction aterial aopte fro Calcls y Frank Ayres Jr

2 RULES OF DIFFERENTIATION. Let e a ifferentiale fnction of ; then. (sin ) = cos 7. (cot ) = csc 5. (cos ) = sin 8. (sec ) = sec tan 6. (tan ) sec = 9. (csc ) = csc cot 8. Differentiation of Inverse trigonoetric fnctions THE INVERSE TRIGONOMETRIC FUNCTIONS. If = sin y, the inverse fnction is written y = arc sin. The oain of efinition of arc sin is, the range of sin y; the range of arc sin is the set if real ners, the oain of efinition of sin y. The oain if efinition an the range of the reaining inverse trigonoetric fnctions ay e estalishe in a siilar anner. The inverse trigonoetric fnctions are lti-vale. In orer that there e agreeent on separating the graph into single-vale arcs, we efine elow one sch arc (calle the principal ranch) for each fnction. In the accopanying graphs, the principal ranch is inicate y a thickening of the line. y = arc sin y = arc cos y = arc tan Fnction y = arc sin y = arc cos y = arc tan y = arc cot y = arc sec y = arc csc Fig. - Principal Branch y π π 0 y π π < y < π 0 < y < π y < π, 0 y < π π < y π, 0 < y π π This instrction aterial aopte fro Calcls y Frank Ayres Jr

3 RULES OF DIFFERENTIATION. Let e a ifferentiale fnction of, then 0. (arc sin ) =. (arc cot ) =. (arc cos ) =. (arc sec ) =. (arc tan ) = 5. (arc csc ) = 9. DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS THE NUMBER e = h k 0 h li = li( k) h / k = K K =.78 8K!! n! NOTATION. If a > 0 an a, an if a y =, then y = log a y = log e = ln y = log 0 = log The oain of efinition is > 0; the range is the set of real ners. y = ln y = e a y = e -a Fig. - Rles of ifferentiation. If is a ifferentiale fnction of, a lna 6. ( log ) =, ( a > 0, a ) 7. (ln ) = 8. (a ) = a ln a, ( a > 0) 9. (e ) = e This instrction aterial aopte fro Calcls y Frank Ayres Jr 5

4 LOGARITHMIC DIFFERENTIATION. If a ifferentiale fnction y = f() is the proct of several factors, the process of ifferentiation ay e siplifie y taking the natral logarith of the fnction efore ifferentiating or, what is the sae thing, y sing the forla 0. (y) = y ( ln y) 0. DIFFERENTIATION OF HYPERBOLIC FUNCTIONS DEFINITIONS OF HYPERBOLIC FUNCTION. For any real ner, ecept where note: e e e e sinh = coth = =, ( 0) tanh e e e e cosh = sec h = = cosh e e sinh e e tanh = = csch = =, ( 0) cosh e e sinh e e DIFFERENTIATION FORMULAS. If is a ifferentiale fnction of,. (sinh ) = cosh. (coth ) = csc h. (cosh ) = sinh 5. (sec h ) = sec h tanh. (tanh ) sec h = 6. (csc h ) = csc h coth DEFINITIONS OF INVERSE HYPERBOLIC FUNCTIONS. sinh cosh tanh = ln( ),all coth = ln, ( > ) = ln( ), ( ) sec h = ln, (0 < ) = ln, ( < ) csc ln h =, ( 0) This instrction aterial aopte fro Calcls y Frank Ayres Jr 6

5 Differentiation forlas. If is a ifferentiale fnction of, 7. (sinh ) = 8. (cosh ) =, ( > ) 9. (tanh ) =, ( < ) 0. (coth ) =, ( > ). (sec h ) =, (0 < < ). (csc h ) =, ( 0). PARAMETRIC REPRESENTATION OF CURVES PARAMETRIC EQUATIONS. If the coorinates (, y) of a point P on a crve are given as fnctions = f(), y = g() of a thir variale or paraeters, the eqations = f(), y = g() are calle paraetric eqations of the crve. Eaple: (a) = cos θ, y = sin θ are paraetric eqations, with paraeter θ, of the paraola y =, since y = cos θ sin θ = () = t, y = t is another paraetric representation, with paraeter t, of the sae crve. (a) Fig. 6- () This instrction aterial aopte fro Calcls y Frank Ayres Jr 7

6 It shol e note that the first set of paraetric eqations represents only a portion of the paraola, whereas the secon represents the entire crve. y THE FIRST DERIVATIVE is given y y y The secon erivative is given y = y y / = / y SOLVED PROBLEMS y y. Fin an, given = θ - sin θ, y = - cosθ y = cosθ, = sinθ, an θ θ y sinθ θ cosθ = = θ cosθ cosθ y y / θ sinθ = = / θ cosθ = ( ) cosθ ( cosθ ). FUNDAMENTAL INTEGRATION FORMULAS IF F() IS A FUNCTION Whose erivative F`()=f() on a certain interval of the - ais, then F() is calle an anti-erivative or inefinite integral of f(). The inefinite integral of a given fnction is not niqe; for eaple,, 5, are inefinite integral of f() = since ( ) = ( 5) = ( ) =. All inefinite integrals of f() = are then incle in C where C, calle the constant of integration, is an aritrary constant. The syol f ( ) is se to inicate that the inefinite integral of f() is to e fon. Ths we write = C FUNDAMENTAL INTEGRATION FORMULAS. A ner of the forlas elow follow ieiately fro the stanar ifferentiation forlas of earlier chapters while Forla 5, for eaple, ay e checke y showing that arcsin a a C = a a Asolte vale signs appear in certain of the forlas. For eaple, we write 5. = ln C This instrction aterial aopte fro Calcls y Frank Ayres Jr 8

7 instea of 5(a) = ln C, > 0 5(). ln( ), = C < 0 an 0. tan = ln sec C instea of 0(a) 0() tan = ln sec C, all sch that sec tan = ln ( sec ) C, all sch that sec - Fnaental Integration Forlas 8. csc cot csc. [ f() ] = f() C. v) = = C ( v 9. = arc tan C a a a a =, a any constant 0. = arc sin C a a. a. = C, 5. = ln C a 6. a = C, a > 0, a lna 7. e = e C, 8. sin = cos C 9. cos = sin C 0. tan = ln sec C. cot = ln sin C. sec = ln sec tan C. csc = ln csc cot C. sec = tan C 5. csc = cot C 6. sec tan = sec C This instrction aterial aopte fro Calcls y Frank Ayres Jr 9

8 . = arc sec C a a a a. = ln C a a a a. = ln C a a a. = ln a C a 5. = ln a a C 6. a = a a arcsin C a 7. a = a a ln a C 8. a = a a ln a C. INTEGRATION BY PARTS INTEGRATION BY PARTS. When an v are ifferentiale fnction of (v) = v v v = (v) v() v = v v (i) To se (i) in effecting a reqire integration, the given integral st e separate into two parts, one part eing an the other part, together with, eing v. (For this reason, integration y the se of (i) is calle integration y parts.) Two general rles can e state: (a) the part selecte as v st e reaily integrale () v st not e ore cople than v Eaple : Fin e Take = an v = e ; then = an v = e e e e = e e = C. Now y the rle This instrction aterial aopte fro Calcls y Frank Ayres Jr 0

9 Eaple : Fin ln( ) Take = ln ( ) an v = ; then = an v =. By the rle. ln( ) = ln( ) = ln( ) = ln ( ) arc tan / C REDUCTION FORMULAS. The laor involve in sccessive applications of integration y parts to evalate an integral ay e aterially rece y the se of rection forlas. In general, a rection forla yiels a new integral of the sae for as the original t with an eponent increase or rece. A rection forla sccees if ltiately it proces an integral which can e evalate. Aong the rection forlas are: (A). =, ( ± ) ( )( ± ) a a a ( a ± ) (B). ( a ± ) ( a ± ) = a ( a ± ), / (C). =, ( ) ( )( ) a a a ( a ) (D). (E). (F). (G). (H). (I). (J). ( a ) ( a ) = e e a sin sin cos cos cos sin n n a a a a = e sin ( a ) = sin = cos sin cos n n n, / n n cos = sin cos sin cos n sin cos = sin cos n n = cos cos = sin sin n, n This instrction aterial aopte fro Calcls y Frank Ayres Jr

10 . TRIGONOMETRIC INTEGRALS THE FOLLOWING IDENTITIES are eploye to fin the trigonoetric integrals of this chapter.. sin cos =. tan = sec. cot = csc. sin = ( cos ) 5. cos = ( cos ) 6. sin cos = sin 7. sin cos y = [ sin( y) sin( y) ] 8. sin sin y = [ cos( y) cos( y) ] 9. cos cos y = [ cos( y) cos( y) ] 0. cos = sin. cos = cos. ± sin = ± cos( π ) SOLVED PROBLEMS SINES AND COSINES sin = C. ( cos ) = sin cos = 6 C. ( cos6 ) = sin = )sin = cos 8 cos C. sin sin sin = ( cos. cos cos cos = ( sin ) cos sin cos = cos = sin cos = sin sin sin C 5. sin cos sin cos cos = sin cos sin 5 = sin ( sin ) cos 5 = cos = sin 5 sin C This instrction aterial aopte fro Calcls y Frank Ayres Jr

11 5. TRIGONOMETRIC SUBSTITUTIONS AN INTEGRAND, which contains one of the fors a, a, or a t no other irrational factor, ay e transfore into another involving trigonoetric fnctions of a new variale as follows: For Use To otain a sin z a tan z a sec z a = a sin z = a cos z a = a tan z = a sec z a = a sec z = a tan z In each case, integration yiels an epression in the variale z. The corresponing epression in the original variale ay e otaine y the se of a right triangle as shown in the solve proles elow.. Fin SOLVED PROBLEMS Let = tan z; then = sec z z an = = = sec z z sec z z ( tan z)( sec z) tan z = sin sec z cos z z = C = C sin z. Fin Let = sec z; then = sec z tan z z an = tan z z = ( sec z tan z z) = sec z tan z = sec z tan z ln sec z tan z C This instrction aterial aopte fro Calcls y Frank Ayres Jr sec = ln C z z

12 9. Fin Let = sin z ; then = cos z z an 9 = 9 cos z cos z sin z = ( cos z z) = cos sin z z z sin z = z = csc z z sin z z sin z = ln csc z cot z cos z C = ln 9 9 C 6. INTEGRATION BY PARTIAL FRACTIONS A POLYNOMIAL IN is a fnction of the for a 0 n a n a n- n a n, where the a s are constants, a 0 0, an n is a positive integer incling zero. If two polynoials of the sae egree are eqal for all vales of the variale, the coefficients of the like powers of the variale in the two polynoials are eqal. Every polynoial with real coefficients can e epresse (at least, theoretically) as a proct of real linear factors of the for a an real irrecile qaratic factors of the for a c. f ( ) A fnction F ( ) =, where f() an g() are polynoials, is calle a rational g( ) fraction. If the egree of f() is less than the egree of g(), F() is calle proper; otherwise. F() is calle iproper. An iproper rational fraction can e epresse as the s of a polynoial an a proper rational fraction. Ths, =. Every proper rational fraction can e epresse (at least, theoretically) as a s of sipler fractions (partial fractions) whose enoinators are of the for (a ) n an (a c) n, n eing a positive integer. For cases, epening pon the natre of the factors of the enoinator, arise. This instrction aterial aopte fro Calcls y Frank Ayres Jr

13 CASE I. DISTINCT LINEAR FACTORS To each linear factor a occrring once in the enoinator of a proper A rational fraction, there correspons a single partial fraction of the for a, where A is a constant to e eterine. CASE II. REPEATED LINEAR FACTORS To each linear factor a occrring n ties in the enoinator of a proper rational fraction, there correspons a s of n partial fractions of the for A A An K a ( a ) ( a ) where the A s are constants to e eterine. n CASE III. DISTINCT QUADRATIC FACTORS To each irrecile qaratic factor a c occrring once in the enoinator of a proper rational fraction, there correspons a single partial fraction of A B the for, where A an B are constants to e eterine. a c CASE IV. REPEATED QUADRATIC FACTORS To each irrecile qaratic factor a c occrring n ties in the enoinator of a proper rational fraction, there correspons a s of n partial fraction of the for A B A B An Bn K a c ( a c) ( a c) where the A s an B s are constants to e eterine. n SOLVED PROBLEMS. Fin (a) Factor the enoinator: = (-)() A B Write = an clear of fraction to otain () = A( ) B( ) or () = (A B) (A B) () Deterine the constants General etho. Eqate coefficients of like powers of in () an solve siltaneosly for the constants. Ths, A B = 0 an A B = ; A =, an B =. This instrction aterial aopte fro Calcls y Frank Ayres Jr 5

14 Short etho. Sstitte in () the vales = an = - to otain = A an = -B; then A = an B =, as efore. (Note that the vales of se are those for which the enoinator of the partial fractions ecoe 0). (c) By either etho: = = = ln ln C = ln an C ( ). Fin 6 (a) 6 = ( )( ). Then = 6 A B C an () = A( )( ) B( ) C ( ) or () = (A B C) (A B C) 6A () General etho. Solve siltaneosly the syste of eqation A B C = 0, A B C =, an -6A = To otain A = -/6, B = /0, an C = -/5. Short etho. Sstitte in () the vales = 0, =, an = - to otain = - 6A or A = -/6, = 0B or B = /0, an = 5C or C = -/5 ( ) (c) = ln ln ln C = ln C / 6 / /0 = 5 ( 5). Fin = ()(-) 5 A B C. Then = ( ) 5 = A( ) B( )( ) C (). an For = -, = A an A = ½. For =, 8 = C an C =. To eterine the reaining constant, se any other vale of, say = 0; for = 0, 5 = A B C an B = -½. Ths ( 5) = = ln ln = ln C - - ( ) - C This instrction aterial aopte fro Calcls y Frank Ayres Jr 6