( ) ( ) ( ) ( u) ( u) = are shown in Figure =, it is reasonable to speculate that. = cos u ) and the inside function ( ( t) du
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1 Porlan Communiy College MTH 51 Lab Manual The Chain Rule Aciviy 38 The funcions f ( = sin ( an k( sin( Since f ( cos( k ( = cos( 3. Bu his woul imply ha k ( f ( = are shown in Figure =, i is reasonable o speculae ha 0 = 0 = 1, an a quick glance of he wo funcions a 0 shoul convince you ha his k 0 > f 0. is no rue; clearly ( ( The funcion k moves hrough hree perios for every one perio Figure 38.1: f an k generae by he funcion f. Since he ampliues of he wo funcions are he same, he only way k can generae perios a a rae of 3:1 (compare o f is if is k = 3cos 3 ; please noe ha 3 is he firs rae of change is hree imes ha of f. In fac, ( ( erivaive of 3. This means ha he formula for k ( he ousie funcion ( sin ( u = cos u an he insie funcion ( ( u ( ( is he prouc of he raes of change of 3 = 3. k is an eample of a composie funcion (as illusrae in Figure 38.. If we efine g by he rule g k = f g. ( = 3, hen ( ( ( 3 sin( 3 ( Noe ha k ( 0 g ( 0 f g( 0 g Figure 38.: g( = 3, f ( u = sin( u, an k( = f ( g( Taking he oupu from g an processing i hrough a secon funcion, f, is he acion ha characerizes k as a composie funcion. =. This las equaion is an eample of wha we call he chain rule for iffereniaion. Loosely, he chain rule ells us ha when fining he rae of change for a composie funcion (a 0, we nee o muliply he rae of change of he ousie funcion, f ( g( 0, wih he rae of change of he insie funcion, g ( 0. This is symbolize for general u = g. values of in Equaion 38.1 where u represens a funcion of (e.g. ( Equaion 38.1 ( ( f u = f ( u ( u This rule is use o fin erivaive formulas in eamples 38.1 an 38.. f Lab Aciviies 57
2 Porlan Communiy College MTH 51 Lab Manual Problem Eample 38.1 Soluion The facor of ( a chain rule facor. is calle Fin y if y sin ( =. y = cos ( ( ( ( = cos = cos Fin f ( Eample 38. Problem Soluion 9 if f ( sec ( =. The facor of ( sec( is calle a chain rule facor. 9 f ( = sec( 8 f ( = 9 sec( ( sec( 8 = 9sec sec an = 9sec 9 ( ( ( ( an( Problem Eample 38.3 Soluion Fin y if y = 4. y ln ( 4 = 4 Please noe ha he chain rule was no applie here because he funcion being iffereniae was no a composie funcion. While you ulimaely wan o perform he chain rule sep in your hea, your insrucor may wan you o illusrae he sep while you are firs pracicing he rule. For his reason, he sep will be eplicily shown in every eample given in his lab. Problem 38.1 Fin he firs erivaive formula for each funcion. In each case ake he erivaive wih respec o he inepenen variable as implie by he epression on he righ sie of he equal sign. Make h. sure ha you use he appropriae name for each erivaive (e.g. ( h ( cos ( ( 3 4 = P sin ( θ 4 z = 7 ln z ( θ sin ( θ 3 = w( α = co ( α 1 = P ( β = an ( β y = sin ( 17 ln( T = y ( = sec ( e 58 L ab Aciviies
3 Porlan Communiy College MTH 51 Lab Manual Problem 38. A funcion, f, is shown in Figure Answer each of he following quesions in reference o his funcion Use he graph o rank he following in ecreasing orer: f 1. f (, f (, f (, f (, an ( f = e e + 3. Fin he formulas for f an f an use he formulas o verify your answer o problem Fin he equaion of he angen line o f a Fin he equaion of he angen line o f a The formula for f is ( Figure 38.3: f There is an anierivaive of f ha passes hrough he poin ( 0,7. Fin he equaion of he angen line o his anierivaive a 0. Aciviy 39 When fining erivaives of comple formulas you nee o apply he rules for iffereniaion in he reverse of orer of operaions. For eample, when fining ( sin( e he firs rule you nee o apply is he erivaive formula for ( nees o be applie is he prouc rule. ( e sin u bu when fining sin( he firs rule ha Problem 39.1 Fin he firs erivaive formula for each funcion. In each case ake he erivaive wih respec o he inepenen variable as implie by he epression on he righ sie of he equal sign. Make f. sure ha you use he appropriae name for each erivaive (e.g. ( f ( = sin( e g( = sin ( e an ( ln ( cos ( y = f ( y ln ( y = sin y z = G = sin 1 ( ln ( 3 Aciviy 40 As always, you wan o simplify an epression before jumping in o ake is erivaive. Never-heless, i can buil confience o see ha he rules work even if you on simplify firs. Lab Aciviies 59
4 Porlan Communiy College MTH 51 Lab Manual Problem 40.1 Consier he funcions f ( g = an ( ( = Assuming ha is no negaive, how oes each of hese formulas simplify? Use he f g. simplifie formula o fin he formulas for ( an ( Use he chain rule (wihou firs simplifying o fin he formulas for f ( an g ( simplify each resul (assuming ha is posiive Are f an g he same funcion? Eplain why or why no. Problem 40. π π So long as falls on he inerval, 1 an ( an ( Problem How oes he formula g( ln ( e g ( an an =. Use he chain rule o fin, 1 ( ( an show ha i simplifies as i shoul. ; = simplify an wha oes his ell you abou he formula for? Afer answering hose quesions use he chain rule o fin he formula for g ( ha i simplifies as i shoul. Problem Consier he funcion g( = ln 3. sec( Fin he formula for g ( wihou firs simplifying he formula for ( g. an show Use he quoien, prouc, an power rules of logarihms o epan he formula for g( ino hree logarihmic erms. Then fin g ( by aking he erivaive of he epane version of g Show ha he wo resulan formulas are in fac he same. Also, reflec upon which process of iffereniaion was less work an easier o clean up. Aciviy 41 So far we have worke wih he chain rule as epresse using funcion noaion. In some applicaions i is easier o hink of he chain rule using Leibniz noaion. Consier he following eample During he 1990s, he amoun of elecriciy use per ay in Eown increase as a funcion of populaion a he rae of 18 kw/person. On July 1, 1997, he populaion of Eown was 100,000 an he populaion was ecreasing a a rae of 6 people/ay. In Equaion 41.1 (page 61 we use hese values o eermine he rae a which elecrical usage was changing (wih respec o ime in Eown on 7/1/1997. Please noe ha in his eremely simplifie eample we are ignoring all facors ha conribue o ciywie elecrical usage oher han populaion (such as emperaure. 60 L ab Aciviies
5 Porlan Communiy College MTH 51 Lab Manual Equaion 41.1 kw people kw 18 6 = 18 person ay ay Le s efine g( as he populaion of Eown years afer January 1, 1990 an f ( u as he aily amoun of elecriciy use in Eown when he populaion was u. From he given informaion, ,000 g 7.5 = 6, an f ( u 18 y = f u where g ( =, ( u g( =, hen we have (from Equaion 41.1: = for all values of u. If we le ( Equaion 41. y u y = u u = 100,000 = 7.5 = 7.5 You shoul noe ha Equaion 41. is an applicaion of he chain rule epresse in Leibniz noaion; f g 7.5 g 7.5. specifically, he epression on he lef sie of he equal sign represens ( ( ( In general, we can epress he chain rule as shown in Equaion Equaion 41.3 y y u = u Problem 41.1 = is Suppose ha Carla is jogging in her swee new running shoes. Suppose furher ha r f ( Carla s pace (mi/hr hours afer 1 pm an y h( r = is Carla s hear rae (bpm when she jogs a a rae of r mi/hr. In his cone we can assume ha all of Carla s moion was in one irecion, so he wors spee an velociy are compleely inerchangeable Wha is he meaning of f (.75 = 7? Wha is he meaning of h ( 7 = 15? Wha is he meaning of ( ( Wha is he meaning of Wha is he meaning of h f.75 = 15? r =.75 y r r = 7 = ? = 8? Assuming ha all of he previous values are for real, wha is he value of wha oes his value ell you abou Carla? y =.75 an Lab Aciviies 61
6 Porlan Communiy College MTH 51 Lab Manual Problem 41. Porions of SW 35 h Avenue are eremely hilly. Suppose ha you are riing your bike along SW 35 h Ave from Vermon Sree o Capiol Highway. Le u = ( be he isance you have ravelle (f where is he number of secons ha have passe since you began your journey. Suppose ha y = eu ( is he elevaion (m of SW 35 h Ave where u is he isance (f from Vermon S heae owars Capial Highway Wha, incluing unis, woul be he meanings of ( 5 = 300, ( ( e ( 5 = 140? e 300 = 140, an 41.. Wha, incluing unis, woul be he meanings of u = 5 = 14 an y = 0.1 u = u Suppose ha he values sae in problem are accurae. Wha, incluing uni, is he y value of? Wha oes his value ell you in he cone of his problem? = 5 Problem 41.3 Accoring o Hooke s Law, he force (lb, F, require o hol a spring in place when is isplacemen from he naural lengh of he spring is (f, is given by he formula F = k where k is calle he spring consan. The value of k varies from spring o spring. Suppose ha i requires 10 lb of force o hol a given spring 1.5 f beyon is naural lengh Fin he spring consan for his spring. Inclue unis when subsiuing he values for F an ino Hooke s Law so ha you know he uni on k Wha, incluing uni, is he consan value of F? Suppose ha he spring is sreche a a consan rae of.03 f/s. If we efine o be he amoun of ime (s ha passes since he sreching begins, wha, incluing uni, is he consan value of? Use he chain rule o fin he consan value (incluing uni of F. Wha is he coneual significance of his value? 6 L ab Aciviies
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