Chaper The Derivaive Applie Calculus 11 Secion 5: Chain Rule There is one more ype of complicae funcion ha we will wan o know how o iffereniae: composiion. The Chain Rule will le us fin he erivaive of a composiion. (This is he las erivaive rule we will learn!) Example 1 y x + 15x. Fin he erivaive of ( ) This is no a simple polynomial, so we can use he basic builing block rules ye. I is a prouc, so we coul wrie i as y ( x + 15x) ( x + 15x)( x + 15x) an use he prouc rule. Or we coul muliply i ou an simply iffereniae he resuling polynomial. I ll o i he secon way: 6 y ( x + 15x) 16x + 10x + 5x 5 y' 6x + 80x + 50x y x + 15x. We coul wrie i as a prouc wih 0 facors an use he prouc rule, or we coul muliply i ou. Bu I on wan o o ha, o you? Now suppose we wan o fin he erivaive of ( ) 0 We nee an easier way, a rule ha will hanle a composiion like his. The Chain Rule is a lile complicae, bu i saves us he much more complicae algebra of muliplying somehing like his ou. I will also hanle composiions where i wouln be possible o muliply i ou. The Chain Rule is he mos common place for suens o make misakes. Par of he reason is ha he noaion akes a lile geing use o. An par of he reason is ha suens ofen forge o use i when hey shoul. When shoul you use he Chain Rule? Almos every ime you ake a erivaive. This chaper is (c) 01. I was remixe by Davi Lippman from Shana Calaway's remix of Conemporary Calculus by Dale Hoffman. I is license uner he Creaive Commons Aribuion license.
Chaper The Derivaive Applie Calculus 115 Derivaive Rules: Chain Rule In wha follows, f an g are iffereniable funcions wih y f ( u) an u g( x) (h) Chain Rule (Leibniz noaion): y y u x u x Noice ha he u s seem o cancel. This is one avanage of he Leibniz noaion; i can remin you of how he chain rule chains ogeher. (h) Chain Rule (using prime noaion): f ' x f ' u g' x f ' g x g' x ( ) ( ) ( ) ( ( )) ( ) (h) Chain Rule (in wors): The erivaive of a composiion is he erivaive of he ousie, wih he insie saying he same, TIMES he erivaive of wha s insie. I recie he version in wors each ime I ake a erivaive, especially if he funcion is complicae. Example y x + 15x. Fin he erivaive of ( ) This is he same one we i before by muliplying ou. This ime, le s use he Chain Rule: The insie funcion is wha appears insie he parenheses: x + 15x. The ousie funcion is he firs hing we fin as we come in from he ousie i s he square funcion, (insie). The erivaive of his ousie funcion is (*insie). Now using he chain rule, he erivaive of our original funcion is: (*insie) TIMES he erivaive of wha s insie (which is 1x + 15 ): y y' ( x + 15x) ( x + 15x) ( 1x + 15) If you muliply his ou, you ge he same answer we go before. Hurray! Algebra works! Example Fin he erivaive of y ( x + 15x) 0 Now we have a way o hanle his one. I s he erivaive of he ousie TIMES he erivaive of wha s insie.
Chaper The Derivaive Applie Calculus 116 The ousie funcion is (insie) 0, which has he erivaive 0(insie) 19. y y' 0 ( x + 15x) 19 0( x + 15x) ( 1x + 15) Example Differeniae x +5 e. This isn a simple exponenial funcion; i s a composiion. Typical calculaor or compuer synax can help you see wha he insie funcion is here. On a TI calculaor, for example, x when you push he e key, i opens up parenheses: e ^ ( This ells you ha he insie of he exponenial funcion is he exponen. Here, he insie is he exponen x + 5. Now we can use he Chain Rule: We wan he erivaive of he ousie TIMES he erivaive of wha s insie. The ousie is he e o he somehing funcion, so is erivaive is he same hing. The erivaive of wha s insie is x. So x + 5 x + 5 ( e ) ( e ) ( x) x Example 5 The able gives values for f, f ', g an g ' a a number of poins. Use hese values o eermine ( f g )(x) an ( f g ) '(x) a x 1 an 0. x f(x) g(x) f'(x) g'(x) ( f g )(x) ( g f )(x) -1 1 0 0-1 1 1 1 0-1 -1 0 1 0-1 ( f g )( 1) f( g( 1) ) f( ) 0 ( f g )(0) f( g(0) ) f( 1 ) 1. ( f g ) '( 1) f '( g( 1) ). g '( 1 ) f '( ). (0) ()(0) 0 an ( f g ) '( 0 ) f '( g( 0 ) ). g '( 0 ) f '( 1 ). ( ) ( 1)().
Chaper The Derivaive Applie Calculus 117 Example 6 If 00 people now have a isease, an he number of people wih he isease appears o ouble every years, hen he number of people expece o have he isease in years is y 00 / (a) How many people are expece o have he isease in years? (b) When are 50,000 people expece o have he isease? (c) How fas is he number of people wih he isease expece o grow now an years from now? (a) In years, y 00. /,810 people. (b) We know y 50,000, an we nee o solve 50,000 00. / for. We coul sar by isolaing he exponenial by iviing boh sies by 00, 50000 / Taking he logarihm of boh sies, 00 50000 / ln ln ( ) Using he exponen propery for logs, 00 50000 ln ln ( ) Solving for, 00 50000 ln 00 1.1 years ln() We expec 50,000 people o have he isease abou 1.1 years from now. (c) This is asking for y/ when 0 an years. Using he chain rule, y 1 ( 00 / ) 00 / ln() 55.5 / Now, a 0, he rae of growh of he isease is approximaely 55.5. 0 55.5 people/year. In years he rae of growh will be approximaely 55.5. / 880 people/year. Derivaives of Complicae Funcions You re now reay o ake he erivaive of some mighy complicae funcions. Bu how o you ell wha rule applies firs? Come in from he ousie wha o you encouner firs? Tha s he firs rule you nee. Use he Prouc, Quoien, an Chain Rules o peel off he layers, one a a ime, unil you re all he way insie.
Chaper The Derivaive Applie Calculus 118 Example 7 x ( ) x Fin e ln( 5x Coming in from he ousie, I see ha his is a prouc of wo (complicae) funcions. So I ll nee he Prouc Rule firs. I ll fill in he pieces I know, an hen I can figure he res as separae seps an subsiue in a he en: x x x ( e ln( 5x ) ( e ) ln( 5x x ( ) + ( e ) ( ln( 5x ) x x Now as separae seps, I ll fin x x x ( e ) e x (using he Chain Rule) an 1 5x + 7 ( ln( 5x ) 5 (also using he Chain Rule). Finally, o subsiue hese in heir places: x x x 1 ( e ln( 5x ) ( e ) ln( 5x + e 5 x 5x + 7 (An please on ry o simplify ha!) ( ) ( ) Example 8 Differeniae e z ( 1) Don panic! As you come in from he ousie, wha s he firs hing you encouner? I s ha h power. Tha ells you ha his is a composiion, a (complicae) funcion raise o he h power. Sep One: Use he Chain Rule. The erivaive of he ousie TIMES he erivaive of wha s insie. z e ( 1) e ( 1) e ( 1)
Chaper The Derivaive Applie Calculus 119 Now we re one sep insie, an we can concenrae on jus he e ( 1) par. Now, as you come in from he ousie, he firs hing you encouner is a quoien his is he quoien of wo (complicae) funcions. Sep Two: Use he Quoien Rule. The erivaive of he numeraor is sraighforwar, so we can jus calculae i. The erivaive of he enominaor is a bi rickier, so we'll leave i for now. ( 9 )( e ( 1) ) ( ) ( ( 1) ) e ( 1) e e 1 ( ( )) ( ) Now we ve gone one more sep insie, an we can concenrae on jus he e ( 1) Now we have a prouc. Sep Three: Use he Prouc Rule: ( e ( 1) ) ( e )( 1) + ( e )( 1) An now we re all he way in no more erivaives o ake. Sep Four: Now i s jus a quesion of subsiuing back be careful now! ( e ( 1) ) ( e )( 1) + ( e )( 1), so e ( 1) ( 9 )( e ( 1) ) ( )( e )( 1) + ( e )( 1) ) ( e ( 1) ), so par. z e ( 1) e ( 1) ( 9 )( e ( 1) ) ( )( e )( 1) + ( e )( 1) ) ( ( )) e 1 Phew!.
Chaper The Derivaive Applie Calculus 10 Wha if he Derivaive Doesn Exis? A funcion is calle iffereniable a a poin if is erivaive exiss a ha poin. We ve been acing as if erivaives exis everywhere for every funcion. This is rue for mos of he funcions ha you will run ino in his class. Bu here are some common places where he erivaive oesn exis. Remember ha he erivaive is he slope of he angen line o he curve. Tha s wha o hink abou. Where can a slope no exis? If he angen line is verical, he erivaive will no exis. Example 9 Show ha f ( x) x 1/ x is no iffereniable a x 0. 1 / 1 Fining he erivaive, f ( x) x. A x 0, his funcion is unefine. From he / x graph, we can see ha he angen line o his curve a x 0 is verical wih unefine slope, which is why he erivaive oes no exis a x 0. Where can a angen line no exis? If here is a sharp corner (cusp) in he graph, he erivaive will no exis a ha poin because here is no well-efine angen line (a eeering angen, if you will). If here is a jump in he graph, he angen line will be ifferen on eiher sie an he erivaive can exis. Example 10 Show ha f ( x) x is no iffereniable a x 0. On he lef sie of he graph, he slope of he line is -1. On he righ sie of he graph, he slope is +1. There is no wellefine angen line a he sharp corner a x 0, so he funcion is no iffereniable a ha poin.
Chaper The Derivaive Applie Calculus 11.5 Exercises 1. The graph of y f(x) is shown. (a) A which inegers is f coninuous? (b) A which inegers is f iffereniable?. The graph of y g(x) is shown. (a) A which inegers is g coninuous? (b) A which inegers is g iffereniable? Problems an refer o he values given in his able: x f(x) g(x) f '(x) g '(x) ( f g )(x) ( f g )' (x) 1 1 1 1 1 0 0 1 1 1 0 1 1 0 1 1. Use he able of values o eermine ( f g )(x) an ( f g )' (x) a x 1 an.. Use he able of values o eermine ( f g )(x) an ( f g )' (x) a x, 1 an 0. 5. Use he graphs o esimae he values of g(x), g '(x), (f g)(x), f '( g(x) ), an ( f g ) '( x ) a x 1. 6. Use he graphs o esimae he values of g(x), g '(x), (f g)(x), f '( g(x) ), an ( f g ) '( x ) for x. In problems 7 1, fin he erivaive of each funcion. 7. f(x) (x 8) 5 8. f(x) (6x x ) 10 9. f(x) x. (x 5 10. f(x) (x + ) 6. (x ) 11. f(x) x + 6x 1 1. f(x) x 5 (x + ) 1. If f is a iffereniable funcion, (a) how are he graphs of y f(x) an y f(x) + k relae? (b) how are he erivaives of f(x) an f(x) + k relae?