3.4 THE DERIVATIVE AS A RATE OF CHANGE

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1 3 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION y y y y y y y = m( ) Figure THE DERIVATIVE AS A RATE OF CHANGE In the cse of liner function y = m + b, the grph is stright line n the slope m mesures the steepness of the line by giving the rte of climb of the line, the rte of chnge of y with respect to. As chnges from to, y chnges m times s much: y y = m( ) (Figure 3.4.) Thus the slope m gives the chnge in y per unit chnge in. In the more generl cse of ifferentible function y = f () the grph is curve. The slope y = f () still gives the rte of chnge of y with respect to, but this rte of chnge cn vry from point to point. At = (see Figure 3.4.) the rte of chnge of y with respect to is f ( ); the steepness of the grph is tht of line of slope f ( ). At =, the rte of chnge of y with respect to is f ( ); the steepness of the grph is tht of line of slope f ( ). At = 3, the rte of chnge of y with respect to is f ( 3 ); the steepness of the grph is tht of line of slope f ( 3 ). m = f '( ) m 3 = f '( 3 ) m = f '( ) 3 Figure 3.4. The erivtive s rte of chnge is one of the funmentl ies of clculus. Keep it in min whenever you see erivtive. This section is only introuctory. We ll evelop the ie further s we go on. Emple The re of squre is given by the formul A = where is the length of sie. As chnges, A chnges. The rte of chnge of A with respect to is the erivtive A = ( ) =. When = 4, this rte of chnge is : the re is chnging t hlf the rte of. When =, the rte of chnge of A with respect to is : the re is chnging t the sme rte s. When =, the rte of chnge of A with respect to is : the re is chnging t twice the rte of. In Figure we hve plotte A ginst. The rte of chnge of A with respect to t ech of the inicte points ppers s the slope of the tngent line. Emple An equilterl tringle of sie hs re A = 4 3. (Check this out.)

2 3.4 THE DERIVATIVE AS A RATE OF CHANGE 3 A m = m = m = 4 A =, > Figure The rte of chnge of A with respect to is the erivtive A = 3. When = 3, the rte of chnge of A with respect to is 3. In other wors, when the sie hs length 3, the re is chnging three times s fst s the length of the sie. Emple 3 Set y =. () Fin the rte of chnge of y with respect to t =. (b) Fin the vlue(s) of t which the rte of chnge of y with respect to is. SOLUTION The rte of chnge of y with respect to is given by the erivtive, y/: y = () ( )() = = 4 3. () At =, (b) Setting y 4 =, we hve 3 with respect to t = 4 is. y = 4 = 3 4. =, n therefore = 4. The rte of chnge of y r Emple 4 Suppose tht we hve right circulr cyliner of chnging imensions. (Figure ) When the bse rius is r n the height is h, the cyliner hs volume h V = πr h. Figure 3.4.4

3 3 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION If r remins constnt while h chnges, then V cn be viewe s function of h. The rte of chnge of V with respect to h is the erivtive V h = πr. If h remins constnt while r chnges, then V cn be viewe s function of r. The rte of chnge of V with respect to r is the erivtive V h = πrh. Suppose now tht r chnges but V is kept constnt. How oes h chnge with respect to r? To nswer this, we epress h in terms of r n V: h = V πr = V π r. Since V is hel constnt, h is now function of r. The rte of chnge of h with respect to r is the erivtive h r = V π r 3 = (πr h) r 3 = h π r. EXERCISES 3.4. Fin the rte of chnge of the re of circle with respect to the rius r. Wht is the rte when r =?. Fin the rte of chnge of the volume of cube with respect to the length s of sie. Wht is the rte when s = 4? 3. Fin the rte of chnge of the re of squre with respect to the length z of igonl. Wht is the rte when z = 4? 4. Fin the rte of chnge of y = / with respect to t =. 5. Fin the rte of chnge of y = [( + )] with respect to t =. 6. Fin the vlues of t which the rte of chnge of y = with respect to is zero. 7. Fin the rte of chnge of the volume of sphere with respect to the rius r. 8. Fin the rte of chnge of the surfce re of sphere with respect to the rius r. Wht is this rte of chnge when r = r? How must r be chosen so tht the rte of chnge is? 9. Fin given tht the rte of chnge of y = + with respect to t = is 4.. Fin the rte of chnge of the re A of circle with respect to () the imeter ; (b) the circumference C.. Fin the rte of chnge of the volume V of cube with respect to () the length w of igonl on one of the fces. (b) the length z of one of the igonls of the cube.. The imensions of rectngle re chnging in such wy tht the re of the rectngle remins constnt. Fin the rte of chnge of the height h with respect to the bse b. 3. The re of sector in circle is given by the formul A = r θ where r is the rius n θ is the centrl ngle mesure in rins. () Fin the rte of chnge of A with respect to θ if r remins constnt. (b) Fin the rte of chnge of A with respect to r if θ remins constnt. (c) Fin the rte of chnge of θ with respect to r if A remins constnt. 4. The totl surfce re of right circulr cyliner is given by the formul A = πr(r + h) where r is the rius n h is the height. () Fin the rte of chnge of A with respect to h if r remins constnt. (b) Fin the rte of chnge of A with respect to r if h remins constnt. (c) Fin the rte of chnge of h with respect to r if A remins constnt. 5. For wht vlue of is the rte of chnge of y = + b + c with respect to the sme s the rte of chnge of z = b + + c with respect to? Assume tht, b, c re constnt with b. 6. Fin the rte of chnge of the prouct f ()g()h() with respect to t = given tht f () =, g() =, h() =, f () =, g () =, h () =

4 3.5 THE CHAIN RULE THE CHAIN RULE In this section we tke up the ifferentition of composite functions. Until we get to Theorem 3.5.6, our pproch is completely intuitive no rel efinitions, no proofs, just informl iscussion. Our purpose is to give you some eperience with the stnr computtionl proceures n some insight into why these proceures work. Theorem puts this ll on soun footing. Suppose tht y is ifferentible function of u n u in turn is ifferentible function of. Then y is composite function of. Does y hve erivtive with respect to? Yes it oes, n y/ is given by formul tht is esy to remember: (3.5.) y = y u u. This formul, known s the chin rule, sys tht the rte of chnge of y with respect to is the rte of chnge of y with respect to u times the rte of chnge of u with respect to. Plusible s ll this souns, remember tht we hve prove nothing. All we hve one is ssert tht the composition of ifferentible functions is ifferentible n given you formul formul tht nees justifiction n is justifie t the en of this section. Before using the chin rule in elborte computtions, let s confirm its vliity in some simple instnces. If y = u n u = 3, then y = 6. Clerly y = 6 = 3 = y u u, n so, in this cse, the chin rule is confirme: y = y u u. If y = u 3 n u =, then y = ( ) 3 = 6. This time y = 6 5, n once gin y u = 3u = 3( ) = 3 4, y = 6 5 = 3 4 = y u u. u = Emple Fin y/ by the chin rule given tht SOLUTION y u so tht y = u u + = (u + )() (u )() (u + ) = y = y [ y u = (u + ) n u =. (u + ) n ] = 4 ( + ). u =

5 34 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION Remrk We woul hve obtine the sme result without the chin rule by first writing y s function of n then ifferentiting: n with y = u u + n u =, we hve y = + y = ( + ) ( ) ( + ) = 4 ( + ). Suppose now tht you were ske to clculte [( ) ]. You coul epn ( ) into polynomil by using the binomil theorem (tht s ssuming tht you re fmilir with the theorem n re ept t pplying it) or you coul try repete multipliction, but in either cse you woul hve terrible mess on your hns: ( ) hs terms. Using the chin rule, we cn erive formul tht will rener such clcultions lmost trivil. By the chin rule, we cn show tht, if u is ifferentible function of n n is positive or negtive integer, then (3.5.) (un n u ) = nu. If n is positive integer, the formul hols without restriction. If n is negtive, the formul is vli ecept t those numbers where u() =. PROOF gives Set y = u n. In this cse, y = y u u (un ) = u (un ) u = nun u. To clculte [( ) ], we set u =. Then by our formul [( ) ] = ( ) 99 ( ) = ( ) 99 = ( ) 99. Remrk While it is cler tht (3.5.) is the only prcticl wy to clculte the erivtive of y = ( ), you o hve choice when ifferentiting similr, but simpler, function such s y = ( ) 4. By (3.5.) [( ) 4 ] = 4( ) 3 ( ) = 4( ) 3 = 8( ) 3. On the other hn, if we were to first epn the epression ( ) 4, we woul get y =

6 3.5 THE CHAIN RULE 35 n then y = As finl nswer, this is correct but somewht unwiely. To reconcile the two results, note tht 8 is fctor of y/: y = 8( ), n the epression in prentheses is ( ) 3 multiplie out. Thus, y = 8( ) 3, s we sw bove. However, (3.5.) gve us this net, compct result much more efficiently. Here re itionl emples of similr sort. Emple [ ( + ) ] 3 ( = 3 + ) 4 ( + ) ( = 3 + ) 4 ( ). Emple 3 [ + ( + 3)5 ] 3 = 3[ + ( + 3) 5 ] [ + ( + 3)5 ]. Since [ + ( + 3)5 ] = 5( + 3) 4 ( + 3) = 5( + 3)4 (3) = 5( + 3) 4, we hve [ + ( + 3)5 ] 3 = 3[ + ( + 3) 5 ] [5( + 3) 4 ] = 45( + 3) 4 [ + ( + 3) 5 ]. Emple 4 Clculte the erivtive of f () = 3 ( 3) 4. SOLUTION Here we nee to use the prouct rule n the chin rule: [ 3 ( 3) 4 ] = 3 [( 3) 4 ] + ( 3) 4 ( 3 ) = 3 [4( 3) 3 ()] + ( 3) 4 (6 ) = 6 4 ( 3) ( 3) 4 = ( 3) 3 ( 9). The formul y = y u u cn be etene to more vribles. For emple, if itself epens on s, then we hve (3.5.3) y s = y u u s.

7 36 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION If, in ition, s epens on t, then (3.5.4) y t = y u s u s t, n so on. Ech new epenence s new link to the chin. Emple 5 Fin y/s given tht y = 3u +, u =, = s. SOLUTION y u = 3, u = 3, s =. Therefore y s = y u u s = (3)( 3 )( ) = 6 3 = 6( s) 3. Emple 6 Fin y/t t t = 9 given tht y = u + u, u = (3s 7), s = t. SOLUTION As you cn check, y u = 3 (u ), u s = 6(3s 7), s At t = 9, we hve s = 3 n u = 4, so tht y u = 3 (4 ) = 3, u s = 6(9 7) =, s Thus, t t = 9, y = y ( u s t u s t = ) () 3 t = t. ( ) = 6 3. t = 9 = 6. Emple 7 Grvel is being poure by conveyor onto conicl pile t the constnt rte of 6π cubic feet per minute. Frictionl forces within the pile re such tht the height is lwys two-thirs of the rius. How fst is the rius of the pile chnging t the instnt the rius is 5 feet? SOLUTION height h is The formul for the volume V of right circulr cone of rius r n V = 3 πr h. However, in this cse we re tol tht h = r, n so we hve 3 ( ) V = 9 πr 3. Since grvel is being poure onto the pile, the volume, n hence the rius, re functions of time t. We re given tht V/t = 6π n we wnt to fin r/t t the

8 3.5 THE CHAIN RULE 37 instnt r = 5. Differentiting ( ) with respect to t by the chin rule, we get V = V r t r t = ( 3 πr ) r t. Solving for r/t n using the fct tht V/t = 6π, we fin tht r t = 8π πr = 9 r. When r = 5, r t = 9 (5) = 9 5 = 3.6. Thus, the rius is incresing t the rte of 3.6 feet per minute t the instnt the rius is 5 feet. So fr we hve worke entirely in Leibniz s nottion. Wht oes the chin rule look like in prime nottion? Let s go bck to the beginning. Once gin, let y be ifferentible function of u: sy Let u be ifferentible function of : sy Then y = f (u). u = g(). y = f (u) = f (g()) = ( f g)() n, ccoring to the chin rule (s yet unprove), y = y u u. Since y = [( f g)()] = ( f y g) (), u = f (u) = f (g()), the chin rule cn be written u = g (), (3.5.5) ( f g) () = f (g()) g (). The chin rule in prime nottion sys tht the erivtive of composition f g t is the erivtive of f t g() times the erivtive of g t. In Leibniz s nottion the chin rule ppers seuctively simple, to some even obvious. After ll, to prove it, ll you hve to o is cncel the u s : y = y u u. Of course, this is just nonsense. Wht woul one cncel from ( f g) () = f (g())g ()? Although Leibniz s nottion is useful for routine clcultions, mthemticins generlly turn to prime nottion where precision is require.

9 38 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION It is time for us to be precise. How o we know tht the composition of ifferentible functions is ifferentible? Wht ssumptions o we nee? Uner wht circumstnces is it true tht ( f g) () = f (g())g ()? The following theorem provies the efinitive nswer. THEOREM THE CHAIN-RULE THEOREM If g is ifferentible t n f is ifferentible t g(), then the composition f g is ifferentible t n ( f g) () = f (g())g (). A proof of this theorem ppers in the supplement to this section. The rgument is not s esy s cnceling the u s. One finl point. The sttement is often written ( f g) () = f (g())g () [ f (g())] = f (g())g (). EXERCISES 3.5 Eercises 6. Differentite the function: () by epning before ifferentition, (b) by using the chin rule. Then reconcile your results.. y = ( + ).. y = ( 3 ). 3. y = ( + ) y = ( + ) y = ( + ). 6. y = (3 ). Eercises 7. Differentite the function. 7. f () = ( ). 8. f () = ( + ) 5. ( 9. f () = ( 5 ).. f () = + ) 3. (. f () = ) 4 ( ) 4.. f (t) =. + t 3. f () = ( ) f (t) = (t t ) 3. ( ) f (t) = (t + t ) f () =. 5 ( ) f () = f () = [( + ) + ( + ) ] 3. ( 3 9. f () = ).. f () = [(6 + 5 ) + ]. Eercises 4. Fin y/ t =.. y =, u = +. + u. y = u + u, u = (3 + )4. 3. y = u 4u, u = (5 + ) y = u 3 u +, u = +. Eercises 5 6. Fin y/t. 5. y = 7u + u, u = +, = t y = + u, u = 7, = 5t +. + Eercises 7 8. Fin y/ t =. 7. y = (s + 3), s = t 3, t =. 8. y = + s s, s = t t, t =. Eercises Evlute the following, given tht f () =, f () =, f () =, f () =, f () =, f () =, g() =, g () =, g() =, g () =, g() =, g () =, h() =, h () =, h() =, h () =, h() =, h () =,

10 3.5 THE CHAIN RULE ( f g) (). 3. ( f g) (). 3. ( f g) (). 3. (g f ) (). 33. (g f ) (). 34. (g f ) (). 35. ( f h) (). 36. ( f h g) (). 37. (g f h) (). 38. (g h f ) (). Eercises Fin f (). 39. f () = ( 3 + ) f () = ( 5 + ). ( ) 3 4. f () =. 4. f () = ( + recll tht [ ] = ). Eercises Epress the erivtive in prime nottion. [ ( )] 43. [ f ( + )]. 44. f. + [ ] 45. [[ f f () ()] + ] f () + Eercises Determine the vlues of for which () f () = ; (b) f () > ; (c) f () <. 47. f () = ( + ). 48. f () = ( ). 49. f () = ( + ). 5. f () = ( ) 3. Eercises Fin formul for the nth erivtive. 5. y =. 5. y = y = ( + b) n ; n positive integer,, b constnts. 54. y = b + c,, b, c constnts. Eercises Fin function y = f () with the given erivtive. Check your nswer by ifferentition. 55. y = 3( + ) (). 56. y = ( ). 57. y = ( 3 )(3 ). 58. y = 3 ( 3 + ). 59. A function L hs the property tht L () = / for. Determine the erivtive with respect to of L( + ). 6. Let f n g be ifferentible functions such tht f () = g() n g () = f (), n let H() = [ f ()] [g()]. Fin H (). 6. Let f n g be ifferentible functions such tht f () = g() n g () = f (), n let T () = [ f ()] + [g()]. Fin T (). 6. Let f be ifferentible function. Use the chin rule to show tht: () if f is even, then f is o. (b) if f is o, then f is even. 63. The number is clle ouble zero (or zero of multiplicity ) of the polynomil P if P() = ( ) q() n q(). Prove tht if is ouble zero of P, then is zero of both P n P, n P (). 64. The number is clle triple zero (or zero of multiplicity 3) of the polynomil P if P() = ( ) 3 q() n q(). Prove tht if is triple zero of P, then is zero of P, P, n P, n P (). 65. The number is clle zero of multiplicity k of the polynomil P if P() = ( ) k q() n q(). Use the results in Eercises 63 n 64 to stte theorem bout zero of multiplicity k. 66. An equilterl tringle of sie length n ltitue h hs re A given by 3 A = 4 where = 3 3 h. Fin the rte of chnge of A with respect to h n etermine this rte of chnge when h = As ir is pumpe into sphericl blloon, the rius increses t the constnt rte of centimeters per secon. Wht is the rte of chnge of the blloon s volume when the rius is centimeters? (The volume V of sphere of rius r is 4 3 πr 3.) 68. Air is pumpe into sphericl blloon t the constnt rte of cubic centimeters per secon. How fst is the surfce re of the blloon chnging when the rius is 5 centimeters? (The surfce re S of sphere of rius r is 4πr.) 69. Newton s lw of grvittionl ttrction sttes tht if two boies re t istnce r prt, then the force F eerte by one boy on the other is given by F(r) = k r where k is positive constnt. Suppose tht, s function of time, the istnce between the two boies is given by r(t) = 49t 4.9t, t. () Fin the rte of chnge of F with respect to t. (b) Show tht (F r) (3) = (F r) (7). 7. Set f () = 3. () Use CAS to fin f (9). Then fin n eqution for the line l tngent to the grph of f t the point (9, f (9)). (b) Use grphing utility to isply l n the grph of f in one figure. (c) Note tht l is goo pproimtion to the grph of f for close to 9. Determine the intervl on which the verticl seprtion between l n the grph of f is of bsolute vlue less thn..

11 4 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION 7. Set f () = +. () Use CAS to fin f (). Then fin n eqution for the line l tngent to the grph of f t the point (, f ()). (b) Use grphing utility to isply l n the grph of f in one figure. (c) Note tht l is goo pproimtion to the grph of f for close to. Determine the intervl on which the verticl seprtion between l n the grph of f is of bsolute vlue less thn.. 7. Use CAS to fin ] [ 4 ( + ) Use CAS to epress the following erivtives in f nottion. [ ( )] [ ( )] () f, (b) f, + [ ] f () (c). + f () 74. Use CAS to fin the following erivtives: () [u (u ())], (b) [u (u (u 3 ()))], (c) [u (u (u 3 (u 4 ())))]. 75. Use CAS to fin formul for [ f (g())]. *SUPPLEMENT TO SECTION 3.5 To prove Theorem 3.5.6, it is convenient to use slightly ifferent formultion of erivtive. THEOREM The function f is ifferentible t iff If this limit eists, it is f (). lim t f (t) f () t eists. PROOF Fi. For ech t in the omin of f, efine f (t) f () G(t) =. t Note tht f ( + h) f () G( + h) = h n therefore f is ifferentible t iff lim G( + h) eists. h The result follows from observing tht lim h G( + h) = L iff lim G(t) = L. t For the equivlence of these two limits we refer you to (..6). PROOF OF THEOREM By Theorem it is enough to show tht lim t f (g(t)) f (g()) t = f (g())g (). We begin by efining n uiliry function F on the omin of f by setting f (y) f (g()), y g() F(y) = y g() f (g()), y = g()

12 3.6 THE CHAIN RULE 4 F is continuous t g() since lim y g() F(y) = lim y g() f (y) f (g()), y g() n the right-hn sie is (by Theorem 3.5.7) f (g()), which is the vlue of F t g(). For t, () f (g(t)) f (g()) t [ g(t) g() = F(g(t)) t To see this we note tht, if g(t) = g(), then both sies re. If g(t) g(), then ]. F(g(t)) = f (g(t)) f (g()), g(t) g() so tht gin we hve equlity. Since g, being ifferentible t, is continuous t n since F is continuous t g(), we know tht the composition F g is continuous t. Thus This, together with (), gives lim t lim F(g(t)) = F(g()) = f (g()). t by our efinition of F f (g(t)) f (g()) t = f (g())g (). PROJECT 3.5 ON THE DERIVATIVE OF u n If n is positive or negtive integer n the function u is ifferentible t, then by the chin rule [u()]n = n[u()] n [u()], ecept tht, if n is negtive, the formul fils t those numbers where u() =. We cn obtin this result without ppeling to the chin rule by using the prouct rule n crrying out n inuction on n. Let u be ifferentible function of. Then [u()] = [u() u()] = u() [u()] + u() [u()] = u() [u()]; [u()]3 = [u() [u()] ] = u() [u()] + [u()] [u()] Problem. = [u()] [u()] + [u()] [u()] = 3[u()] [u()]. Show tht [u()]4 = 4[u()] 3 [u()]. Problem. Show by inuction tht [u()]n = n[u()] n [u()] for ll positive integers n. Problem 3. Show tht if n is negtive integer, then [u()]n = n[u()] n [u()] ecept t those numbers where u() =. HINT: Problem n the reciprocl rule.

13 4 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION 3.6 DIFFERENTIATING THE TRIGONOMETRIC FUNCTIONS An outline review of trigonometry efinitions, ientities, n grphs ppers in Chpter. As inicte there, the clculus of the trigonometric functions is simplifie by the use of rin mesure. We will use rin mesure throughout our work n refer to egree mesure only in pssing. The erivtive of the sine function is the cosine function: (3.6.) (sin ) = cos. PROOF Fi ny number. For h, sin( + h) sin h Now, s shown in Section.5 cos h lim h h = [sin cos h + cos sin h] [sin ] h = sin cos h h + cos sin h h. sin h = n lim h h =. Since is fie, sin n cos remin constnt s h pproches zero. It follows tht ( sin( + h) sin lim = lim sin cos h + cos sin h ) h h h h h ( ) ( ) cos h sin h = sin lim + cos lim. h h h h Thus sin( + h) sin lim = (sin )() + (cos )() = cos. h h The erivtive of the cosine function is the negtive of the sine function: (3.6.) (cos ) = sin. PROOF Fi ny number. For h, cos( + h) = cos cos h sin sin h. Therefore cos( + h) cos lim h h [cos cos h sin sin h] [cos ] = lim h h ( ) ( cos h = cos lim sin lim h h h = sin. ) sin h h Emple To ifferentite f () = cos sin, we use the prouct rule: f () = cos (sin ) + sin (cos ) = cos (cos ) + sin ( sin ) = cos sin.

14 We come now to the tngent function. Since tn = sin / cos, we hve cos (sin ) sin (cos ) (tn ) = = cos + sin = cos cos cos = sec. The erivtive of the tngent function is the secnt squre: 3.6 DIFFERENTIATING THE TRIGONOMETRIC FUNCTIONS 43 (3.6.3) (tn ) = sec. The erivtives of the other trigonometric functions re s follows: (3.6.4) (cot ) = csc, (sec ) = sec tn, (csc ) = csc cot. The verifiction of these formuls is left s n eercise. It is time for some smple problems. Emple Fin f (π/4) for f () = cot. SOLUTION We first fin f (). By the prouct rule, f () = (cot ) + cot () = csc + cot. Now we evlute f t π/4: f (π/4) = π 4 ( ) + = π. Emple 3 Fin [ sec tn ]. SOLUTION By the quotient rule, [ ] sec = tn (sec tn = ) tn ( sec ) ( sec ) tn (tn ) = tn ( sec tn ) ( sec )(sec ) tn = sec (sec tn ) sec tn = sec sec tn = sec ( sec ) tn. Emple 4 Fin n eqution for the line tngent to the curve y = cos t the point where = π/3.

15 44 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION SOLUTION Since cos π/3 = /, the point of tngency is (π/3, /). To fin the slope of the tngent line, we evlute the erivtive y = sin t = π/3. This gives m = 3/. The eqution for the tngent line cn be written y 3 ( = π ). 3 Emple 5 Set f () = + sin. Fin the numbers in the open intervl (, π) t which () f () =, (b) f () >, (c) f () <. SOLUTION The erivtive of f is the function f () = + cos. The only numbers in (, π) t which f () = re the numbers t which cos = : = π/3 n = 4π/3. These numbers seprte the intervl (, π) into three open subintervls (, π/3), (π/3, 4π/3), (4π/3, π). On ech of these subintervls f keeps constnt sign. The sign of f is recore below: sign of f π/3 4π/3 Answers: () f () = t = π/3 n = 4π/3. (b) f () > on (, π/3) (4π/3, π). (c) f () < on (π/3, 4π/3). π The Chin Rule Applie to the Trigonometric Functions If f is ifferentible function of u n u is ifferentible function of, then, s you sw in Section 3.5, [ f ()] = u [ f (u)]u = f (u) u. Written in this form, the erivtives of the si trigonometric functions pper s follows: (3.6.5) u (sin u) = cos u, (tn u) = sec u u, u (sec u) = sec u tn u, u (cos u) = sin u, (cot u) = csc u u, u (csc u) = csc u cot u. Emple 6 Emple 7 (cos ) = sin () = sin. [sec( + )] = sec( + ) tn( + ) ( + ) = sec( + ) tn( + ).

16 Emple 8 (sin3 π ) = (sin π )3 = 3(sin π ) (sin π ) = 3(sin π ) cos π (π ) = 3(sin π ) cos π (π) = 3π sin π cos π. Our tretment of the trigonometric functions hs been bse entirely on rin mesure. When egrees re use, the erivtives of the trigonometric functions contin the etr fctor 8 π =.75. Emple 9 Fin (sin ). SOLUTION Since = 8 π rins, (sin ) = (sin 8 π ) = 8 π cos 8 π = 8 π cos. The etr fctor π is isvntge, prticulrly in problems where it occurs 8 repetely. This tens to iscourge the use of egree mesure in theoreticl work. 3.6 DIFFERENTIATING THE TRIGONOMETRIC FUNCTIONS 45 EXERCISES 3.6 Eercises. Differentite the function.. y = 3 cos 4 sec.. y = sec. 3. y = 3 csc. 4. y = sin. 5. y = cos t. 6. y = 3t tn t. 7. y = sin 4 u. 8. y = u csc u. 9. y = tn.. y = cos.. y = [ + cot π ] 4.. y = [ sec ] 3. Eercises 3 4. Fin the secon erivtive. 3. y = sin. 4. y = cos. 5. y = cos + sin. 6. y = tn3 π. 7. y = cos 3 u. 8. y = sin 5 3t. 9. y = tn t.. y = cot 4u.. y = sin 3.. y = sin cos. 3. y = sin + cos. 4. y = sec tn. Eercises 5 3. Fin the inicte erivtive (sin ) (cos ). 4 4 ] [ 7. [t (t cos 3t). 8. t ] t t t t (cos t ). 9. [ f (sin 3)]. 3. [sin( f (3))]. Eercises Fin n eqution for the line tngent to the curve t the point with coorinte. 3. y = sin ; =. 3. y = tn ; = π/ y = cot ; = π/ y = cos ; =. 35. y = sec ; = π/ y = csc ; = π/3. Eercises Determine the numbers between n π where the line tngent to the curve is horizontl. 37. y = cos. 38. y = sin. 39. y = sin + 3 cos. 4. y = cos 3 sin. 4. y = sin. 4. y = cos. 43. y = tn. 44. y = 3 cot y = sec + tn. 46. y = cot csc. Eercises Fin ll in (, π) t which () f () = ; (b) f () > ; (c) f () <. 47. f () = + cos. 48. f () = sin. 49. f () = sin + cos. 5. f () = sin cos. Eercises Fin y/t () by the chin rule n (b) by writing y s function of t n then ifferentiting. 5. y = u, u = sec, = πt. 5. y = [ ( + u)]3, u = cos, = t. 53. y = [ ( u)]4, u = cos, = t. 54. y = u, u = csc, = 3t.

17 46 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION 55. It cn be shown by inuction tht the nth erivtive of the sine function is given by the formul n { ( ) (sin ) = (n )/ cos, n o n ( ) n/ sin, n even. Persue yourself tht this formul is correct n obtin similr formul for the nth erivtive of the cosine function. 56. Verify the following ifferentition formuls: () (cot ) = csc. (b) (sec ) = sec tn. (c) (csc ) = csc cot. 57. Use the ientities ( π ) cos = sin n ( π ) sin = cos to give n lterntive proof of (3.6.). 58. The ouble-ngle formul for the sine function tkes the form: sin = sin cos. Differentite this formul to obtin ouble-ngle formul for the cosine function. 59. Set f () = sin. Show tht fining f () from the efinition of erivtive mounts to fining lim sin. (see Section.5) 6. Set f () = cos. Show tht fining f () from the efinition of erivtive mounts to fining lim cos. Eercises Fin function f with the given erivtive. Check your nswer by ifferentition. 6. f () = cos 3 sin. 6. f () = sec csc. 63. f () = cos + sec tn. 64. f () = sin 3 csc cot. 65. f () = cos( ) sin. 66. f () = sec ( 3 ) + sec tn. { sin(/), 67. Set f () = n g() = f()., =. In Eercise 6, Section 3., you were ske to show tht f is continuous t but not ifferentible there, n tht g is ifferentible t. Both f n g re ifferentible t ech. () Fin f () n g () for. (b) Show tht g is not continuous t. { cos, 68. Set f () = + b, <. () For wht vlues of n b is f ifferentible t? (b) Using the vlues of n b you foun in prt (), sketch the grph of f. { sin, π/3 69. Set g() = + b, π/3 < π. () For wht vlues of n b is g ifferentible t π/3? (b) Using the vlues of n b you foun in prt (), sketch the grph of g. { + cos, π/3 7. Set f () = b + sin(/), > π/3. () For wht vlues of n b is f ifferentible t π/3? (b) Using the vlues of n b you foun in prt (), sketch the grph of f. 7. Let y = A sin ωt + B cos ωt where A, B,ω re constnts. Show tht y stisfies the eqution y t + ω y =. 7. A simple penulum consists of mss m swinging t the en of ro or wire of negligible mss. The figure shows simple penulum of length L. The ngulr isplcement θ t time t is given by trigonometric epression: where A,ω,φ re constnts. θ(t) = A sin(ωt + φ) () Show tht the function θ stisfies the eqution θ t + ω θ =. (Ecept for nottion, this is the eqution of Eercise 7.) (b) Show tht θ cn be written in the form θ(t) = A sin ωt + B cos ωt where A, B,ωre constnts. 73. An isosceles tringle hs two sies of length c. The ngle between them is rins. Epress the re A of the tringle s function of n fin the rte of chnge of A with respect to. 74. A tringle hs sies of length n b, n the ngle between them is rins. Given tht n b re kept constnt, fin the rte of chnge of the thir sie c with respect to. HINT: Use the lw of cosines. 75. Let f () = cos k, k positive integer. Use CAS to fin n () [ f ()], n (b) ll positive integers m for which y = f () is solution of the eqution y + my =. 76. Use CAS to show tht y = A cos + B sin is solution of the eqution y + y =. Fin A n B given tht y() = n y () = 3. Verify your results nlyticlly. θ L

18 3.7 IMPLICIT DIFFERENTIATION; RATIONAL POWERS Let f () = sin cos for π. () Use grphing utility to estimte the points on the grph where the tngent is horizontl. (b) Use CAS to estimte the numbers t which f () =. (c) Reconcile your results in () n (b). 78. Eercise 77 with f () = sin sin for π. Eercises Fin n eqution for the line l tngent to the grph of f t the point with -coorinte c. Use grphing utility to isply l n the grph of f in one figure. Note tht l is goo pproimtion to the grph of f for close to c. Determine the intervl on which the verticl seprtion between l n the grph of f is of bsolute vlue less thn f () = sin ; c =. 8. f () = tn ; c = π/ IMPLICIT DIFFERENTIATION; RATIONAL POWERS Up to this point we hve been ifferentiting functions efine eplicitly in terms of n inepenent vrible. We cn lso ifferentite functions not eplicitly given in terms of n inepenent vrible. Suppose we know tht y is ifferentible function of n stisfies prticulr eqution in n y. If we fin it ifficult to obtin the erivtive of y, either becuse the clcultions re burensome or becuse we re unble to epress y eplicitly in terms of, we my still be ble to obtin y/ by process clle implicit ifferentition. This process is bse on ifferentiting both sies of the eqution stisfie by n y. Emple eqution We know tht the function y = (Figure 3.7.) stisfies the + y =. (Figure 3.7.) We cn obtin y/ by crrying out the ifferentition in the usul mnner, or we cn o it more simply by working with the eqution + y =. Differentiting both sies of the eqution with respect to (remembering tht y is ifferentible function of ), we hve ( ) + (y ) = () + y y = }{{ } (by the chin rule) y = y. We hve obtine y/ in terms of n y. Usully this is s fr s we cn go. Here we cn go further since we hve y eplicitly in terms of. The reltion y = gives y =. Verify this result by ifferentiting y =. in the usul mnner. Emple Assume tht y is ifferentible function of which stisfies the given eqution. Use implicit ifferentition to epress y/ in terms of n y. () y y 3 + = + y. (b) cos( y) = ( + ) 3 y. SOLUTION () Differentiting both sies of the eqution with respect to, we hve y + 4y 3y y = + y }{{}}{{ } (by the prouct rule) (by the chin rule) ( 3y ) y = 4y. y y = Figure 3.7. y (, ) (, ) + y = Figure 3.7.

19 48 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION y Therefore y = 4y 3y. (b) We ifferentite both sies of the eqution with respect to : [ sin( y) y ] = ( + ) 3 y + 3( + ) ()y }{{} (by the chin rule) [sin( y) ( + ) 3 ] y = 6( + ) y + sin( y). Thus y = 6( + ) y + sin( y) sin( y) ( + ). 3 (, ) Emple 3 Figure shows the curve 3 + y 3 = 9y n the tngent line t the point (, ). Wht is the slope of the tngent line t tht point? y 3 = 9y Figure We wnt y/ where = n y =. We procee by implicit iffer- SOLUTION entition: 6 + 6y y + y y Setting = n y =, we hve + 8 y = 3 y + 6, = 9 y + 9y = 3 y + 3y. 5 y = 4, The slope of the tngent line t the point (, ) is 4/5. y = 4 5. We cn lso fin higher erivtives by implicit ifferentition. Emple 4 The function y = (4 + ) /3 stisfies the eqution y 3 = 4. Use implicit ifferentition to epress y/ in terms of n y. SOLUTION Differentition with respect to gives ( ) 3y y =. Differentiting gin, we hve 3y ( ) ( ) y y + (3y ) = }{{} (by the prouct rule) ( ) 3y y y + 6y =. Since ( ) gives y = 3y,

20 we hve ( ) 3y y + 6y =. 3y As you cn check, this gives y = 6y3 8. 9y 5 Remrk If we ifferentite + y = implicitly, we fin tht + y y = n therefore y = y. However, the result is meningless. It is meningless becuse there is no rel-vlue function y of tht stisfies the eqution + y =. Implicit ifferentition cn be pplie meningfully to n eqution in n y only if there is ifferentible function y of tht stisfies the eqution. Rtionl Powers You hve seen tht the formul ( n ) = n n hols for ll rel if n is positive integer n for ll if n is negtive integer. For, we cn stretch the formul to n = (n it is bit of stretch) by writing ( ) = () = =. The formul cn then be etene to ll rtionl eponents p/q: 3.7 IMPLICIT DIFFERENTIATION; RATIONAL POWERS 49 (3.7.) ( p/q ) = p q (p/q). The formul pplies to ll where p/q is efine. DERIVATION OF (3.7.) We operte uner the ssumption tht the function y = /q is ifferentible t ll where /q is efine. (This ssumption is reily verifie from consiertions epline in Section 7..) From y = /q we get y q =. Implicit ifferentition with respect to gives q y qy = n therefore y = q y q = q ( q)/q = q (/q). So fr we hve shown tht ( /q ) = q (/q). The function y = p/q is composite function: y = p/q = ( /q ) p.

21 5 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION Applying the chin rule, we hve y = p( /q ) p ( /q ) = p (p )/q q (/q) = p q (p/q) s sserte. Here re some simple emples: ( /3 ) = 3 /3, ( 5/ ) = 5 3/, If u is ifferentible function of, then, by the chin rule ( 7/9 ) = 7 9 6/9. (3.7.) (u p/q ) = p u u(p/q) q. The verifiction of this is left to you. The result hols on every open -intervl where u (p/q) is efine. Emple 5 () [( + ) /5 ] = 5 ( + ) 4/5 () = 5 ( + ) 4/5. (b) [( ) /3 ] = 3 ( ) /3 ( ) = 4 3 ( ) /3. (c) [( ) /4 ] = 4 ( ) 3/4 ( ) = ( ) 3/4. The first sttement hols for ll rel, the secon for ll ±, n the thir only for (, ). Emple 6 [ ( ) ] / = ( + + ) / ( ) + ( ) / ( + )() () = + ( + ) ( + ) / The result hols for ll >. = ( + ) = / ( + ). 3/ EXERCISES 3.7 Preliminry note. In mny of the eercises below you re ske to use implicit ifferentition. We ssure you tht in ech cse there is function y = y() tht stisfies the inicte eqution n hs the requisite erivtive(s). Eercises. Use implicit ifferentition to epress y/ in terms of n y.. + y = y 3 3y = y = y = y + y 4 =. 6. y + y + y =. 7. ( y) y =. 8. (y + 3) 4 =. 9. sin ( + y) = y.. tn y = y.

22 3.7 IMPLICIT DIFFERENTIATION; RATIONAL POWERS 5 Eercises 6. Epress y/ in terms of n y.. y + y = 6.. y + 4y = y + y = y = tn y = sin + cos y =. Eercises 7. Evlute y/ n y/ t the point inicte. 7. 4y = 9; (5, ) y + y = ; (, ). 9. cos( + y) = ; (π/6,π/6).. = sin y; (,π/4). Eercises 6. Fin equtions for the tngent n norml lines t the point inicte.. + 3y = 5; (, 3) y = 7; (, 3) y + y = 8; (, 3) y + 3y = 3 3 ; (, ). ( 5. = cos y;, π ) 3. ( ) 6. tn y = ;, π 4. Eercises 7 3. Fin y/. 7. y = ( 3 + ) /. 8. y = ( + ) /3. 9. y = y = ( + ) /3 ( + ) /3. 3. y = y = ( 4 + ) 3. Eercises Crry out the ifferentition. ( ( ) ) ( ) ( ) (Importnt) Show the generl form of the grph. () f () = /n, n positive even integer. (b) f () = /n, n positive o integer. (c) f () = /n, n n o integer greter thn. Eercises Fin the secon erivtive. 38. y = y = 3 + b. 4. y =. 4. y = tn. 4. y = sin. 43. Show tht ll normls to the circle + y = r pss through the center of the circle. 44. Determine the -intercept of the tngent to the prbol y = t the point where =. The ngle between two curves is the ngle between their tngent lines t the point of intersection. If the slopes re m n m, then the ngle of intersection α cn be obtine from the formul tn α = m m + m m. 45. At wht ngles o the prbols y = p + p n y = p p intersect? 46. At wht ngles oes the line y = intersect the curve y + y = 8? 47. The curves y = n = y 3 intersect t the points (, ) n (, ). Fin the ngle between the curves t ech of these points. 48. Fin the ngles t which the circles ( ) + y = n + (y ) = 5 intersect. Two curves re si to be orthogonl iff, t ech point of intersection, the ngle between them is right ngle. Show tht the curves given in Eercises 49 n 5 re orthogonl. 49. The hyperbol y = 5 n the ellipse 4 + 9y = The ellipse 3 + y = 5 n y 3 =. HINT: The curves intersect t (, ) n (, ) Two fmilies of curves re si to be orthogonl trjectories (of ech other) if ech member of one fmily is orthogonl to ech member of the other fmily. Show tht the fmilies of curves given in Eercises 5 n 5 re orthogonl trjectories. 5. The fmily of circles + y = r n the fmily of lines y = m. 5. The fmily of prbols = y n the fmily of ellipses + y = b. 53. Fin equtions for the lines tngent to the ellipse 4 + y = 7 tht re perpeniculr to the line + y + 3 =. 54. Fin equtions for the lines norml to the hyperbol 4 y = 36 tht re prllel to the line + 5y 4 =. 55. The curve ( + y ) = y is clle lemniscte. The curve is shown in the figure. Fin the four points of the curve t which the tngent line is horizontl. y (, ) (, ) 56. The curve /3 + y /3 = /3 is clle n stroi. The curve is shown in the figure. (, ) y (, ) (, ) (, )

23 5 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION () Fin the slope of the grph t n rbitrry point (, y ), which is not verte. (b) At wht points of the curve is the slope of the tngent line,,? 57. Show tht the sum of the - n y-intercepts of ny line tngent to the grph of / + y / = c / is constnt n equl to c. 58. A circle of rius with center on the y-is is inscribe in the prbol y =. See the figure. Fin the points of contct. y (, ) y = 59. Set f () = 3 3. Use CAS to f (h) f () () Fin (h) =. h (b) Fin lim (h) n lim (h). h h + (c) Is there tngent line t (, )? Eplin. () Use grphing utility to rw the grph of f on [, ]. 6. Eercise 59 with f () = 3 3. Eercises 6 n 6. Use grphing utility to etermine where () f () = ; (b) f () > ; (c) f () <, 6. f () = f () = A grphing utility in prmetric moe cn be use to grph some equtions in n y. Drw the grph of the eqution + y = 4 first by setting = t, y = 4 t n then by setting = t, y = 4 t. Eercises Use CAS to fin the slope of the line tngent to the curve t the given point. Use grphing utility to rw the curve n the tngent line together in one figure y = 6; P(, ) y = ; P(3, 4). 66. sin y cos = ; P(,π/6) y = 4; P(, 3 3). 68. () Use grphing utility to rw the grph of the eqution 3 + y 3 = 6y. (b) Use CAS to fin equtions for the lines tngent to the curve t the points where = 3. (c) Drw the grph of the eqution n the tngent lines in one figure. 69. () Use grphing utility to rw the figure-eight curve 4 = y. (b) Fin the -coorintes of the points of the grph where the tngent line is horizontl. 7. Use grphing utility to rw the curve ( )y = 3. Such curve is clle cissoi. CHAPTER 3. REVIEW EXERCISES Eercises 4. Differentite by tking the limit of the pproprite ifference quotient.. f () = f () = g() =. 4. F() = sin. Eercises 5. Fin the erivtive. 5. y = /3 7 /3. 6. y = 3/4 4 /4. 7. y = f (t) = ( 3t ) 3. 3 ( 9. f () =.. y = b ). (. y = + b ) 3.. y = y = tn g() = cos( ). 5. F() = ( + ) y = h(t) = t sec t + t 3 sin. 8. y = + cos. 9. s = 3 3t + 3t.. r = θ 3 4θ. sin. f (θ) = cot(3θ + π).. y = +. Eercises 3 6. Fin f (c). 3. f () = 3 + ; c = f () = 8 ; c =. 5. f () = sin π ; c = f () = cot 3; c = 9 π. Eercises 7 3. Fin equtions for the lines tngent n norml to the grph of f t the point inicte. 7. f () = 3 + 3; (, 4). 8. f () = ; (, 5).

24 3.7 IMPLICIT DIFFERENTIATION; RATIONAL POWERS f () = ( + ) sin ; (, ). 3. f () = + ; (, ). Eercises Fin the secon erivtive. 3. f () = cos ( ). 3. f () = ( + 4) 3/. 33. y = sin. 34. g(u) = tn u. Eercises Fin formul for the n th erivtive. 35. y = ( b) n. 36. y = b + c. Eercises Use implicit ifferentition to epress y/ in terms of n y y + y 3 =. 38. tn ( + y) = y cos y = y y = + /y. Eercises 4 4. Fin equtions for the lines tngent n norml to the curve t the point inicte y 3y = 9; (3, ). 4. y sin sin y = 4 π; ( 4 π, π). Eercises Fin ll t which () f () = ; (b) f () > ; (c) f () <. 43. f () = f () = +. Eercises Fin ll in (, π) t which () f () = ; (b) f () > ; (c) f () <. 45. f () = + sin 46. f () = 3 cos. 47. Fin the points on the curve y = 3 3/ where the inclintion of the tngent line is () π/4, (b) 6, (c) π/ Fin equtions for ll tngents to the curve y = 3 tht pss through the point (, ). 49. Fin equtions for ll tngents to the curve y = 3 tht pss through the point (, ). 5. Fin A, B, C given tht the curve y = A + B + C psses through the point (, 3) n is tngent to the line y + = t the point (, 3). 5. Fin A,B,C,D given tht the curve y = A 3 + B + C + D is tngent to the line y = 5 4 t the point (, ) n is tngent to the line y = 9 t the point (, 9). 5. Show tht /( n ) = n/ n+ for ll positive integers n by showing tht [ lim h h ( + h) ] = n n n. n+ Eercises Evlute the following limits. HINT: Apply either Definition 3.. or (3..5). ( + h) ( + h) lim. h h 9 + h 3 sin ( lim. 55. lim π + h). h h h h lim. sin 57. lim π π. 58. The figure is intene to epict function f which is continuous on [, ) n ifferentible on (, ). b c e For ech (, ) efine M() = mimum vlue of f on [, ]. m() = minimum vlue of f on [, ].. Sketch the grph of M n specify the number(s) t which M fils to be ifferentible. b. Sketch the grph of m n specify the number(s) t which m fils to be ifferentible. f

25 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES 4. THE MEAN-VALUE THEOREM We come now to the men-vlue theorem. From this theorem flow most of the results tht give power to the process of ifferentition. THEOREM 4.. THE MEAN-VALUE THEOREM If f is ifferentible on the open intervl (, b) n continuous on the close intervl [, b], then there is t lest one number c in (, b) for which f f (b) f () (c) =. b Note tht for this number f (b) f () = f (c)(b ). The quotient f (b) f () b is the slope of the line l tht psses through the points (, f ()) n (b, f (b)). To sy tht there is t lest one number c for which f f (b) f () (c) = b is to sy tht the grph of f hs t lest one point (c, f (c)) t which the tngent line is prllel to the line l. See Figure The theorem ws first stte n prove by the French mthemticin Joseph-Louis Lgrnge (736 83).

26 4. THE MEAN-VALUE THEOREM 55 tngent l (b, f (b)) (, f ()) c b Figure 4.. We will prove the men-vlue theorem in steps. First we will show tht if function f hs nonzero erivtive t some point, then, for close to, f () is greter thn f ( ) on one sie of n less thn f ( ) on the other sie of. THEOREM 4.. Suppose tht f is ifferentible t. If f ( ) >, then f ( h) < f ( ) < f ( + h) for ll positive h sufficiently smll. If f ( ) <, then f ( h) > f ( ) > f ( + h) for ll positive h sufficiently smll. PROOF We tke the cse f ( ) > n leve the other cse to you. By the efinition of the erivtive, lim k f ( + k) f ( ) k = f ( ). With f ( ) > we cn use f ( ) itself s ɛ n conclue tht there eists δ>such tht if < k < δ, then f ( + k) f ( ) f ( ) k < f ( ). For such k we hve n thus In prticulr, f ( ) < f ( + k) f ( ) k < f ( o + k) f ( ) k f ( ) < f ( ) < f ( ). (Why?) f ( + k) f ( ) ( ) >. k We hve shown tht ( ) hols for ll numbers k which stisfy the conition < k <δ.if< h <δ, then < h <δn < h <δ. Consequently, f ( + h) f ( ) h > n f ( h) f ( ) h >.

27 56 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES The first inequlity shows tht f ( + h) f ( ) > n therefore f ( ) < f ( + h). The secon inequlity shows tht f ( h) f ( ) < n therefore f ( h) < f ( ). c Figure 4.. tngent b Net we prove specil cse of the men-vlue theorem, known s Rolle s theorem [fter the French mthemticin Michel Rolle (65 79), who first nnounce the result in 69]. In Rolle s theorem we mke the itionl ssumption tht f () n f (b) re both. (See Figure 4...) In this cse the line through (, f ()) n (b, f (b)) is horizontl. (It is the -is.) The conclusion is tht there is point (c, f (c)) t which the tngent line is horizontl. THEOREM 4..3 ROLLE'S THEOREM Suppose tht f is ifferentible on the open intervl (, b) n continuous on the close intervl [, b]. If f () n f (b) re both, then there is t lest one number c in (, b) for which f (c) =. PROOF If f is constntly on [, b], then f (c) = for ll c in (, b). If f is not constntly on [, b], then f tkes on either some positive vlues or some negtive vlues. We ssume the former n leve the other cse to you. Since f is continuous on [, b], f must tke on mimum vlue t some point c of [, b] (Theorem.6). This mimum vlue, f (c), must be positive. Since f () n f (b) re both, c cnnot be n it cnnot be b. This mens tht c must lie in the open intervl (, b) n therefore f (c) eists. Now f (c) cnnot be greter thn n it cnnot be less thn becuse in either cse f woul hve to tke on vlues greter thn f (c). (This follows from Theorem 4...) We cn conclue therefore tht f (c) =. Remrk Rolle s theorem is sometimes formulte s follows: Suppose tht g is ifferentible on the open intervl (, b) n continuous on the close intervl [, b]. If g() = g(b), then there is t lest one number c in (, b) for which g (c) =. Tht these two formultions re equivlent is reily seen by setting f () = g() g() (Eercise 44). Rolle s theorem is not just stepping stone towr the men-vlue theorem. It is in itself useful tool. Emple one rel zero. We use Rolle s theorem to show tht p() = hs ectly

28 SOLUTION Since p is cubic, we know tht p hs t lest one rel zero (Eercise 9, Section.6). Suppose tht p hs more thn one rel zero. In prticulr, suppose tht p() = p(b) = where n b re rel numbers n b. Without loss of generlity, we cn ssume tht < b. Since every polynomil is everywhere ifferentible, p is ifferentible on (, b) n continuous on [, b]. Thus, by Rolle s theorem, there is number c in (, b) for which p (c) =. But p () = for ll, n p (c) cnnot be. The ssumption tht p hs more thn one rel zero hs le to contriction. We cn conclue therefore tht p hs only one rel zero. We re now rey to give proof of the men-vlue theorem. 4. THE MEAN-VALUE THEOREM 57 PROOF OF THE MEAN-VALUE THEOREM We crete function g tht stisfies the conitions of Rolle s theorem n is so relte to f tht the conclusion g (c) = les to the conclusion f f (b) f () (c) =. b The function [ ] f (b) f () g() = f () ( ) + f () b (, f ()) f g() (b, f (b)) b is ectly such function. A geometric view of g() is given in Figure The line tht psses through (, f ()) n (b, f (b)) hs eqution f (b) f () y = ( ) + f (). b [This is not hr to verify. The slope is right, n, t =, y = f ().] The ifference [ ] f (b) f () g() = f () ( ) + f () b is simply the verticl seprtion between the grph of f n the line feture in the figure. If f is ifferentible on (, b) n continuous on [, b], then so is g. As you cn check, g() n g(b) re both. Therefore, by Rolle s theorem, there is t lest one number c in (, b) for which g (c) =. Since g () = f f (b) f () (), b we hve g (c) = f f (b) f () (c). b Since g (c) =, f f (b) f () (c) =. b Figure 4..3 Emple The function f () =,

29 58 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES stisfies the conitions of the men-vlue theorem: it is ifferentible on (, ) n continuous on [, ]. Thus, we know tht there eists number c between n t which f (c) = f () f ( ) ( ) =. f () =, f ( ) = Wht is c in this cse? To nswer this, we ifferentite f. By the chin rule, f () =. The conition f (c) = gives c =. Solve this eqution for c n you ll fin tht c =. The tngent line t (, f ( )) = (, ) is prllel to the secnt line tht psses through the enpoints of the grph. (Figure 4..4) y Figure 4..4 Emple 3 Suppose tht f is ifferentible on (, 4), continuous on [, 4], n f () =. Given tht f () 3 for ll in (,4), wht is the lest vlue tht f cn tke on t 4? Wht is the gretest vlue tht f cn tke on t 4? SOLUTION By the men-vlue theorem, there is t lest one number c between n 4 t which f (4) f () = f (c)(4 ) = 3 f (c). Solving this eqution for f (4), we hve f (4) = f () + 3 f (c). Since f () for every in (, 4), we know tht f (c). It follows tht f (4) + 3() = 8. Similrly, since f () 3 for every in (, 4), we know tht f (c) 3, n therefore f (4) + 3(3) =. We hve shown tht f (4) is t lest 8 n no more thn. Functions which o not stisfy the hypotheses of the men-vlue theorem (ifferentibility on (, b), continuity on [, b]) my fil to stisfy the conclusion of the theorem. This is emonstrte in the Eercises. EXERCISES 4. Eercises 4. Show tht f stisfies the conitions of Rolle s theorem on the inicte intervl n fin ll numbers c on the intervl for which f (c) =.. f () = 3 ; [, ].. f () = 4 8; [, ]. 3. f () = sin ; [, π]. 4. f () = /3 /3 ; [, 8]. Eercises 5. Verify tht f stisfies the conitions of the men-vlue theorem on the inicte intervl n fin ll numbers c tht stisfy the conclusion of the theorem. 5. f () = ; [, ]. 6. f () = 3 4; [, 4]. 7. f () = 3 ; [, 3]. 8. f () = /3 ; [, 8].

30 4. THE MEAN-VALUE THEOREM f () = ; [, ].. f () = 3 3; [, ].. Determine whether the function f () = /(3 + ) stisfies the conitions of Rolle s theorem on the intervl [, ]. If so, fin the numbers c for which f (c) =.. The function f () = /3 hs zeros t = n t =. () Show tht f hs no zeros in (, ). (b) Show tht this oes not contrict Rolle s theorem. 3. Does there eist ifferentible function f with f () =, f () = 5, n f () for ll in (, )? If not, why not? 4. Does there eist ifferentible function f with f () = only t =,, 3, n f () = only t =, 3/4, 3/? If not, why not? 5. Suppose tht f is ifferentible on (, 6) n continuous on [, 6]. Given tht f () 3 for ll in (, 6), show tht 4 f (6) f (). 6. Fin point on the grph of f () = + + 3, between n, where the tngent line is prllel to the line through (, 3) n (, 9). 7. Sketch the grph of { +, f () = 3, > n fin the erivtive. Determine whether f stisfies the conitions of the men-vlue theorem on the intervl [ 3, ] n, if so, fin the numbers c tht stisfy the conclusion of the theorem. 8. Sketch the grph of { + f () = 3, 3, > n fin the erivtive. Determine whether f stisfies the conitions of the men-vlue theorem on the intervl [, ] n, if so, fin the numbers c tht stisfy the conclusion of the theorem. 9. Set f () = A + B + C. Show tht, for ny intervl [, b], the number c tht stisfies the conclusion of the menvlue theorem is ( + b)/, the mipoint of the intervl.. Set f () =, =, b =. Verify tht there is no number c for which f (c) = f (b) f (). b Eplin how this oes not violte the men-vlue theorem.. Eercise with f () =.. Grph the function f () = 3. Verify tht f ( ) = = f () n yet f () is never. Eplin how this oes not violte Rolle s theorem. 3. Show tht the eqution = oes not hve more thn two istinct rel roots. 4. Show tht the eqution = hs ectly one rel root. 5. Show tht the eqution = hs ectly one rel root. 6. () Let f be ifferentible on (, b). Prove tht if f () for ech (, b), then f hs t most one zero in (, b). (b) Let f be twice ifferentible on (, b). Prove tht if f () for ech (, b), then f hs t most two zeros in (, b). 7. Let P() = n n be nonconstnt polynomil. Show tht between ny two consecutive roots of the eqution P () = there is t most one root of the eqution P() =. 8. Let f be twice ifferentible. Show tht, if the eqution f () = hs n istinct rel roots, then the eqution f () = hs t lest n istinct rel roots n the eqution f () = hs t lest n istinct rel roots. 9. A number c is clle fie point of f if f (c) = c. Prove tht if f is ifferentible on n intervl I n f () < for ll I, then f hs t most one fie point in I. HINT: Form g() = f (). 3. Show tht the eqution b = hs ectly one rel root if n t most one rel root between 3 3 n 3 3 if <. 3. Set f () = b. () Show tht f () = for t most one number in [, ]. (b) Determine the vlues of b which gurntee tht f () = for some number in [, ]. 3. Set f () = b, >. Show tht f () = for t most one number in [, ]. 33. Show tht the eqution n + + b =, n n even positive integer, hs t most two istinct rel roots. 34. Show tht the eqution n + + b =, n n o positive integer, hs t most three istinct rel roots. 35. Given tht f () for ll rel numbers, show tht f ( ) f ( ) for ll rel numbers n. 36. Let f be ifferentible on n open intervl I. Prove tht, if f () = for ll in I, then f is constnt on I. 37. Let f be ifferentible on (, b) with f () = f (b) = n f (c) = for some c in (, b). Show by emple tht f nee not be continuous on [, b]. 38. Prove tht for ll rel n y () cos cos y y. (b) sin sin y y. 39. Let f be ifferentible on (, b) n continuous on [, b]. () Prove tht if there is constnt M such tht f () M for ll (, b), then f (b) f () + M(b ).