# Chapter 6: Transcendental functions: Table of Contents: 6.3 The natural exponential function. 6.2 Inverse functions and their derivatives

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1 Chpter 6: Trnscendentl functions: In this chpter we will lern differentition nd integrtion formuls for few new functions, which include the nturl nd generl eponentil nd the nturl nd generl logrithmic function We will lso look t simple differentil equtions tht involve the nturl eponentil function Tble of Contents: 63 The nturl eponentil function 6 Inverse functions nd their derivtives 6 The nturl logrithm function 64 Generl Eponentil nd Logrithmic Functions 65 Eponentil Growth nd Dec 66 First order liner differentil equtions 67 Approimtions for differentil equtions (direction fields nd Euler s method) (optionl) 68 The inverse trigonometric functions nd their derivtives 69 The hperbolic functions, their inverses nd their derivtives

2 63 The nturl eponentil function: Consider the function :, here nd tht we hve fied positive bse Define We define : s follows: f R, f ( ), R,, note tht the rgument is n eponent N: n, if n times, then define (P) ; Similrl, n, for ever n N(P) Note tht (P) nd (P) follow s nturl properties for n m m n m n powers ( mens multiplied with itself times, nd since mn, then (cross n multipl), which genertes P) Therefore, if Z, then either, then tke the definition bove, or, nd use (P) If p Q( generic rtionl number), then define q the propert m n m n 3 Emple: 9, 8 p q, such tht q p (here we keep in mind However, if R Q, wht is? For emple, let monotonicit (continuit) of, strt with 3 73? In order to mintin the nd we cn improve this bound of between to rtionl eponents s close s we wnt In other words, strt with the grph of which is defined onl for rtionl eponents until now, nd fill the holes with to irrtionl powers to fill the holes (lws respecting tht the function is incresing (for is unique vlue of 3, nd we cn get s close s we wnt to 3 ) This definition of, is unique, since there 3 this w Properties of : Bsed on these definitions, we collect the properties of eponentil (strt to prove these with, N nd continue with the other sets):

3 (PE),, R,, R,, R *, b b,, b R, R Also note wh we hve the necessr restriction, since is ssumed rel here, from where we find tht the domin nd rnges of f ( ) 3 Grphs of : re :, f R Consider the grphs of f ( ) for few vlues of : Figure 63 Grphs of,,,, 3, We see tht the function f ( ) is strictl decresing for, constnt for, nd strictl incresing for Moreover, (<) increses more rpidl for lrger (>), nd lso decreses more rpidl for smller 3

4 4 Define the number e : We consider prticulr eponentil function ( prticulr bse ) with prticulr propert of More precisel, we clculte the slopes of the tngent lines to for for (we do not hve formul for ' et, so we need to drw tngent lines nd mesure slopes) We find tht the slope of the tngent line to for when is m 7 nd the sme slope for 3is m3 We consider the number e, e 3, s the bse of tht eponentil function for which the slope of the tngent line to e for is We will see tht this number e, s the bse of this prticulr eponentil function (with this propert), is ver importnt, nd it will be used etensivel Alternte properties (definitions) cn be given For now, from the bove propert, we hve tht: ( P ) f h f ( h) f () e '() lim lim h h h h We cn plug in different vlues for e to clculte this limit, nd we find tht e 788 (irrtionl number) From ( P ) we cn non-rigorousl derive the more direct reltion: P ) elim h h ( h (we will give rigorous proof of this result lter, let s keep this in mind for now) Two importnt properties of the eponentil follow from ( P ): 5 Find e ' nd First, e d : P ) h h ( h h e e e e ' lim e lim e h h Therefore, we hve found e ' for ever, bsed on e ' for (dd this to formuls of derivtives nd memorize, since we ll use it often) It is remrkble tht the eponentil is itself However, remember tht e d becuse of ( P ), we hve tht (the nti-derivtive of e is the onl function whose derivtive e ) is function whose derivtive is e, so in this cse, ( P 3 ) e d e C Add this to the tble with formuls for integrls nd memorize s it will be used mn times 4

5 d e B ppling the Chin Rule, obtin tht d 6 Emples: Find d e d u( ) u( ) e u '( ) Let f ( ) e Find where f is incresing nd decresing nd where it is concve upwrd nd concve downwrd Also, identif ll etreme vlues nd points of inflection Then sketch grph of f( ) Evlute 4 e d Evlute e 3 d Evlute 3 3 e d Evlute 6e d Find the derivtive of: f ( ) e () t t g t e e g() t cost e f ( ) sin e e e 5

6 e Evlute the following indefinite integrls: e 3 d e e d ep e d (86 AB5, BC) Let A() be the re of the rectngle inscribed under the curve nd (, ), > e with vertices t (-, ) () Find A() (b) Wht is the gretest vlue of A()? Justif our nswer (c) Wht is the verge vlue of A() on the intervl [, ]? Homework: Choose from 4,9,,,4,6,39,44,46,47,5,53 from the end of section 63 + from bove 6 Inverse functions nd their derivtives: Definition nd min theorem: Let f : D R be one to one function on D (psses the horizontl line test: ever horizontl line intersects the grph of f( ) t most once) We define g : R D to be the inverse of the function f if it stisfies: (DI) g f ( ) D f g( ) R Theorem : A (surjective) function f : D R is invertible (it hs n inverse) if nd onl if f is one to one (ie it psses the horizontl line test) 6

7 Proof: (tr the proof) Consider s homework too Defn: f : D R is function if (PF) D! R s t f ( ) (note the uniqueness here, otherwise the reltion would be mbiguous (not D f R function), Figure 6: Emple of non-surjective function Emple: is f : R R, f ( ), function?, Wht bout f : R R, f ( ),,? Wht bout f : R R, f ( ),? Defn: A function f : D R is surjective (onto) function if (PS) R D s t f ( ) D f R f : R,, f ( ) surjective? Figure 6: Emple of surjective function Emple: Is,, Wht bout f : R R, f ( )? Wht bout f : R R, f ( ),, Defn: A function f : D R is one to (injective) function if (PI) f ( ) f ( ),, D, tht is the in (PS) bove is unique D f R Figure 63: Emple of one to one surjective (tht is, bijective) function 7

8 Emple: Is f : R [, ), f ( ) bijective function?, Wht bout f : R R, f ( )? (solution b discussion fter nd ), Note: Think lso bout these definitions in terms of the grph of f Bsed on these definitions, let s prove first, tht is f bijective, implies f invertible Use (PS) with unique (since injective), tke g( ) Due to surjectivit nd injectivit of f, g is function: R! D s t f ( ) g( ) (b choice of ) Since f ( ) Let us ppl g to f ( ) invertible then f ( g( )), R to get g( f ( )) g( ), D Therefore, when f is bijective, f is Now, tht is, we hve to show tht if f is n invertible function, then f is bijective, tht is (PS) with unique Since g is function, then R! D s t g( ) f ( ) (b ppling f nd using tht f nd g re inverses to ech other) QED Method to find the inverse nd emples: To find the inverse of function f, solve the eqution () f ( ) This is obvious from the proof bove: If f is given b f ( ) for, then to find the inverse g( ), we need to reverse the reltion bove, tht is, we need to find from which cme This reltion is cler in Figure 3: we hve to reverse the rrow (the effect of f ) in order to undo its effect over (give numbers here) Therefore, in order to find the inverse of function f, solve the eqution () f ( ) Emple : for Find the inverse of f( ) First, ou need to find the domin nd rnge of f, nd mke sure tht f is one to one (bijective) over its domin 8

9 Note tht the grphs of f ( ) nd set of points, nd, (This is becuse f ( ) re smmetric versus the line, since the represent the f ( ) f ( ) re the sme grph, but we needed to switch the roles of nd to plot the inverse) This is n importnt propert which gives n es w to plot the inverse of n function A simpler criterion (thn one-to-one) for deciding if function is invertible: Theorem : If f : D R is strictl monotonic on D, then f is invertible on D Proof: Here, like usul, we ssume tht f is surjective on D But f strictl monotonic on D implies tht f is one to one on D (es to prove tht f ( ) f ( ) ) Therefore f is invertible on D from Theorem Emple : Show tht 5 f ( ) is invertible Emple 3: ) Let f( ) Find the domin nd the rnge of f( ), show tht f( ) is invertible over its domin, find formul for f ( ), nd verif (DI) nd the propert of the grphs b) Let f ( ) 4,, f :[, ) [ 4, ) Show tht f( ) is invertible over its domin, find formul for f ( ), nd verif (DI) nd the propert of the grphs 3 The inverse function theorem: Theorem 3 (inverse function theorem): Let f : D Rbe differentible one to one (or strictl monotonic) function on D Then: () d f ( ), R d d f ( ) d Note tht the derivtives bove re with respect to different vribles, first with respect to, the second with respect to Also, to find this useful ( formul for of, fter we found the derivtive) d f ( ) d, we need to epress the right hnd side in terms 9

10 Proof: d f ( ) chin rule d f ( ) f f ( ) () d d Note: This inverse function theorem bove will be useful in mn instnces, when we wnt to clculte the derivtive of n inverse of function, nd we cn find the derivtive of the originl function, nd epress somehow in terms of fter tht Emples: Problems 6,,, 37 tetbook Emple: Let f Homework: 5 ( ) Find f '(4) Choose from:, 5,,,,, 5, 9, 3, 35, 38, 4, 43 from end of section 6 Etr Credit: 45 6 The nturl logrithm function: Definition: Consider f ( ) e, :, f R, e s defined before Then f '( ) e, R, nd therefore f( ) is strictl incresing f( ) is one to one so invertible (cn esil check tht the function is lso surjective) Therefore, there eists function f :, f ( ) ln( ), such tht: R, which we cll the nturl logrithm function, () ln e, R ln( ) e,, Emple: ln(3)? (from wht, =3 cme?), solve e 3 nd so on ln( e)?

11 Figure 6 Grph of e (red), ln( ) (mgent) nd their line of smmetr (blue) From the grph, we note ln(), ln(), ln( ) Properties of ln : Remember the properties of e : () e e e e,, R e e,, R e e e,, R (3) these produce ln() ln( ) ln( b) ln( b),, b(, ) ln( ) ln( b) ln,, b(, ) b ln b b ln( ),, b(, ) Prove some of these First, b definition: ln() is number such tht e For the second, cll e nd e bin nd ppl ln using the inverse propert Similr for 3 For 4 let 3 e nd b ln( ) ' nd d : b, then ln ln( ) b ln( ) Let us look now t ln( ) '

12 Using the inverse function theorem for derivtives for f ( ) e : d Therefore, we hve found ln( ) d d e e ln( ) ', for d This cn be generlized to: ln ', for (use chin rule when ) Therefore, we hve found new integrl: d ln C 4 Logrithmic differentition: Logrithmic differentition is ver useful (nd sometimes indispensble tool) It mkes use of the properties of ln to simplif function first, fter which we clculte the derivtive Emples: ) f ( ) ln 3, : introductor emple, works both ws but suggests log differentition much esier b) f( ) /3 (cn compre methods nd get domin nd rnge on the w) c) f ( ) e Therefore, log differentition is useful when there re lrge products or frction, or powers, which reduce to sums, differences, coefficients, fter ppling ln Don t forget to multipl bck b f ( ) t end However, there re other cses in which log differentition is ver useful (nd prett much indispensble): d) f ( ) find domin, rnge nd derivtive e) f ( ) find domin, rnge nd derivtive Other emples: ) Find d ln d b) Find d ln d

13 c) Find 5 7 d d) Evlute 3 d e) tn d f) Given the function f defined b f ( ) ln( 9) ) Determine the smmetr of the grph of f b) Find the domin of f c) Find ll vlues of such tht f() = d) Grph the function g) Find the derivtive of f ( ) ln( ) ln( ) (discuss the domin nd the rnge first) -4 h) Find '( ) if e i) Find '( ) if sin e j) Sketch the grph of ln f ( ), - - k) Evlute: e e d l) Evlute: ep e d 3

14 m) Evlute: d (85 BC5) Let f be the function defined b f ( ) ln for nd let R be the region between the grph of f nd the -is () Determine whether the region R hs finite re Justif our nswer (b) Determine whether the solid generted b revolving the region R bout the -is hs finite volume Justif our nswer (9 AB) Let f be the function given b f( ) ln () Wht is the domin of f? (b) Find the vlue of the derivtive of f t (c) Write n epression for f ( ) where f denotes the inverse function of f Homework: Choose from problems 3,5,,6,,,33,4,45 /end of section 6 nd from bove Etr credit: 55,56 64 Generl Eponentil nd Logrithmic Functions ' nd d We wnt to generlize bit the formuls for derivtives nd integrls which we found to generl eponentil nd logrithmic functions Consider f ( ),, f : (, ) In order to find ' bsed on wht formuls we lred hve, write e ln( ) ln( ) ln( ) ' e ' ln( ) ' e ln( ) Then Therefore () ' ln( ), for (dd this to tble of derivtives nd memorize) Cn we then find d? From bove, () d = ln( ) C ( dd to tble of integrls nd memorize) log ( ) nd its derivtive 4

15 Let s consider now f ( ),, f : (, ), remember the grphs of Note tht ll these functions re strictl monotonic, therefore f ( ) f( ),,,,3,5 5 3 is strictl monotonic if Therefore, f ( ),,, f : R (, ) is invertible nd we cll its inverse f R f :,, ( ) log ( ) with properties (3) log, R,, log ( ),,,, For emple, log 3 such tht 3 New properties for f ( ) log ( ) follow, which correspond nd cn be shown in similr w with the properties (3) in 6, usull b replcing e (4) log log blog Proof: b Let log nd log z Then nd z b which shows () b replcing b b A new propert is: z z Therefore b log b z log b (this needs to be proven s indicted bove), nd z The derivtive log ( ) ' follows from the inverse function theorem, like before: For f ( ), log ( ) ' ( )' ln( ) ln( ) You cn re-write this s (5) log ( ) ', for ln( ), dd to tble of derivtives, memorize nd use in this form No new integrl forms, since (5) still involves just the function on the right hnd side (cn write d ln( )log ( ) C but this is known formul, using (4)) 5

16 Emples: ) Find d 3 d d if 5 d 4 b) Find 5 4 c) 3, 4find d d 3 4 d d) If log e) Compre the grphs of,, nd is eponentil function, is n power function D ln but D 6 d C nd d C ln Find D, f) 5 d g) Find if sin d d h) Find the derivtive of ech of the following: f( ) 3 cos f( ) 7 f ( ) ep log f 3 f( ) 4 f ( ) 3 ln f( ) f( ) log3 ( ) log i) Find d log Homework: From bove nd from problems 5,4,6,7,,6,7,33,34(EC),38,4,49 6

17 65 Eponentil Growth nd Dec: Eponentil growth nd dec: Since the derivtive of function represents its instntneous rte of chnge, mn mthemticl nd phsicl models for different lws in nture involve equtions which contin the derivtive of function (these re clled differentil equtions) The eponentil function is ver importnt nd omnipresent in mthemtics lso becuse it ppers often in mthemticl models We often her the term of quntit growing (or decing) eponentill, but wht does it men? For n eponentil growth of popultion, we ssume tht the rte of growth (births minus deths) of given popultion t n certin fied time is proportionl with the size of the current popultion (no eponentil in our model et!) Although idelistic (neglects limittion problems) this model is bsed on resonble ssumption If t () is the size of the popultion t time t, this model gives: () d k dt, where k is just proportionlit constnt nd it cn be considered popultion dependent Note tht for k we hve popultion growth nd for k we hve popultion dec Solving () for t t with ( t) gives kt () () t e, from where the nme of eponentil growth (nd dec) Therefore, the ssumption () produced solution of eponentil growth/dec Emple: If the popultion of the world ws 64 billion t the beginning of 4 nd k for the world popultion is pproimtel 3 (ssuming t is mesured in ers), how long will it tke for the world popultion to double? The time intervl needed for popultion (growing eponentill) to double is clled the doubling time Note tht if n eponentill growing quntit doubles from to in n initil intervl of length T, it will double in n intervl of length T ktt t T e e () t e kt kt Logistic model: As mentioned before, the eponentil growth model is unrelistic, since it does not include n limittion (due to resources, etc), nd it projects fster nd fster growth indefinitel into the future A more relistic model, clled the 7

18 logistic model, ssumes tht the rte of growth is proportionl to both the popultion size nd to the difference L, where L is the mimum popultion tht cn be supported This leds to the differentil eqution d dt (3) k L Note tht for smll, d kl (second fctor not essentil model-wise) which suggests eponentil-tpe growth But dt s pproches L, the second fctor becomes importnt, growth is curtiled, nd d gets smller nd smller (until it dt gets ver close to zero so there is lmost no chnge in t ()), producing growth curve tht flttens out Eqution (3) cn be solved ectl to obtin this tpe of solution (indicte steps) 3 Eponentil dec: Some quntities dec t rte proportionl to their current size, one such dec model is the rdioctive dec Decing popultions re modeled b () but with k Emple: Problem 6 or 7 t end of section 4 Newton s lw of cooling: Newton s lw of cooling ssumes tht the rte t which n object cools (or wrms) is proportionl with the difference in temperture between the objects nd the temperture of the surrounding medium: dt (4) k( T T ) dt (note tht k is negtive in this model) This is similr model with (), the onl difference is tht the solution increses (or decreses) to T (s T gets close to T the rte of chnge gets close to ) Wht is the ect solution of (4)? kt T() t T T T e 5 A better definition of e : We hinted in 63 tht (5) elim h h h 8

19 but we defined e from (*) h e lim h h Now we hve ll tools (properties of eponentil nd logrithms) to prove (5) As mentioned, (5) is preferble method nd we wnt to memorize it, s it will be useful for mn limits (nd test problems!!!) Proof: Remember tht for (*) we strted with the slope of the tngent line to e t (,) is We see tht we need some form of n inverse, in order to subtrct (solve for e) in (*) Therefore, we consider now the slope of the tngent line t the corresponding point ((,)) for the inverse of e : ln( ) From the inverse function theorem, or using ln( ) '/ Using this, tr to find (5) formul: Let f ( ) ln( ) Since '() Therefore: h h ln( h) ln() ln( h) h lim lim lim ln h h h h h f, then h / / h ln lim ln( e) (since f ( ) ln( ) is continuous function, wh?, since it is the inverse of continuous function) But f ( ) ln( ) is one to one function, which produces the result Let us memorize (5) s it is the preferred w to define e From (5) we cn derive tht: (6) lim n n n e b mking chnge of vrible in limit Using (5) or (6), solve the following limits: lim h h h n lim n 3 n lim n n lim n n n n h h lim h lim n n n lim n n lim n n n3 lim n n n lim n n n lim n n n lim n n n 3 lim n n 3n lim n n n3 Homework: Problems 3, 8,, 6, 7, 8, 9, 36, 37 (importnt) 9

20 66 First order liner differentil equtions: d Method for solving P( ) Q( ) d : The differentil eqution () d k dt which we solved in the previous section is one of the simplest tpes of differentil equtions Let us look now t the more generl differentil eqution: d () P( ) Q( ) This eqution is clled liner (becuse, where is the eqution d opertor), first order with vrible coefficients differentil eqution (usull for liner differentil eqution the rhs is of the liner form P( ) Q( ) ) Eqution () is not seprble nmore, so we cnnot seprte the vribles s we did for () Insted, we ppl ver prticulr method for this eqution (mke sure tht ou hve ectl this form before solving the eqution, especill mke sure tht d d hs coefficient nd tht the term P( ) is on the sme side with d d ): In order to obtin seprble eqution, we tr to get rid of the P( ) term b multipling both sides of () b P( ) d e After this, () becomes: (3) d e Q ( ) e d P( ) d P( ) d such tht P( ) d P( ) d e Q( ) e d C Which gives tht: e Q( ) e d Ce P( ) d P( ) d P( ) d (creful with the plce t which C pops up in the problem) As with ll first order differentil equtions, n initil condition should be given (quntit chnges, but t wht vlue does it strt?) The vlue of C will be determined from this initil condition Emple: Problem 3/pp 358: Solve the differentil eqution: (4) Is this of the form ()? Not et, we need to divide first b ', < After this, we hve:

21 ' (5) P( ) nd Q( ) which is of the form () with (note tht the division is llowed in domin) First, P( ) d d ln C, P ( ) d so e Multipling (5) b this integrtion fctor, we obtin: ' d d 3/ 3/ 3/ This is the seprble form, which, fter integrtion, gives: d C 3/ (ver importnt to put the C here!) Therefore: C (ou should check this b plugging it bck into (4)) 3 Emple 3/356 (tnk problems): Initil: gllons brine 75 lbs slt Rte in = gls brine/min with lbs of slt/gll Rte out=gls brine /min with / slt/gll Find () - the totl mount of slt in tnk fter hour (6 mins) Let (t) the totl quntit of slt in tnk t time t d d Then rte in - rte out gl/min slt/gl gl / min slt/gl dt dt d So 4 dt, where time is mesured in minutes Solving, we find 6 ( t) 44 69e 6 t

22 So (6) 44 69e 86 lbs Red other problems with tnk Cn lso do problem 8/358 (we hve to solve two eqution there, for (t): d 3 gl / min lbs / gl gl / min lbs / gl dt z nd for z(t): dz 3 dt ) Homework: Problems,6,8,3,4,6,8, 68 The inverse trigonometric functions nd their derivtives: Let us turn now to the si trigonometric functions: sin( ), cos( ), tn( ), cot( ), sec( ), csc( ) Gols: sin ( ) rcsin( ), cos ( ) rccos( ) Remember the inverses tn ( ) rctn( ), cot ( ) rc cot( ) d(rcsin( )) d(rccos( )), d d Find nd memorize the formuls for d(rctn( )) d( rc cot( )), d d Use the formuls bove to find, memorize nd use formuls for new integrls: d d d d

23 The si trigonometric functions listed bove re not es to work with, with respect to their inverses Since the re ll periodic, the re not one to one (do not pss the horizontl line test) Therefore we will need to restrict the domin for ech of them, choosing the lrgest possible restricted domin such tht the function is one-to-one (invertible) on this restricted domin Some freedom is possible with this choice of the restricted domin, nd we will then choose the most nturl restricted domin Let us strt with the inverse of sin( ) : sin : R [,] is not one-to-one, restrict the domin to, (look t the grph) Therefore, sin :, [,] is one-to-one (invertible) So, there eists () rc sin :, [, ] such tht () rcsin(sin( )),, sin(rcsin( )),, Plot sin( ) nd sin ( ) rcsin( ) on the sme grph: Figure 68: Grph of sin( ) (green) (blue) nd sin ( ) (red) Emple: Clculte ) sin b) sin c) sin 3 d) sin sin e) sin 3

24 Similrl, cos : [,] is not one-to-one, nd to mke it one to one we restrict its domin to, (there re more choices here but this is the most nturl ) Therefore cos :,, is invertible, so there eists cos rccos :, [, ] such tht (4) (3) rc rccos cos( ),, cos cos( ),, Plot cos( ) nd cos ( ) rccos( ) on the sme grph Figure 68: Grph of cos( ) (green) (blue) nd cos ( ) (red) (ignore the left to zero prt for cos( ) ) Creful with the domin of the restricted cos, nd the rnge of rccos, the will be importnt when finding vlues Emple: Clculte: ) rccos 3 b) rccos 3 c) rccos d) 7 rccos cos 6 e) rccos cos 3 Similrl, is not one-to-one, nd we restrict its domin to, to mke it one- to-one (k ) tn : R \, k Z R So, there eists 4

25 (5) tn rctn : R, such tht (6) rctn tn( ),, tn rctn( ), R Plot tn( ) nd tn ( ) rctn( ) on sme grph: Figure 683: Grph of tn( ) (green) (blue) nd tn ( ) (red) Emple: Clculte tn ( ) Similrl, cot : R \ k, k Z R is not one-to-one, nd we restrict its domin to Eercise: Prove the following identities (eventull using the fundmentl trigonometric identit sin ( ) cos ( ) :, to mke it one-to-one ) sin cos b) cos sin c) sectn d) Let us find now the derivtives of the four inverse trigonometric function bove tn sec Using the inverse function theorem (note tht ll these restricted trigonometric functions re strictl monotonic nd differentible), we hve: For sin( ), d sin d dsin( ) cos( ) sin ( ) d (the sign in the identit before the lst comes from,, for which cos( ) ) 5

26 Therefore, the new formul is: (7) sin ( ) ' (dd to tble of derivtives nd memorize) Similrl: (8) cos ( ) ' derivtives nd memorize) (the minus comes becuse cos( ) ' sin( ) now, dd to tble of Also, for tn( ), d tn cos ( ) d d tn( ) sec ( ) tn ( ) d, which gives tht: (9) tn ( ) ' You cn lso derive (on our own) in similr mnner tht: cot ( ) ' () Like for ll the previous derivtives tht we obtined in section 6, 63 nd 64, formuls (7) () produce new integrls: () d rcsin( ) C, which generlizes to () rcsin of the right side or tr to obtin the integrl in the left with chnge of vribles ) d C (tke the derivtive nd: (3) rctn( ) d C, which generlizes to (4) rctn d C The other two functions do not give new integrls Add the boed formuls to tble of integrls nd memorize These new formuls for derivtives nd integrls m pper comple, but the will be ver useful in solving more complicted integrls (frctions nd rdicls) Eercises: 6

27 Let s remember (or clculte) the following integrls: ) sec ( ) d b) csc ( ) d c) tn( ) d d) cot( ) d Lter on (net section) we ll lern three more importnt formuls for integrls: d, d nd d which gives us bsic set of formuls, which, if we memorize, help us solve ll integrl problems in the tet Eercises: Find: sin 3 ' tn ' d A mn is stnding on top of verticl cliff feet bove lke As he wtches, motorbot moves directl w from the foot of the cliff t rte of 5 feet per second How fst is the ngle of depression of his line of sight chnging when the bot is 5 feet from the foot of the cliff? 3 Differentite: f ( ) sin sec e f( ) sin sin t P, 6 4 Evlute: f ( ) tn ln d d 9 d 4 d 3 d 6 5 Homework: From bove nd problems 4,7,8,4,7,,4,5,3,3,36,46,48,57,6,67,7,7,7, nd 77(EC) 7

28 69 The hperbolic functions, their inverses nd their derivtives: We wnt to prove, memorize nd use the following formuls for integrls, which will complete our tble of integrtion for now: () ln d C () d ln C (3) d ln C These formuls will help us solve integrls with qudrtics in the denomintor, under or not under squre root Tking the derivtive of the right side we obtin the left side, which is enough for proof Wh? Solve few integrl problems here: ) 5 d b) 4 c) 4 d d) 3 t dt rctn( ) d e) d f) 4 6 sec ( ) tn ( ) d However, these functions bove cn be relted to new set of functions clled hperbolic functions Therefore, we define: e e e e (4) sinh( ), cosh( ), sinh( ) e e tnh( ) cosh( ) e e, cosh( ) e e coth( ) sinh( ) e e These functions re clled hperbolic since prmetriztion : cosh( t), sinh( t) describes the right brnch of the unit hperbol: Therefore, we hve : (5) cosh ( ) sinh ( ) Other identities, similr with the trigonometric identities, cn be derived 8

29 For emple: (6) d sinh( ) d cosh( ), however: (7) d cosh( ) d sinh( ) sinh : R Ris one to one (invertible) (plot or tr to prove injective directl) Also, the restriction cosh :, [, ) is one to one (invertible) (plot) We cn esil find tht (give this s n eercise in clss): sinh : R R is sinh ( ) ln Also, cosh :[, ) [, ) is cosh ( ) ln We lso hve tht tnh :[,) R is tnh ( ) ln We cn find the derivtive of these functions either directl or using the inverse function theorem to find the prticulr form of the formuls () (3) A generliztion to get () (3) follows b chnge of vribles in the corresponding integrls Homework: Problems,8,6,8,5,4,4,5 t the end of section Preprtion for test: All problems t the end of chpter, strting with the true nd flse tpe questions 9

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