5.4 Exponential Functions: Differentiation and Integration TOOTLIFTST:


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1 .4 Eponntial Functions: Diffrntiation an Intgration TOOTLIFTST: Eponntial functions ar of th form f ( ) Ab. W will, in this sction, look at a spcific typ of ponntial function whr th bas, b, is This function is call th Natural Eponntial Function f ( ) Lik all ponntial functions, it is ontoon, thrfor, it has an invrs. It has a y intrcpt of, a Horizontal Asymptot on th ais, an is monotonic incrasing. This invrs of f ( ) is th natural log function g( ) ln. This mans that y if an only if ln y. Also, w gt th following rlationships ( ) ln an ln Hr ar a coupl of ampls that utiliz ths proprtis. Rmmbr from Prcalculus, that to solv ponntial quations, all you ha to o was tak th natural log of both sis, rgarlss of what th ponntial bas was. For ampl: Solv 7 + Th first thing you wr train to o was to look if you coul us th GTBTS mtho (Gtting th bass th sam mtho.) In this cas, it is not going to b asy to writ 7 as to th somthingth powr, so w tak th natural log of both sis. ln 7 ln ln 7 ln 7 + ( + ) ln ( + )( ) ln7.946 Hr s anothr ampl of solving an ponntial quation. Solv ln( 3)
2 Notic hr, thr is alray a natural log in th problm, an w CANNOT istribut it. In orr to gt at th insi th log, w must first gt ri of it by ponntiating both sis. ( 3) ln (This is lik th invrs of th GTBTS mtho) Now for th calculus part of. On of th most intriguing, bizarr, an usful charactristics of f ( ) is th fact that it is its own rivativ!! Plas rra th last sntnc, thn ra on. f () is th fact that it is its own rivativ!! That s lik somon giving birth to himslf! Prhaps lik asual rprouction by buing, who knows? Prhaps th rally stans for asy.?! Hr s th official rul: Th Drivativ of th Natural Eponntial Function Lt u b a iffrntial function of. Thn. u u u an. (this is th goo ol chain rul again.) Graph f ( ) an thn graph th rivativ by analyzing th slops. You will quickly convinc yourslf of th abov. But why osn t it work for othr bass? Lt s try som quick rivativs: a. ( ) ( ) itslf chain rul 3/ 3/ 3 3 b. ( ) 3/ simplifi W can now us this rul in conjunction with othr ruls to fin usful information:
3 c. Fin th Rlativ Etrma of f ( ) f ( ) + by th prouct rul f ( ) ()( ) ( )( ) or ( + ) + bcaus will nvr qual zro y ), th only solution is, which is a rlativ min by th first rivativ tst. Th actual point is (, ). S how asy it was going forwar, that is, iffrntiating? It s almost as asy going backwars, or intgrating. All it rquirs you to o is to notic th sam pattrns you hav grown accustom to looking for. Hr s th official ruls: Intgration ruls for Natural Eponntial Functions Lt u b a iffrntiabl function of. Thn. an. u u u Ha! So it s not only its own rivativ, but its own intgral as wll. Prtty crpy. Th pattrn you ar looking for now will involv th function u that is th ponnt of th factor. You will likly call this th insi function an chcking to s if its rivativ is th othr factor on th outsi. Por jmplo: 3 Th rivativ of th ponnt of is 3. W o not hav a scalar multipl of 3 on th outsi, so w will hav to mak a corrction of /3 in our guss (which will, or cours, b corrct.) 3 3 Pic of π!
4 Now, rmmbr that you can only mak corrctions of Scalar Multipls, not of variabls. If th intgral ha bn, th answr woul NOT b! U substitution woul not work hr ithr. This intgral, to you, right now, is not possibl. Hr s anothr ampl: First, rwrit th intgral, pulling all isting scalar multipls out front. Notic that from th rivativ of th ponnt of, w ar only of by a ngativ two, so our guss has th corrction in it, an ONLY involvs th part (rmmbr, th part coms from th chain rul upon iffrntiation.) () ( ) Hr ar som mor ampl of both infinit an finit intgrals. Try to o thm quickly on your own bfor, that s BEFORE, you look at th answr, rwriting first whn you n to. Sinc th answrs will b right nt to th problms, this will not only b a tst of your intgral abilitis, but also of your Emotional Intllignc! Don t forgt: All th prvious ruls you hav larn can b us too!! / / / a. ( ) cos cos b. sin + c. ln+ ( ln( + ) ln ). 6 (that on was fun). cos sin sin sin.48
5 Di you gt thm corrct? Th last two wr kina tricky, an th last on was vry triggy. Notic that in ach of thm, th insi function was not th ponnt of, bcaus was not th Main or outr function. This is asy to spot if you ha u troubl: th function appar MORE THAN ONCE in th intgran, maning, that it HAD to b th innr function itslf, so that it can gnrat itslf upon iffrntiating with th chain rul. Dos that mak sns? Rra, until it os. Hr s th gran final, finally. sc tan OUCH, it s vn painful to look at. This on rquirs som clvr rwriting using th commutativ an associativ proprtis of multiplication. ( sc )( sc tan )( )( ) Notic th rivativ of th first factor is th scon, an th thir factor is th rivativ of th angl AND th last factor is th rivativ of th angl s ponnt (off by only a.) This is th chain rul insi of th chain rul which will rquir th Powr rul. Hr s our corrct guss (with th corrction) corrction powr rul sc ( sc ) 4 If you on t s it, try usubstitution, ltting u sc, or by taking th rivativ of th answr until you s th pattrn. Happy intgrating.
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