Lecture for Week 10 (Secs. 4.5–8)Derivative Miscellany III1Growth and decay problemsI already discussed the theory of these prob-lems two weeks ago, so let’s just do an ex ampl e.2Exercise 4.5.3A culture s t arts with 500 bacteria, and after 3hours there are 8000.(a) Find t he formula f or the number after thours.(b) Find th e number after 4 hours.(c) When will t he popul at i on reach 30,000?3The b as i c assumption is th at the numberof new bact eria is proportional to t he numberalready there (p are nts) . SoB(t) = B(0)ektfor some constant k. So according to the data,8000 = 500e3k,or 3k = ln805= ln 16, or k =13ln 16.B(t) = 500e13t ln 16= 500(16)t/3.4ThenB(4) = 500(16)4/3= (c al culator output).For the last part,30, 000 = 500(16)t/3⇒t3= log163005⇒ t = 3 log1660= 3ln 60ln 16.5Inverse trigonometric functionsThere are two aspect s of inverse trig func -tions that need to be st udied:• the definitions (especially branch ch oi ces);• their derivatives.The most important inverse t rig functionsare sin−1and tan−1.6Both of the problems we encountered for t hesquare root function also appear for the inversesine.1. sin θ is alway s between −1 and 1, s o the re isno way to define sin−1z if | z| > 1 (un l ess wego i nto comple x numbers, which we won’t).2. For −1 ≤ z ≤ 1 there is more than oneθ with z as i t s sine. (In fact , there are in-finitel y many.) S o, we have to choose a prin-7cipal value (or branch) of the inverse func-tion. The standard choice is to pick θ sothat −π2≤ θ ≤π2.Please refer to the book for the graphs (p. 276and p. 278 in this case). Recall that to get thegraph of an inverse function, you can plot the or ig-inal function on a tr ansparent sheet and flip it overso that the horizontal and vertical axes are inter-changed.8Thus sin(sin−1z) = z always , butsin−1(sin θ) = θ is false if θ is not in the prin cipalinterval.Recall al s o t hat sin−1z does not mean(sin z)−1(that is, 1/ sin z), alth ough sin2z doesmean (si n z)2. This n ot ational i nconsistency isunfortunat e, but we’re s t uck with it. (Let’ s noteven ask what sin−2z means.) Another notationfor the inverse is arcsin z.9The i nverse tange nt is easier (see graphsp. 279), because it is defin ed for all z and all thebranches look the same (have positive slope).But there are still infinitely many branches, andthe standard choice is −π2< θ <π2. (Why is it“<” here bu t “≤” for the inverse sine?)This tan−1z is a very n i ce function. It i n-creases smoothly between horizontal asy mptotesat θ = −π2and θ = +π2.10The u s ual technique for differentiat i ng animplicit or inverse fu nction yiel ds the formulasddxsin−1x =1√1 − x2,ddxtan−1x =11 + x2.These are ordinary algebraic f unctions ! (All trac eof trig seems to have disappeared.) One reasoninverse trig fun ctions are important is that th eyhelp provid e th e antiderivatives of certain alge-braic functi ons.11Exercise 4.6.51Find t he derivative of g(x) = sin−1(3x + 1) andstate the domains of g and g′.ExerciseFind an antiderivative off(x) =3√4 − 4x2−10x2+ 1.12g(x) = sin−1(3x + 1).g′(x) =3p1 − (3x + 1)2(which could be simplified). For g to be definedwe need |3x + 1| ≤ 1.Case 1: 3x + 1 ≥ 0. The n3x + 1 ≤ 1 ⇒ x ≤ 0,3x + 1 ≥ 0 ⇒ x ≥ −13.13Case 2: 3x + 1 < 0. The n−3x − 1 ≤ 1 ⇒ x ≥ −23,3x + 1 < 0 ⇒ x < −13.So th e domain of g consists of the two intervals−13≤ x ≤ 0 and −23≤ x < −13,which fit toget her to give−23≤ x ≤ 0.14For g′to be defined we also need 3x + 1 6= 0,hence the interval shrinks to −23< x ≤ 0. (Seethe vertical tangents at the end s of th e graph,Fig. 4 on p. 278.)Alternative solution of the inequality:|3x + 1| ≤ 1 ⇐⇒x +13≤13.This clearl y describes the numbers w hose dis-tance from −13is at most13— namely, the interval−23, 0 .15F′(x) =3√4 − 4x2−10x2+ 1;what is F ?F′(x) =321√1 − x2−10x2+ 1,so the obvious choic e isF (x) =32sin−1x − 10 tan−1x.16Soon we will reach the proof that the only otherantiderivatives are eq ual to this one p lus a con-stant.(What if the two numbers inside the squareroot were not the same? Look forward t o t he ex-citement of Chapter 8 in Math. 152!)17Hyperbolic functionsThis topic is not in the syllabus for Math.151 at TAMU. To see why it should be, read mypaper i n College Math. Journal 36 (2005) 381–387. It al s o explains why I d on’t tal k about cot,csc, sec−1, et c.18Indeterminate forms (l’Hospital’s rule)(The name is pronounced “Loap -it-ALL”(more or less) and sometimes spell ed “l’Hˆopital”.)In my opinion, th e two most importantthings to learn about l’Hosp i t al’s rule are• when not to use it• what it te ach es us about limits of exp andln at infinity.19Suppose we want t o calculate th e li mit off(x)g(x)as x → a (a may be ∞), and suppose thatboth f(x) and g(x) approach 0 in that limit, orboth ap proach ∞. L’Hospital’ s rule states thatthat l imit is the same as the limit off′(x)g′(x)(whichmay be easier to calculate).Please don’t confuse this formula with the“limit law” for a quotient, or wi th the formul a forthe derivative of a quotient. They are thr ee differ-ent thi ngs!20Here is an e x ampl e of the correct use of therule:limx→0sin(5x)7x=“00”= limx→05 cos(5x)7=57.However, you didn’t re al ly need th e rule to dothis problem, did you? You alre ady know thatsin(5x) ≈ 5x when x ≈ 0 (or can appeal tolimu→0sin uu= 1).21After st udying Taylor series (Chapter 10)you will know many ot her sit uations where thebehavior of the func t ions f and g near a i s obvi-ous, s o l’Hospi t al is unnecessary. Many studentsoveruse l’Hospital’s rule, rel y ing on it as a “bl ackbox” when the y would learn much more (an dsolve th e problems equal l y fast) by just taking aclose look at, and comparing, the behavior of thenumerator an d denominator as x → a.22Here is an e x ampl e where using the ru le isabsolutel y wrong: We know that limx→0+cos xx=+∞, be cause the numerator approaches 1 whi l ethe denominator approach es 0. If you incor-rectly applied l’Hospital’s rule, you would getlimx→0+− sin x1= 0. This fraction did not satisfy thehypotheses of l’Hospital’s rule, because the limitof th e numerator, cos x, i s not 0.23Finally, consider limx→0(x ln x). This is an in-determinate form of th e type “ 0 × ∞”. To app l yl’Hospital’s
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