Lecture for Week 5 (Secs. 3.4–5)Trig Functions and the Chain Rule1The important differenti ation formulas fortrigonometric funct ions areddxsin x = cos x,ddxcos x = −sin x.Memorize them! To evaluate any other trig deriva-tive, you just combine these with the product (andquotient) rule and chain rule and the definitions ofthe other trig functions, of which the most impor-tant istan x =sin xcos x.2Exercise 3.4.19Prove thatddxcot x = −csc2x.Exercise 3.4.23Find the derivative of y = csc x cot x.3What isddxcot x ? Well, the de finition of thecotangent iscot x =cos xsin x.So, by the quotient rule, its derivative issin x(−sin x) − cos x(cos x)sin2x= −1sin2x≡ −csc2x(since sin2x + cos2x = 1).4To differentiate csc x cot x u s e th e productrule:dydx=d csc xdxcot x + csc xd cot xdx.The second derivative is the one we ju s t cal-culated, and the other one is found similarly(Ex. 3.4.17):ddxcsc x = −csc x cot x.5Sodydx= −csc x cot2x − csc3x.This could be rewritten using trig identities, butthe other versions are no simpler.Another method:y = csc x cot x =cos xsin2x.Now use the quotient rule (and cancel some extrafactors of sin x as the last step).6Exercise 5.7.11 (p. 353)Find the antiderivatives ofh(x) = sin x − 2 c os x.7We want to find the func tions whose deriva-tive is h(x) = sin x − 2 cos x. If we know twofunct i ons whose derivati ves are (respectively)sin x and cos x, we’re home free. But we do!d(−cos x)dx= sin x,d sin xdx= cos x.So we let H(x) = −cos x − 2 sin x and check thatH′(x) = h(x). The most general antiderivative ofh is H(x) + C where C is an arbitrary constant .8Now let’s drop back to see where the trigderivatives came from. On pp . 180–181 we’reoffered 4 trigonometric limits, but they are not ofequal profundity.The first two just say that the sine and co-sine functions are continuous at θ = 0. (But youalready knew that, didn’t you?)9The third limit is the important one:limθ→0sin θθ= 1.It can be prove d from the inequ al itiessin θ < θ < tan θ (for 0 < θ < π/2),which are made obvious by drawing some pic-tures.It says that sin θ “behaves like” θ when θ issmall.10In contrast, the fourth limit formula,limθ→0cos θ − 1θ= 0,says that cos θ “behaves like” 1 and that the dif-ferenc e from 1 vanishes faster than θ as θ goesto 0.In fact, later we will see thatcos θ ≈ 1 −θ22.11Exercise 3.4.15limx→0tan 3x3 tan 2x.12Split the functi on into a prod uct of functionswhose limits we know:tan 3x3 tan 2x=13sin 3xcos 3xcos 2xsin 2x=sin 3x3x2xsin 2xcos 2x2 cos 3x.As x → 0, 2x and 3x ap proach 0 as well. There-fore, the two si ne q uotients approach 1. Eachcosine also goe s to 1. So the limit is12.13The chain rule is the most important andpowerful theorem about derivatives. For a fi rstlook at it, let’s approach the last example of lastwe ek’s lecture in a different way :Exercise 3.3.11 (revisited and shortened)A stone is d ropped into a lake, creating a circu-lar ripple that travel s outward at a speed of 60cm/s. Find the rate at wh ich the circle’s area isincreasing afte r 3 s .14In applied problems it’s usually easier to usethe “Leibniz notation”, such as df/dx, instead ofthe “prime” notation for derivatives (which is es-sentially Newton’s notation).The area of a circ le of radius r isA = πr2.SodAdr= 2πr.15(Notice that dA/dr = 2πr is the circumference.That makes sense, since when the radius changesby ∆r, the region enclosed changes by a thin cir-cular strip of length 2πr and width ∆r, hence area∆A = 2πr ∆r.)From A′(r) = 2πr we could compute therate of change of area with respect to radiusby plugging in the appropriate value of r(namely, 60 × 3 = 180). But the question asks forthe rate with respect to time and tells u s that16drdt= 60 cm/s .Common sense says that we should just mult iplyA′(r) by 60, gettingdAdt=dAdrdrdt= 2π(60)2t, t = 3.Of course, this is t he same re s ult we got lastwe ek.17Now consider a slight variation on the prob-lem:Exercise 3.3.11 (modified)A stone is d ropped into a lake, creating a circu-lar ripple that travel s outward at a speed of 60cm/s. Find the rate at wh i ch the area is increas-ing when the radius is 180 cm .18To answer this quest i on by the method oflast week, using A(t) = 3600πt2, we would ne edto calculate the time when r = 180. That ’s easyenough in this case (t = 3), but it carries the ar-gument through an unnecessary loop through aninverse function. It is more natural and simplerto use this wee k ’s formula,dAdt=dAdrdrdt.It givesdAdt= (2πr)60 = 21600π.19In both ve rsions of the exercise we dealtwith a function of the ty pe A(r(t)), where theoutput of one function is plugged in as the i n-put to a different on e. The composite functionis sometimes denoted A ◦ r (or (A ◦ r)(t)). Thechain rule says that(A ◦ r)′(t) = A′(r(t))r′(t),ordAdt=dAdrdrdt.20Now we can use the chain rule to differenti-ate particular funct i ons.Exercise 3.5.7Differenti ateG(x) = (3x − 2)10(5x2− x + 1)12.21G(x) = (3x − 2)10(5x2− x + 1)12.It would be f ool i s h to multiply out the pow-ers when we can use the chain rule instead. Ofcourse, the first step is a product rule.G′(x) = 10(3x − 2)9ddx(3x − 2) (5x2− x + 1)12+12(3x − 2)10(5x2− x + 1)11ddx(5x2− x + 1)22= 30(3x − 2)9(5x2− x + 1)12+ 12(10x − 1)(3x − 2)10(5x2− x + 1)11.(The book’s answer combines some terms at theexpense of factoring out a messy polynomial.)23Note that it is not smart to use the quotientrule on a probl em likeddxx + 1(x2+ 1)3.You’ll find yourself cancell ing extra fact ors ofx2+ 1. It’ s much better to us e th e product ru l eon(x + 1)(x2+ 1)−3,getting only 4 factors of (x2+ 1)−1instead of 6.24Exercise 3.5.51Find the tangent line to the graph ofy =8√4 + 3xat the point (4, 2).25y = 8(4 + 3x)−1/2.dydx= −4(4 + 3x)−3/2(3).When x = 4, y′= −12(16)−3/2= −3/16 (andy = 8(16)−1/2= 2 as the problem claims). So thetangent isy = 2 −316(x − 4).26The book simplifies the equationy = 2 −316(x − 4)to 3x + 16y = 44, but I think it is better to leavesuch equations in the form that emphasizes thedependence on ∆x (= x − 4 in this case). Thepoint of a tangent line is that is the best linearapproximation to the curve in the region where ∆xis small.27Exercise 3.5.59Suppose F (x) = f(g(x)) andg(3) = 6, g′(3) = 4, f′(3) = 2, f′(6) = 7.Find F′(3).28F′(x) = f′(g(x))g′(x)= f′(g(3))g′(3) = f′(6) × 4 = 28.f′(3)
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