Lecture for Week 5 Secs 3 4 5 Trig Functions and the Chain Rule 1 The important differentiation formulas for trigonometric functions are d d sin x cos x cos x sin x dx dx Memorize them To evaluate any tive you just combine these with quotient rule and chain rule and the other trig functions of which tant is sin x tan x cos x 2 other trig derivathe product and the definitions of the most impor Exercise 3 4 19 Prove that d cot x csc2 x dx Exercise 3 4 23 Find the derivative of y csc x cot x 3 What is cotangent is d dx cot x Well the definition of the cos x sin x So by the quotient rule its derivative is cot x sin x sin x cos x cos x sin2 x 1 2 csc x 2 sin x since sin2 x cos2 x 1 4 To differentiate csc x cot x use the product rule dy d csc x d cot x cot x csc x dx dx dx The second derivative is the one we just calculated and the other one is found similarly Ex 3 4 17 d csc x csc x cot x dx 5 So dy csc x cot2 x csc3 x dx This could be rewritten using trig identities but the other versions are no simpler Another method y csc x cot x cos x sin2 x Now use the quotient rule and cancel some extra factors of sin x as the last step 6 Exercise 5 7 11 p 353 Find the antiderivatives of h x sin x 2 cos x 7 We want to find the functions whose derivative is h x sin x 2 cos x If we know two functions whose derivatives are respectively sin x and cos x we re home free But we do d cos x sin x dx d sin x cos x dx So we let H x cos x 2 sin x and check that H x h x The most general antiderivative of h is H x C where C is an arbitrary constant 8 Now let s drop back to see where the trig derivatives came from On pp 180 181 we re offered 4 trigonometric limits but they are not of equal profundity The first two just say that the sine and cosine functions are continuous at 0 But you already knew that didn t you 9 The third limit is the important one sin 1 0 It can be proved from the inequalities lim sin tan for 0 2 which are made obvious by drawing some pictures It says that sin behaves like when is small 10 In contrast the fourth limit formula cos 1 0 0 says that cos behaves like 1 and that the difference from 1 vanishes faster than as goes to 0 lim In fact later we will see that 2 cos 1 2 11 Exercise 3 4 15 tan 3x lim x 0 3 tan 2x 12 Split the function into a product of functions whose limits we know tan 3x 1 sin 3x cos 2x 3 tan 2x 3 cos 3x sin 2x sin 3x 2x cos 2x 3x sin 2x 2 cos 3x As x 0 2x and 3x approach 0 as well Therefore the two sine quotients approach 1 Each cosine also goes to 1 So the limit is 12 13 The chain rule is the most important and powerful theorem about derivatives For a first look at it let s approach the last example of last week s lecture in a different way Exercise 3 3 11 revisited and shortened A stone is dropped into a lake creating a circular ripple that travels outward at a speed of 60 cm s Find the rate at which the circle is increasing after 3 s 14 In applied problems it s usually easier to use the Leibniz notation such as df dx instead of the prime notation for derivatives which is essentially Newton s notation The area of a circle of radius r is A r 2 So dA 2 r dr 15 Notice that dA dr 2 r is the circumference That makes sense since when the radius changes by r the region enclosed changes by a thin circular strip of length 2 r and width r hence area A 2 r r From A r 2 r we could compute the rate of change of area with respect to radius by plugging in the appropriate value of r namely 60 3 180 But the question asks for the rate with respect to time and tells us that 16 dr 60 cm s dt Common sense says that we should just multiply A r by 60 getting dA dA dr 2 60 2 t dt dr dt t 3 Of course this is the same result we got last week 17 Now consider a slight variation on the problem Exercise 3 3 11 modified A stone is dropped into a lake creating a circular ripple that travels outward at a speed of 60 cm s Find the rate at which the circle is increasing when the radius is 180 cm 18 To answer this question by the method of last week using A t 3600 t2 we would need to calculate the time when r 180 That s easy enough in this case t 3 but it carries the argument through an unnecessary loop through an inverse function It is more natural and simpler to use this week s formula dA dA dr dt dr dt dA 2 r 60 21600 It gives dt 19 In both versions of the exercise we dealt with a function of the type A r t where the output of one function is plugged in as the input to a different one The composite function is sometimes denoted A r or A r t The chain rule says that A r t A r t r t or dA dA dr dt dr dt 20 Now we can use the chain rule to differentiate particular functions Exercise 3 5 7 Differentiate G x 3x 2 10 5x2 x 1 12 21 G x 3x 2 10 5x2 x 1 12 It would be foolish to multiply out the powers when we can use the chain rule instead Of course the first step is a product rule d 2 12 3x 2 5x x 1 dx 2 10 2 11 d 5x x 1 12 3x 2 5x x 1 dx G x 10 3x 2 9 22 30 3x 2 9 5x2 x 1 12 12 10x 1 3x 2 10 2 5x x 1 11 The book s answer combines some terms at the expense of factoring out a messy polynomial 23 Note that it is not smart to use the quotient rule on a problem like d x 1 2 3 dx x 1 You ll find yourself cancelling extra factors of x2 1 It s much better to use the product rule on x 1 x2 1 3 getting only 4 factors of x2 …
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