Transcendental Functions

Transcription

1 4 Transcenental Functions So far we have use only algebraic functions as examples when fining erivatives, that is, functions that can be built up by the usual algebraic operations of aition, subtraction, multiplication, ivision, an raising to constant powers Both in theory an practice there are other functions, calle transcenental, that are very useful Most important among these are the trigonometric functions, the inverse trigonometric functions, exponential functions, an logarithms º½ ÌÖ ÓÒÓÑ ØÖ ÙÒØ ÓÒ When you first encountere the trigonometric functions it was probably in the context of triangle trigonometry, efining, for example, the sine of an angle as the sie opposite over the hypotenuse While this will still be useful in an informal way, we nee to use a more expansive efinition of the trigonometric functions First an important note: while egree measure of angles is sometimes convenient because it is so familiar, it turns out to be ill-suite to mathematical calculation, so (almost) everything we o will be in terms of raian measure of angles 7

2 72 Chapter 4 Transcenental Functions To efine the raian measurement system, we consier the unit circle in the xy-plane: A (cosx,sinx) x y (,0) B An angle, x, at the center of the circle is associate with an arc of the circle which is sai to subten the angle In the figure, this arc is the portion of the circle from point (,0) to point A The length of this arc is the raian measure of the angle x; the fact that the raian measure is an actual geometric length is largely responsible for the usefulness of raian measure The circumference of the unit circle is 2πr = 2π() = 2π, so the raian measure of the full circular angle (that is, of the 360 egree angle) is 2π While an angle with a particular measure can appear anywhere aroun the circle, we nee a fixe, conventional location so that we can use the coorinate system to efine properties of the angle The stanar convention is to place the starting raius for the angle on the positive x-axis, an to measure positive angles counterclockwise aroun the circle In the figure, x is the stanar location of the angle π/6, that is, the length of the arc from (,0) to A is π/6 The angle y in the picture is π/6, because the istance from (,0) to B along the circle is also π/6, but in a clockwise irection Now the funamental trigonometric efinitions are: the cosine of x an the sine of x are the first an secon coorinates of the point A, as inicate in the figure The angle x shown can be viewe as an angle of a right triangle, meaning the usual triangle efinitions of the sine an cosine also make sense Since the hypotenuse of the triangle is, the sie opposite over hypotenuse efinition of the sine is the secon coorinate of point A over, which is just the secon coorinate; in other wors, both methos give the same value for the sine The simple triangle efinitions work only for angles that can fit in a right triangle, namely, angles between 0 an π/2 The coorinate efinitions, on the other han, apply

3 4 Trigonometric Functions 73 to any angles, as inicate in this figure: x (cosx,sinx) A The angle x is subtene by the heavy arc in the figure, that is, x = 7π/6 Both coorinates of point A in this figure are negative, so the sine an cosine of 7π/6 are both negative The remaining trigonometric functions can be most easily efine in terms of the sine an cosine, as usual: tanx = sinx cosx cotx = cosx sinx secx = cosx cscx = sinx an they can also be efine as the corresponing ratios of coorinates Although the trigonometric functions are efine in terms of the unit circle, the unit circle iagram is not what we normally consier the graph of a trigonometric function (The unit circle is the graph of, well, the circle) We can easily get a qualitatively correct iea of the graphs of the trigonometric functions from the unit circle iagram Consier the sine function, y = sinx As x increases from 0 in the unit circle iagram, the secon coorinate of the point A goes from 0 to a maximum of, then back to 0, then to a minimum of, then back to 0, an then it obviously repeats itself So the graph of y = sinx must look something like this: 2π 3π/2 π π/2 π/2 π 3π/2 2π

4 74 Chapter 4 Transcenental Functions Similarly, as angle x increases from 0 in the unit circle iagram, the first coorinate of the point A goes from to 0 then to, back to 0 an back to, so the graph of y = cosx must look something like this: 2π 3π/2 π π/2 π/2 π 3π/2 2π Exercises 4 Some useful trigonometric ientities are in appenix B Fin all values of θ such that sin(θ) = ; give your answer in raians 2 Fin all values of θ such that cos(2θ) = /2; give your answer in raians 3 Use an angle sum ientity to compute cos(π/2) 4 Use an angle sum ientity to compute tan(5π/2) 5 Verify the ientity cos 2 (t)/( sin(t)) = +sin(t) 6 Verify the ientity 2 csc(2θ) = sec(θ) csc(θ) 7 Verify the ientity sin(3θ) sin(θ) = 2cos(2θ)sin(θ) 8 Sketch y = 2sin(x) 9 Sketch y = sin(3x) 0 Sketch y = sin( x) Fin all of the solutions of 2sin(t) sin 2 (t) = 0 in the interval [0,2π] º¾ Ì Ö Ú Ø Ú Ó sinx What about the erivative of the sine function? The rules for erivatives that we have are no help, since sinx is not an algebraic function We nee to return to the efinition of the erivative, set up a it, an try to compute it Here s the efinition: sinx = x x 0 sin(x+ x) sinx x Using some trigonometric ientities, we can make a little progress on the quotient: sin(x+ x) sinx x = sinxcos x+sin xcosx sinx x = sinx cos x x +cosx sin x x

5 43 A har it 75 This isolates the ifficult bits in the two its cos x x 0 x an sin x x 0 x Here we get a little lucky: it turns out that once we know the secon it the first is quite easy The secon is quite tricky, however Inee, it is the harest it we will actually compute, an we evote a section to it º Ö Ð Ñ Ø We want to compute this it: sin x x 0 x Equivalently, to make the notation a bit simpler, we can compute sinx x 0 x In the original context we nee to keep x an x separate, but here it oesn t hurt to rename x to something more convenient To o this we nee to be quite clever, an to employ some inirect reasoning The inirect reasoning is emboie in a theorem, frequently calle the squeeze theorem THEOREM 43 Squeeze Theorem Suppose that g(x) f(x) h(x) for all x close to a but not equal to a If x a g(x) = L = x a h(x), then x a f(x) = L This theorem can be prove using the official efinition of it We won t prove it here, but point out that it is easy to unerstan an believe graphically The conition says that f(x) is trappe between g(x) below an h(x) above, an that at x = a, both g an h approach the same value This means the situation looks something like figure 43 The wiggly curve is x 2 sin(π/x), the upper an lower curves are x 2 an x 2 Since the sine function is always between an, x 2 x 2 sin(π/x) x 2, an it is easy to see that x 0 x 2 = 0 = x 0 x 2 It is not so easy to see irectly, that is algebraically, that x 0 x 2 sin(π/x) = 0, because the π/x prevents us from simply plugging in x = 0 The squeeze theorem makes this har it as easy as the trivial its involving x 2 To o the har it that we want, x 0 (sinx)/x, we will fin two simpler functions g an h so that g(x) (sinx)/x h(x), an so that x 0 g(x) = x 0 h(x) Not too surprisingly, this will require some trigonometry an geometry Referring to figure 432, x is the measure of the angle in raians Since the circle has raius, the coorinates of

6 76 Chapter 4 Transcenental Functions Figure 43 The squeeze theorem point A are (cosx,sinx), an the area of the small triangle is (cosxsinx)/2 This triangle is completely containe within the circular wege-shape region borere by two lines an the circle from (,0) to point A Comparing the areas of the triangle an the wege we see (cosxsinx)/2 x/2, since the area of a circular region with angle θ an raius r is θr 2 /2 With a little algebra this turns into (sinx)/x /cosx, giving us the h we seek A B 0 x Figure 432 Visualizing sin x/x To fin g, we note that the circular wege is completely containe insie the larger triangle The height of the triangle, from (,0) to point B, is tanx, so comparing areas we get x/2 (tanx)/2 = sinx/(2cosx) With a little algebra this becomes cosx (sinx)/x So now we have cosx sinx x cosx

7 43 A har it 77 Finally, the two its x 0 cosx an x 0 /cosx are easy, because cos(0) = By the squeeze theorem, x 0 (sinx)/x = as well Before we can complete the calculation of the erivative of the sine, we nee one other it: x 0 cosx x This it is just as har as sinx/x, but closely relate to it, so that we on t have to o a similar calculation; instea we can o a bit of tricky algebra cosx x = cosx x cosx+ cosx+ = cos2 x x(cosx+) = sin2 x x(cosx+) = sinx x sinx cosx+ To compute the esire it it is sufficient to compute the its of the two final fractions, as x goes to 0 The first of these is the har it we ve just one, namely The secon turns out to be simple, because the enominator presents no problem: x 0 sinx cosx+ = sin0 cos0+ = 0 2 = 0 Thus, cosx x 0 x = 0 Exercises 43 Compute x 0 sin(5x) x 3 Compute x 0 cot(4x) csc(3x) 5 Compute x π/4 sinx cosx cos(2x) 2 Compute x 0 sin(7x) sin(2x) tanx 4 Compute x 0 x 6 For all x 0, 4x 9 f(x) x 2 4x+7 Fin x 4 f(x) 7 For all x, 2x g(x) x 4 x 2 +2 Fin x g(x) 8 Use the Squeeze Theorem to show that x 0 x 4 cos(2/x) = 0

8 78 Chapter 4 Transcenental Functions º Ì Ö Ú Ø Ú Ó sinx ÓÒØ ÒÙ Now we can complete the calculation of the erivative of the sine: sin(x+ x) sinx sinx = x x 0 x = sinxcos x +cosx sin x x 0 x x = sinx 0+cosx = cosx The erivative of a function measures the slope or steepness of the function; if we examine the graphs of the sine an cosine sie by sie, it shoul be that the latter appears to accurately escribe the slope of the former, an inee this is true: π/2 π 3π/2 2π sinx π/2 π 3π/2 2π cosx Notice that where the cosine is zero the sine oes appear to have a horizontal tangent line, an that the sine appears to be steepest where the cosine takes on its extreme values of an Of course, now that we know the erivative of the sine, we can compute erivatives of more complicate functions involving the sine EXAMPLE 44 Compute the erivative of sin(x 2 ) x sin(x2 ) = cos(x 2 ) 2x = 2xcos(x 2 ) EXAMPLE 442 Compute the erivative of sin 2 (x 3 5x) x sin2 (x 3 5x) = x (sin(x3 5x)) 2 = 2(sin(x 3 5x)) cos(x 3 5x)(3x 2 5) = 2(3x 2 5)cos(x 3 5x)sin(x 3 5x)

9 45 Derivatives of the Trigonometric Functions 79 Exercises 44 Fin the erivatives of the following functions sin 2 ( x) xsinx sinx 4 x 2 +x sinx sin 2 x º Ö Ú Ø Ú Ó Ø ÌÖ ÓÒÓÑ ØÖ ÙÒØ ÓÒ All of the other trigonometric functions can be expresse in terms of the sine, an so their erivatives can easily be calculate using the rules we alreay have For the cosine we nee to use two ientities, Now: cosx = sin(x+ π 2 ), sinx = cos(x+ π 2 ) x cosx = x sin(x+ π 2 ) = cos(x+ π ) = sinx 2 x tanx = sinx xcosx = cos2 x+sin 2 x cos 2 x = cos 2 x = sec2 x x secx = x (cosx) = (cosx) 2 ( sinx) = sinx cos 2 x = secxtanx The erivatives of the cotangent an cosecant are similar an left as exercises Exercises 45 Fin the erivatives of the following functions sinxcosx 2 sin(cosx) 3 xtanx 4 tanx/(+sinx) 5 cotx 6 cscx 7 x 3 sin(23x 2 ) 8 sin 2 x+cos 2 x 9 sin(cos(6x)) 0 Compute secθ θ +secθ Compute t t5 cos(6t) 2 Compute t t 3 sin(3t) cos(2t) 3 Fin all points on the graph of f(x) = sin 2 (x) at which the tangent line is horizontal

10 80 Chapter 4 Transcenental Functions 4 Finallpointsonthegraphoff(x) = 2sin(x) sin 2 (x)atwhichthetangent lineishorizontal 5 Fin an equation for the tangent line to sin 2 (x) at x = π/3 6 Fin an equation for the tangent line to sec 2 x at x = π/3 7 Fin an equation for the tangent line to cos 2 x sin 2 (4x) at x = π/6 8 Fin the points on the curve y = x+2cosx that have a horizontal tangent line 9 Let C be a circle of raius r Let A be an arc on C subtening a central angle θ Let B be the chor of C whose enpoints are the enpoints of A (Hence, B also subtens θ) Let s be the length of A an let be the length of B Sketch a iagram of the situation an compute θ 0 +s/ º ÜÔÓÒ ÒØ Ð Ò ÄÓ Ö Ø Ñ ÙÒØ ÓÒ An exponential function has the form a x, where a is a constant; examples are 2 x, 0 x, e x The logarithmic functions are the inverses of the exponential functions, that is, functions that uno the exponential functions, just as, for example, the cube root function unoes the cube function: 23 = 2 Note that the original function also unoes the 3 inverse function: ( 3 8) 3 = 8 Let f(x) = 2 x The inverse of this function is calle the logarithm base 2, enote log 2 (x) or (especially in computer science circles) lg(x) What oes this really mean? The logarithm must uno the action of the exponential function, so for example it must be that lg(2 3 ) = 3 starting with 3, the exponential function prouces 2 3 = 8, an the logarithm of 8 must get us back to 3 A little thought shows that it is not a coincience that lg(2 3 ) simply gives the exponent the exponent is the original value that we must get back to In other wors, the logarithm is the exponent Remember this catchphrase, an what it means, an you won t go wrong (You o have to remember what it means Like any goo mnemonic, the logarithm is the exponent leaves out a lot of etail, like Which exponent? an Exponent of what? ) EXAMPLE 46 What is the value of log 0 (000)? The 0 tells us the appropriate number to use for the base of the exponential function The logarithm is the exponent, so the question is, what exponent E makes 0 E = 000? If we can fin such an E, then log 0 (000) = log 0 (0 E ) = E; fining the appropriate exponent is the same as fining the logarithm In this case, of course, it is easy: E = 3 so log 0 (000) = 3 Let s review some laws of exponents an logarithms; let a be a positive number Since a 5 = a a a a aan a 3 = a a a, it s clear that a 5 a 3 = a a a a a a a a= a 8 = a 5+3, an in general that a m a n = a m+n Since the logarithm is the exponent, it s no surprise that this translates irectly into a fact about the logarithm function Here are three facts

11 46 Exponential an Logarithmic functions 8 from the example: log a (a 5 ) = 5, log a (a 3 ) = 3, log a (a 8 ) = 8 So log a (a 5 a 3 ) = log a (a 8 ) = 8 = 5+3 = log a (a 5 )+log a (a 3 ) Now let s make this a bit more general Suppose A an B are two numbers, A = a x, an B = a y Then log a (AB) = log a (a x a y ) = log a (a x+y ) = x+y = log a (A)+log a (B) Now consier (a 5 ) 3 = a 5 a 5 a 5 = a = a 5 3 = a 5 Again it s clear that more generally (a m ) n = a mn, an again this gives us a fact about logarithms If A = a x then A y = (a x ) y = a xy, so log a (A y ) = xy = ylog a (A) the exponent can be pulle out in front Wehavecheateabitintheprevioustwoparagraphs Itisobviousthata 5 = a a a a a an a 3 = a a a an that the rest of the example follows; likewise for the secon example But when we consier an exponential function a x we can t be ite to substituting integers for x What oes a 25 or a 3 or a π mean? An is it really true that a 25 a 3 = a 25 3? The answer to the first question is actually quite ifficult, so we will evae it; the answer to the secon question is yes We ll evae the full answer to the har question, but we have to know something about exponential functions You nee first to unerstan that since it s not obvious what 2 x shoul mean, we are really free to make it mean whatever we want, so long as we keep the behavior that is obvious, namely, when x is a positive integer What else o we want to be true about 2 x? We want the properties of the previous two paragraphs to be true for all exponents: 2 x 2 y = 2 x+y an (2 x ) y = 2 xy After the positive integers, the next easiest number to unerstan is 0: 2 0 = You have presumably learne this fact in the past; why is it true? It is true precisely because we want 2 a 2 b = 2 a+b to be true about the function 2 x We nee it to be true that x = 2 0+x = 2 x, an this only works if 2 0 = The same argument implies that a 0 = for any a The next easiest set of numbers to unerstan is the negative integers: for example, 2 3 = /2 3 We know that whatever 2 3 means it must be that = = 2 0 =, which means that 2 3 must be /2 3 In fact, by the same argument, once we know what 2 x means for some value of x, 2 x must be /2 x an more generally a x = /a x Next, consier an exponent /q, where q is a positive integer We want it to be true that (2 x ) y = 2 xy, so (2 /q ) q = 2 This means that 2 /q is a q-th root of 2, 2 /q = q 2 This is all we nee to unerstan that 2 p/q = (2 /q ) p = ( q 2 ) p an a p/q = (a /q ) p = ( q a ) p What s left is the har part: what oes 2 x mean when x cannot be written as a fraction, like x = 2 or x = π? What we know so far is how to assign meaning to 2 x

12 82 Chapter 4 Transcenental Functions whenever x = p/q; if we were to graph this we see something like this: But this is a poor picture, because you can t see that the curve is really a whole lot of iniviual points, above the rational numbers on the x-axis There are really a lot of holes in the curve, above x = π, for example But (this is the har part) it is possible to prove that the holes can be fille in, an that the resulting function, calle 2 x, really oes have the properties we want, namely that 2 x 2 y = 2 x+y an (2 x ) y = 2 xy Exercises 46 Expan log 0 ((x+45) 7 (x 2)) 2 Expan log 2 x 3 3x 5+(7/x) 3 Write log 2 3x+7log 2 (x 2) 2log 2 (x 2 +4x+) as a single logarithm 4 Solve log 2 (+ x) = 6 for x 5 Solve 2 x2 = 8 for x 6 Solve log 2 (log 3 (x)) = for x º Ö Ú Ø Ú Ó Ø ÜÔÓÒ ÒØ Ð Ò ÐÓ Ö Ø Ñ ÙÒØ ÓÒ As with the sine, we on t know anything about erivatives that allows us to compute the erivatives of the exponential an logarithmic functions without going back to basics Let s o a little work with the efinition again: a x+ x a x x ax = x 0 x a x a x a x = x 0 x = x 0 axa x x = a x x 0 a x x

13 47 Derivatives of the exponential an logarithmic functions 83 There are two interesting things to note here: As in the case of the sine function we are left with a it that involves x but not x, which means that whatever x 0 (a x )/ x is, we know that it is a number, that is, a constant This means that a x has a remarkable property: its erivative is a constant times itself We earlier remarke that the harest it we woul compute is sinx/x = ; we x 0 now have a it that is just a bit too har to inclue here In fact the har part is to see that x 0 (a x )/ x even exists oes this fraction really get closer an closer to some fixe value? Yes it oes, but we will not prove this fact We can look at some examples Consier (2 x )/x for some small values of x:, , , , , when x is, /2, /4, /8, /6, /32, respectively It looks like this is settling in aroun 07, which turns out to be true (but the it is not exactly 07) Consier next (3 x )/x: 2, , , , , , at the same values of x It turns out to be true that in the it this is about Two examples on t establish a pattern, but if you o more examples you will fin that the it varies irectly with the value of a: bigger a, bigger it; smaller a, smaller it As we can alreay see, some of these its will be less than an some larger than Somewhere between a = 2 an a = 3 the it will be exactly ; the value at which this happens is calle e, so that e x x 0 x As you might guess from our two examples, e is closer to 3 than to 2, an in fact e 278 Now we see that the function e x has a truly remarkable property: = e x+ x e x x ex = x 0 x e x e x e x = x 0 x = x 0 exe x x = e x e x x 0 x = e x That is, e x is its own erivative, or in other wors the slope of e x is the same as its height, or the same as its secon coorinate: The function f(x) = e x goes through the point (z,e z ) an has slope e z there, no matter what z is It is sometimes convenient to express the function e x without an exponent, since complicate exponents can be har to rea In such cases we use exp(x), eg, exp(+x 2 ) instea of e +x2

14 84 Chapter 4 Transcenental Functions What about the logarithm function? This too is har, but as the cosine function was easier to o once the sine was one, so the logarithm is easier to o now that we know the erivative of the exponential function Let s start with log e x, which as you probably know is often abbreviate ln x an calle the natural logarithm function Consier the relationship between the two functions, namely, that they are inverses, that one unoes the other Graphically this means that they have the same graph except that one is flippe or reflecte through the line y = x, as shown in figure 47 Figure 47 The exponential an logarithm functions This means that the slopes of these two functions are closely relate as well: For example, the slope of e x is e at x = ; at the corresponing point on the ln(x) curve, the slope must be /e, because the rise an the run have been interchange Since the slope of e x is e at the point (,e), the slope of ln(x) is /e at the point (e,) Figure 472 Slope of the exponential an logarithm functions More generally, we know that the slope of e x is e z at the point (z,e z ), so the slope of ln(x) is /e z at (e z,z), as inicate in figure 472 In other wors, the slope of lnx is the reciprocal of the first coorinate at any point; this means that the slope of lnx at (x,lnx) is /x The upshot is: x lnx = x

15 47 Derivatives of the exponential an logarithmic functions 85 We have iscusse this from the point of view of the graphs, which is easy to unerstan but is not normally consiere a rigorous proof it is too easy to be le astray by pictures that seem reasonable but that miss some har point It is possible to o this erivation without resorting to pictures, an inee we will see an alternate approach soon Note that lnx is efine only for x > 0 It is sometimes useful to consier the function ln x, a function efine for x 0 When x < 0, ln x = ln( x) an x ln x = x ln( x) = x ( ) = x Thus whether x is positive or negative, the erivative is the same What about the functions a x an log a x? We know that the erivative of a x is some constant times a x itself, but what constant? Remember that the logarithm is the exponent an you will see that a = e lna Then a x = (e lna ) x = e xlna, an we can compute the erivative using the chain rule: x ax = x (elna ) x = x exlna = (lna)e xlna = (lna)a x The constant is simply lna Likewise we can compute the erivative of the logarithm function log a x Since x = e lnx we can take the logarithm base a of both sies to get Then log a (x) = log a (e lnx ) = lnxlog a e x log ax = x log ae This is a perfectly goo answer, but we can improve it slightly Since a = e lna log a (a) = log a (e lna ) = lnalog a e = lnalog a e lna = log a e,

16 86 Chapter 4 Transcenental Functions we can replace log a e to get x log a x = xlna You may if you wish memorize the formulas x ax = (lna)a x an x log ax = xlna Because the trick a = e lna is often useful, an sometimes essential, it may be better to remember the trick, not the formula EXAMPLE 47 Compute the erivative of f(x) = 2 x x 2x = x (eln2 ) x = = x exln2 ( ) x xln2 e xln2 = (ln2)e xln2 = 2 x ln2 EXAMPLE 472 Compute the erivative of f(x) = 2 x2 = 2 (x2) = x 2x2 = x ex2 ln2 ( ) x x2 ln2 = (2ln2)xe x2 ln2 = (2ln2)x2 x2 e x2 ln2 EXAMPLE 473 Compute the erivative of f(x) = x x At first this appears to be a new kin of function: it is not a constant power of x, an it oes not seem to be an exponential function, since the base is not constant But in fact it is no harer than the previous example x xx = = x exlnx ( ) x xlnx = (x x +lnx)xx = (+lnx)x x e xlnx

17 48 Implicit Differentiation 87 EXAMPLE 474 Recall that we have not justifie the power rule except when the exponent is a positive or negative integer We can use the exponential function to take care of other exponents x xr = = x erlnx ( ) x rlnx = (r x )xr = rx r e rlnx Exercises 47 In 9, fin the erivatives of the functions 3 x2 2 sinx 3 (e x ) 2 4 sin(e x ) 5 e sinx 6 x sinx 7 x 3 e x 8 x+2 x 9 (/3) x2 0 e 4x /x ln(x 3 +3x) 2 ln(cos(x)) 3 ln(x2 )/x 4 ln(sec(x)+tan(x)) 5 x cos(x) 6 xlnx 7 ln(ln(3x)) +ln(3x 2 ) 8 +ln(4x) 9 x 8 (x 23) /2 27x 6 (4x 6) 8 20 Fin the value of a so that the tangent line to y = ln(x) at x = a is a line through the origin Sketch the resulting situation 2 If f(x) = ln(x 3 +2) compute f (e /3 ) e x º ÁÑÔÐ Ø Ö ÒØ Ø ÓÒ As we have seen, there is a close relationship between the erivatives of e x an lnx because these functions are inverses Rather than relying on pictures for our unerstaning, we woul like to be able to exploit this relationship computationally In fact this technique can help us fin erivatives in many situations, not just when we seek the erivative of an inverse function

18 88 Chapter 4 Transcenental Functions We will begin by illustrating the technique to fin what we alreay know, the erivative of lnx Let s write y = lnx an then x = e lnx = e y, that is, x = e y We say that this equation efines the function y = lnx implicitly because while it is not an explicit expression y =, it is true that if x = e y then y is in fact the natural logarithm function Now, for the time being, preten that all we know of y is that x = e y ; what can we say about erivatives? We can take the erivative of both sies of the equation: x x = x ey Then using the chain rule on the right han sie: = ( ) x y e y = y e y Then we can solve for y : y = e y = x There is one little ifficulty here To use the chain rule to compute /x(e y ) = y e y we nee to know that the function y has a erivative All we have shown is that if it has a erivative then that erivative must be /x When using this metho we will always have to assume that the esire erivative exists, but fortunately this is a safe assumption for most such problems The example y = lnx involve an inverse function efine implicitly, but other functions can be efine implicitly, an sometimes a single equation can be use to implicitly efine more than one function Here s a familiar example The equation r 2 = x 2 + y 2 escribes a circle of raius r The circle is not a function y = f(x) because for some values of x there are two corresponing values of y If we want to work with a function, we can break thecircleintotwo pieces, theupper anlowersemicircles, eachofwhich isafunction Let s call these y = U(x) an y = L(x); in fact this is a fairly simple example, an it s possible to give explicit expressions for these: U(x) = r 2 x 2 an L(x) = r 2 x 2 But it s somewhat easier, an quite useful, to view both functions as given implicitly by r 2 = x 2 + y 2 : both r 2 = x 2 +U(x) 2 an r 2 = x 2 + L(x) 2 are true, an we can think of r 2 = x 2 +y 2 as efining both U(x) an L(x) Now we can take the erivative of both sies as before, remembering that y is not simply a variable but a function in this case, y is either U(x) or L(x) but we re not yet specifying which one When we take the erivative we just have to remember to apply the

19 chain rule where y appears x r2 = x (x2 +y 2 ) 0 = 2x+2yy y = 2x 2y = x y 48 Implicit Differentiation 89 Now we have an expression for y, but it contains y as well as x This means that if we want to compute y for some particular value of x we ll have to know or compute y at that value of x as well It is at this point that we will nee to know whether y is U(x) or L(x) Occasionally it will turn out that we can avoi explicit use of U(x) or L(x) by the nature of the problem EXAMPLE 48 Fin the slope of the circle 4 = x 2 +y 2 at the point (, 3) Since we know both the x an y coorinates of the point of interest, we o not nee to explicitly recognize that this point is on L(x), an we o not nee to use L(x) to compute y but we coul Using the calculation of y from above, y = x y = 3 = 3 It is instructive to compare this approach to others We might have recognize at the start that (, 3) is on the function y = L(x) = 4 x 2 We coul then take the erivative of L(x), using the power rule an the chain rule, to get L (x) = 2 (4 x2 ) /2 x ( 2x) = 4 x 2 Then we coul compute L () = / 3 by substituting x = Alternately, we coulrealize that the point is onl(x), but use the fact that y = x/y Since the point is on L(x) we can replace y by L(x) to get y = x L(x) = x 4 x 2, without computing the erivative of L(x) explicitly Then we substitute x = an get the same answer as before In the case of the circle it is possible to fin the functions U(x) an L(x) explicitly, but there are potential avantages to using implicit ifferentiation anyway In some cases it is more ifficult or impossible to fin an explicit formula for y an implicit ifferentiation is the only way to fin the erivative

20 90 Chapter 4 Transcenental Functions EXAMPLE 482 Fin theerivativeofany function efine implicitlyby yx 2 +e y = x We treat y as an unspecifie function an use the chain rule: x (yx2 +e y ) = x x (y 2x+y x 2 )+y e y = y x 2 +y e y = 2xy y (x 2 +e y ) = 2xy y = 2xy x 2 +e y You might think that the step in which we solve for y coul sometimes be ifficult after all, we re using implicit ifferentiation here because we can t solve the equation yx 2 +e y = x for y, so maybe after taking the erivative we get something that is har to solve for y In fact, this never happens All occurrences y come from applying the chain rule, an whenever the chain rule is use it eposits a single y multiplie by some other expression So it will always be possible to group the terms containing y together an factor out the y, just as in the previous example If you ever get anything more ifficult you have mae a mistake an shoul fix it before trying to continue It is sometimes the case that a situation leas naturally to an equation that efines a function implicitly EXAMPLE 483 Consier all the points(x, y) that have the property that the istance from (x,y) to (x,y ) plus the istance from (x,y) to (x 2,y 2 ) is 2a (a is some constant) These points form an ellipse, which like a circle is not a function but can viewe as two functions paste together Because we know how to write own the istance between two points, we can write own an implicit equation for the ellipse: (x x ) 2 +(y y ) 2 + (x x 2 ) 2 +(y y 2 ) 2 = 2a Then we can use implicit ifferentiation to fin the slope of the ellipse at any point, though the computation is rather messy EXAMPLE 484 We have alreay justifie the power rule by using the exponential function, but we coul also o it for rational exponents by using implicit ifferentiation Suppose that y = x m/n, where m an n are positive integers We can write this implicitly asy n = x m, then becausewejustifie thepower ruleforintegers, wecantaketheerivative

21 48 Implicit Differentiation 9 of each sie: ny n y = mx m y = m n y = m n x m y n x m (x m/n ) n y = m n xm (m/n)(n ) y = m n xm m+(m/n) y = m n x(m/n) Exercises 48 In exercises 8, fin a formula for the erivative y at the point (x,y): y 2 = +x 2 2 x 2 +xy +y 2 = 7 3 x 3 +xy 2 = y 3 +yx 2 4 4cosxsiny = 5 x+ y = 9 6 tan(x/y) = x+y 7 sin(x+y) = xy 8 x + y = 7 9 A hyperbola passing through (8,6) consists of all points whose istance from the origin is a constant more than its istance from the point (5,2) Fin the slope of the tangent line to the hyperbola at (8,6) 0 Compute y for the ellipse of example 483 If y = log a x then a y = x Use implicit ifferentiation to fin y 2 The graph of the equation x 2 xy+y 2 = 9 is an ellipse Fin the lines tangent to this curve at the two points where it intersects the x-axis Show that these lines are parallel 3 Repeat the previous problem for the points at which the ellipse intersects the y-axis 4 Fin the points on the ellipse from the previous two problems where the slope is horizontal an where it is vertical 5 Fin an equation for the tangent line to x 4 = y 2 +x 2 at (2, 2) (This curve is the kampyle of Euoxus) 6 Fin an equation for the tangent line to x 2/3 +y 2/3 = a 2/3 at a point (x,y ) on the curve, with x 0 an y 0 (This curve is an astroi) 7 Fin an equation for the tangent line to (x 2 +y 2 ) 2 = x 2 y 2 at a point (x,y ) on the curve, when y 0 (This curve is a lemniscate)

22 92 Chapter 4 Transcenental Functions Definition Two curves are orthogonal if at each point of intersection, the angle between their tangent lines is π/2 Two families of curves, A an B, are orthogonal trajectories of each other if given any curve C in A an any curve D in B the curves C an D are orthogonal For example, the family of horizontal lines in the plane is orthogonal to the family of vertical lines in the plane 8 Show that x 2 y 2 = 5 is orthogonal to 4x 2 + 9y 2 = 72 (Hint: You nee to fin the intersection points of the two curves an then show that the prouct of the erivatives at each intersection point is ) 9 Show that x 2 +y 2 = r 2 is orthogonal to y = mx Conclue that the family of circles centere at the origin is an orthogonal trajectory of the family of lines that pass through the origin Note that there is a technical issue when m = 0 The circles fail to be ifferentiable when they cross the x-axis However, the circles are orthogonal to the x-axis Explain why Likewise, the vertical line through the origin requires a separate argument 20 For k 0 an c 0 show that y 2 x 2 = k is orthogonal to yx = c In the case where k an c are both zero, the curves intersect at the origin Are the curves y 2 x 2 = 0 an yx = 0 orthogonal to each other? 2 Suppose that m 0 Show that the family of curves {y = mx+b b R} is orthogonal to the family of curves {y = (x/m)+c c R} º ÁÒÚ Ö ÌÖ ÓÒÓÑ ØÖ ÙÒØ ÓÒ The trigonometric functions frequently arise in problems, an often it is necessary to invert the functions, for example, to fin an angle with a specifie sine Of course, there are many angles with the same sine, so the sine function oesn t actually have an inverse that reliably unoes the sine function If you know that sinx = 05, you can t reverse this to iscover x, that is, you can t solve for x, as there are infinitely many angles with sine 05 Nevertheless, it is useful to have something like an inverse to the sine, however imperfect The usual approach is to pick out some collection of angles that prouce all possible values of the sine exactly once If we iscar all other angles, the resulting function oes have a proper inverse The sine takes on all values between an exactly once on the interval [ π/2,π/2] If we truncate the sine, keeping only the interval [ π/2,π/2], as shown in figure 49, then this truncate sine has an inverse function We call this the inverse sine or the arcsine, an write y = arcsin(x) Recall that a function an its inverse uno each other in either orer, for example, ( 3 x) 3 = x an 3 x 3 = x This oes not work with the sine an the inverse sine because the inverse sine is the inverse of the truncate sine function, not the real sine function It is true that sin(arcsin(x)) = x, that is, the sine unoes the arcsine It is not true that the arcsine unoes the sine, for example, sin(5π/6) = /2 an arcsin(/2) = π/6, so oing first the sine then the arcsine oes not get us back where we starte This is because 5π/6

23 49 Inverse Trigonometric Functions 93 2π 3π/2 π π/2 π/2 π 3π/2 2π π/2 π/2 π/2 π/2 Figure 49 The sine, the truncate sine, the inverse sine is not in the omain of the truncate sine If we start with an angle between π/2 an π/2 then the arcsine oes reverse the sine: sin(π/6) = /2 an arcsin(/2) = π/6 What is the erivative of the arcsine? Since this is an inverse function, we can iscover the erivative by using implicit ifferentiation Suppose y = arcsin(x) Then sin(y) = sin(arcsin(x)) = x Now taking the erivative of both sies, we get y cosy = y = cosy As we expect when using implicit ifferentiation, y appears on the right han sie here We woul certainly prefer to have y written in terms of x, an as in the case of lnx we can actually o that here Since sin 2 y + cos 2 y =, cos 2 y = sin 2 y = x 2 So cosy = ± x 2, but which is it plus or minus? It coul in general be either, but this isn t in general : since y = arcsin(x) we know that π/2 y π/2, an the cosine of an angle in this interval is always positive Thus cosy = x 2 an x arcsin(x) = x 2 Note that this agrees with figure 49: the graph of the arcsine has positive slope everywhere We can o something similar for the cosine As with the sine, we must first truncate the cosine so that it can be inverte, as shown in figure 492 Then we use implicit

24 94 Chapter 4 Transcenental Functions π π/2 π/2 π Figure 492 The truncate cosine, the inverse cosine ifferentiation to fin that x arccos(x) = x 2 Note that the truncate cosine uses a ifferent interval than the truncate sine, so that if y = arccos(x) we know that 0 y π The computation of the erivative of the arccosine is left as an exercise Finally we look at the tangent; the other trigonometric functions also have partial inverses but the sine, cosine an tangent are enough for most purposes The tangent, truncate tangent an inverse tangent are shown in figure 493; the erivative of the arctangent is left as an exercise π/2 π/2 π/2 π/2 π/2 π/2 Figure 493 The tangent, the truncate tangent, the inverse tangent Exercises 49 Show that the erivative of arccosx is x 2 2 Show that the erivative of arctanx is +x 2

25 40 Limits revisite 95 3 The inverse of cot is usually efine so that the range of arccot is (0,π) Sketch the graph of y = arccotx In the process you will make it clear what the omain of arccot is Fin the erivative of the arccotangent 4 Show that arccotx+arctanx = π/2 5 Fin the erivative of arcsin(x 2 ) 6 Fin the erivative of arctan(e x ) 7 Fin the erivative of arccos(sinx 3 ) 8 Fin the erivative of ln((arcsinx) 2 ) 9 Fin the erivative of arccose x 0 Fin the erivative of arcsinx+arccosx Fin the erivative of log 5 (arctan(x x )) º½¼ Ä Ñ Ø Ö Ú Ø We have efine an use the concept of it, primarily in our evelopment of the erivative Recall that f(x) = L is true if, in a precise sense, f(x) gets closer an closer to x a L as x gets closer an closer to a While some its are easy to see, others take some ingenuity; in particular, the its that efine erivatives are always ifficult on their face, since in f(x+ x) f(x) x 0 x both the numerator an enominator approach zero Typically this ifficulty can be resolve when f is a nice function an we are trying to compute a erivative Occasionally such its are interesting for other reasons, an the it of a fraction in which both numerator an enominator approach zero can be ifficult to analyze Now that we have the erivative available, there is another technique that can sometimes be helpful in such circumstances Before we introuce the technique, we will also expan our concept of it, in two ways When the it of f(x) as x approaches a oes not exist, it may be useful to note in what way it oes not exist We have alreay talke about one such case: one-sie its Another case is when f goes to infinity We also will occasionally want to know what happens to f when x goes to infinity EXAMPLE 40 What happens to /x as x goes to 0? From the right, /x gets bigger an bigger, or goes to infinity From the left it goes to negative infinity EXAMPLE 402 What happens to the function cos(/x) as x goes to infinity? It seems clear that as x gets larger an larger, /x gets closer an closer to zero, so cos(/x) shoul be getting closer an closer to cos(0) =

26 96 Chapter 4 Transcenental Functions x a As with orinary its, these concepts can be mae precise Roughly, we want f(x) = to mean that we can make f(x) arbitrarily large by making x close enough to a, an x f(x) = L shoul mean we can make f(x) as close as we want to L by making x large enough Compare this efinition to the efinition of it in section 23, efinition 232 DEFINITION 403 If f is a function, we say that f(x) = if for every N > 0 x a there is a δ > 0 such that whenever x a < δ, f(x) > N We can exten this in the obvious ways to efine f(x) =, x a f(x) = ±, an +f(x) = ± x a DEFINITION 404 Limit at infinity Iff isafunction, wesaythat f(x) = L x if for every ǫ > 0 there is an N > 0 so that whenever x > N, f(x) L < ǫ We may similarly efine f(x) = L, an using the iea of the previous efinition, we may x efine f(x) = ± x ± We inclue these efinitions for completeness, but we will not explore them in etail Suffice it to say that such its behave in much the same way that orinary its o; in particular there are some analogs of theorem 236 Now consier this it: x 2 π 2 x π sinx As x approaches π, both the numerator an enominator approach zero, so it is not obvious what, if anything, the quotient approaches We can often compute such its by application of the following theorem THEOREM 405 L Hôpital s Rule For sufficiently nice functions f(x) an g(x), if f(x) = 0 = g(x) or both f(x) = ± an x a g(x) = ±, an if x a x a x a f (x) f(x) g exists, then (x) x a g(x) = f (x) x a g This remains true if x a is replace by (x) x or x x a This theorem is somewhat ifficult to prove, in part because it incorporates so many ifferent possibilities, so we will not prove it here We also will not nee to worry about the precise efinition of sufficiently nice, as the functions we encounter will be suitable x 2 π 2 EXAMPLE 406 Compute x π sinx in two ways x a

27 40 Limits revisite 97 First we use L Hôpital s Rule: Since the numerator an enominator both approach zero, x 2 π 2 x π sinx = 2x x π cosx, provie the latter exists But in fact this is an easy it, since the enominator now approaches, so x 2 π 2 x π sinx = 2π = 2π We on t really nee L Hôpital s Rule to o this it Rewrite it as an note that (x+π)x π x π sinx x π x π sinx = x π x π sin(x π) = x x 0 sinx since x π approaches zero as x approaches π Now as before (x+π)x π x π sinx = (x+π) x = 2π( ) = 2π x π x 0 sinx 2x 2 3x+7 EXAMPLE 407 Compute x x 2 +47x+ in two ways As x goes to infinity both the numerator an enominator go to infinity, so we may apply L Hôpital s Rule: 2x 2 3x+7 x x 2 +47x+ = 4x 3 x 2x+47 In the secon quotient, it is still the case that the numerator an enominator both go to infinity, so we are allowe to use L Hôpital s Rule again: x 4x 3 2x+47 = 4 x 2 = 2 So the original it is 2 as well Again, we on t really nee L Hôpital s Rule, an in fact a more elementary approach is easier we ivie the numerator an enominator by x 2 : 2x 2 3x+7 x x 2 +47x+ = 2x 2 3x+7 x x 2 +47x+ x 2 x 2 = x 2 3 x + 7 x x + x 2 Now as x approaches infinity, all the quotients with some power of x in the enominator approach zero, leaving 2 in the numerator an in the enominator, so the it again is 2

28 98 Chapter 4 Transcenental Functions EXAMPLE 408 Compute x 0 secx sinx Both the numerator an enominator approach zero, so applying L Hôpital s Rule: secx x 0 sinx = x 0 secxtanx cosx = 0 = 0 EXAMPLE 409 Compute x 0 + xlnx This oesn t appear to be suitable for L Hôpital s Rule, but it also is not obvious As x approaches zero, lnx goes to, so the prouct looks like (something very small) (something very large an negative) But this coul be anything: it epens on how small an how large For example, consier (x 2 )(/x), (x)(/x), an (x)(/x 2 ) As x approaches zero, each of these is (something very small) (something very large), yet the its are respectively zero,, an We can in fact turn this into a L Hôpital s Rule problem: xlnx = lnx /x = lnx x Now as x approaches zero, both the numerator an enominator approach infinity (one an one +, but only the size is important) Using L Hôpital s Rule: lnx = x 0 + x x 0 + /x = x 2 x 0 + x ( x2 ) = x = 0 x 0 + One way to interpret this is that since x 0 + xlnx = 0, the x approaches zero much faster than the lnx approaches Exercises 40 Compute the its cosx x 0 sinx 3 x2 +x x 2 x x lnx 5 x x e x 2 x x 3 4 x lnx x e x +e x 6 x e x e x 9+x 3 (/t) 7 8 x 0 x t + t 2 2t+ 2 x+2 t+5 2/t /t x 2 4 x 2 t 3t+2 /t 2 y ++ y x 2 y y x 3 x

29 ( x) /4 3 x 0 x ( 5 t 0 + t + t (u ) 3 7 u (/u) u 2 +3u 3 +5/ x 9 x / x 40 Limits revisite 99 ( 4 t+ ) ((4 t) 3/2 8) t 0 t ) ( x 2 t+ ) 6 x 0 2x+ 2+(/x) 8 x 0 3 (2/x) 3+x /2 +x 20 x x /2 x+x /2 +x / x x 2/3 +x /4 23 t t t t t cosx 25 x π/2 (π/2) x x 2 27 x 0 e x x 29 x 0 ln(x 2 +) x 2 t 24 x sin(2x) 3 x 0 ln(x+) x 33 x + x x +x /2 35 x 5+x 37 x +2x 4 x 0 e x 26 x 0 x t t+ 4t+ t+2 x+x + x lnx 28 x x 30 x xlnx x 2 x /4 32 x x 34 x x+x /2 36 x 3x 2 +x+2 39 x 0 x 4 x+ x x 0 38 x2 + x+ 45 x 0 +(x+5) x 3 6x 2 47 x 2 x x + x 3 +4x+8 2x 3 2 ( 2x + ) x+2 x 40 x 0 42 x 0 + x x x+x 2 2x+x 2 4x 2x2 + x+ x+4 2 x++ x+ ( 44 (x+5) x 2x + ) x+2 ( 46 (x+5) x 2x + ) x+2 48 x 2 x 3 6x 2 x 3 4x

30 00 Chapter 4 Transcenental Functions x 50 The function f(x) = has two horizontal asymptotes Fin them an give a rough x2 + sketch of f with its horizontal asymptotes º½½ ÀÝÔ Ö ÓÐ ÙÒØ ÓÒ The hyperbolic functions appear with some frequency in applications, an are quite similar in many respects to the trigonometric functions This is a bit surprising given our initial efinitions DEFINITION 4 The hyperbolic cosine is the function an the hyperbolic sine is the function coshx = ex +e x, 2 sinhx = ex e x 2 Notice that cosh is even (that is, cosh( x) = cosh(x)) while sinh is o (sinh( x) = sinh(x)), an coshx +sinhx = e x Also, for all x, coshx > 0, while sinhx = 0 if an only if e x e x = 0, which is true precisely when x = 0 LEMMA 42 The range of coshx is [, ) Proof Let y = coshx We solve for x: y = ex +e x 2 2y = e x +e x 2ye x = e 2x + 0 = e 2x 2ye x + e x = 2y ± 4y e x = y ± y 2 From the last equation, we see y 2, an since y 0, it follows that y Now suppose y, so y± y 2 > 0 Then x = ln(y± y 2 ) is a real number, an y = coshx, so y is in the range of cosh(x)

31 DEFINITION 43 The other hyperbolic functions are tanhx = sinhx coshx cothx = coshx sinhx sechx = coshx cschx = sinhx 4 Hyperbolic Functions 0 The omain of coth an csch is x 0 while the omain of the other hyperbolic functions is all real numbers Graphs are shown in figure Figure 4 The hyperbolic functions: cosh, sinh, tanh, sech, csch, coth Certainly the hyperbolic functions o not closely resemble the trigonometric functions graphically But they o have analogous properties, beginning with the following ientity THEOREM 44 For all x in R, cosh 2 x sinh 2 x = Proof The proof is a straightforwar computation: cosh 2 x sinh 2 x = (ex +e x ) 2 4 (ex e x ) 2 4 = e2x +2+e 2x e 2x +2 e 2x 4 = 4 4 = This immeiately gives two aitional ientities: tanh 2 x = sech 2 x an coth 2 x = csch 2 x The ientity of the theorem also helps to provie a geometric motivation Recall that the graph of x 2 y 2 = is a hyperbola with asymptotes x = ±y whose x-intercepts are

32 02 Chapter 4 Transcenental Functions ± If (x,y) is a point on the right half of the hyperbola, an if we let x = cosht, then y = ± x 2 = ± cosh 2 t = ±sinht So for some suitable t, cosht an sinht are the coorinates of a typical point on the hyperbola In fact, it turns out that t is twice the area shown in the first graph of figure 42 Even this is analogous to trigonometry; cost an sint are the coorinates of a typical point on the unit circle, an t is twice the area shown in the secon graph of figure (cosht,sinht) 2 3 (cost,sint) Figure 42 Geometric efinitions of sin, cos, sinh, cosh: t is twice the shae area in each figure Given the efinitions of the hyperbolic functions, fining their erivatives is straightforwar Here again we see similarities to the trigonometric functions THEOREM 45 Proof e x +e x 2 x coshx = e x +e x x 2 = coshx coshx = sinhx an sinhx = coshx x x = ex e x 2 = sinhx, an x sinhx = e x e x x 2 Since coshx > 0, sinhx is increasing an hence injective, so sinhx has an inverse, arcsinhx Also, sinhx > 0 when x > 0, so coshx is injective on [0, ) an has a (partial) inverse, arccosh x The other hyperbolic functions have inverses as well, though arcsech x is only a partial inverse We may compute the erivatives of these functions as we have other inverse functions THEOREM 46 x arcsinhx = +x 2 Proof Let y = arcsinhx, so sinhy = x Then x sinhy = cosh(y) y =, an so y = coshy = +sinh 2 y = +x 2 =

33 4 Hyperbolic Functions 03 The other erivatives are left to the exercises Exercises 4 Show that the range of sinhx is all real numbers (Hint: show that if y = sinhx then x = ln(y + y 2 +)) 2 Compute the following its: a coshx x b sinhx x c tanhx x (coshx sinhx) x 3 Show that the range of tanhx is (,) What are the ranges of coth, sech, an csch? (Use the fact that they are reciprocal functions) 4 Prove that for every x,y R, sinh(x + y) = sinhxcoshy + coshxsinhy Obtain a similar ientity for sinh(x y) 5 Prove that for every x,y R, cosh(x + y) = coshxcoshy + sinhxsinhy Obtain a similar ientity for cosh(x y) 6 Use exercises 4 an 5 to show that sinh(2x) = 2sinhxcoshx an cosh(2x) = cosh 2 x+sinh 2 x for every x Conclue also that (cosh(2x) )/2 = sinh 2 x 7 Show that x (tanhx) = sech2 x Compute the erivatives of the remaining hyperbolic functions as well 8 What are the omains of the six inverse hyperbolic functions? 9 Sketch the graphs of all six inverse hyperbolic functions

34