Some functions and their derivatives

Transcription

1 Chapter Some functions an their erivatives. Derivative of x n for integer n Recall, from eqn (.6), for y = f (x), Also recall that, for integer n, Hence, if y = x n then y x = lim δx 0 (a + b) n = a n + na n b + y x = lim (x + δx) n x n δx 0 δx = lim δx 0 f(x + δx) f(x). δx n (n ) a n b b n.! ³ x n + nx n δx + n(n ) x n δx +... δx nx n δx + n(n ) x n δx +... = lim δx 0 δx nx n + = lim δx 0 y x = nxn.. Trigonometric Functions See Appenix C for further etails. n (n ) x n δx +... x n 0

2 .. Definition an Graphs of Sine an Cosine The two basic trigonometric functions are the sine an cosine. From an angle θ, we may efine cos θ an sin θ geometrically. Given the point P in the (x, y) plane such that OP is of length an makes an anticlockwise angle θ with the positive x axis, then cos θ = x-coorinate of P. sin θ = y-coorinate of P. In the figure, 0 < θ < π an, for the right-angle triangle OAP, sin θ = length of sie opposite to θ length of hypotenuse, cos θ = length of sie ajactent to θ. length of hypotenuse In calculus, an THROUGHOUT THIS COURSE, all angles will be measure in RADIANS. In the picture, the angle θ inraiansisefine to be the length of the arc from the point (, 0) to P of the circle of raius centre O. Our geometrical efinitions of cos θ an sin θ make sense not just for 0 < θ < π/ as in the picture but for all real values of θ of either sign. The graphs of these two functions in the range ( 3π, 3π) are:

3 By consiering the limiting cases of the previously illustrate right-angle triangle when θ =0an when θ = π/, we immeiately euce (in accorance with the graphs just rawn) that.. Inverse Trigonometric Functions sin 0 = 0, (.) cos π = 0, (.) cos 0 =, (.3) sin π =. (.4) From the form of the graph of the sine function, we see that it cannot have an inverse as it stans, since each horizontal line y = c (with <c<) cuts the graph in infinitely many points. We restrict it to an interval on which it is an increasing function, an the conventional choice for that interval is [ π/, π/]. The increasing restriction

4 sin : π, π! [, ] has increasing inverse sin =arcsin:[, ]! π, π. By efinition, given any y in [, ], sin y istheuniqueanglebetween π/ an π/ whose sine is y. NB. It is totally ifferent from (sin y) =cosec y. NB The graph of sin y has vertical tangents at y =, sotheslope x y these two points. is infinite at Question: Over what restriction will cos have an inverse, arccos? Answer: We restrict cos to be either increasing or ecreasing. It is usual to choose the ecreasing restriction cos : [0, π]! [, ], with inverse arccos : [, ]! [0, π]. 3

5 As for arcsin,givenanyy in [, ], cos (y) = arccos y istheuniqueanglebetween0 an π whose cosine is y. tan an arctan 4

6 The function tan is increasing over many intervals, but we conventionally take the interval containing x =0: tan : ( π/, π/)! R with inverse arctan : R! ( π/, π/). 5

7 ..3 Some ientities an a useful limit NB.AnientityisanequationwhichholsforALLpermittevaluesofitsvariables. See Appenix C for proofs of many trig. ientities. Three which will be use in the next section are sin(x + y) = sinx cos y +cosxsin y, (.5) cos x = sin x (.6) cos x +sin x =. (.7) Appenix C4 also contains the proof of a useful limit: sin x lim =, x 0 x (.8) i.e. sin x ¼ x for small angles x...4 Derivatives of the trigonometric functions By efinition (see section.4), the erivative of sin x is (sin x) = lim x δx 0 sin(x + δx) sin x δx sin x cos δx +cosxsin δx sin x = lim δx 0 δx sin x (cos δx ) + cos x sin δx = lim δx 0 δx sinx sin (δx/) + cos x sin δx = lim δx 0 δx sin δx/ δx = sin x lim δx 0 δx/ = cosx using (.8). +cosx lim δx 0 sin δx δx using (.5) using (.6) This is eqn B.4 of Appenix B. There is a similar proof that (cos x) = sin x. x Derivatives of other trig functions may be erive from those of sin an cos. e.g, using the quotient rule: µ sin x (tan x) = x x cos x cos x (cos x) sin x( sin x) = cos x = cos x +sin x cos x = using (.7) cos x = sec x. One similarly euces that x (cot x) = cosec x. The remaining erivatives are all given in Appenix B. 6

8 ..5 Derivatives of the inverse trigonometric functions Derivatives of inverse functions To fin the erivatives of inverse functions, we use the result y x = ³, provie x x y 6=0 which follows irectly from the efinition of the erivative. y Inverse Sine Set y =sin x an x =siny (with π/ y π/ an <x< ) cos y 0). Using sin y +cos y =,wehavecos y =+ p sin y =+ p x. Hence x (sin x) = y x = = = ³ ³ x y y cos y (sin y) an so x (sin x)= Note that this! + as x!. p x ( <x<). Inverse Cosine Using an exactly similar argument we fin that x (cos x)= p x ( <x<), which also! as x!. Inverse Tan x =tany ) x y =sec y =+tan y =+x,so x (tan x)= x y = for any real x. +x.3 The Exponential an Logarithm Functions.3. The Exponential Function Consiering the unique function y(x) (efine for all real x) satisfying the conition y = y(x) for all real x, (.9) x [NBthisisaifferential equation cf MATH970 next semester.] 7

9 Repeately ifferentiating (.9), we see that y x = y x = y, an more generally n y = y for all positive integers n. xn It follows (from Taylor s Theorem - which you haven t met yet!) that a function satisfies conition (.9) if an only if y(x) =y(0) +x + x! + x3 3! + = y(0) X n=0 x n n! for all real x, (.0) where n! =n(n )(n ) is the prouct of the first n positive integers (calle factorial n ) an y(0) is the value of y(x) at x =0. For now, you can satisfy yourself that (.0) works by ifferentiating it: y x = y(0) +x + x x! + x3 3! + = y(0) 0++ x! + 3x + 4x3 3! 4! = y(0) +x + x! + x3 3! + = y(x). It turns out that the sum of infinitely many terms is nevertheless finite in value for any real x. [It is an example of a Taylor series, ofwhichmorelateroninthis course.] Definition of the exponential function We efine the exponential function exp(x) to be the unique function (efine for all real x) satisfying the two conitions [exp(x)] x = exp(x) (for all real x), (.) exp (0) =. (.) The unique function exp(x) satisfying these is given explicitly by This efinition: exp(x) =+x + x! + x3 3! + = X n=0 x n n!. (.3) ² allows the numerical evaluation of exp (x) for all real x. ² allows one to show (see Appenix D.) that: exp (x + y) = exp(x)exp(y), exp (x) > 0, exp ( x) = exp (x) Given exp (x) > 0 it follows from (.) that [exp(x)] > 0 an so x exp (x) is a strictly increasing function, i.e. x > y) exp(x) > exp(y). (.4) 8

10 From.3 we can see that an so, given exp ( x) = exp(x), So, the graph of the function exp(x) looks like: exp(x)! + as x! +, (.5) exp(x)! 0 as x!. (.6).3. The Logarithmic Function Since exp : R! (0, ) is a strictly increasing function, it must have a (unique) strictly increasing inverse ln : (0, )! R calle the logarithmic function or natural logarithm. For x R, y (0, ), y =exp(x) if an only if x =lny. (.7) Taking x =0in (.7) an using exp (0) =, weseethat Similarly it follows from (.5), (.6) an (.7) that ln = 0. (.8) ln x! + as x! +, (.9) ln x! as x! 0. (.0) Properties (.8) (.0) all show up clearly on the graph of ln: 9

11 .3.3 Definition of an Arbitrary Real Power of a Positive Real Number We know so we see exp(x + x )=exp(x )exp(x ) (.) exp(x + x + + x n )=exp(x )exp(x )...exp(x n ), (.) where n is any positive integer an x,x,...,x n any real numbers. Taking x = x = = x n =lnx, wherex>0, an using (.7), we get exp(n ln x) =[exp(lnx)] n = x n, for any positive integer n. We can efine an arbitrary real power x α of a positive real number x by equivalent [by (.7)] to x α =exp(α ln x) (x>0, α real), (.3) ln(x α )=α ln x (x >0, α real). (.4) The efinition (.3) satisfies the stanar rules for multiplication an ivision of powers: for all real α an β. x α x β = x α+β, x α x β = x α β, 0

12 .3.4 The Number e, an the Exponential Function as a Power We efine the positive real number e (whose value is approximately.78) by e =exp, (.5) from which Using (.3), i.e..3.5 Derivatives The exponential function ln e =. (.6) e x =exp(xln e) =exp(x) exp x = e x for all real x. (.7) By efinition. [exp(x)] = exp(x). (.8) x The logarithmic function y =lnx, ansox =exp(y). Then x [ln(x)] = x y = exp (y) = x (for x>0). (.9) Arbitrary real powers From the above efinition of x α we have x (xα ) = (exp (α ln x)) x = exp(α ln x) α x (using the chain rule) = αx α x = αx α. Thus x (xα )=αx α (α real, x>0), (.30) NBthereisasimilartrickfor x (ax )= (exp (x ln a)) =... x.4 The Hyperbolic Functions These are constructe from the exponential function, but obey ientities closely analogous to those satisfie by the corresponing trigonometric functions (obtaine by ropping the final h ).

13 .4. Definitions cosh x = (ex + e x ), (.3) sinh x = (ex e x ), (.3) tanh x = sinh x cosh x = ex e x e x = e x + e x +e = ex x e x +, (.33) coth x = tanh x = cosh x sinh x = ex + e x +e x = e x e x e = ex + x e x (x 6= 0),(.34) sech x = cosh x =, e x + e x (.35) cosech x = sinh x = e x e x (x 6= 0). (.36).4. The Funamental Ientity Corresponing to the well-known ientity cos x +sin x =for the trigonometric functions, we have for the hyperbolic functions the ientity cosh x sinh x = (.37) (holing for all real x). NB the minus sign! To prove (.37), just substitute the efinitions (.3) an (.3) into the LHS, to get 4 (ex +e x ) 4 (ex e x ) = 4 (ex +e x +e x e x ) 4 (ex +e x e x e x )=e x e x = e x x = e 0 =..4.3 Further Ientities tanh x = sech x, (.38) coth x = cosech x, (.39) sinh(x + y) = sinhx cosh y +coshxsinh y, (.40) sinh(x y) = sinhx cosh y cosh x sinh y, (.4) cosh(x + y) = coshx cosh y +sinhxsinh y, (.4) cosh(x y) = coshx cosh y sinh x sinh y, (.43) sinh x = sinhx cosh x, (.44) cosh x = sinh x +cosh x, (.45) cosh x = (cosh x +), (.46) sinh x = (cosh x ), (.47) sinh x cosh y = [sinh (x + y)+sinh(x y)], (.48) cosh x cosh y = [cosh (x + y)+cosh(x y)], (.49) sinh x sinh y = [cosh (x + y) cosh (x y)]. (.50)

14 Each of these has a trigonometric counterpart (see Appenix C), but there are some important ifferences of sign, which shoul be carefully note. The ientities are all easy to prove from the efinitions above..4.4 The Graphs of cosh, sinh an tanh Using the efinitions of the hyperbolic functions, an our knowlege of the exponential function, we can sketch the following graphs an euce some properties for the hyperbolic functions. cosh 0 =, cosh x! + as x!, cosh ( x) = cosh(x) (i.e. it is an EVEN function). 3

15 sinh 0 = 0, sinh x! + as x! +, sinh x! as x!, sinh ( x) = sinh (x) (i.e. it is an ODD function). 4

16 tanh 0 = 0, tanh x! + as x! +, tanh x! as x!, tanh ( x) = tanh (x) (i.e. it is an ODD function). Like all even functions, cosh has a a graph which is symmetrical about Oy, i.e. invariant uner reflection in Oy. It has a minimum value of at x =0, where its erivative sinh x is zero. On the other han, the graphs of the functions sinh an tanh, like those of all o functions, are invariant uner rotation through 80 about O. Notethatsinhan tanh are both increasing functions an that the graph of tanh has horizontal asymptotes y = approache as x! respectively..4.5 Symmetry an Asymptotic Properties of other hyperbolic functions Similarly, we can euce the following properties of coth, sech an cosech: coth( x) = coth x, sech ( x) = sechx, cosech ( x) = cosechx. coth x! as x! +, coth x! as x!, coth x! + as x! 0 from above, coth x! as x! 0, from below, sech x! 0 as x!, cosech x! 0 as x!, sech 0= cosech x! + as x! 0 from above, cosech x! as x! 0 from below..4.6 Inverse Hyperbolic Functions cosh We restrict cosh x to the interval [0, ), on which it is an increasing function with increasing inverse cosh :[, )! [0, ). Fory 0, x, wehavey =cosh x, x =coshy. cosh x may be expresse in terms of logarithms as follows. If y =cosh x (x ) then y is the positive solution of the equation x =coshy, that is x = e y + e y. Multiplying through by e y gives a quaratic for e y,withroots xe y = e y + e y xe y + = 0 (e y ) xe y + = 0 e y = x p 4x 4 = x p x. ³ so, y = ln x p x. 5

17 But we want y>0, so we choose the + sign, i.e. ³ y =cosh x =ln x + p x for x. (.5) sinh sinh: R! R is an increasing function, so there is no nee for restriction. By similar arguments to those use for cosh, one shows that, for any x, sinh x =ln ³x + p +x. (.5) tanh tanh : R! (, ) is an increasing function, so there is no nee for restriction. It is easy to show that tanh x = µ +x ln for <x<. (.53) x.4.7 Derivatives The Derivatives of the Hyperbolic Functions We know: Hence x (ex )=e x, (cosh x) = x x (sinh x) = x x x (e x )= e x x e x + e = e x e x =sinhx, x e x e = e x + e x =coshx. Bythequotientruleweeucethat µ sinh x cosh x cosh x sinh x sinh x (tanh x) = = x x cosh x cosh x = cosh x = sech x, (sech x) = x x (cosh x) = (cosh x) sinh x = cosh x an one may similarly prove (provie x 6= 0)that sinh x = sech x tanh x, cosh x (see Appenix B.) x (coth x) = cosech x, (cosech x) x = cosech x coth x. Derivatives of Inverse Hyperbolic Functions 6

18 Inverse Cosh Recall y =cosh x, x =coshy with y 0, x. So,sinh y 0 an hence Then i.e. sinh y = cosh y q sinh y = + cosh y =+ p x. y x = x y x (cosh x)= = sinh y =+ p x p for x>. x You can also prove this by ifferentiating eqn (.5) irectly. Inverse Sinh Similarly x (sinh x)= p +x. Inverse Tanh y =tanh x ) x =tanhy, sowehave so x y = sech x = tanh x = y, x (tanh x)= x y = for <x<. x 7