CHAPTER 7. Di erentiation

CHAPTER 7 Di erentiation 1. Te Derivative at a Point Definition 7.1. Let f be a function defined on a neigborood of x 0. f is di erentiable at x 0, if te following it exists: f 0 fx 0 + ) fx 0 ) x 0 )=. Define Df) ={x : f 0 x) exists}. Te standard notations for te derivative will be used; e. g., f 0 x), etc. df x) dx, Dfx), An equivalent way of stating tis definition is to note tat if x 0 2 Df), ten f 0 x 0 )= x!x 0 fx) fx 0 ) x x 0. See Figure 1.) Tis can be interpreted in te standard way as te iting slope of te secant line as te points of intersection approac eac oter. Example 7.1. If fx) =c for all x and some c 2 R, ten So, f 0 x) =0everywere. Example 7.2. If fx) =x, ten fx 0 + ) fx 0 ) c c = =0. fx 0 + ) fx 0 ) x 0 + x 0 = = =1. So, f 0 x) =1everywere. Teorem 7.2. For any function f, Df) Cf). Proof. Suppose x 0 2 Df). Ten fx) x!x 0 fx 0 ) = x!x 0 fx) fx 0 ) x x 0 x x 0 ) = f 0 x 0 )0=0. Tis sows x!x0 fx) =fx 0 ), and x 0 2 Cf). Of course, te converse of Teorem 7.2 is not true. 1 c Lee Larson LLarson@Louisville.edu) 7-1

7-2 CHAPTER 7. DIFFERENTIATION Figure 1. Tese graps illustrate tat te two standard ways of writing te di erence quotient are equivalent. Example 7.3. Te function fx) = x is continuous on R, but so f 0 0) fails to exist. f0 + ) f0) f0 + ) f0) =1=, #0 "0 Teorem 7.2 and Example 7.3 sow tat di erentiability is a strictly stronger condition tan continuity. For a long time most matematicians believed tat every continuous function must certainly be di erentiable at some point. In te nineteent century, several researcers, most notably Bolzano and Weierstrass, presented examples of functions continuous everywere and di erentiable nowere. 2 It as since been proved tat, in a tecnical sense, te typical continuous function is nowere di erentiable [4]. So, contrary to te impression left by many beginning calculus classes, di erentiability is te exception rater tan te rule, even for continuous functions.. 2. Di erentiation Rules Following are te standard rules for di erentiation learned in every beginning calculus course. Teorem 7.3. Suppose f and g are functions suc tat x 0 2 Df) \ Dg). a) x 0 2 Df + g) and f + g) 0 x 0 )=f 0 x 0 )+g 0 x 0 ). b) If a 2 R, ten x 0 2 Daf) and af) 0 x 0 )=af 0 x 0 ). c) x 0 2 Dfg) and fg) 0 x 0 )=f 0 x 0 )gx 0 )+fx 0 )g 0 x 0 ). d) If gx 0 ) 6= 0, ten x 0 2 Df/g) and f g 0 x 0 )= f 0 x 0 )gx 0 ) fx 0 )g 0 x 0 ) gx 0 )) 2. 2 Bolzano presented is example in 1834, but it was little noticed. Te 1872 example of Weierstrass is more well-known [2]. A translation of Weierstrass original paper [16] is presented by Edgar [8]. Weierstrass example is not very transparent because it depends on trigonometric series. Many more elementary constructions ave since been made. One suc will be presented in Example 9.9.

2. DIFFERENTIATION RULES 7-3 b) c) Proof. a) f + g)x 0 + ) f + g)x 0 )!0 fx 0 + )+gx 0 + ) fx 0 ) gx 0 ) = fx0 + ) fx 0 ) = + gx 0 + ) gx 0 ) = f 0 x 0 )+g 0 x 0 ) af)x 0 + ) af)x 0 ) fx 0 + ) fx 0 ) = a = af 0 x 0 )!0 fg)x 0 + ) fg)x 0 ) fx 0 + )gx 0 + ) fx 0 )gx 0 ) =!0 Now, slip a 0 into te numerator and factor te fraction. fx 0 + )gx 0 + ) fx 0 )gx 0 + )+fx 0 )gx 0 + ) fx 0 )gx 0 ) = fx0 + ) fx 0 ) = gx 0 + )+fx 0 ) gx 0 + ) gx 0 ) Finally, use te definition of te derivative and te continuity of f and g at x 0. = f 0 x 0 )gx 0 )+fx 0 )g 0 x 0 ) d) It will be proved tat if gx 0 ) 6= 0, ten 1/g) 0 x 0 )= g 0 x 0 )/gx 0 )) 2. Tis statement, combined wit c), yields d). 1 1 1/g)x 0 + ) 1/g)x 0 ) gx = 0 + ) gx 0 )!0 gx 0 ) gx 0 + ) 1 = gx 0 + )gx 0 ) = g0 x 0 ) gx 0 ) 2 Plug tis into c) to see 0 f x 0 )= f 1 0 x 0 ) g g = f 0 1 x 0 ) gx 0 ) + fx g 0 x 0 ) 0) gx 0 )) 2 = f 0 x 0 )gx 0 ) fx 0 )g 0 x 0 ) gx 0 )) 2. Combining Examples 7.1 and 7.2 wit Teorem 7.3, te following teorem is easy to prove. Corollary 7.4. A rational function is di erentiable at every point of its domain.

7-4 CHAPTER 7. DIFFERENTIATION Teorem 7.5 Cain Rule). If f and g are functions suc tat x 0 2 Df) and fx 0 ) 2 Dg), ten x 0 2 Dg f) and g f) 0 x 0 )=g 0 fx 0 )f 0 x 0 ). Proof. Let y 0 = fx 0 ). By assumption, tere is an open interval J containing fx 0 ) suc tat g is defined on J. SinceJ is open and x 0 2 Cf), tere is an open interval I containing x 0 suc tat fi) J. Define : J! R by 8 < gy) gy 0 ) g 0 y 0 ), y 6= y 0 y) = y y 0. : 0, y = y 0 Since y 0 2 Dg), we see gy) gy 0 ) y) = g 0 y 0 )=g 0 y 0 ) g 0 y 0 )=0=y 0 ), y!y 0 y!y 0 y y 0 so y 0 2 C). Now, x 0 2 Cf) and fx 0 )=y 0 2 C), so Teorem 6.15 implies x 0 2 C f). In particular 53) x!x 0 fx) =0. From te definition of f for x 2 I wit fx) 6= fx 0 ), we can solve for 54) g fx) g fx 0 )= fx)+g 0 fx 0 ))fx) fx 0 )). Notice tat 54) is also true wen fx) =fx 0 ). Divide bot sides of 54) by x x 0, and use 53) to obtain g fx) g fx 0 ) = fx)+g 0 fx 0 )) fx) fx 0) x!x 0 x x 0 x!x 0 x x 0 =0+g 0 fx 0 ))f 0 x 0 ) = g 0 fx 0 )f 0 x 0 ). Teorem 7.6. Suppose f :[a, b]! [c, d] is continuous and invertible. If x 0 2 Df) and f 0 x 0 ) 6= 0 for some x 0 2 a, b), ten fx 0 ) 2 Df 1 ) and f 1 0 fx 0 )) = 1/f 0 x 0 ). Proof. Let y 0 = fx 0 ) and suppose y n is any sequence in f[a, b]) \{y 0 } converging to y 0 and x n = f 1 y n ). By Teorem 6.24, f 1 is continuous, so Terefore, x 0 = f 1 y 0 )= n!1 f 1 y n )= n!1 x n. f 1 y n ) f 1 y 0 ) x n x 0 = n!1 y n y 0 n!1 fx n ) fx 0 ) = 1 f 0 x 0 ). Example 7.4. It follows easily from Teorem 7.3 tat fx) =x 3 is di erentiable everywere wit f 0 x) =3x 2. Define gx) = 3p x. Ten gx) =f 1 x). Suppose gy 0 )=x 0 for some y 0 2 R. According to Teorem 7.6, g 0 y 0 )= 1 f 0 x 0 ) = 1 1 3x 2 = 0 3gy 0 )) 2 = 1 3 3p y 0 ) 2 = 1 3y 2/3 0.

3. DERIVATIVES AND EXTREME POINTS 7-5 In te same manner as Example 7.4, te following corollary can be proved. Corollary 7.7. Suppose q 2 Q, fx) =x q and D is te domain of f. Ten f 0 x) =qx q 1 on te set D, wen q 1 D \{0}, wen q<1. 3. Derivatives and Extreme Points As is learned in calculus, te derivative is a powerful tool for determining te beavior of functions. Te following teorems form te basis for muc of di erential calculus. First, we state a few familiar definitions. Definition 7.8. Suppose f : D! R and x 0 2 D. f is said to ave a relative maximum at x 0 if tere is a >0 suc tat fx) apple fx 0 ) for all x 2 x 0, x 0 + )\D. f as a relative minimum at x 0 if f as a relative maximum at x 0.Iff as eiter a relative maximum or a relative minimum at x 0, ten it is said tat f as a relative extreme value at x 0. Te absolute maximum of f occurs at x 0 if fx 0 ) fx) for all x 2 D. Te definitions of absolute minimum and absolute extreme are analogous. Examples like fx) =x on 0, 1) sow tat even te nicest functions need not ave relative extrema. Corollary 6.23 sows tat if D is compact, ten any continuous function defined on D assumes bot an absolute maximum and an absolute minimum on D. Teorem 7.9. Suppose f :a, b)! R. Iff as a relative extreme value at x 0 and x 0 2 Df), ten f 0 x 0 )=0. Proof. Suppose fx 0 ) is a relative maximum value of f. Tenteremustbe a >0 suc tat fx) apple fx 0 )weneverx 2 x 0, x 0 + ). Since f 0 x 0 )exists, 55) x 2 x 0, x 0 ) =) fx) fx 0) x x 0 0 =) f 0 x 0 )= x"x0 fx) fx 0 ) x x 0 0 and 56) x 2 x 0,x 0 + ) =) fx) fx 0) x x 0 apple 0 =) f 0 x 0 )= x#x0 fx) fx 0 ) x x 0 apple 0. Combining 55) and 56) sows f 0 x 0 ) = 0. If fx 0 ) is a relative minimum value of f, apply te previous argument to f. Teorem 7.9 is, of course, te basis for muc of a beginning calculus course. If f :[a, b]! R, ten te extreme values of f occur at points of te set C = {x 2 a, b) :f 0 x) =0}[{x 2 [a, b] :f 0 x) does not exist}. Te elements of C are often called te critical points or critical numbers of f on [a, b]. To find te maximum and minimum values of f on [a, b], it su ces to find its maximum and minimum on te smaller set C, wic is finite in elementary calculus courses.

7-6 CHAPTER 7. DIFFERENTIATION 4. Di erentiable Functions Di erentiation becomes most useful wen a function as a derivative at eac point of an interval. Definition 7.10. Te function f is di erentiable on an open interval I if I Df). If f is di erentiable on its domain, ten it is said to be di erentiable. In tis case, te function f 0 is called te derivative of f. Te fundamental teorem about di erentiable functions is te Mean Value Teorem. Following is its simplest form. Lemma 7.11 Rolle s Teorem). If f : [a, b]! R is continuous on [a, b], di erentiable on a, b) and fa) =0=fb), ten tere is a c 2 a, b) suc tat f 0 c) =0. Proof. Since [a, b] is compact, Corollary 6.23 implies te existence of x m,x M 2 [a, b] suc tat fx m ) apple fx) apple fx M ) for all x 2 [a, b]. If fx m )=fx M ), ten f is constant on [a, b] and any c 2 a, b) satisfies te lemma. Oterwise, eiter fx m ) < 0 or fx M ) > 0. If fx m ) < 0, ten x m 2 a, b) and Teorem 7.9 implies f 0 x m ) = 0. If fx M ) > 0, ten x M 2 a, b) and Teorem 7.9 implies f 0 x M ) = 0. Rolle s Teorem is just a stepping-stone on te pat to te Mean Value Teorem. Two versions of te Mean Value Teorem follow. Te first is a version more general tan te one given in most calculus courses. Te second is te usual version. 4 Teorem 7.12 Caucy Mean Value Teorem). If f :[a, b]! R and g :[a, b]! R are bot continuous on [a, b] and di erentiable on a, b), ten tere is a c 2 a, b) suc tat g 0 c)fb) fa)) = f 0 c)gb) ga)). Proof. Let x) =gb) ga))fa) fx)) + gx) ga))fb) fa)). Because of te assumptions on f and g, is continuous on [a, b] and di erentiable on a, b) wita) =b) = 0. Teorem 7.11 yields a c 2 a, b) suc tat 0 c) = 0. Ten 0= 0 c) = gb) ga))f 0 c)+g 0 c)fb) fa)) =) g 0 c)fb) fa)) = f 0 c)gb) ga)). Corollary 7.13 Mean Value Teorem). If f :[a, b]! R is continuous on [a, b] and di erentiable on a, b), ten tere is a c 2 a, b) suc tat fb) fa) = f 0 c)b a). Proof. Let gx) = x in Teorem 7.12. Many of te standard teorems of beginning calculus are easy consequences of te Mean Value Teorem. For example, following are te usual teorems about monotonicity. 3 c Lee Larson LLarson@Louisville.edu) 4 Teorem 7.12 is also often called te Generalized Mean Value Teorem.

Section 4: Di erentiable Functions 7-7 Figure 2. Tis is a picture proof of Corollary 7.13. Teorem 7.14. Suppose f :a, b)! R is a di erentiable function. f is increasing on a, b) i f 0 x) 0 for all x 2 a, b). f is decreasing on a, b) i f 0 x) apple 0 for all x 2 a, b). Proof. Only te first assertion is proved because te proof of te second is pretty muc te same wit all te inequalities reversed. )) Ifx, y 2 a, b) witx 6= y, ten te assumption tat f is increasing gives fy) y fx) x 0 =) f 0 fy) x) = y!x y fx) x ) Let x, y 2 a, b) witx<y. According to Teorem 7.13, tere is a c 2 x, y) suc tat fy) fx) =f 0 c)y x) 0. Tis sows fx) apple fy), so f is increasing on a, b). Corollary 7.15. Let f :a, b)! R be a di erentiable function. f is constant i f 0 x) =0for all x 2 a, b). It follows from Teorem 7.2 tat every di erentiable function is continuous. But, it s not true tat a derivative need be continuous. Example 7.5. Let fx) = x 2 sin 1 x, x 6= 0 0, x =0. We claim f is di erentiable everywere, but f 0 is not continuous. To see tis, first note tat wen x 6= 0, te standard di erentiation formulas give tat f 0 x) =2x sin1/x) cos1/x). To calculate f 0 0), coose any 6= 0. Ten f) = 2 sin1/) apple 2 = and it easily follows from te definition of te derivative and te Squeeze Teorem Teorem 6.3) tat f 0 0) = 0. Let x n =1/2 n for n 2 N. Tenx n! 0 and f 0 x n )=2x n sin1/x n ) cos1/x n )= 1 for all n. Terefore, f 0 x n )! 1 6= 0=f 0 0), and f 0 is not continuous at 0. 0.

7-8 CHAPTER 7. DIFFERENTIATION But, derivatives do sare one useful property wit continuous functions; tey satisfy an intermediate value property. Compare te following teorem wit Corollary 6.26. Teorem 7.16 Darboux s Teorem). If f is di erentiable on an open set containing [a, b] and is between f 0 a) and f 0 b), ten tere is a c 2 [a, b] suc tat f 0 c) =. Proof. If f 0 a) =f 0 b), ten c = a satisfies te teorem. So, we may as well assume f 0 a) 6= f 0 b). Tere is no generality lost in assuming f 0 a) <f 0 b), for, oterwise, we just replace f wit g = f. Figure 3. Tis could be te function of Teorem 7.16. Let x) =fx) x so tat Df) =D) and 0 x) =f 0 x). In particular, tis implies 0 a) < 0 < 0 b). Because of tis, tere must be an ">0 small enoug so tat a + ") a) < 0 =) a + ") <a) " and b) b ") > 0 =) b ") <b). " See Figure 3.) In ligt of tese two inequalities and Teorem 6.23, tere must be a c 2 a, b) suc tat c) =glb {x) :x 2 [a, b]}. Now Teorem 7.9 gives 0= 0 c) =f 0 c), and te teorem follows. Here s an example sowing a possible use of Teorem 7.16. Example 7.6. Let fx) = Teorem 7.16 implies f is not a derivative. Since A more striking example is te following 0, x 6= 0 1, x =0. Example 7.7. Define sin 1 fx) = x, x 6= 0 and gx) = 1, x =0 fx) gx) = 0, x 6= 0 2, x =0 sin 1 x, x 6= 0 1, x =0. does not ave te intermediate value property, at least one of f or g is not a derivative. Actually, neiter is a derivative because fx) = g x).)

5. APPLICATIONS OF THE MEAN VALUE THEOREM 7-9 5. Applications of te Mean Value Teorem In te following sections, te standard notion of iger order derivatives is used. To make tis precise, suppose f is defined on an interval I. Te function f itself can be written f 0).Iff is di erentiable, ten f 0 is written f 1). Continuing inductively, if n 2!, f n) exists on I and x 0 2 Df n) ), ten f n+1) x 0 )=df n) x 0 )/dx. 5.1. Taylor s Teorem. Te motivation beind Taylor s teorem is te attempt to approximate a function f near a number a by a polynomial. Te polynomial of degree 0 wic does te best job is clearly p 0 x) =fa). Te best polynomial of degree 1 is te tangent line to te grap of te function p 1 x) =fa)+f 0 a)x a). Continuing in tis way, we approximate f near a by te polynomial p n of degree n suc tat f k) a) =p k) n a) for k =0, 1,...,n. A simple induction argument sows tat nx f k) a) 57) p n x) = x a) k. k! k=0 Tis is te well-known Taylor polynomial of f at a. Many students leave calculus wit te mistaken impression tat 57) is te important part of Taylor s teorem. But, te important part of Taylor s teorem is te fact tat in many cases it is possible to determine ow large n must be to acieve a desired accuracy in te approximation of f; i. e., te error term is te important part. Teorem 7.17 Taylor s Teorem). If f is a function suc tat f,f 0,...,f n) are continuous on [a, b] and f n+1) exists on a, b), ten tere is a c 2 a, b) suc tat nx f k) a) fb) = b a) k + f n+1) c) k! n + 1)! b a)n+1. k=0 Proof. Let te constant be defined by 58) nx f k) a) fb) = b a) k + b k! n + 1)! a)n+1 and define F x) =fb) k=0 nx k=0! f k) x) b x) k + b x)n+1. k! n + 1)! From 58) we see tat F a) = 0. Direct substitution in te definition of F sows tat F b) = 0. From te assumptions in te statement of te teorem, it is easy to see tat F is continuous on [a, b] and di erentiable on a, b). An application of Rolle s Teorem yields a c 2 a, b) suc tat f 0=F 0 n+1) c) c) = b c) n b c)n =) = f n+1) c), n! n! as desired. 5 c Lee Larson LLarson@Louisville.edu)

7-10 CHAPTER 7. DIFFERENTIATION 4 n = 4 n = 8 2 n = 20 2 4 6 8 y = cosx) -2-4 n = 2 n = 6 n = 10 Figure 4. Here are several of te Taylor polynomials for te function cosx) graped along wit cosx). Now, suppose f is defined on an open interval I wit a, x 2 I. If f is n +1 times di erentiable on I, ten Teorem 7.17 implies tere is a c between a and x suc tat fx) =p n x)+r f n, x, a), were R f n, x, a) = f n+1) c) n+1)! x a) n+1 is te error in te approximation. 6 Example 7.8. Let fx) =cos x. Suppose we want to approximate f2) to 5 decimal places of accuracy. Since it s an easy point to work wit, we ll coose a = 0. Ten, for some c 2 0, 2), 59) R f n, 2, 0) = f n+1) c) 2 n+1 apple 2n+1 n + 1)! n + 1)!. A bit of experimentation wit a calculator sows tat n = 12 is te smallest n suc tat te rigt-and side of 59) is less tan 5 10 6. After doing some aritmetic, it follows tat p 12 2) = 1 2 2 2! + 24 4! 2 6 6! + 28 8! is a 5 decimal place approximation to cos2). 2 10 10! + 212 12! = 27809 66825 0.41614. But, tings don t always work out te way we migt like. Consider te following example. Example 7.9. Suppose fx) = e 1/x2, x 6= 0 0, x =0. 6 Tere are several di erent formulas for te error. Te one given ere is sometimes called te Lagrange form of te remainder. In Example 8.4 a form of te remainder using integration instead of di erentiation is derived.

5. APPLICATIONS OF THE MEAN VALUE THEOREM 7-11 Figure 5 below as a grap of tis function. In Example 7.11 below it is sown tat f is di erentiable to all orders everywere and f n) 0) = 0 for all n 0. Wit tis function te Taylor polynomial centered at 0 gives a useless approximation. 5.2. L Hospital s Rules and Indeterminate Forms. According to Teorem 6.4, fx) x!a gx) = x!a fx) x!a gx) wenever x!a fx) and x!a gx) bot exist and x!a gx) 6= 0. But, it is easy to find examples were bot x!a fx) = 0 and x!a gx) = 0 and x!a fx)/gx) exists, as well as similar examples were x!a fx)/gx) fails to exist. Because of tis, suc a it problem is said to be in te indeterminate form 0/0. Te following teorem allows us to determine many suc its. Teorem 7.18 Easy L Hospital s Rule). Suppose f and g are eac continuous on [a, b], di erentiable on a, b) and fb) =gb) =0. If g 0 x) 6= 0on a, b) and x"b f 0 x)/g 0 x) =L, were L could be infinite, ten x"b fx)/gx) =L. Proof. Let x 2 [a, b), so f and g are continuous on [x, b] and di erentiable on x, b). Caucy s Mean Value Teorem, Teorem 7.12, implies tere is a cx) 2 x, b) suc tat f 0 cx))gx) =g 0 cx))fx) =) fx) gx) = f 0 cx)) g 0 cx)). Since x<cx) <b, it follows tat x"b cx) =b. Tis sows tat f 0 x) L = x"b g 0 x) = f 0 cx)) x"b g 0 cx)) = fx) x"b gx). Several tings sould be noted about tis proof. First, tere is noting special about te left-and it used in te statement of te teorem. It could just as easily be written in terms of te rigt-and it. Second, if x!a fx)/gx) is not of te indeterminate form 0/0, ten applying L Hospital s rule will usually give a wrong answer. To see tis, consider x 1 =06= 1= x!0 x +1 x!0 1. Anoter case were te indeterminate form 0/0 occurs is in te it at infinity. Tat L Hôpital s rule works in tis case can easily be deduced from Teorem 7.18. Corollary 7.19. Suppose f and g are di erentiable on a, 1) and fx) = gx) =0. x!1 x!1 If g 0 x) 6= 0on a, 1) and x!1 f 0 x)/g 0 x) =L, were L could be infinite, ten x!1 fx)/gx) =L. Ten Proof. Tere is no generality lost by assuming a>0. Let f1/x), x 2 0, 1/a] g1/x), x 2 0, 1/a] F x) = and Gx) = 0, x =0 0, x =0 F x) = fx) =0= gx) = Gx), x#0 x!1 x!1 x#0.

7-12 CHAPTER 7. DIFFERENTIATION so bot F and G are continuous at 0. It follows tat bot F and G are continuous on [0, 1/a] and di erentiable on 0, 1/a) witg 0 x) = g 0 x)/x 2 6= 0 on 0, 1/a) and x#0 F 0 x)/g 0 x) = x!1 f 0 x)/g 0 x) =L. Te rest follows from Teorem 7.18. Te oter standard indeterminate form arises wen fx) =1 = gx). x!1 x!1 Tis is called an 1/1 indeterminate form. It is often andled by te following teorem. Teorem 7.20 Hard L Hospital s Rule). Suppose tat f and g are di erentiable on a, 1) and g 0 x) 6= 0on a, 1). If ten f 0 x) fx) = gx) =1 and x!1 x!1 x!1 g 0 = L 2 R [ { 1, 1}, x) fx) x!1 gx) = L. Proof. First, suppose L 2 R and let ">0. Coose a 1 >alarge enoug so tat f 0 x) g 0 L <", 8x >a 1. x) Since x!1 fx) =1 = x!1 gx), we can assume tere is an a 2 >a 1 suc tat bot fx) > 0 and gx) > 0wenx>a 2. Finally, coose a 3 >a 2 suc tat wenever x>a 3,tenfx) >fa 2 ) and gx) >ga 2 ). Let x>a 3 and apply Caucy s Mean Value Teorem, Teorem 7.12, to f and g on [a 2,x]tofindacx) 2 a 2,x) suc tat 60) If f 0 cx)) g 0 cx)) = fx) fa fx) 2) gx) ga 2 ) = gx) x) = 1 ga 2) gx), 1 fa 2) fx) 1 1 fa 2) fx) ga 2) gx) ten 60) implies fx) gx) = f 0 cx)) g 0 cx)) x). Since x!1 x) = 1, tere is an a 4 >a 3 suc tat wenever x>a 4,ten x) 1 <".Ifx>a 4,ten fx) gx) L = f 0 cx)) g 0 cx)) x) L. = f 0 cx)) g 0 x) Lx)+Lx) L cx)) apple f 0 cx)) g 0 L x) + L x) 1 cx)) <"1 + ")+ L " =1+ L + ")"

5. APPLICATIONS OF THE MEAN VALUE THEOREM 7-13 1!" 3 2 1 1 2 3 Figure 5. Tis is a plot of fx) =exp flattens out near te origin. 1/x 2 ). Notice ow te grap can be made arbitrarily small troug a proper coice of ". Terefore fx)/gx) =L. x!1 Te case wen L = 1 is done similarly by first coosing a B>0 and adjusting 60) so tat f 0 x)/g 0 x) >Bwen x>a 1. A similar adjustment is necessary wen L = 1. Tere is a companion corollary to Teorem 7.20 wic is proved in te same way as Corollary 7.19. Corollary 7.21. Suppose tat f and g are continuous on [a, b] and di erentiable on a, b) wit g 0 x) 6= 0on a, b). If ten f 0 x) fx) =gx) =1 and x#a x#a x#a g 0 = L 2 R [ { 1, 1}, x) fx) x#a gx) = L. Example 7.10. If >0, ten x!1 ln x/x is of te indeterminate form 1/1. Taking derivatives of te numerator and denominator yields 1/x x!1 x 1 = 1 x!1 x =0. Teorem 7.20 now implies x!1 ln x/x = 0, and terefore ln x increases more slowly tan any positive power of x. Example 7.11. Let f be as in Example 7.9. See Figure 5.) It is clear f n) x) exists wenever n 2! and x 6= 0. We claim f n) 0) = 0. To see tis, we first prove tat 61) e 1/x2 x!0 x n =0, 8n 2 Z.

7-14 CHAPTER 7. DIFFERENTIATION Wen n apple 0, 61) is obvious. So, suppose 61) is true wenever m apple n for some n 2!. Making te substitution u =1/x, wesee 62) Since e 1/x 2 x#0 x n+1 = u n+1. u!1 e u2 e 1/x 2 x#0 n + 1)u n n + 1)u n 1 = = n +1 u!1 2ue u2 u!1 2e u2 2 x n 1 =0 by te inductive ypotesis, Teorem 7.20 gives 62) in te case of te rigt-and it. Te left-and it is andled similarly. Finally, 61) follows by induction. Wen x 6= 0, a bit of experimentation can convince te reader tat f n) x) is of te form p n 1/x)e 1/x2,werep n is a polynomial. Induction and repeated applications of 61) establis tat f n) 0) = 0 for n 2!. 7.1. If fx) = ten sow Df) ={0} and find f 0 0). 6. Exercises x 2, x 2 Q 0, oterwise, 7.2. Let f be a function defined on some neigborood of x = a wit fa) = 0. Prove f 0 a) = 0 if and only if a 2 D f ). 7.3. If f is defined on an open set containing x 0,tesymmetric derivative of f at x 0 is defined as f s fx 0 + ) fx 0 ) x 0 )=.!0 2 Prove tat if f 0 x) exists,tensodoesf s x). Is te converse true? 7.4. Let G be an open set and f 2 DG). If tere is an a 2 G suc tat x!a f 0 x) exists,ten x!a f 0 x) =f 0 a). 7.5. Suppose f is continuous on [a, b] and f 00 exists on a, b). If tere is an x 0 2 a, b) suc tat te line segment between a, fa)) and b, fb)) contains te point x 0,fx 0 )), ten tere is a c 2 a, b) suc tat f 00 c) = 0. 7.6. If = {f : f = F 0 for some F : R! R}, ten is closed under addition and scalar multiplication. Tis sows te di erentiable functions form a vector space.) 7.7. If and f 1 x) = f 2 x) = ten at least one of f 1 and f 2 is not in. 1/2, x =0 sin1/x), x 6= 0 1/2, x =0 sin 1/x), x 6= 0,

6. EXERCISES 7-15 7.8. Prove or give a counter example: If f is continuous on R and di erentiable on R \{0} wit x!0 f 0 x) =L, tenf is di erentiable on R. 7.9. Suppose f is di erentiable everywere and fx+y) =fx)fy) for all x, y 2 R. Sow tat f 0 x) =f 0 0)fx) and determine te value of f 0 0). 7.10. If I is an open interval, f is di erentiable on I and a 2 I, tentereisa sequence a n 2 I \{a} suc tat a n! a and f 0 a n )! f 0 a). 7.11. Use te definition of te derivative to find d p x. dx 7.12. Let f be continuous on [0, 1) and di erentiable on 0, 1). If f0) = 0 and f 0 x) < fx) for all x>0, ten fx) = 0 for all x 0. 7.13. Suppose f : R! R is suc tat f 0 is continuous on [a, b]. If tere is a c 2 a, b) suc tat f 0 c) = 0 and f 00 c) > 0, ten f as a local minimum at c. 7.14. Prove or give a counter example: If f is continuous on R and di erentiable on R \{0} wit x!0 f 0 x) =L, tenf is di erentiable on R. 7.15. Let f be continuous on [a, b] and di erentiable on a, b). If fa) = and f 0 x) < for all x 2 a, b), ten calculate a bound for fb). 7.16. Suppose tat f :a, b)! R is di erentiable and f 0 is bounded. If x n is a sequence from a, b) suc tat x n! a, tenfx n ) converges. 7.17. Let G be an open set and f 2 DG). If tere is an a 2 G suc tat x!a f 0 x) exists,ten x!a f 0 x) =f 0 a). 7.18. Prove or give a counter example: If f 2 Da, b)) suc tat f 0 is bounded, ten tere is an F 2 C[a, b]) suc tat f = F on a, b). 7.19. Sow tat fx) =x 3 +2x + 1 is invertible on R and, if g = f 1,tenfind g 0 1). 7.20. Suppose tat I is an open interval and tat f 00 x) 0 for all x 2 I. Ifa 2 I, ten sow tat te part of te grap of f on I is never below te tangent line to te grap at a, fa)). 7.21. Suppose f is continuous on [a, b] and f 00 exists on a, b). If tere is an x 0 2 a, b) suc tat te line segment between a, fa)) and b, fb)) contains te point x 0,fx 0 )), ten tere is a c 2 a, b) suc tat f 00 c) = 0. 7.22. Let f be defined on a neigborood of x. a) If f 00 x) exists,ten fx ) 2fx)+fx + )!0 2 = f 00 x). b) Find a function f were tis it exists, but f 00 x) does not exist.

7-16 CHAPTER 7. DIFFERENTIATION 7.23. If f : R! R is di erentiable everywere and is even, ten f 0 is odd. If f : R! R is di erentiable everywere and is odd, ten f 0 is even. 7 7.24. Prove tat wen x apple1. sin x x x 3 6 + x5 120 < 1 5040 7.25. Te exponential function e x is not a polynomial. 7 Afunctiong is even if g x) =gx) foreveryx and it is odd if g x) = gx) forevery x. Tetermsareevenandoddbecausetisisowgx) =x n beaves wen n is an even or odd integer, respectively.