Differentiation: Our First View

Transcription

1 Capter 6 Differentiation: Our First View We are now ready to reàect on a particular application of its of functions namely, te derivative of a function. Tis view will focus on te derivative of real-valued functions on subsets of U 1. Looking at derivatives of functions in U k requires a different enoug perspective to necessitate separate treatment tis is done wit Capter 9 of our text. Except for te last section, our discussion is restricted to aspects of differential calculus of one variable. You sould ave seen most of te results in your rst exposure to calculus MAT21A on tis campus. However, some of te results proved in tis capter were only stated wen you rst saw tem and some of te results are more general tan te earlier versions tat you migt ave seen. Te good news is tat te presentation ere isn t dependent on previous exposure to te topic on te oter and, reàecting back on prior work tat you did wit te derivative can enance your understanding and foster a deeper level of appreciation. 6.1 Te Derivative De nition Areal-valued function f on a subset P of U is differentiable at a point? + P if and only if f is de ned in an open interval containing? and *? f * f? *? (6.1) exists. Te value of te it is denoted by f )?. Te function is said to be differentiable on P if and only if it is differentiable at eac? + P. 229

2 230 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW Remark For a function f and a xed point?, te expression M * f * f? *? is one form of wat is often referred to as a difference quotient. Sometimes it is written as f * were te Greek letter is being offered as a reminder tat difference starts wit a d. It is te latter form tat motivates use of te notation df for te rst derivative d* of f as a function of *. Oter commonly used notations are D * and D 1 tese only become useful alternatives wen we explore functions over several real variables. Tere is an alternative form of (6.1) tat is often more useful in terms of computation and formatting of proofs. Namely, if we let *?, (6.1) can be written as f? f?. (6.2) Remark Wit te form given in (6.2), te difference quotient can be abbreviated as f. De nition A real-valued function f on a subset P of U is rigt-and differentiable at a point? + P if and only if f is de ned in a alf open interval in te form [?? = for some = 0 and te one-sided derivative from te rigt, denoted by D f?, f? f? Å exists te function f is left-and differentiable at a point? + P if and only if fisde ned in a alf open interval in te form? =?] for some = 0 and te one-sided derivative from te left, denoted by D f?, f? f? exists.

3 6.1. THE DERIVATIVE 231 De nition Areal-valued function f is differentiable on a closed interval [a b] if and only if f is differentiable in a b, rigt-and differentiable at x a and left-and differentiable at x b. Example Use te de nition to prove tat f x x 2 is differentiable at x 1 x 2. Note tat f is de ned in te open interval 1 3 wic contains * 2. Furtermore, t u * 2 4 f * f 2 * * 2 *2 * 2 *2 * 2 *2 * 2 Hence, f is differentiable at * 2 and f ) 2 3. * Example Use te de nition to prove tat gx x 2 is not differentiable at x 2. Since dom g U, te function g is de ned in any open interval tat contains x 2. Hence, g is differentiable at x 2 if and only if exists. Let M Å Å g 2 g 2 for / 0. Note tat 1 and 1 Tus, M 0 / M 0 from wic we conclude tat M does not exist. Terefore, g is not differentiable at x 2. Remark Because te function g given in Example is left-and differentiable at x 2 and rigt-and differentiable at x 2, we ave tat g is differentiable in eac of * 2] and [2 *. Example Discuss te differentiability of eac of te following at x 0.! x sin 1, for x / 0 1. G x x! 0, for x 0

4 232 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW! x 3 sin 1, for x / 0 2. F x x! 0, for x 0 First of all, notice tat, toug te directions did not specify appeal to te de - nition, making use of te de nition is te only viable option because of te way te function is de ned. Discussing te differentiability of functions tat are de ned in pieces requires consideration of te pieces. On segments were te functions are realized as simple algebraic combinations of nice functions, te functions can be declared as differentiable based on noting te appropriate nice properties. If te function is de ned one way at a point and a different way to te left and/or rigt, ten appeal to te difference quotient is mandated. For (1), we note tat G is de ned for all reals, consequently, it is de ned in every interval tat contains 0. Tus, G is differentiable at 0 if and only if G 0 G 0 sin 1 0 t sin 1 u exists. For / 0, letm sin 1. For eac n + M, letp 2 n H 2n 1. Now, p n * n1 converges to 0 as n approaces in nity but Mp n * n1 j 1 n1k * n1 diverges. From te Sequences Caracterization for Limits of Functions (Teorem ), we conclude tat Mdoes not exist. Terefore, G is not differentiable at x 0. Te function F given in (2) is also de ned in every interval tat contains 0. Hence, F is differentiable at 0 if and only if 1 F 0 F 0 3 sin 0 t 2 sin 1 u exists. Now we know tat, for / 0, n sin 1 n n 1 and 2 0 it follows from a simple t modi cation of wat was proved in Exercise #6 of Problem Set D tat 2 sin 1 u 0. Terefore, F is differentiable at x 0 and F ) 0 0.

5 6.1. THE DERIVATIVE 233 Excursion In te space provided, sketc graps of G and F on two different representations of te Cartesian coordinate system in intervals containing 0. ***For te sketc of G using te curves y x and y x as guides to stay witin sould ave elped give a nice sense for te appearance of te grap te guiding (or bounding) curves for F are y x 3 and y x 3.*** Remark Te two problems done in te last example illustrate wat is sometimesreferredtoasasmootingeffect.inourtext,itissowntat! K x! x 2 sin 1 x, for x / 0 0, for x 0 is also differentiable at x 0. Te function! sin 1, for x / 0 L x x! 0, for x 0 is not continuous at x 0 wit te discontinuity being of te second kind. Te niceness of te function is improving wit te increase in exponent of te smooting function x n.

6 234 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW In te space provided, sketc graps of K and L on two different representations of te Cartesian coordinate system in intervals containing 0. Te function L is not continuous at x 0 wile G is continuous at x 0 but not differentiable tere. Now we know tat K and F are bot differentiable at x 0 in fact, it can be sown tat F can be de ned to be differentiable at x 0 wile at most continuity at x 0 can be gained for te derivative of K at x 0. Our rst teorem in tis section will justify te claim tat being differentiable is a stronger condition tan being continuous tis offers one sense in wic we claim tat F is a nicer function in intervals containing 0 tan K is tere. Excursion Fill in wat is missing in order to complete te following proof tat te function f x T x is differentiable in U 0 *. Proof. Let f x T x and suppose tat a + U.Tenfis r a s in te segment 2 2a tat contains x a. Hence, f is differentiable at x aif and only if 1 2 3

7 6.1. THE DERIVATIVE 235 exists. Now 3 bt T cbt T c a a a a bt a T a c Consequently, f is differentiable at x a and f ) a was arbitrary, we conclude tat 1x 7 v bx + U F f x T x c " f ) x 1 2 T x. Since a + U w. ***Acceptable responses are: (1) de ned, (2) d f a f a b 1ce,(3) dbt T cb a a 1 ce a a, (4) bt a T a c, (5) bt a T a c 1, (6) b 2 T a c 1 1,and(7) 2 T a.*** Te next result tells us tat differentiability of a function at a point is a stronger condition tan continuity at te point. Teorem If a function is differentiable at? + U, ten it is continuous tere. Excursion Make use of te following observations and your understanding of properties of its of functions to prove Teorem Some observations to ponder: Te function f being differentiable at? assures te existence of a = tat f is de ned in te segment? =? = 0 suc

8 236 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW GivenafunctionGde ned in a segment a b, we know tat G is continuous at any point p + a b if and only if G x G p wic is equivalent d e xp to aving G x G p 0. xp Space for scratc work. Proof. ***Once you tink of te possibility of writing d G x G p e as d G x G p x p 1 e x p for x / p te it of te product teorem does te rest of te work.*** Remark We ave already seen two examples of functions tat are continuous at a point witout being differentiable at te point namely, g x x 2 at x 2 and, for x 0,! x sin 1, for x / 0 G x x.! 0, for x 0

9 6.1. THE DERIVATIVE 237 To see tat G is continuous at x 0, note tat n sin 1 x n n 1 for x / 0 and x 0 x0 t t implies tat x sin 1 uu 0. Alternatively, for 0, let= ten x0 x 0 x 0 =implies tat n x sin 1 n nnn x 0 x n sin 1 x n n x =. t Hence, x sin 1 u 0 G 0. Eiter example is suf cient to justify tat te x0 x converse of Teorem is not true. Because te derivative is de ned as te it of te difference quotient, it sould come as no surprise tat we ave a set of properties involving te derivatives of functions tat follow directly and simply from te de nition and application of our it teorems. Te set of basic properties is all tat is needed in order to make a transition from nding derivatives using te de nition to nding derivatives using simple algebraic manipulations. Teorem (Properties of Derivatives) c ) x 0. (a) If c is a constant function, ten (b) If f is differentiable at? and k is a constant, ten x kfx is differentiable at? and )? kf )?. (c) If f and g are differentiable at?,tenfx f gx is differentiable at? and F )? f )? g )?. (d) If u and ) are differentiable at?, tengx u)x is differentiable at? and G )? u? ) )? )?u )?. (e) If f is differentiable at? and f? / 0, tenhx [ f x] 1 is differentiable at? and H )? f )? [ f? ] 2. (f) If px x n for n an integer, p is differentiable werever it is de ned and p ) x nx n1.

10 238 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW Te proofs of (a) and (b) are about as easy as it gets wile te straigtforward proofs of (c) and (f) are left as exercises. Completing te next two excursions will provide proofs for (d) and (e). Excursion Fill is wat is missing in order to complete te following proof tat, if u and ) are differentiable at?,tengx u)x is differentiable at? and G )? u?) )? )?u )?. Proof. Suppose u, ), and G are as described in te ypotesis. Because u and ) are differentiable at?, tey are de ned in a segment containing?. Hence, G x u x ) x is de ned in a segment containing?. Hence, G is differentiable at? if and only if exists. Note tat )? [u? u? ] u? [)? )? ] v t u t uw u? u? )?)? )? u?. Since ) is differentiable at? it is continuous tere tus, )?. 3 Now te differentiability of u and ) wit te it of te product and it of te sum teorems yield tat 1 Terefore, G is differentiable at?. 4. ***Acceptable responses are: (1) d G? G? 1e, (2) d u? )? u? )? 1e,(3))?,and(4))?u )? u? ) )?.***

11 6.1. THE DERIVATIVE 239 Excursion Fill is wat is missing in order to complete te following proof tat, if f is differentiable at? and f? / 0, tenhx [ f x] 1 is differentiable at? and H )? f )? [ f?] 2. Proof. Suppose tat te function f is differentiable at? and f? / 0. From Teorem , f is at?.hence. f x.since x? 1 2 f? 0, it follows tat tere exists = 0 suc tat 2 3 implies tat f x f? f?. Te (oter) triangular inequality, yields 2 f? tat, for, f? f x from wic 2 3 f? we conclude tat f x in te segment. Terefore, te 2 4 function H x [ f x] 1 is de ned in a segment tat contains? and it is differentiable at? if and only if exists. Now simple algebraic def H? H? manipulations yield tat H? H? vt f? f? ut uw 1. f? f? From te of f at?, it follows tat f?. 5 6 In view of te differentiability of f and te it of te product teorem, we ave tat H? H? 7. ***Acceptable responses are: (1) continuous, (2) f?, (3) x? =,(4)? =? =, (5) continuity, (6) f?,and(7) b f )? c [ f? ] 2.*** Te next result offers a different way to tink of te difference quotient.

12 240 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW Teorem (Fundamental Lemma of Differentiation) Suppose tat f is differentiable at x 0. Ten tere exists a de ned on an open interval containing 0 for 0 and is continuous at 0. f x 0 f x 0 [ f ) x (6.3) Before looking at te proof take a few moments to reàect on wat you can say about for 0. f x 0 f x 0 f ) x 0 Proof. Suppose tat = 0 is suc tat f is de ned in x x 0 =and let 1 d! f x0 f x 0 e f ) x 0,if0 0, if 0 Because f is differentiable at x 0, it follows from te it of te sum teorem , we conclude is continuous at 0. Finally, 1 d f x0 f x 0 e f ) x 0 for f x 0 f x 0 yields (6.3). Remark If f is differentiable at x 0, ten f x 0 s f x 0 f ) x 0 for very small i.e., te function near to x 0 is approximated by a linear function wose slope is f ) x 0. Next, we will use te Fundamental Lemma of Differentiation to obtain te derivative of te composition of differentiable functions.

13 6.1. THE DERIVATIVE 241 Teorem (Cain Rule) Suppose tat g and u are functions on U and tat f x gux. If u is differentiable at x 0 and g is differentiable at ux 0,ten f is differentiable at x 0 and f ) x 0 g ) ux 0 u ) x 0 *************************** Before reviewing te offered proof, look at te following and tink about wat prompted te indicated rearrangement Wat sould be put in te boxes to enable us to relate to te given information? We want to consider f x 0 f x 0 g ux 0 g ux 0 ux 0 g ux 0 % #g & $ *************************** Proof. Let f f x 0 f x 0, u ux 0 ux 0 and u 0 u x 0. Ten f gux 0 gux 0 gu 0 u gu 0. Because u is continuous at x 0,weknowtat u 0. By te Fundamental Lemma of Differentiation, tere exists a 0, tat is continuous at 0 and is suc tat f [g ) u Hence, f t [g ) u u u g ) u 0 u ) x 0 from te it of te sum and it of te product teorems.

14 242 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW Formulas for Differentiation As a consequence of te results in tis section, we can justify te differentiation of all polynomials and rational functions. From Excursion , we know tat te formula given in te Properties of Derivatives Teorem (f) is valid for n 1 2. In fact, it is valid for all nonzero real numbers. Prior to te Cain Rule, te only r way to nd te derivative of f x x 3 b 3x 2 7 c s 12 8, oter tan appeal to tede nition, was to expand te expression and apply te Properties of Derivatives Teorem, parts (a), (b), (c) and (f) in view of te Cain Rule and te Properties of Derivatives Teorem, we ave t r s u 12 7 v r s w 11 f ) x 8 x 3 3x 2 7 3x 2 72x 3x 2 7. Wat we don t ave yet is te derivatives of functions tat are not realized as algebraic combinations of polynomials most notably tis includes te trigonometric functions, te inverse trig functions, : x for any xed positive real number :, and te logaritm functions. For any x + U, we know tat sin x sin x sin cos x cos sin x sin x v t u t uw sin cos 1 cos x sin x and cos x cos x cos cos x sin sin x cos x v t u t uw cos 1 sin cos x sin x. Consequently, in view of te it of te sum and it of te product teorems, nding t te derivatives u of tte sine anducosine functions depends on te existence of sin cos 1 and. Using elementary geometry and trigonometry, it can be sown tat te values of tese its are 1 and 0, respectively. An outline for te proofs of tese two its, wic is a review of wat is sown in an elementary calculus course, is given as an exercise. Te formulas for te derivatives

15 6.1. THE DERIVATIVE 243 of te oter trigonometric functions follow as simple applications of te Properties of Derivatives. Recall tat e 1? 1? and y ln x % x e y. Wit tese in addition? 0 to basic properties of logaritms, for x apositivereal, ln x ln x v 1 ln ln Keepinginmindtatx is a constant, it follows tat ln x ln x t 1 uw x u 1 t 1 x t ln 1 u x 1x x t 1 x ln 1 u x x t Because 1 u x e as 0andlne 1, te same argument tat x was used for te proof of Teorem allows us to conclude tat ln x ln x 1 x. Formulas for te derivatives of te inverse trigonometric functions and : x,for any xed positive real number :, will follow from te teorem on te derivative of te inverses of a function tat is proved at te end of tis capter Revisiting A Geometric Interpretation for te Derivative Completing te following gure sould serve as a nice reminder of one of te common interpretations and applications of te derivative of a function at te point. On te x-axis, label te x-coordinate of te common point of intersection of te curve, f x, and te tree indicated lines as c.

16 244 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW Corresponding to eac line ( 1, ( 2,and( 3,ontex-axis label te x-coordinate of te common point of intersection of te curve, f x, wit te line as c 1, c 2,andc 3 in ascending order. Note tat 1, 2 and 3 are negative in te set-up tat is sown. Eac of te lines ( 1, ( 2,and( 3 are called secant lines. Find te slopes m 1, m 2,andm 3, respectively, of te tree lines. Excursion Using terminology associated wit te derivative, give a brief description tat applies to eac of te slopes m j for j Excursion Give a concise well-written description of te geometric interpretation for te derivative of f at x c, if it exists. 6.2 Te Derivative and Function Beavior Te difference quotient is te ratio of te cange in function values to te cange in arguments. Consequently, it sould come as no surprise tat te derivative provides information related to monotonicity of functions.

17 6.2. THE DERIVATIVE AND FUNCTION BEHAVIOR 245 In te following, continuity on an interval I [a b] is equivalent to aving continuity on a b, rigt-and continuity at x a and left-and continuity at x b. For rigt-and continuity at x a f a f a, wile left-and continuity at x b requires tat f b f b. De nition A real valued function f on a metric space X d X as a local maximum at a point p + X if and only if 2= 0 d 1qq + N = p " f q n f p e te function as a local minimum at a point p + X if and only if 2= 0 d 1qq + N = p " f p n f q e. De nition A real valued function f on a metric space X d X as a (global) maximum at a point p + X if and only if d 1xx + X " f x n f p e te function as a (global) minimum at a point p + X if and only if d 1xq + X " f p n f x e Teorem (Interior Extrema Teorem) Suppose tat f is a function tat is de nedonanintervali [a b]. If f as a local maximum or local minimum at a point x 0 + a b and f is differentiable at x 0,ten f ) x 0 0. Space for scratc work or motivational picture. Proof. Suppose tat te function f is de ned in te interval I [a b], as a local maximum at x 0 + a b, and is differentiable at x 0. Because f as a local maximum at x 0, tere exists a positive real number = suc tat x 0 = x 0 = t a b and 1t d t + x 0 = x 0 = " f t n f x 0 e. Tus, for t + x 0 = x 0, f t f x 0 t x 0 o 0 (6.4)

18 246 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW wile t + x 0 x 0 = implies tat f t f x 0 t x 0 n 0. (6.5) f t f x 0 Because f is differential at x 0, exists and is equal to f ) x 0. tx 0 t x 0 From (6.4) and (6.5), we know tat f ) x 0 o 0and f ) x 0 n 0, respectively. Te Tricotomy Law yields tat f ) x 0 0. Te Generalized Mean-Value Teorem tat follows te next two results contains Rolle s Teorem and te Mean-Value Teorem as special cases. We offer te results in tis order because it is easier to appreciate te generalized result after reàecting upon te geometric perspective tat is offered by te two lemmas. Lemma (Rolle s Teorem) Suppose tat f is a function tat is continuous on te interval I [a b] and differentiable on te segment I i a b. If fa f b, tentereisanumberx 0 + I i suc tat f ) x 0 0. Space for scratc work or building intuition via a typical picture. Proof. If f is constant, we are done. Tus, we assume tat f is not constant in te interval a b. Sincef is continuous on I, by te Extreme Value Teorem, tere exists points? 0 and? 1 in I suc tat f? 0 n f x n f? 1 for all x + I. Because f is not constant, at least one of x + I : f x f a and x + I : f x f a is nonempty. If x + I : f x f a a b, ten f? 0 f a f b and, by te Interior Extrema Teorem,? 1 + a b is suc tat f )? 1 0. If x + I : f x f a a b, ten f? 1 f a f b,? 0 + a b, and te Interior Extrema Teorem implies tat f )? 0 0. Finally, if x + I : f x f a / a b and x + I : f x f a / a b, ten bot? 0 and? 1 are in a b and f )? 0 f )? 1 0. Lemma (Mean-Value Teorem) Suppose tat f is a function tat is continuous on te interval I [a b] and differentiable on te segment I i a b. Ten

19 6.2. THE DERIVATIVE AND FUNCTION BEHAVIOR 247 tere exists a number G + I i suc tat f ) G f b f a b a. Excursion Use te space provided to complete te proof of te Mean-Value Teorem. Proof. Consider te function F de ned by Fx f x f b f a b a x a f a as a candidate for application of Rolle s Teorem. Teorem (Generalized Mean-Value Teorem) Suppose tat f and F are functions tat are continuous on te interval I [a b] and differentiable on te segment I i.iff ) x / 0 on I i, ten (a) Fb Fa / 0, and t (b) 2G G + I i F f b f a Fb Fa f ) u G F ). G Excursion Fill in te indicated steps in order to complete te proof of te Generalized Mean-Value Teorem.

20 248 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW Proof. To complete a proof of (a), apply te Mean-Value Teorem to F. For (b), for x + I,de ne te function by f b f a x f x f a [Fx Fa]. Fb Fa It follows directly tat a b 0. Teorem (Monotonicity Test) Suppose tat a function f is differentiable in te segment a b. (a) If f ) x o 0 for all x + a b, ten f is monotonically increasing in a b. (b) If f ) x 0 for all x + a b, ten f is constant in a b. (c) If f ) x n 0 for all x + a b, ten f is monotonically decreasing in a b. Excursion Fill in wat is missing in order to complete te following proof of te Monotonicity Test. Proof. Suppose tat f is differentiable in te segment a b and x 1 x 2 + a b are suc tat x 1 x 2. Ten f is continuous in [x 1 x 2 ] and 1

21 6.2. THE DERIVATIVE AND FUNCTION BEHAVIOR 249 in x 1 x 2.Fromte tat 2, tere exists G + x 1 x 2 suc f ) G f x 1 f x 2. x 1 x 2 If f ) x o 0 for all x + a b, ten f ) G o 0. Since x 1 x 2 0, it follows tat i.e., f x 1 n f x 2.Sincex 1 and x 2 3 were arbitrary, we ave tat 1x 1 1x 2 x 1 x 2 + a b F Hence, f is 5 If f ) x 0 for all x + a b, ten 4 in a b. " f x 1 n f x Finally, if f ) x n 0 for all x + a b, 7 ***Acceptable responses are: (1) differentiable, (2) Mean-Value Teorem, (3) f x 1 f x 2 n 0, (4) x 1 x 2, (5) monotonically increasing, (6) f x 1 f x 2 0 i.e., f x 1 f x 2.Sincex 1 and x 2 were arbitrary, we ave tat f is constant trougout a b., (7) ten f ) G n 0andx 1 x 2 0 implies tat f x 1 f x 2 o 0 i.e., f x 1 o f x 2. Because x 1 and x 2 were arbitrary we conclude tat f is monotonically decreasing in a b.*** Example Discuss te monotonicity of f x 2x 3 3x 2 36x 7. For x + U, f ) x 6x 2 6x 36 6 x 3x 2. Since f ) is positive in * 3 and 2 *, f is monotonically increasing tere, wile f ) negative in 3 2 yields tat f is monotonically decreasing in tat segment.

22 250 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW Remark Actually, in eac of open intervals * 3, 2 *, and 3 2 tat were found in Example , we ave strict monotonicity i.e., for x 1 x 2 + * 3 or x 1 x *, x 1 x 2 implies tat f x 1 f x 2, wile x 1 x and x 1 x 2 yields tat f x 1 f x Continuity (or Discontinuity) of Derivatives Given a real-valued function f tat is differentiable on a subset P of U, te derivative F f ) is a function wit domain P. We ave already seen tat F need not be continuous. It is natural to ask if tere are any nice properties tat can be associated wit te derivative. Te next teorem tells us tat te derivative of a real function tat is differentiable on an interval satis es te intermediate value property tere. Teorem Suppose tat f is a real valued function tat is differentiable on [a b] and f ) a f ) b. Ten for any D + U suc tat f ) a D f ) b,tere exists a point x + a b suc te f ) x D. Proof. Suppose tat f is a real valued function tat is differentiable on [a b] and D + U is suc tat f ) a D f ) b. Let Gt f t Dt. From te Properties of Derivatives, G is differentiable on [a b]. By Teorem , G is continuous on [a b] from wic te Extreme Value Teorem yields tat G as a minimum at some x + [a b]. Since G ) a f ) t D0and G ) b f ) t D 0, tereexistsat 1 + a b and t 2 + a b suc tat G t 1 G a and G t 2 G b. It follows tat neiter a G a nor b G b is a minimum of G in [a b]. Tus, a x b. In view of te Interior Extrema Teorem, we ave tat G ) x 0 wic is equivalent to f ) x D Remark Wit te obvious algebraic modi cations, it can be sown tat te same result olds if te real valued function tat is differentiable on [a b] satis es f ) a f ) b. Corollary If f is a real valued function tat is differentiable on [a b], ten f ) cannot ave any simple( rst kind) discontinuities on [a b]. Remark Te corollary tells us tat any discontinuities of real valued functions tat are differentiable on an interval will ave only discontinuities of te second kind.

23 6.3. THE DERIVATIVE AND FINDING LIMITS Te Derivative and Finding Limits Te next result allows us to make use of derivatives to obtain some its: It can be used to nd its in te situations for wic we ave been using te Limit of Almost Equal Functions and to nd some its tat we ave not ad an easy means of nding. Teorem (L Hôpital s Rule I) Suppose tat f and F are functions suc tat f ) and F ) existonasegmenti a b and F ) / 0 on I. t f ) u t u f (a) If f a Fa 0 and F ) a L, ten a L. F t f ) u t u f (b) If f a Fa *and F ) a L, ten a L. F Excursion Fill in wat is missing in order to compete te following proof of part (a). Proof. Suppose tat f and F are differentiable on a segment I a b, F ) / 0 on I, and f a Fa 0. Setting f a f a and F a Fa extends f and F to functions tat are in [a b. Wit tis, F a 0 and F ) x / 0 in I yields tat F x. 2 t f ) u Suppose tat 0 is given. Since F ) a L, tere exists = 0 suc tat a *a = implies tat 1 2 From te Generalized Mean-Value Teorem and te fact tat F a f a 0,it follows tat 4 ]`_^ f x L F x F a L L _^]` n 3 n n n n _ ^] ` n 5

24 252 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW for some G satisfying a Ga =. Hence, f x n F x L n.since arbitrary, we conclude tat. 6 0 was ***Acceptable responses are: (1) continuous, (2) / 0, (3) f ) * n F ) * L n,(4) F x,(5) f x f a, (6) f ) G f x F ),and(7) G xa Å F x L.*** Proof. Proof of (b). Suppose tat f and F are functions suc tat f ) and F ) existonanopenintervali t x : a x b, F ) / 0onI, f a Fa * f ) u and a L. Ten f and F are continuous on I and tere exists 0 F ) suc tat F ) / 0inI x : a x a. For> 0, tere exists a = wit 0 = suc tat f ) G n F ) G L n > 2 for all G in I = x : a x a = Let x and c be suc tat x c and x c + I =. By te Generalized Mean-Value f x f c Teorem, tere exists a G in I = suc tat Fx Fc f ) G F ) G. Hence, f x f c n Fx Fc L n > 2. In particular, for >1, we ave tat n f x f c nnn n Fx Fcn f x f c Fx Fc L L n L1 2. Wit a certain amount of playing around we claim tat n f x f x f c nnn n Fx Fx Fcn f c Fx Fc f x f x f c Fx Fcn n t n f c nnn n Fxn Fc f xn L 1 u. 2 For c xed, suc tat f c Fx 0and Fc f x 0asx a. Hence, tere exists = 1,0= 1 =, f c n Fxn > n nnn 4 and Fc f xn 1 4L12.

25 6.3. THE DERIVATIVE AND FINDING LIMITS 253 Combining te inequalities leads to n n f x n Fx L nnn n n f x f x f c nnn Fx Fx Fcn f x f c Fx Fc L n > wenever a x a = 1.Since> 0 was arbitrary, we conclude tat t u f f x a F xa Å Fx L. Remark Te two statements given in L Hôpital s Rule are illustrative of te set of suc results. For example, te x a can be replaced wit x b, x *,x *, and x *, wit some appropriate modi cations in te statements. Te following statement is te one tat is given as Teorem 5.13 in our text. Teorem (L Hôpital s Rule II) Suppose f and g are real and differentiable in a b,were* n a b n*,g ) f ) x x / 0 for all x + a b, and xa g ) x f x A. If f x 0 F g x 0 or g x *,ten xa xa xa xa g x A. Excursion Use an appropriate form of L Hôpital s Rule to nd x 2 5x 6 7sinx x3 2x 6

26 254 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW t u * ** * 1 ***Hopefully, you got 3 ande, respectively.*** 6.4 Inverse Functions Recall tat for a relation, S, onu, te inverse relation of S, denoted by S 1,iste set of all ordered pairs y x suc tat x y + S. Wile a function is a relation tat is single-valued, its inverse need not be single-valued. Consequently, we cannot automatically apply te tools of differential calculus to inverses of functions. Wat follows if some criteria tat enables us to talk about inverse functions. Te rst result tells us tat were a function is increasing, it as an inverse tat is a function. Remark If u and ) are monotonic functions wit te same monotonicity, ten teir composition (if de ned) is increasing. If u and ) are monotonic functions wit te opposite monotonicity, ten teir composition (if de ned) is decreasing. Teorem (Inverse Function Teorem) Suppose tat f is a continuous function tat is strictly monotone on an interval I wit f I J. Ten (a) J is an interval (b) te inverse relation g of f is a function wit domain J tat is continuous and strictly monotone on J and (c) we ave g f x x for x + I and f gy y for y + J. Proof. Because te continuous image of a connected set is connected and f is strictly monotone, J is an interval. Witout loss of generality, we take f to be

27 6.4. INVERSE FUNCTIONS 255 decreasing in te interval I. Ten f x 1 / f x 2 implies tat x 1 / x 2 and we conclude tat, for eac * 0 in J, tere exists one and only one? 0 + I suc tat * 0 f? 0. Hence, te inverse of f is a function and te formulas given in (c) old. It follows from te remark above and (c) tat g is strictly decreasing. To see tat g is continuous on J, let* 0 be an interior point of J and suppose tat g* 0 x 0 i.e., f x 0 * 0. Coose points * 1 and * 2 in J suc tat * 1 * 0 * 2. Ten tere exist points x 1 and x 2 in I, suc tat x 1 x 0 x 2, f x 1 * 2 and f x 2 * 1. Hence, x 0 is an interior point of I. Now, witout loss of generality, take > 0 small enoug tat te interval x 0 > x 0 > is contained in I and de ne *`1 f x 0 > and *`2 f x 0 > so *`1 *`2. Since g is decreasing, x 0 > g*`1 o g* o g*`2 x 0 > for * suc tat *`1 n * n *`2 Hence, g* 0 > o g* o g* 0 > for * suc tat *`1 n * n *`2 Now taking = to be te minimum of *`2 * 0 and * 0 *`1 leads to g* g* 0 >wenever * * 0 = Remark Wile we ave stated te Inverse Function Teorem in terms of intervals, please note tat te term intervals can be replaced by segments a b were a can be * and/or b can be *. In view of te Inverse Function Teorem, wen we ave strictly monotone continuous functions, it is natural to tink about differentiating teir inverses. For a proof of te general result concerning te derivatives of inverse functions, we will make use wit te following partial converse of te Cain Rule. Lemma Suppose te real valued functions F, G, and u are suc tat F x G u x, u is continuous at x 0 + U, F ) x 0 exists, and G ) u x 0 exists and differs from zero. Ten u ) x 0 is de ned and F ) x 0 G ) u x 0 u ) x 0. Excursion Fill in wat is missing to complete te following proof of te Lemma.

28 256 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW Proof. Let F Fx 0 Fx 0, u ux 0 ux 0 and u 0 u x 0. Ten F 1 Gu 0 u Gu 0. Since u is continuous at x 0, we know tat u 0. By te Fundamental Lemma of Differentiation, tere exists a wit, tat is continuous at 0 and is suc tat F 2 3. Hence, u F [G ) u From u 0, it follows 0 as 0. Because G ) u 0 exists and is nonzero, u ) x 0 ux 0 ux 0 F [G ) u F) x 0 G ) u 0. Terefore, u ) x 0 exists and 4. ***Acceptable responses are: (1) Gux 0 Gux 0, (2)@0 0, (3) [G ) u and (4) F ) x 0 G ) u 0 u ) x 0.*** Teorem (Inverse Differentiation Teorem) Suppose tat f satis es te ypoteses of te Inverse Function Teorem. If x 0 is a point of J suc tat f ) gx 0 is de ned and is different from zero, ten g ) x 0 exists and g ) x 0 1 f ) g x 0. (6.6) Proof. From te Inverse Function Teorem, f g x x. Takingu g and G f in Lemma yields tat g ) x 0 exists and f ) g x g ) x 1. Since f ) gx 0 / 0, it follows tat g ) 1 x 0 f ) gx 0 as needed.

29 6.4. INVERSE FUNCTIONS 257 Corollary For a xed nonnegative real number :, letgx : x. Ten dom g U and, for all x + U, g ) x : x ln :. Proof. We know tat g x : x is te inverse of f x log : x were f is a strictly increasing function wit domain 0 * and range * *. Because A log : B % : A B % A ln : ln B, it follows tat Hence log : B ln B ln :. f ) x b log : x c t u ) ln x ) 1 ln : x ln :. From te Inverse Differentiation Teorem, we ave tat g ) x g x ln : : x ln :. 1 f ) g x Remark Taking : e in te Corollary yields tat e x ) e x. In practice, nding particular inverses is usually carried out by working directly wit te functions given rater tan by making a sequence of substitutions. Example Derive a formula, in terms of x, for te derivative of y arctan x, H 2 x H 2. We know tat te inverse of u tan ) is a relation tat is not a function consequently we need to restrict ourselves to a subset ofrte domain. Because u is strictly increasing and continuous in te segment I H 2 H s te corresponding segment is * *. We denote te inverse tat corresponds to tis segment by 2 y f x arctan x. From y arctan x if and only if x tan y, it follows directly tat b sec 2 y c dy dx 1 or dy dx 1 sec 2 y. On te oter and, tan2 y 1 sec 2 ywit x tan y implies tat sec 2 y x 2 1. Terefore, dy dx f ) x 1 x 2 1. Excursion Use f x x to verify te Inverse Differentiation Teorem 1 x on te segment 2 4 i.e., sow tat te teorem applies, nd te inverse g and

30 258 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW its derivative by te usual algebraic manipulations, and ten verify te derivative satis es equation (6.6) ***Hopefully, you tougt to use te Monotonicity Test tat f is strictly increasing in I 2 4 follows immediately upon t noting tat f ) x 1 x 2 0in I. Te corresponding segment J 2 4 u is te domain for te inverse g 3 tat we seek. Te usual algebraic manipulations for nding inverses leads us to solving x y 1 y 1 for y. Ten application of te quotient rule sould ave led to g ) x 1 x 2. Finally, to verify agreement wit wat is claimed wit equation (6.6), substitute g into f ) x 1 x 2 and simplify.*** 6.5 Derivatives of Higer Order If f is a differentiable function on a set P ten corresponding to eac x + P, tere is a uniquely determined f ) x. Consequently, f ) is also a function on P. Weave already seen tat f ) need not be continuous on P. However,iff ) is differentiable on a set t P, ten its derivative is a function on wic can also be considered for differentiability. Wen tey exist, te subsequent derivatives are called iger order derivatives. Tis process can be continued inde nitely on te oter and, we could arrive at a function tat is not differentiable or, in te case of polynomials, we ll eventually obtain a iger order derivative tat is zero everywere. Note tat we can speak of iger order derivatives only after we ave isolated te set on wic te previous derivative exists. De nition If f is differentiable on a set P and f ) is differentiable on a set P 1 t P, ten te derivative of f ) is denoted by f )) or d2 f and is called te second dx2

31 6.5. DERIVATIVES OF HIGHER ORDER 259 derivative of f if te second derivative of f is differentiable on a set P 2 t P 1, ten te derivative of f )), denoted by f ))) or f 3 or d3 f,iscalledtetird derivative dx3 of f. Continuing in tis manner, wen it exists, f n denotes te n t derivative of f and is given by b f n1c ). Remark Te statement f k exists at a point x 0 asserts tat f k1 t is de ned in a segment containing x 0 (or in a alf-open interval aving x 0 as te included endpoint in cases of one-sided differentiability) and differentiable at x 0.If k 2, ten te same two claims are true for f k2. In general, f k exists at a point x 0 implies tat eac of f j, for j 1 2 k 1, isde nedinasegment containing x 0 and is differentiable at x 0. Example Given f x 3 5 2x 2 in U 5 }, nd a general formula 2 for f n. From f x 3 5 2x 2, it follows tat f ) x 325 2x 3 2, f )) x x 4 b 2 2c,f 3 x x 5 b 2 3c, and f 4 x x 6 b 2 4c. Basic pattern recognition suggests tat f n x 1 n 3 2 n n 1! 5 2x n2. (6.7) Remark Equation (6.7) was not proved to be te case. Wile it can be proved by Matematical Induction, te set-up of te situation is direct enoug tat claiming te formula from a suf cient number of carefully illustrated cases is suf cient for our purposes. Teorem (Taylor s Approximating Polynomials) Suppose f is a real function on [a b] suc tat tere exists n + M for wic f n1 is continuous on [a b] and f n exists for every t + a b. For< + [a b], let P n1 < t n1 ; k0 f k < k! t < k. Ten, for : and ; distinct points in [a b], tere exists a point x between : and ; suc tat f ; P n1 : ; f n x ; : n. (6.8) n!

32 260 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW Excursion Fill in wat is missing to complete te following proof of Taylor s Approximating Polynomials Teorem. Proof. Since P n1 : ;, ; : n and f ; are xed, we ave tat for some M + U. Let f ; P n1 : ; M ; : n g t def f t P n1 : t M t : n. Ten g is a real function on [a b] for wic g n exists in a b because te Properties of Derivatives, for t + a b, we ave tat 2 1 is continuous and. From g ) t f ) t ; n1 k1 f k : k 1! t :k1 nm t : n1, and g )) t 3. In general, for j suc tat 1 n j n n 1 and t + a b, it follows tat Finally, g j t f j t ; n1 k j f k : k j! t :k j n! n j! M t :n j. g n t 4 (6.9) Direct substitution yields tat g : 0. Furtermore, for eac j, 1 n j n n 1, t : implies tat 3 n1 f k : j k j t :k f k : consequently, k j! g : g j : 0 for eac j, 1 n j n n 1.

33 6.5. DERIVATIVES OF HIGHER ORDER 261 In view of te coice of M, we ave tat g ; 0. Because g is differentiable in a b, continuous in [a b],andg: g ; 0,by 5 tere exists x 1 between : and ; suc tat g ) x 1 0. Assuming tat n 1, from Rolle s Teorem, g ) differentiable in a b and in [a b] wit g ) : g ) x 1 0 for : x 1 + a b yields te existence of x 2 between : and x 1 suc tat. If n 2, Rolle s Teorem can be applied to g )) to obtain x 3 between 7 8 suc tat g 3 x 3 0. We can repeat tis process troug g n, te last iger order derivative tat we are assured exists. After n steps, we ave tat tere is an x n between : and x n1 suc tat g n x n 0. Substituting x n into equation (6.9) yields tat 6, 0 g n x n 9. Hence, tere exists a real number x x n tat is between : and ; suc tat f n f n x x n!m i.e., M. Te de nition of M yields equation (6.8). n! ***Acceptable responses are: (1) g n1,(2)g is te sum of functions aving tose properties, (3) f )) t 3 n1 f k : k2 k 2! t :k2 n n 1 M t : n2,(4) f n t n!m, (5) Rolle s Teorem or te Mean-Value Teorem, (6) continuous, (7) g )) x 2 0, (8) : and x 2, and (9) f n x n n!m.*** Remark For n 1, Taylor s Approximating Polynomials Teorem is te Mean-Value Teorem. In te general case, te error from using P n1 : ; instead of f ; is f n x ; : n for some x between : and ; consequently, we ave n! an approximation of tis error wenever we ave bounds on n f n x n. v Example Let f x 1 x 1 in n! 1 x n1 is continuous in v w. Ten, for eac n + M, f n x w 8. Consequently, te ypoteses for Tay-

34 262 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW lor s Approximating Polynomial Teorem are met for eac n + M. For n 2, P n1 < t P 1 < t 1 1 < 1 t <. 2 1 < If : 1 t 4 and ; 1 2, te Teorem claims te existence of x u suc 4 tat t f 1 u t u 1 P f 2 t x 1 2 2! u Since t u 1 P t t u u t we wis to nd x u suc tat 2 t u x 3 te only real 16 solution to te last equation is x T wic is approximately equal to Because x 0 is between : 1 4 and ; 1, tis veri es te Teorem for te 2 speci ed coices. 6.6 Differentiation of Vector-Valued Functions In te case of its and continuity we ave already justi ed tat for functions from U into U k, properties are ascribed if and only if te property applies to eac coordinate. Consequently, it will come as no surprise tat te same by co-ordinate property assignment carries over to differentiability. De nition A vector-valued function f from a subset P of U into U k is differentiable at a point? + P if and only if f is de ned in a segment containing? and tere exists an element of U k, denoted by f )?, suctat t? n f t f? t? were denotes te Euclidean k-metric. f )? n 0

35 6.6. DIFFERENTIATION OF VECTOR-VALUED FUNCTIONS 263 Lemma Suppose tat f 1 f 2 f k are real functions on a subset P of U and f x f 1 x f 2 x f k x for x + P. Ten f is differentiable at? + P wit derivative f )? if and only if eac of te functions f 1 f 2 f k is differentiable at? and f )? b f 1 )? f 2 )? f ) k? c. Proof. For t and? in U, we ave tat t f t f? f ) f1 t f 1?? t? t? f ) 1? f k t f k? t? f ) k? u. Consequently, te result follows immediately from Lemma and te Limit of Sequences Caracterization for te Limits of Functions. Lemma If f is a vector-valued function from P t U into U k tat is differentiable at a point? + P, tenf is continuous at?. Proof. Suppose tat f is a vector-valued function from P t U into U k tat is differentiable at a point? + P. Ten f is de ned in a segment I containing? and, for t + I,weavetat t f1 t f 1? f t f? t? f k t f k? t? u t? t? b f 1 )? 0 f 2 ) )? 0 f k? 0c as t?. Hence, for eac j + M, 1n j n k, t? f j t f j? i.e., eac f j is continuous at?. From Teorem (a), it follows tat f is continuous at?. We note tat an alternative approac to proving Lemma simply uses Lemma In particular, from Lemma 6.6.2, f x f 1 x f 2 x f k x differentiable at? implies tat f j is differentiable at? for eac j, 1n j n k. By Teorem , f j is continuous at? for eac j,1n j n k, from wic Teorem (a) allows us to conclude tat f x f 1 x f 2 x f k x is continuous at?. Lemma If f and g are vector-valued functions from P t U into U k tat are differentiable at a point? + P, ten te sum and inner product are also differentiable at?. Proof. Suppose tat f x f 1 x f 2 x f k x and g x g 1 xg 2 x g k x are vector-valued functions from P t U into U k tat are differentiable at a point? + P. Ten f gx f 1 g 1 x f 2 g 2 x f k g k x

36 264 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW and f gx f 1 g 1 x f 2 g 2 x f k g k x. b From te c Properties b c of Derivatives (c) and (d), b for eac c j + M, 1n j n k, ) f j g j and f j g j are differentiable at? wit f j g j? f ) j? g ) j? and b c ) f j g j? f ) j? g j? f j? g ) j?. From Lemma 6.6.2, it follows tat f g is differentiable at? wit f g? b f ) g )c? bb f 1 ) b 1c g)? f ) 2 g 2 ) c b? f ) k g k ) c c? and f g is differentiable at? wit f g )? b f ) g c? b f g )c?. Te tree lemmas migt prompt an unwarranted leap to te conclusion tat all of te properties tat we ave found for real-valued differentiable functions on subsets of U carry over to vector-valued functions on subsets of U. A closer scrutiny reveals tat we ave not discussed any results for wic te ypoteses or conclusions eiter made use of or relied on te linear ordering on U. Since we loose te existence of a linear ordering wen we go to U 2, it souldn t be a sock tat te Mean-Value Teorem does not extend to te vector-valued functions from subsets of U to U 2. Example For x + U, letf x cos x sin x. Sow tat tere exists an interval [a b] suc tat f satis es te ypoteses of te Mean-Value Teorem witout yielding te conclusion. From Lemma and Lemma 6.6.3, we ave tat f is differentiable in a b and continuous in [a b] for any a b + U suc tat a b. Since f 0 f 2H 1 0, f 2Hf Becausef ) x sin x cos x, n n f ) x n n 1 for eac x + 0 2H. In particular, 1x + 0 2H b f ) x / 0 0 c from wic we see tat 1x + 0 2H b f 2H f 0 / 2H 0 f ) x c i.e., 2x d x + 0 2H F b f 2H f 0 2H 0 f ) x ce Remark Example 5.18 in our text justi es tat L Hôpital s Rule is also not valid for functions from U into F.

37 6.6. DIFFERENTIATION OF VECTOR-VALUED FUNCTIONS 265 Wen we justify tat a result known for real-valued differentiable functions on subsets of U does not carry over to vector-valued functions on subsets of U, itis natural to seek modi cations of te original results in terms of properties tat migt carry over to te different situation. In te case of te Mean-Value Teorem, success in acieved wit an inequality tat follows directly from te teorem. From te Mean-Value Teorem, if f is a function tat is continuous on te interval I [a b] and differentiable on te segment I i a b, ten tere exists a number G + I i suc tat f b f a f ) G b a. Since G + I i, n f ) G n n n sup f ) x n. x+i n i Tis leads to te weaker statement tat f b f a n b a sup f ) x n.on x+i i te oter and, tis statement as a natural candidate for generalization because te absolute value or Euclidean 1-metric can be replaced wit te Euclidean k-metric. We end tis section wit a proof of a vector-valued adjustment of te Mean-Value Teorem. Teorem Suppose tat f is a continuous mapping of [a b] into U k tat is differentiableina b. Ten tere exists x + a b suc tat f b f a n b a n n f ) x n n (6.10) Proof. Suppose tat f f 1 f 2 is a continuous mapping of [a b] into U k tat is differentiable in a b and let z f b f a. Equation 6.10 certainly olds if z 0 0 consequently, we suppose tat z / 0 0. By Teorem (b) and Lemma 6.6.4, te real-valued function M t z f t for t + [a b] is continuous in [a b] and differentiable in a b. Applying te Mean-Value Teorem to M,weavetattereexistsx + a b suc tat Now, Mb Ma M ) xb a. (6.11) MbMa z f b z f a f b f a f b f b f a f a f b f a f b f a z z z 2.

38 266 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW For z 1 f 1 b f 1 a and z 2 f 2 b f 2 a, M x n n z f ) x n n n nz1 f ) 1 x z 2 f ) 2 xn n n S z 1 z 2 T nn f ) 1 xn n n n f ) 2 x n n z n nf ) x n n by Scwarz s Inequality. Substituting into equation (6.11) yields z 2 b a n n z f ) x n n n b a z n nf ) x n n wic implies z n b a n n f ) x n n because z / Problem Set F 1. Use te de nition to determine weter or not te given function is differentiable at te speci ed point. Wen it is differentiable, give te value of te derivative. (a) f x x 3 x 0 x 3, for 0 n x n 1 (b) f x x 1 T x,for x 1 T 1! x sin,for x / 0 (c) f x x! 0, for x 0 x 0 9 (d) f x 2x 2 1 x 2 2. Prove tat, if f and g are differentiable at?, ten Fx f gx is differentiable at? and F )? f )? g )?. 3. Use te de nition of te derivative to prove tat f x x n is differentiable on U for eac n + M. x 2,for x + T 4. Let f x. 0, for x + T Is f differentiable at x 0? Carefully justify your position.

39 6.7. PROBLEM SET F If f is differentiable at?, provetat f? : f? ; : ; f )?. 6. Discuss te differentiability of te following functions on U. (a) f x x x 1 (b) f x x x 7. Suppose tat f : U U is differentiable at a point c + U. Given any two sequences a n * n1 and b n * n1 suc tat a n / b n for eac n + M and a n b n c, is it true tat n* n* f b n f a n f ) c? n* b n a n State your position and carefully justify it. 8. Use te Principle of Matematical Induction to prove te Leibnitz Rule for te n t derivative of a product: n; t u n fg n x f nk x g k x k k0 were b nc n! k n k! k! and f 0 x f x. 9. Use derivative formulas to nd f ) x for eac of te following. Do only te obvious simpli cations. (a) f x 4x 6 3x 1 rx 5 4x 2 b 5x 3 7x 4c 7 s (b) f x 4x 2 b 1 2x 3 b4x c 9 3x 2 10 c 2 2 x 2 3 b % 2x 2 3x 5c 3 7 (c) f x # r 14 4 T x 2 3 s 4 & $ 15

40 268 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW (d) f x (e) f x t 3x 5 1 x 5 T 3 S 2 T 1 x 10. Complete te following steps to prove tat sin A A0 A u 10 r 4 T 12 7x 4 3 5x 2s 5 1 and A0 cos A 1 A 0. (a) Draw a gure tat will serve as an aid towards completion of a proof sin A tat 1. A0 A i. On a copy of a Cartesian coordinate system, draw a circle aving radius 1 tat is centered at te origin. Ten pick an arbitrary point on te part of te circle tat is in te rst quadrant and label it P. ii. Label te origin, te point 1 0, and te point were te line x P would intersect te x-axis, O B and A, respectively. iii. Suppose tat te argument of te point P, inradianmeasure,isa. Indicate te coordinates of te point P and sow te line segment joining P to A in your diagram. iv. If your completed diagram is correctly labelled, it sould illustrate tat sin A A n npa n n lengt of ' PB were n n PA denotes te lengt of te line segment joining points P and A and PB ' denotes te arc of te unit circle from te point B to te point P. v. Finally, te circle aving radius n n OA and centered at te origin will pass troug te point A and a point and a point on te ray OP. Label te point of intersection wit OP wit te letter C and sow te arc CAon ' your diagram.

41 6.7. PROBLEM SET F 269 (b) Recall tat, for a circle of radius r, te area of a sector subtended by A radians is given by Ar 2. Prove tat 2 A cos 2 A 2 cos A sin A 2 A 2 for A satisfying te set-up from part (a). sin A (c) Prove tat 1. A0 A (d) Recall tat sin 2 A cos 2 cos A 1 A 1. Prove tat 0. A0 A 11. Te result of Problem 10 in conjunction wit te discussion tat was offered in te section on Formulas for Derivatives justi es te claim tat, for any x + U, sin x ) cos x and cos x ) sin x, werex is interpreted as radians. Use our Properties of Derivatives and trig identities to prove eac of te following. (a) tan x ) sec 2 x (b) sec x ) sec x tan x (c) csc x ) csc x cot x (d) ln sec x tan x ) sec x (e) ln csc x cot x ) csc x 12. Use derivative formulas to nd f ) x for eac of te following. Do only te obvious simpli cations. r r (a) f x sin 5 3x 4 cos 2 2x 2 T ss x 4 7 (b) f x tan3 b 4x 3x 2c 1 cos 2 b 4x 5c (c) f x b 1 sec 3 3x c 4 t x 3 r (d) f x cos 3 x 4 4 T s 4 1 sec 4 x 3 2x 2 1 tan x u 2

42 270 CHAPTER 6. DIFFERENTIATION: OUR FIRST VIEW 13. Find eac of te following. Use L Hôpital s Rule wen it applies. (a) x H 2 tan x x H2 (b) x0 tan 5 x tan 3 x 1 cos x (c) x 3 x* e 2x 4x 3 2x 2 x (d) x* 5x 3 3x 2 2x tan x x (e) x0 x 3 (f) x 2 ln x 2 x2 Å 14. For f x x 3 and x 0 2 in te Fundamental Lemma of Differentiation, sow For f x x 1 2x 1 and x 0 1 in te Fundamental Lemma of Differentiation, nd te 16. Suppose tat f, g,and are tree real-valued functions on U and c is a xed real number suc tat f c g c c and f ) c g ) c ) c. If A 1 A 2 A 3 is a partition of U, and f x,for x + A 1 L x g x,for x + A 2, x,for x + A 3 prove tat L is differentiable at x c. 17. If te second derivative for a function f exists at x 0 + U, sow tat f x 0 2 f x 0 f x 0 2 f )) x For eac of te following, nd formulas for f n in terms of n + M.