MATH 27 Lecture Guide UNIT 1 for AY 2018-19 Revised.pdf -...

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Unformatted text preview: MATH 27 LECTURE GUIDE UNIT 1. DERIVATIVES OF AND INTEGRALS YIELDING TRANSCENDENTAL FUNCTIONS In your basic calculus, you were already introduced with limits, derivatives, and integrals. However, you may observe that the functions that are usually considered are algebraic in nature, i.e., polynomial functions, square root functions, rational functions, etc. In this section, we will deal with a class of functions different from algebraic functions called transcendental functions. These includes trigonometric and inverse trigonometric functions, exponential and logarithmic functions, among others. For the first Unit, derivatives of and integrals yielding this type of functions will be discussed. Moreover, a technique in solving derivatives of “convoluted” functions, called logarithmic differentiation, will also be introduced. We will also look into some application problems such as optimization and related rates problem that involves derivatives of transcendental functions. Lastly, we evaluate limits of functions of indeterminate forms using L’Hopital’s Rule. Our goals for this unit are as follows. By the end of the unit, you should be able to ✓ ✓ ✓ ✓ ✓ ✓ evaluate limit of functions using limit theorems find derivatives of algebraic and transcendental functions; solve integrals of and integrals yielding algebraic and transcendental functions; apply logarithmic differentiation appropriately; apply derivatives of and integrals yielding transcendental functions to real-world problems in various fields; and evaluate limits of functions using L'Hopital's rule. 1.1 Limit of a Function The limit of a function is a fundamental concept in calculus. It describes the behavior of a function as its variable gets closer to a specific point. The derivatives and the definite integral are formally defined in terms of limits. Intuitively, when we say “the limit of () as approaches is ”, it means that as gets closer and closer to , the function values of () gets closer and closer to . Notation: () = ⟶ read as “the limit of () as approaches is ” For example, consider the function defined by () = . Note that as gets closer and closer to , the value of () gets closer and closer to . One can verify this fact by either looking at the graph of or by investigating the values of () for values of near like . or . . Then we write, = ⟶ Definition. (Limit of a Function) Let be a function defined at every number in some open interval containing , except possibly at itself. Then, () = if for every > (however small), there exists > such ⟶ that if < | − | < , then |() − | < . The inequalities < | − | < and |() − | < describe the “closeness” that is referred to when we say “ approaches ” or “() approaches . MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) We state here its formal definition. 1 1.1.1 Limit Theorems Instead of using this formal definition of limit, we compute for limits using limit theorems. MUST REMEMBER!!! Limit theorems 1. If () = and () = , then = . ⟶ (uniqueness) ⟶ 2. If is a constant, then = . ⟶ = and for ∈ ℕ, = . 3. ⟶ ⟶ 4. If and are any constants, then ( + ) = + . ⟶ 5. If () = and () = , then ⟶ ⟶ • (() ± ()) = ± ⟶ • (() ∙ ()) = ∙ ⟶ • ( () ⟶ () )= provided ≠ 6. If () = , > and ∈ ℕ is even, then √() = √. ⟶ ⟶ If () = and ∈ ℕ is odd, then √() = √. ⟶ ⟶ From the theorems, it is immediate that for polynomial functions, say (), computing a limit is simply evaluating () at = , in other words, () = (). ⟶ ILLUSTRATION: 1. To compute ( − ⟶− + ), we simply evaluate the limit of the numerator and the denominator. Note ( − ) = − and ( + ) = . ⟶− Hence, ( − ⟶− + ⟶− )=− . 2. To compute (√ + − ⟶ Hence, (√ + − ⟶ − − + ), we simply compute for the limit term-by-term. + ) = − + = + . TRY THIS! Compute the following limits. 2. 3. 4. 5. ( + − + ) ⟶− ⟶ − + √ − ⟶ − − − ⟶− ⟶− MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) 1. 2 1.1.2 One-sided limits We next consider one-side limit. Consider a real number and a function . Limit from the right Let be defined from the right of , i.e. on some open interval of the form (, ). The limit of from the right of is if the function values () approaches as approaches from the right. Notation: + () = ⟶ Limit from the left Let be defined from the left of , i.e. on some open interval of the form (, ). The limit of from the left of is if the function values () approaches as approaches from the left. Notation: − () = ⟶ MUST REMEMBER!!! Theorem. () exists and is equal to if and only if + () and − () both exist and both ⟶ ⟶ ⟶ are equal to . ILLUSTRATION: 1. Consider the function () = √ − with domain = [, +∞). Note that is defined from the right of = but is undefined from the left of = . Since is undefined from the left of = , it is immediate that − √ − does not exist. ⟶ Now, from the right of = , the function values approaches . Hence, + √ − = . ⟶ Even if + √ − exists but − √ − does not exist, the “two-sided limit” √ − does ⟶ ⟶ ⟶ not exist. 2. Consider the function () = { + ≤ with domain = ℝ. − > To obtain the limit from the right, we consider the function values from the right of = . This is given by the condition > . So, + () = +( − ) = . ⟶ ⟶ To obtain the limit from the left, we consider the function values from the left of = . This is given by the condition ≤ . So, − () = −( + ) = . ⟶ ⟶ Since + () and − () both exist and are both equal to , the theorem tells us that ⟶ ⟶ () also exists and that () = . ⟶ ⟶ MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) We focus on the behavior of the function as approaches = . 3 1.1.3 Infinite limits It is also possible that () can increase or decrease without bound. We discuss this case in the next section. Consider the function () = whose graph was given on the right. Note that ⟶ does not exist. The function values do not approach any particular values as ⟶ . However, we can still establish the behavior of the function as ⟶ + and as ⟶ − . As ⟶ + , the values bound. Note that + ⟶ increases without is of the form + meaning the denominator approaches 0 through positive values. Hence, we write + = +∞ . ⟶ As ⟶ − , the values decreases without bound. Note that − ⟶ is of the form − meaning the denominator approaches 0 through negative values. Hence, we write + = −∞. ⟶ Remark: Infinite limits of the form () = +∞ or () = −∞ describe the behavior of the ⟶ ⟶ function. The limit () does not exist but the behavior increases without bound (+∞) or ⟶ decreases without bound (−∞). MUST REMEMBER!!! Theorems on Infinite Limits Theorem. Let be a positive integer. i. ii. ⟶+ ⟶− = +∞ ={ +∞ −∞ Moreover, if () = and () = , then ⟶ If > and () ⟶ + , then ii. If > and () ⟶ − , then () ⟶ () () + ⟶ () () − ⟶ () () iii. If < and () ⟶ , then iv. If < and () ⟶ , then ⟶ () + = +∞. ( +) = −∞. ( −) = −∞. ( +) = +∞. ( −) + − − ILLUSTRATION: − To obtain + , note that +( − ) = − and + ( + ) = + , i.e. + ⟶− + ⟶− ⟶− approaches through positive values. Hence, + ⟶− − + − = −∞ ( + ). Similarly, − ⟶− − + − = +∞ ( − ). MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) ⟶ i. 4 1.1.4 Limits at Infinity Next, we look at the case when increases or decreases without bound. Let be a function defined on the interval (, +∞). The limit of () as increases without bound is if the function values () approaches as ⟶ +∞. Notation: () = ⟶+∞ Let be a function defined on the interval (−∞, ). The limit of () as decreases without bound is if the function values () approaches as ⟶ −∞. Notation: () = ⟶−∞ MUST REMEMBER!!! Theorem on Limits at Infinity Theorem. Let be a positive integer. Then ⟶+∞ = and ⟶−∞ = . We discuss more of infinite limits and limits at infinity in Section ___ , which is about indeterminate forms. TRY THIS! Compute the following limits. If the limit does not exist, write “The limit does not exist.” 1. 2. 3. − ⟶ + − ⟶+∞ + ⟶−∞ + − _________________________ For the following conditional functions, properly implement one-sided limits to determine whether the limit at the indicated point exists or does not exist. () = { − < − > () () ⟶− ⟶+ 2. () = { − − () ⟶−+ () ⟶ < − − ≤ < ≥ () ⟶−− () ⟶− REMINDER: For a thorough discussion of limits, students are advised to refer to Module III. Limits of MATH 27 BASIC CALCULUS Preparatory Modules. MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) 1. 5 1.2 Derivatives and Integrals of Functions We first talk about basic differentiation of functions. Differentiation is the process of finding the derivative of a function. The derivative of a function is defined in terms of limit of a function which was discussed in the previous section. REMEMBER THIS!!! Definition of Derivative The derivative of the function , denoted by ′, is defined by ′ () = → if this limit exists. ( + ) − () Further, if is a particular number in the domain of , then the derivative of at may also be defined as () − ( ) ′ ( ) = → − if this limit exists. Alternative notations: For a function = (), the derivative ′ is also denoted by , , or . If ′ ( ) exists, then we say that is differentiable at . Instead of using limits, we will compute for derivatives using the following rules of differentiation. REMEMBER THIS!!! Basic Rules on Differentiation 1. If is a given constant, then [] = . 2. Power Rule If is a given real number, then [ ] = − . 3. If is a given constant and is a given function, then [ ⋅ ()] = ⋅ [()]. 4. Sum/Difference Rule If and are given functions, then [() ± ()] = [()] ± [()]. In general, if , , … , are given functions, then [ () ± () ± ⋯ ± ()] = [ ()] ± [ ()] ± ⋯ ± [ ()]. 5. Product Rule If and are given functions, then Mnemonic: left de-right plus right de-left 6. Quotient Rule If and are given functions, then [ Mnemonic: () () ⋅ [()] − () ⋅ [()]= . [()] () low de-high minus high de-low all over low-squared MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) [() ⋅ ()] = () ⋅ [()] + () ⋅ [()]. 6 1.2.1 Algebraic Functions We first apply the theorems in algebraic functions such as polynomial, radical and rational functions. ILLUSTRATION: 1. For the function () = − √ + , express it in the form () = − − + . Then, implement the Power Rule to each term. Hence, the derivative is given by ′ () = −− − − + . 2. To obtain the derivative of () = ( + )( − ), implement the Product Rule. So, ′() = ( + ) ( − ) + ( − ) ( + ) = ( + )() + ( − )(). 3. To obtain the derivative of = − + − , implement the Quotient Rule. So, ( − ) ( − + ) − ( − + ) ( − ) = ( − ) = ( − )( − ) − ( − + )( ) . ( − ) There are functions wherein the given basic differentiation rules do not apply. We introduce here a rule that can be used in these cases. Chain Rule Chain Rule is a very tool in differentiating composition of functions. A composition is of the form () = ( ∘ )() = (()). REMEMBER THIS! Chain Rule If () = (()), then ′() = ′ (()) ⋅ ′ (). In other notation, if = () and = (), then = ⋅ . ILLUSTRATION: Applying Chain Rule, ′ () = ′ (()) ⋅ ′ () = [()] ⋅ ( − ) = ( − + ) ⋅ ( − ) 2. To compute for [( − − ) ], apply the Power Rule. Then, by Chain Rule, apply the Quotient Rule to the innermost expression. Hence, − − − − ( − )() − ( − )(−) [( ) ] = ( ) ⋅ ( ) = ( ) ( ). ( − ) − − − − MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) 1. The function () = ( − + ) is a composition () = (()), where () = and () = − + . 7 REMINDER: For a thorough discussion of derivatives of algebraic functions, students are advised to refer to Module I. Derivatives of MATH 27 BASIC CALCULUS Preparatory Modules to better understand the concept of derivative and the use of chain rule in solving derivatives involving algebraic functions. The given rules on differentiation will also hold for transcendental functions, which is the concern of this course. TRY THIS! Compute for the derivatives of the following functions using rules for differentiation. 1. () = − − + − 2. () = ⁄ − √ + √ 3. () = + − + 4. () = (√ + )( + + √) 5. () = − ++ 6. ( √ − + − ) 7. [( + + ) − ( − ) ] 8. If = √ + and = + , then = Recall that if ′ () = (), then ∫ () = () + , where is an arbitrary constant. In this case, we call () an antiderivative of (). We also say that the indefinite integral of () with respect to is (). The symbol () is referred to as the integrand. ∫ is called the integral sign and MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) We now talk about the inverse process of differentiation. Antidifferentiation is the process of finding all the antiderivatives of a given function. 8 REMEMBER THIS!!! Basic Rules of Antidifferentiation + + 1. (Power rule) If is any rational number, then ∫ = 2. ∫ = + 3. ∫ () = ∫ () 4. ∫[ () ± () ± ⋯ ± ()] = ∫ () ± ∫ () ± ⋯ ± ∫ () + , where ≠ −. CAUTION!!! There is NO immediate rule for antiderivative of product nor of quotient. Thus, the following may not hold true: ∫ ()() ≠ ∫ () ∫ () ∫ () ∫ () ≠ () ∫ () ILLUSTRATION: 1. To compute for ∫( − + ) , simply take the antiderivative of each term. Hence, ∫( − + ) = − + + . 2. There is NO quotient rule for antiderivatives. Some quotients can be written in the form where Power Rule is applicable. So, − ∫ = ∫ − = + = − + . − 3. There is NO product rule for antiderivatives. Whenever possible, first take the product of the expressions. Then, compute for the term-by-term antiderivatives. Hence, ∫( + )( − ) = ∫( + − ) = + − + = + − + . Substitution Technique In general, given a function and its derivative ’, then [()]+ ∫[()] ∙ ′() = +. + The technique here is to let = () so that = ′(). By substitution, the left side can then be written as + ∫ = +. + MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) Simple substitution techniques can be used to compute some integral forms. This is best used in solving integrals of some products or quotients. 9 ILLUSTRATION: 1. To solve for ∫ √ + , we let = + so that = and = . Hence, ∫ √ + = ∫ √ = ∫ = ⋅ + = ⋅ + = ⋅ ( + ) + . + 2. To solve for ∫ , we let = + + so that = ( + ) and ( + +) ( + ) = . Hence, + − − ∫ = ∫ = ∫ = ⋅ + ( + + ) − =− +=− + ( + + ) REMINDER: For a thorough discussion of antidrivatives of algebraic functions, it is recommended that students should refer to Module II. INTEGRALS of MATH 27 BASIC CALCULUS Preparatory Modules. The given rules on anti-differention is also valid for thranscendental functions. TRY THIS! Compute the following antiderivatives. 1. ∫( − + + ) 2. ∫ ( √ + − + ) ∫( + )( + ) − + 4. ∫ 5. 6. 7. ∫ ( + ) ∫( + )( + + ) ∫ √ − MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) 3. 10 1.2.2 Trigonometric Functions First on the list is trigonometric function. We first determine the derivatives of this type of function. Integrals yielding trigonometric functions will directly follow. In the later sections, integrals of trigonometric functions will also be considred. Recall the following facts for basic trigonometric functions: Function () = () = () = tan Domain ℝ ℝ ℝ − { | is an odd integer} ℝ − {| is an odd integer} ℝ − { | is an odd integer} ℝ − {| is an odd integer} () = () = () = Morever, the given basic trigonometric functions are continuous over their respective domains. (To know more about continuity of a function, please refer to Module III. Limits of MATH 27 BASIC CALCULUS Preparatory Modules.) Using the definition of a derivative, it can be shown that ( ) = and ( ) = − . Now, formulate ( ) and ( ) using the result above. The first few steps are already written. TO DO: ( ) = ( ) (HINT: Use quotient rule for differentiation) ( ) = ( ) = [( )− ] Following the same procedure, it can also be shown that ( ) = − and ( ) = MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) TO DO: 11 MUST REMEMBER!!! Derivatives of Trigonometric Functions ( ) = ( ) = ( ) = ( ) = − ( ) = − ( ) = − It is a fact that the previously discussed trigonometric functions are differentiable over their respective domains. To aid in remembering things, we can think of the trigonometric functions as pairs, such that the derivative of the “co-function” is always negative. In particular, we can pair sine with cosine, tangent with cotangent, and secant with cosecant. This pairing will be very useful as we go through the succeeding sections. If the argument of a trigonometric function is a differentiable function of the variable , then we can use Chain Rule to obtain its corresponding derivative. MUST REMEMBER!!! CHAIN RULE: Derivatives of trigonometric functions Let be a differentiable function of . ( ) = ⋅ ( ) = ⋅ ( ) = ⋅ ( ) = − ⋅ ( ) = − ⋅ ( ) = − ⋅ ILLUSTRATION: 1. Determine ( ). Solution: ( ) = ⋅ ( ) + ⋅ ( ) by Product Rule = (− ) + = − ILLUSTRATION: Solution: ( − ) = ⋅ ( ) − ( ) = ⋅ () − ⋅ () = ⋅ () − ⋅ () = − by Sum/Difference Rule by Chain Rule MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) 2. Evaluate ( − ). 12 TRY THIS! Evaluate the following. 1. ( ) 2. ( ) 3. ( + √ ) 4. ( ) 5. ( √ − √ ) 6. ( ) 7. ( ( )) 8. (√( )) For more exercises, you can refer to: Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage Learning Asia Pte. Ltd., pp. 125,136 For an online tutorial, follow these links: We now consider integrals yielding trigonometric functions. From the previous section, the following results are immediate. MUST REMEMBER!!! Integrals Yielding Trigonometric Functions ∫ = + ∫ = + ∫ = + ∫ = − + ∫ = − + ∫ = − + MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) Let be a differentiable function of . 13 ILLUSTRATION: 1. Evaluate ∫ . Solution: Note that ( ) = . Hence, we can let = , so that = and the given integral can now be rewritten as ∫ . Then, ∫ = + ⇒ ∫ = + To verify that the answer is correct, we show that ( ( + ) = . + ) = ( ⋅ ( )) = TO DO: . 1. Rework the Illustration 1 using an alternate substitute = . Compare your answer to the answer above. Explain why the two answers are both antiderivatives of . Are the two answers equal? 2. Employ the technique similar to Illustration 1 to evaluate ∫ . There would be instances that solving integrals involving trigonometric functions would require us to use identities. We recall here some known trigonometric identities that may be helpful in the succeeding discussions. RECALL: Some Trigonometric Identities + = + = + = ILLUSTRATION: 2. Evaluate By inspecting the list of integrals that we discussed previously, we will see that there is no result whose integrand is . However, there is a result that involves is . Hence, we can use the identity + = to be able to solve the given problem. ∫ = ∫( − ) = ∫ − ∫ = − − + To verify that the answer is correct, we show that (− − + ) = . (− − + ) = −(− ) − = − = MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) ∫ . 14 TO DO: Employ the technique similar to Illustration 2 to evaluate ∫ . TRY THIS! Evaluate the following. 1. ∫ ( ) 2. ∫ 3. (√) √ ∫ ( ) 4. ∫ 5. ∫ 6. ∫ 7. ∫ 8. ∫ For more exercises, you can refer to: Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage Learning Asia Pte. Ltd., p. 334 For an online tutorial, follow these links: Inverse Trigonometric Functions Now, we formulate derivatives of inverse trigonometric functions. Then, we look into integrals that yields inverse trigonometric function later. We first recall the six basic inverse trigonometric functions and their respective domain. Function () = () = () = tan () = () = () = Domain [−, ] [−, ] ℝ ℝ (−∞, −] ∪ [, +∞) (−∞, −] ∪ [, +∞) MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) 1.2.3 15 We note that other references may use the notation − for , and similar notation for other inverse trigonometric functions. For more details about inverse trigonometric functions, we refer the student to Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7th sub-edition. Harper Collins Publishers. pp. 491-503 The given inverse trigonometric functions are continuous over their respective domains except for the endpoints of the interval if they exist. Hence, derivatives of the said class of transcendental functions exist at some points. We first formulate the derivative of the inverse sine function using the derivative of the sine function and implicit differentiation. Recall that ( ) = . TO DO: Formulate the derivative of . Let = . Hence, = . Getting the derivative of both sides of the equation implicitly, we have, Using the same technique, we can show the following results. MUST REMEMBER!!! Derivatives of Inverse Trigonometric Functions ( ...
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