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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 ✓ ✓ ✓ ✓ ✓ find derivatives of transcendental functions; solve integrals of and integrals yielding 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. REMINDER: Before proceeding to the first section of this unit, students are advised to refer to pages 1-4 of 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. 1.1 Derivatives of and Integrals Yielding 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: 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. Using the definition of a derivative, ′ () = → it can be shown that ( + ) − () ( ) = and ( ) = − . MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) Function () = () = () = tan 1 Now, formulate ( ) and ( ) using the result above. The first few steps are already written. TO DO: ( ) = ( ) (HINT: Use quotient rule for differentiation) TO DO: ( ) = ( ) = [( )− ] Following the same procedure, it can also be shown that ( ) = − and ( ) = ( ) = ( ) = ( ) = ( ) = − ( ) = − ( ) = − 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. MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) MUST REMEMBER!!! Derivatives of Trigonometric Functions 2 MUST REMEMBER!!! CHAIN RULE: Derivatives of trigonometric functions Let be a differentiable function of . ( ) = ⋅ ( ) = ⋅ ( ) = ⋅ ( ) = − ⋅ ( ) = − ⋅ ( ) = − ⋅ ILLUSTRATION: 1. Determine ( ). Solution: ( ) = ⋅ ( ) + ⋅ ( ) by Product Rule = (− ) + = − ILLUSTRATION: 2. Evaluate ( − ). Solution: ( − ) = ⋅ ( ) − ( ) = ⋅ () − ⋅ () by Sum/Difference Rule by Chain Rule = ⋅ () − ⋅ () = − TRY THIS! Evaluate the following. 1. ( ) 2. ( ) 3. ( + √ ) 5. ( √ − √ ) 6. ( ) 7. ( ( )) 8. (√( )) MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) 4. ( ) 3 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. We first recall the process of antidifferentiation given in basic calculus. Recall that ′ () = () ⇒ ∫ () = () + In this case, we call () an antiderivative of (). We also say that the indefinite integral of () with respect to is () REMINDER: For a more thorough discussion of antidrivatives of algebraic functions, it is recommended that students should refer to pages 1-3 of Module II. INTEGRALS of MATH 27 BASIC CALCULUS Preparatory Modules. The given rules on anti-differention is also valid for thranscendental functions. From the previous section, the following results are immediate. MUST REMEMBER!!! Integrals Yielding Trigonometric Functions Let be a differentiable function of . ∫ = + ∫ = + ∫ = + ∫ = − + ∫ = − + ∫ = − + 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 ( ( . + ) = . + ) = ( ⋅ ( )) = MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) ∫ . 4 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 (− − + ) = . (− − + ) = −(− ) − = − = TO DO: ∫ . MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) Employ the technique similar to Illustration 2 to evaluate 5 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: 1.2 Derivatives of and Integrals Yielding Inverse Trigonometric Functions (TC7 491-503 / TCWAG 503-513) Now, we formulate derivatives of inverse trigonometric functions. Then, we look into integrals that yields inverse trigonometric function later. Function () = () = () = tan () = () = () = Domain [−, ] [−, ] ℝ ℝ (−∞, −] ∪ [, +∞) (−∞, −] ∪ [, +∞) MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) We first recall the six basic inverse trigonometric functions and their respective domain. 6 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 ( ) = √ − ( ) = − √ − ( ) = + ( ) = − + ( ) = √ − ( ) = − √ − Now, applying the Chain Rule to the previous results, we have the following corollary. MUST REMEMBER!!! CHAIN RULE: Derivatives of inverse trigonometric functions Let be a differentiable function of . Then, ( ) = √ − ( ) = − ⋅ √ − ⋅ ( ) = ⋅ + ( ) = − ⋅ + ( ) = ( ) = √ − √ − ⋅ ⋅ MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) As you will observe, we can still use the pairing of the “co-functions” that we made earlier to easily remember that the derivative of one is the negative of the other. For example, the derivative of is the negative of the derivative of , i.e., ( ) = − ( ) = − . √ − 7 ILLUSTRATION: 1. Solve for if = ( − ) Solution: = ( − ) ⇒ =− ⋅ ( − ) √ − ( − ) =− =− √ − ( − + ) ⋅ √ − ILLUSTRATION: 2. Solve for if = [ ( )] Solution: = [ ( )] ⇒ = [ ( )] ⋅ ( ( ) ) = [ ( )] ⋅ = [ ( )] ⋅ √ − √ − ⋅ ( ) ⋅ TRY THIS! Solve for if 1. = ( ) 2. = 3. = ( ) 4. = √ 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. 372 MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) 5. = ( ) + ( ) 8 Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7th sub-edition. Harper Collins Publishers. p. 505 For an online tutorial, follow these links: We now proceed with integrals yielding inverse trigonometric functions. Using our list of derivatives, we can obtain the following results. MUST REMEMBER!!! Integrals Yielding Inverse Trigonometric Function Let be a differentiable function of , and be a constant. Then , ∫ ∫ ∫ √ − = + = + + √ − = + The given facts can be verified by getting the derivative of the integrand and showing that it is equal to the right hand side of the equation. Moreover, it can be shown that ∫ √ − = − + . The same is true for the other integrals on the list. However, as a matter of convention, we will use the former result. You might also observe that integrals yielding inverse trigonometric functions involve polynomials in the form of sum or difference of two squares. So, whenever you see a sum of two squares or a difference two squares as a radicand, more often than not, that integral will yield inverse trigonometric function. ILLUSTRATION: 1. Evaluate √ − . Solution: We first identify and to be able to transform the problem to one of the forms in the list of integrals given above. Let = and = , so that = and ∫ √ − = ∫ =⋅ = √ − + + MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) ∫ 9 TO DO: Verify that the answer in Ilustration 1 is correct by showing that ( + ) = √ − ILLUSTRATION: 2. Evaluate ∫ . + Solution: We first identify and to be able to transform the problem to one of the forms in the list of integrals given above. Let = √ and = , so that = − ⇒ − = , and ∫ = − ∫ + + =− =− √ + ( √ )+ Some problems would require us to complete the square of some polynomial first so that its form would become a sum or difference of two squares. This scenario is demonstrated below. ILLUSTRATION: 3. Evaluate ∫ √ − . Solution: Observe that the expression inside the radical, − , is not a sum or difference of two squares. However, we can do some manipulations to transform the given polynomial to our desired form. Recall first that to complete the square of a polynomial + , we add the constant ( ) since + + ( ) = ( + ) . Hence, to manipulate the expression − Next, we identify and to be able to transform the problem to one of the forms in the list of integrals given above. Let = and = − , so that = , and ∫ = ∫ = ∫ √ − √ − √ − ( − ) = − + = ( )+ MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) , we add − , which is equal to . In this way, we are not changing the given problem. Now, − = − + − = − ( − + ) = − ( − ) Thus, ∫ = ∫ . √ − √ − ( − ) 10 TO DO: Verify that the answer in Illustration 3 is correct by showing that − ( ( ) + ) = √ − TRY THIS!!! Evaluate the following. 1. 2. 3. 4. 5. 6. 7. 8. ∫ √ − ∫ √ − ∫ ∫ ∫ ∫ ∫ ∫ √ − + + √ (Assume that ( ) = .) − √ − + √ − For more exercises, you can refer to: Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage Learning Asia Pte. Ltd., p. 380 Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7th sub-edition. Harper Collins Publishers. p. 511 ___________ Evaluate , and MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) For an online tutorial, follow this link: 11 1.3 Derivatives of and Integrals Yielding Logarithmic Functions We now consider logarithmic functions. For a review of definition and properties of logarithmic functions, particularly the natural logarithmic function, refer to Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7 th sub-edition. Harper Collins Publishers. pp. 439-447, 472- 473. The natural logarithmic function defined by () = is continuous over (, +∞). Moreover, lim+ = −∞ and lim = +∞ . →0 →+∞ Using an alternative definition for in terms of definite integral and the First Fundamental Theorem of Calculus (FTC 1), we will now formulate the derivative of . But first, we recall FTC 1. RECALL!!! First Fundamental Theorem of Calculus Let be a continuous real-valued function defined on a closed interval [, ]. Let be the function defined, for all in [, ], by () = ∫ () . Then, is uniformly continuous on [, ], differentiable on the open interval (, ), and ′() = () for all in (, ). We now formulate ( ) . The natural logarithmic function defined by () = can defined alternatively as = ∫ . Then, by FTC 1, ( ) = (∫ ) = MUST REMEMBER!!! ( ) = and if is a differentiable function of , then ⋅ ILLUSTRATION: 1. Solve for ′ () if () = ( + ). Solution: () = ( + ) ⇒ ′ () = = ⋅ ( + ) + ( + ) ⋅ ( + ) = = + + MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) ( ) = 12 ILLUSTRATION: 2. Solve for ′ () if () = (√ ⋅ ). Solution: () = (√ ⋅ ) ⇒ ′ () = = = √ ⋅ √ ⋅ ⋅ (√ ⋅ ) ⋅ (√ ⋅ + ⋅ √ ) + = + TO DO: 3. Solve for ′ () if () = , where > , ≠ . Solution: Note that = and that is a constant. MUST REMEMBER!!! Derivative of Logarithmic Function If is a differentiable function of , then ( ) = ⋅ ⋅ ILLUSTRATION: 4. Evaluate [( )] . Solution: [( )] = ⋅ ⋅ ( ) = ⋅ ⋅ ⋅ () = ⋅ ⋅ ⋅ ⋅ MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) Note that the given result above is for a general logarithmic function. Hence, the said result will also hold for the natural logarithmic function. Can you show that ( ) = ⋅ by using the result above? Thus, all you need to remember when it comes to derivative of logarithmic function is the said result. 13 TRY THIS! Solve for if 1. = ( ) 2. = ( ) 3. = ( + ) 4. = ( − ) 5. = ( ) 6. = ( √) 7. = ( ( − + )) 8. = , where is a constant For more exercises, you can refer to: Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage Learning Asia Pte. Ltd., p. 362 Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7th sub-edition. Harper Collins Publishers. pp.449, 457, 476 For an online tutorial, follow these links: MUST REMEMBER!!! If is a differentiable function of , then ∫ = || + . Observe that ∫ = || + instead of . Why is that so? Review the respective domains of and . Moreover, you can solve for (||) by using the definition of || as a conditional function. MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) The following result directly follows from the previous discussions. 14 ILLUSTRATION: 1. Evaluate ∫ . + Solution: At first look, we might think that this integral will yield inverse trigonometric function because of the presence of sum of two squares in the denominator. However, upon careful inspection, we will realize that the integral is solvable by simple substitution 1 and will yield logarithmic function. We let = 2 + 4 so that = 2 and = . 2 Hence, ∫ = ∫ + = || + = | + | + = ( + ) + (Why?) ILLUSTRATION 2. Evaluate ∫ . Solution: Recall that = . Then, we let = so that = − and = −. Hence, ∫ = ∫ = − ∫ = − || + = − | | + = |( )− | + = | | + TO DO: Using similar procedure, show that Integrals of other trigonometric functions are given below. MUST REMEMBER!!! Integrals of the “Other” Trigonometric Functions If is a differentiable function of , then ∫ = | | + ∫ = | + | + ∫ = | | + ∫ = | − | + MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) ∫ = | | + . 15 TRY THIS!!! Evaluate the following. 1. 2. 3. 4. 5. 6. 7. 8. ∫ + ∫ + + ∫ ∫ ∫ − ∫ √ √ where and are constants ∫ ∫ ( ) For more exercises, you can refer to: Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7th sub-edition. Harper Collins Publishers. p. 457, 476 For an online tutorial, follow this link: ___________ REMINDER: Before proceeding to the next section, students are advised to refer to pages 4-5 of Module I. Derivatives of MATH Preparatory Modules to have an idea of Evaluate , 27 BASIC CALCULUS and implicit differentiation since this technique will be used in the following discussions. Differentiate the following function: () = ⋅ ⋅ √ + How many times did you use Quotient Rule? How about Product Rule? By using only basic theorems on differentiation to solve the given problem, it is undeniable that the process is quite tedious. In addition, the probability of committing errors in computation is also high since it involves multiple processes. MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) 1.4 Logarithmic Differentiation 16 In this section, we will introduce an alternative way of solving problems of the same nature as the one given earlier. This method, called logarithmic differentiation, makes it easier to deal with differentiating “super products,” “super quotients,” and composite functions in the form of function raised to another function like () = ( ) . MUST REMEMBER! How to do logarithmic differentiation? Given = (). 1. Consider || = |()|. Get the natural logarithms of both sides of the equation, i.e. || = |()|. Note that (||) = . 2. Use properties of logarithms to express |()| as sums instead of products, as difference instead of quotients and products instead of exponentiations. 3. Get the derivatives of both sides of || = |()| implicitly. Hence, ⋅ = (|()|). 4. Solve for by cross-multiplying and expressing in terms of . ILLUSTRATION: 1. Use logarithmic differentiation to solve for = if ⋅ ⋅ √ + Solution: ⋅ ⋅ √ + ⋅ ⋅ √ + ⋅ ⋅ √ + = ⇒ || = | | ⇒ || = | | ⇒ || = | ⋅ ⋅ √ + | − | | ⇒ || = || + | | + |( + ) | − | | ⇒ ⋅ = + ⋅ ( ) + ⋅ ⋅ ( + ) − ⋅ ( ) + ⇒ ⋅ = + ⋅ + ⋅ − ⋅ ( + ) + ⇒ = ( + + − ⋅ ) + + ⇒ ⋅ ⋅ √ + = ( + + − ⋅ ) + + Compare your first solution to the same problem with the one above. Which one is less complicated? MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) ⇒ || = || + | | + | + | − | | 17 ___________ ILLUSTRATION: 2. Use logarithmic differentiation to solve for if = . Solution: = ⇒ || = | | ⇒ || = | | ⇒ || = ⋅ || ⇒ ⋅ = ⋅ ( ||) + || ⋅ ( ) ⇒ ⋅ = ⋅ + || ⋅ ⇒ = ( ⋅ + || ⋅ ) ⇒ = ( ⋅ + || ⋅ ) by Product Rule Note that in the previous illustration, the only way to do it is by using logarithmic differentiation since we cannot use Power Rule for = . Can you explain why? TRY THIS! Use logarithmic differentiation to solve for 1. = if ⋅ ___________ 2. = 3. = ⋅ ( ) 4. = 5. = ( ) √ ⋅ Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage Learning Asia Pte. Ltd., p. 326 Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7th sub-edition. Harper Collins Publishers. p. 457 MATH 27 Lecture Guide UNIT 1 (IMSP,UPLB) For more exercises, you can refer to: 18 1.5 Derivatives of and Integrals of Exponential Functions We now focus our attention to exponential functions. For a quick review of concepts and properties regarding exponential function, particularly the natural exponential function, refer to Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7th sub-edition. Harper Collins Publishers. pp. 458-462. The natural exponential function defined by () = is continuous at every real number. Also, = an...
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