**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. 1.1 Derivatives of and Integrals Yielding Trigonometric Functions MATH 27 Lecture Guide UNIT 1 1 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
( )
( )
( ) Domain ( )
( )
( ) { } * + { } * + Morever, the given basic trigonometric functions are continuous over their respective domains.
Using the definition of a derivative,
( ( ) ) ( ) it can be shown that
( ) ( Now, formulate
already written. and
) and ( ( ) . ) using the result above. The first few steps are TO DO:
( ) . / (HINT: Use quotient rule for differentiation) TO DO:
) (
,( )
) - MATH 27 Lecture Guide UNIT 1 ( 2 Following the same procedure, it can also be shown that
( ) and ( ) 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 ) by Sum/Difference Rule ) ILLUSTRATION: Solution:
( ( ).
) (
(
( ) ) ( ) (
( ) ) by Chain Rule MATH 27 Lecture Guide UNIT 1 2. Evaluate 3 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, 10 th 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 your basic calculus.
Recall that
( ) ( ) ∫ ( ) ( ) In this case, we call ( ) an antiderivative of ( ). We also say that the indefinite integral of ( )
with respect to is ( )
.
From the previous discussions, we have the following results. Let be a differentiable function of .
∫ ∫ ∫ ∫ ∫ ∫ MATH 27 Lecture Guide UNIT 1 MUST REMEMBER!!! Integrals Yielding Trigonometric Functions 4 ILLUSTRATION:
1. Evaluate
∫ .
Solution:
Note that ( ) . Hence, we can let and the given integral can now be rewritten as , so that ∫ .
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 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.
∫ ∫( ∫ )
∫ MATH 27 Lecture Guide UNIT 1 ∫ . 5 ILLUSTRATION: (cont’n)
To verify that the answer is correct, we show that ( ( ) ( ) ) . 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, 10 th Edition. Philippines: Cengage
Learning Asia Pte. Ltd., p. 334
MATH 27 Lecture Guide UNIT 1 For an online tutorial, follow these links: 6 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.
We first recall the six basic inverse trigonometric functions and their respective domain.
Function
)
)
)
)
)
) (
(
(
(
(
( Domain
,
,
- (
( - ,
, )
) 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, 7 th 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, implicitly, we have, . Getting the derivative of both sides of the equation Using the same technique, we can show the following results. ( ) ( ) ( ) ( ) ( ) ( ) MATH 27 Lecture Guide UNIT 1 MUST REMEMBER!!! Derivatives of Inverse Trigonometric Functions 7 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.,
( ) ( ) . 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, ( ) ( ) ILLUSTRATION: 1. Solve for if ( ) ) ( ) ( ) ( ) ( ) ( Solution: ( ) ( ) ( ) ILLUSTRATION: 2. Solve for if , ( )- Solution: , ( )- , ( )- ( ( ) ) , ( )- , ( )- ( ) MATH 27 Lecture Guide UNIT 1 8 TRY THIS! Solve for if 1. ./ 2. 3. ( ) 4. 5. ( ) 6. ( ) 7. ( ) 8. ( ) For more exercises, you can refer to:
Ron Larson & Bruce H. Edwards. (2016). Calculus, 10 th Edition. Philippines: Cengage
Learning Asia Pte. Ltd., p. 372
Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7 th 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. ∫ ∫
∫ MATH 27 Lecture Guide UNIT 1 MUST REMEMBER!!! Integrals Yielding Inverse Trigonometric Function
Let be a differentiable function of , and be a constant. Then , 9 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
∫ ∫ TO DO:
Verify that the answer is correct by showing that ( ) ILLUSTRATION:
2. Evaluate
∫ . ∫ ∫ ( ) MATH 27 Lecture Guide UNIT 1 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 10 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 , we add , which is equal to . In this way, we are not changing the given
problem.
Now, ∫ ( ) ( ) Thus, ∫ ( ) . 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 ) ( ( ∫ ) TO DO:
Verify that the answer is correct by showing that ) ) MATH 27 Lecture Guide UNIT 1 ( ( 11 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, 10 th Edition. Philippines: Cengage
Learning Asia Pte. Ltd., p. 380
Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7 th sub-edition. Harper
Collins Publishers. p. 511 For an online tutorial, follow this link: MATH 27 Lecture Guide UNIT 1 12 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
Moreover,
and is continuous over ( ). 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 ( ) 13 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 ( ) 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.
ILLUSTRATION:
4. Evaluate ,( )- . ,( )- ( ) () MATH 27 Lecture Guide UNIT 1 Solution: 14 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, 10 th Edition. Philippines: Cengage
Learning Asia Pte. Ltd., p. 362
Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7 th sub-edition. Harper
Collins Publishers. pp.449, 457, 476
For an online tutorial, follow these links:
The following result directly follows from the previous discussions.
MUST REMEMBER!!!
If is a differentiable function of , then Observe that ∫ instead of domains of and
. Moreover, you can solve for
conditional function. . . Why is that so? Review the respective
( ) by using the definition of as a MATH 27 Lecture Guide UNIT 1 ∫ 15 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 . Hence, . Then, we let ∫ ∫ so that ∫ and ( ) TO DO: Using similar procedure, show that
∫ . Integrals of other trigonometric functions are given below. If is a differentiable function of , then
∫ ∫ ∫ ∫ MATH 27 Lecture Guide UNIT 1 MUST REMEMBER!!! Integrals of the “Other” Trigonometric Functions 16 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, 7 th sub-edition. Harper
Collins Publishers. p. 457, 476
For an online tutorial, follow this link: MATH 27 Lecture Guide UNIT 1 17 1.4 Logarithmic Differentiation
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.
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 () . Hence, ( () ). 4. Solve for by cross-multiplying and expressing in terms of .
ILLUSTRATION: 1. Use logarithmic differentiation to solve for if ) ( ( ( ) ( ) ( ( ) ( ) ) ) MATH 27 Lecture Guide UNIT 1 Solution: 18 Compare your first solution to the same problem with the one above. Which one is less
complicated?
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 if 1. 2. 3. ( ) 4. 5. ( ) Ron Larson & Bruce H. Edwards. (2016). Calculus, 10 th Edition. Philippines: Cengage
Learning Asia Pte. Ltd., p. 326
Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7 th sub-edition. Harper
Collins Publishers. p. 457 MATH 27 Lecture Guide UNIT 1 For more exercises, you can refer to: 19 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, 7 th sub-edition. Harper
Collins Publishers. pp. 458-462. Also, The natural exponential function defined by ( )
is continuous at every real number.
and
. In general, for a constant , if
> , then
and
. However, if
, then
and
. Next, we formulate the derivative of . Recall that by definition,
implicit differentiation to solve for ( ). The first few lines are already given. . Use TO DO: . Determine if .
Solution: By using implicit differentiation, MUST REMEMBER!!! Derivative of Natural Exponential Function
If is a differentiable function of , then ( ) . In a similar manner, and using the fact that
result for a general exponential function. , we can also show the next MUST REMEMBER!!! Derivative of Exponential Function
If is a differentiable function of , then Note that if we let
exponential function since . in the preceding statement, we have the result for the natural
. Thus, all you need to remember is the one given above. MATH 27 Lecture Guide UNIT 1 ( ) 20 ILLUSTRATION: 1. Evaluate . /. Solution: . / ( ( ) ) ILLUSTRATION:
2. Evaluate .
Solution: ( ) ( ) by Product Rule ( ) ( ) 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, 10 th Edition. Philippines: Cengage
Learning Asia Pte. Ltd., pp. 352, 362
Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7 th sub-edition. Harper
Collins Publishers. pp. 467, 476 MATH 27 Lecture Guide UNIT 1 21 For an online tutorial, follow these links:
From the earlier discussions, it is easy to see the following facts.
MUST REMEMBER!!! Integrals of Exponential Functions
If is a differentiable function of , then
∫ and ∫ . Similar to our previous remark, you can just remember the second integral only since the first
can be obtained from the second by letting
. ILLUSTRATION 1. Evaluate
∫ Solution:
We can let so that ∫ . and . Hence, ∫ TO DO: . MATH 27 Lecture Guide UNIT 1 Verify that the answer is correct by showing 22 ILLUSTRATION
2. Evaluate
∫ Solution:
We can let . and so that ∫ . Hence, ∫ TO DO:
Verify that the answer is correct by showing . TRY THIS!!!
Evaluate the following. 1. ∫ 2. ∫ 3. ∫ 5. 6. ∫ ∫ 7. 8. ∫ ∫ ( ) MATH 27 Lecture Guide UNIT 1 4. ∫ 23 For more exercises, you can refer to:
Ron Larson & Bruce H. Edwards. (2016). Calculus, 10 th Edition. Philippines: Cengage
Learning Asia Pte. Ltd., pp. 354, 363
Louis Leithold.(1997). The Calculus 7 of a Single Variable, 7 th sub-edition. Harper
Collins Publishers. pp. 467, 476 For an online tutorial, follow this link: MATH 27 Lecture Guide UNIT 1 24 1.6 Some Application on Optimization, Related Rates and Laws of Natural Growth and Decay
In this section, we consider some applications of derivatives of transcendental functions to
real- world problems. In particular, we look into problems involving optimization, related rates and
laws of natural growth and decay.
We first recall from our basic calculus the steps in solving optimization problems.
RECALL: How to solve maximization/min...

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