Differentiate algebraic and transcendental functions
MODULE 3 - DERIVATIVES
OVERVIEW
This module focuses on how to solve derivatives of a function using constant rule, power rule, constant multiple rule, and sum and difference rule, product rule and quotient rule.
This module focuses on how to solve derivatives of a function using constant rule, power rule, constant multiple rule, and sum and difference rule, product rule and quotient rule.
The derivative is an important tool in calculus that represents an infinitesimal change in a function with respect to one of its variables. Given a function f(x), there are many ways to denote the derivative of f with respect to x. Shown on the right is the common notations for derivatives. Personally, I prefer using the y prime notation because among the three, it is the simplest and not so complicated notation. Meanwhile, derivative is significant because many physical phenomena, such as velocity, acceleration, and force, are defined as instantaneous rates of change of some other quantity. The derivative can provide you with an exact instantaneous number for that rate of change, allowing you to model the required quantity precisely.
THE DERIVATIVE AS A FUNCTION
The derivative of the function f is that function, denoted by f′, such that its value at any number x in the domain of f is given by the expression on the left, only if the limit exists.
The derivative of the function f is that function, denoted by f′, such that its value at any number x in the domain of f is given by the expression on the left, only if the limit exists.
BASIC DERIVATIVE RULES
EXAMPLE 1. FINDING THE DERIVATIVE OF A SUM OF FUNCTIONS
This example was taken from my Quiz 2. Note that the Sum and Difference Rule mentioned in the table above allows us to calculate the derivative of a function with multiple terms by adding together the derivatives of each of the terms. For instance. in order to find the derivative of the given function on the left, it is imperative to find first the derivative of each of the term then proceed to adding or subtracting them together accordingly. For example, I got the derivative of 6x-2 by applying both the constant multiple rule and the power rule. 6x-2 is composed of a constant 6 multiplied by a function x-2. As per the constant multiple rule, we need to factor out the constant 6 then take the derivative of the function x-2 to be multiplied to the factored-out constant. Taking the derivative of the function x-2 allows us to apply the power rule in which we need to multiply the exponent to the variable then subtract one from the exponent. By applying these all, the derivative of x-2 is -2x-3 that when multiplied to 6 will be equal to -12x-3.
This example was taken from my Quiz 2. Note that the Sum and Difference Rule mentioned in the table above allows us to calculate the derivative of a function with multiple terms by adding together the derivatives of each of the terms. For instance. in order to find the derivative of the given function on the left, it is imperative to find first the derivative of each of the term then proceed to adding or subtracting them together accordingly. For example, I got the derivative of 6x-2 by applying both the constant multiple rule and the power rule. 6x-2 is composed of a constant 6 multiplied by a function x-2. As per the constant multiple rule, we need to factor out the constant 6 then take the derivative of the function x-2 to be multiplied to the factored-out constant. Taking the derivative of the function x-2 allows us to apply the power rule in which we need to multiply the exponent to the variable then subtract one from the exponent. By applying these all, the derivative of x-2 is -2x-3 that when multiplied to 6 will be equal to -12x-3.
EXAMPLE 2. FINDING THE DERIVATIVE OF A PRODUCT OF FUNCTIONS
The Product Rule is applied in finding the derivative of a product of two functions. When we translate the formula of the product rule above, it is as simple as this: the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. Note that it can be interchanged such as like this: the second function times the derivative of the first function plus the first function times the derivative of the second function because of the commutative property of addition which states that changing the order of the addends does not affect the sum. Nonetheless, I still followed the stated formula on the table above. For instance, 5x2 times the derivative of (x + 47) plus (x + 47) times the derivative of 5x2 is equal to 15x2 + 470x. By the way, this was taken from my Quiz 2.
The Product Rule is applied in finding the derivative of a product of two functions. When we translate the formula of the product rule above, it is as simple as this: the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. Note that it can be interchanged such as like this: the second function times the derivative of the first function plus the first function times the derivative of the second function because of the commutative property of addition which states that changing the order of the addends does not affect the sum. Nonetheless, I still followed the stated formula on the table above. For instance, 5x2 times the derivative of (x + 47) plus (x + 47) times the derivative of 5x2 is equal to 15x2 + 470x. By the way, this was taken from my Quiz 2.
EXAMPLE 3. FINDING THE DERIVATIVE OF A QUOTIENT OF FUNCTIONS
This example was again taken from my Quiz 2. As per the formula above, a Quotient Rule is stated as the ratio of the quantity of the denominator times the derivative of the numerator function minus the numerator times the derivative of the denominator function to the square of the denominator function. To easily memorize this formula, our instructor taught us simple terms. The numerator must be called "high," the denominator must be called "low," and the derivative must be called "de." As such, the formula would be like this: "low de-high minus high de-low, all over low squared." For instance, as per the taken example it would be x2 + 6x times the derivative of x2 + 5 minus x2 + 5 times the derivative of x2 + 6x all over the square of x2 + 6x. I got the derivative of x2 + 5 through the power rule and the constant rule, respectively. The power rule will give us the derivative of x2 which is 2x by multiplying the exponent 2 to the variable x then subtracting one from the exponent. The constant rule will give us the derivative of 5 which is 0 because the rule states that the derivative of a constant function is always zero. 2x + 0 is still 2x. Furthermore, I got the derivative of x2 + 6x through the power rule and the constant multiple rule, respectively. Again, the derivative of x2 is 2x by the power rule. The constant multiple rule will give us the derivative of 6x which is 6 because the rule tells us to factor out the constant 6 and get the derivative of x which is 1 to be multiplied to 6, so 6 times 1 is 6. Thus, the derivative of x2 + 6x is 2x + 6. Finally, I arrived to the final answer by algebraic manipulation specifically by multiplying terms accordingly, combining like terms, and factoring.
This example was again taken from my Quiz 2. As per the formula above, a Quotient Rule is stated as the ratio of the quantity of the denominator times the derivative of the numerator function minus the numerator times the derivative of the denominator function to the square of the denominator function. To easily memorize this formula, our instructor taught us simple terms. The numerator must be called "high," the denominator must be called "low," and the derivative must be called "de." As such, the formula would be like this: "low de-high minus high de-low, all over low squared." For instance, as per the taken example it would be x2 + 6x times the derivative of x2 + 5 minus x2 + 5 times the derivative of x2 + 6x all over the square of x2 + 6x. I got the derivative of x2 + 5 through the power rule and the constant rule, respectively. The power rule will give us the derivative of x2 which is 2x by multiplying the exponent 2 to the variable x then subtracting one from the exponent. The constant rule will give us the derivative of 5 which is 0 because the rule states that the derivative of a constant function is always zero. 2x + 0 is still 2x. Furthermore, I got the derivative of x2 + 6x through the power rule and the constant multiple rule, respectively. Again, the derivative of x2 is 2x by the power rule. The constant multiple rule will give us the derivative of 6x which is 6 because the rule tells us to factor out the constant 6 and get the derivative of x which is 1 to be multiplied to 6, so 6 times 1 is 6. Thus, the derivative of x2 + 6x is 2x + 6. Finally, I arrived to the final answer by algebraic manipulation specifically by multiplying terms accordingly, combining like terms, and factoring.
EXAMPLE 4. FINDING THE DERIVATIVE USING THE CHAIN RULE
The Chain Rule is used whenever we have to find the derivative of a composition of functions. In my own opinion, thinking the chain rule formula in terms of function composition somewhat gives confusion on my part. That is why I only associate the chain rule formula with the power rule with an additional derivative. For instance, it goes like this: Using the power rule formula, multiply the exponent to the function inside the parenthesis, then subtract one from the exponent. The additional derivative is simply the derivative of the inside function. Take a look at my example on the right taken from my Quiz 2. The square root of 4x + 2 is equal to (4x + 2)1/2. Thus, we can now use the chain rule. By using the power rule, I arrived at 1/2(4x + 2)-1/2 to be multiplied to the derivative of 4x + 2 which is 4 (I used constant multiple rule and constant rule). Then to get rid of the negative exponent, I put the function in the denominator and did some cancellation to arrive to the final answer. Note that my answer was again in square root form because any function raised to 1/2 can be transformed to radical form specifically square root form.
The Chain Rule is used whenever we have to find the derivative of a composition of functions. In my own opinion, thinking the chain rule formula in terms of function composition somewhat gives confusion on my part. That is why I only associate the chain rule formula with the power rule with an additional derivative. For instance, it goes like this: Using the power rule formula, multiply the exponent to the function inside the parenthesis, then subtract one from the exponent. The additional derivative is simply the derivative of the inside function. Take a look at my example on the right taken from my Quiz 2. The square root of 4x + 2 is equal to (4x + 2)1/2. Thus, we can now use the chain rule. By using the power rule, I arrived at 1/2(4x + 2)-1/2 to be multiplied to the derivative of 4x + 2 which is 4 (I used constant multiple rule and constant rule). Then to get rid of the negative exponent, I put the function in the denominator and did some cancellation to arrive to the final answer. Note that my answer was again in square root form because any function raised to 1/2 can be transformed to radical form specifically square root form.
MODULE 4 - IMPLICIT DIFFERENTIATION AND HIGHER ORDER DERIVATIVES
OVERVIEW
The intent of this section is to give you the skills necessary to work the related- rate problems. We will show you how to use implicit differentiation and how to find higher order derivatives. Knowing how to do implicit differentiation is a key skill that is necessary for our discussion of related rates.
The intent of this section is to give you the skills necessary to work the related- rate problems. We will show you how to use implicit differentiation and how to find higher order derivatives. Knowing how to do implicit differentiation is a key skill that is necessary for our discussion of related rates.
IMPLICIT DIFFERENTIATION
The technique of implicit differentiation allows us to find the derivative of y with respect to x without having to solve the given equation for y. The chain rule must be used whenever the function y is being differentiated because of the assumption that y may be expressed as a function of x. There are five simple steps to follow use implicit differentiation. See the picture on the right taken from onlinemathlearning.com.
The technique of implicit differentiation allows us to find the derivative of y with respect to x without having to solve the given equation for y. The chain rule must be used whenever the function y is being differentiated because of the assumption that y may be expressed as a function of x. There are five simple steps to follow use implicit differentiation. See the picture on the right taken from onlinemathlearning.com.
EXAMPLE 5. FINDING DERIVATIVE BY IMPLICIT DIFFERENTIATION
By following the five steps mentioned above, I was able to easily get the correct answer of this item from my Quiz 2. First thing I did is differentiate the both sides of the equation with respect to x (1) by of course following the basic derivative rules such as product rule, power rule, and constant rule while also including dy/dx every time an expression involves y (2). Then, I did some algebraic manipulation to isolate the dy/dx terms in only one side of the equation (3), then I solved for dy/dx by factoring it out (4) and dividing the terms beside it(5). Note that I did cancellation to get rid of the negative sign and make my answer look more appealing.
By following the five steps mentioned above, I was able to easily get the correct answer of this item from my Quiz 2. First thing I did is differentiate the both sides of the equation with respect to x (1) by of course following the basic derivative rules such as product rule, power rule, and constant rule while also including dy/dx every time an expression involves y (2). Then, I did some algebraic manipulation to isolate the dy/dx terms in only one side of the equation (3), then I solved for dy/dx by factoring it out (4) and dividing the terms beside it(5). Note that I did cancellation to get rid of the negative sign and make my answer look more appealing.
HIGHER ORDER DERIVATIVES
Since the derivative of a function y = f(x) is itself a function y′ = f′(x), the derivative of f′(x), also referred as the second derivative of f(x), can be expressed as f"(x) or f2(x). This differentiation process can be repeated to obtain the third, fourth, and subsequent derivatives of f(x), also known as higher order derivatives of f(x). Note that there is a so called numerical notation f(n)(x) = y(n) to denote the nth derivate of f(x) such as the fourth derivative on the picture on the right. This can help us avoid confusions of how many lines have we drawn above the y or f. After all, our solutions would look messy whenever we are asked to find the tenth derivative of a certain function. Furthermore, notice the parenthesis used to express the fourth derivative. The presence of parenthesis in the exponent denotes differentiation while the absence of parenthesis denotes exponentiation.
Since the derivative of a function y = f(x) is itself a function y′ = f′(x), the derivative of f′(x), also referred as the second derivative of f(x), can be expressed as f"(x) or f2(x). This differentiation process can be repeated to obtain the third, fourth, and subsequent derivatives of f(x), also known as higher order derivatives of f(x). Note that there is a so called numerical notation f(n)(x) = y(n) to denote the nth derivate of f(x) such as the fourth derivative on the picture on the right. This can help us avoid confusions of how many lines have we drawn above the y or f. After all, our solutions would look messy whenever we are asked to find the tenth derivative of a certain function. Furthermore, notice the parenthesis used to express the fourth derivative. The presence of parenthesis in the exponent denotes differentiation while the absence of parenthesis denotes exponentiation.
EXAMPLE 6. FINDING THE NTH DERIVATIVE USING THE HIGHER ORDER DERIVATIVES
This example was again taken from my Quiz 2. Note that I was asked to find the second derivative only of the given function f(x) = 5x2 (x + 47). Before I arrive to the second derivative, of course I need to find the first derivative of the given function. By observing, we can say that the function is a product of two functions, thus the need to apply the product rule. The product rule states that the first function times the derivative of the second function plus the second function times the derivative of the second function. Through this as well as the application of other basic derivative rules such as the constant rule, constant multiple rule, and power rule, I was able to arrive to the first derivative f′(x) which is equal to 15x2 + 470x. Finally, when we get the derivative of this function, we would arrive to the second derivative prior to what is asked in the problem. It was easy for me to get the second derivative since it only applies the sum rule and the other basic rules such as the constant multiple rule and the power rule. At last the second derivative f"(x) is equal to 30x + 470.
This example was again taken from my Quiz 2. Note that I was asked to find the second derivative only of the given function f(x) = 5x2 (x + 47). Before I arrive to the second derivative, of course I need to find the first derivative of the given function. By observing, we can say that the function is a product of two functions, thus the need to apply the product rule. The product rule states that the first function times the derivative of the second function plus the second function times the derivative of the second function. Through this as well as the application of other basic derivative rules such as the constant rule, constant multiple rule, and power rule, I was able to arrive to the first derivative f′(x) which is equal to 15x2 + 470x. Finally, when we get the derivative of this function, we would arrive to the second derivative prior to what is asked in the problem. It was easy for me to get the second derivative since it only applies the sum rule and the other basic rules such as the constant multiple rule and the power rule. At last the second derivative f"(x) is equal to 30x + 470.
Salazar (BSCE 1B) - Assignment No. 3.pdfASSIGNMENT NO. 3
Quiz 2 - Salazar (BSCE1B).pdfPROBLEM SET 2 (30/30)
Salazar (BSCE 1B) - Problem Set 2.pdfQUIZ 2 SOLUTIONS (40/40)
REFLECTION
In general, Calculus 1 helped me to review some concepts and at the same time taught me new lessons that I believe would be valuable in future computations in the civil engineering course. The topics contained in the first and second learning outcomes (CLO1 and CLO2) were already discussed to us during my Senior High School through the subject Basic Calculus. Thanks to Calculus 1 because I was able to review some useful topics such as functions in Module 1, limits in Module 2, derivatives in Module 3, implicit differentiation and higher order derivatives in Module 4, and derivatives of trigonometric functions in Module 5. Relearning the aforementioned topics gave me satisfaction in this subject since I had already forgotten some of them. The subject somewhat allowed me to reminisce my last days in school because I remember I was already reviewing my notes back then in Basic Calculus in preparation of the final exam, but unfortunately, it was cancelled due to the start of the nationwide lockdown. Going back, I found new topics in this subject such as the derivatives of inverse trigonometric functions of Module 6; derivatives of logarithmic and exponential functions in module 7; tangents and normals, extrema and first derivative test (FDT), concavity, points of inflection, second derivative test, increasing/decreasing intervals, and ultimately graphing functions. Furthermore, I found the last topics to be a bit challenging especially when it comes to graphing of functions that I encountered some confusions. Yet, I am grateful to our instructor for addressing my concerns regarding the topics. Nevertheless, I still enjoyed the whole subject because it is aligned on my interest of solving math problems. I just cannot explain the feeling of satisfaction after finally arriving to every correct answer of the problem sets, assignments, and quizzes given to us. A big thank you to our instructor for discussing the course in a manner that everybody could understand. Probably, I wouldn't enjoy the subject without the pure efforts and dedication of our instructor.
In general, Calculus 1 helped me to review some concepts and at the same time taught me new lessons that I believe would be valuable in future computations in the civil engineering course. The topics contained in the first and second learning outcomes (CLO1 and CLO2) were already discussed to us during my Senior High School through the subject Basic Calculus. Thanks to Calculus 1 because I was able to review some useful topics such as functions in Module 1, limits in Module 2, derivatives in Module 3, implicit differentiation and higher order derivatives in Module 4, and derivatives of trigonometric functions in Module 5. Relearning the aforementioned topics gave me satisfaction in this subject since I had already forgotten some of them. The subject somewhat allowed me to reminisce my last days in school because I remember I was already reviewing my notes back then in Basic Calculus in preparation of the final exam, but unfortunately, it was cancelled due to the start of the nationwide lockdown. Going back, I found new topics in this subject such as the derivatives of inverse trigonometric functions of Module 6; derivatives of logarithmic and exponential functions in module 7; tangents and normals, extrema and first derivative test (FDT), concavity, points of inflection, second derivative test, increasing/decreasing intervals, and ultimately graphing functions. Furthermore, I found the last topics to be a bit challenging especially when it comes to graphing of functions that I encountered some confusions. Yet, I am grateful to our instructor for addressing my concerns regarding the topics. Nevertheless, I still enjoyed the whole subject because it is aligned on my interest of solving math problems. I just cannot explain the feeling of satisfaction after finally arriving to every correct answer of the problem sets, assignments, and quizzes given to us. A big thank you to our instructor for discussing the course in a manner that everybody could understand. Probably, I wouldn't enjoy the subject without the pure efforts and dedication of our instructor.
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