Apply the concept of differentiation in solving word problems involving
optimization, related rates, and approximation
MODULE 5 - DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
According to cuemath.com, the process of finding the derivatives of trigonometric functions is called the differentiation of trigonometric functions. Differentiation of trigonometric functions is a mathematical procedure that determines the rate of change of trigonometric functions with respect to a variable angle in trigonometry. The six basic trigonometric functions namely sine (sin x), cosine (cos x), tangent (tan x), cotangent (cot x), secant (sec x) and cosecant (cosec x) have differentiation formulas shown on the right that can be used in different fields such as electronics, computer engineering, and modeling different cyclic functions. Furthermore, they can also be used in finding the slope of the tangent line and normal line to a trigonometric curve y = f(x) which further determine the equations of the said lines. Lastly, they can also be used in finding the maximum and minimum values of various functions. Meanwhile, after following these guide formulas in the process of differentiation, it is imperative to multiply it to the derivative of the x same as in the chain rule. I will provide examples below.
Pythagorean identities are three of the eight fundamental trigonometric identities which are derived from Pythagoras' theorem. These identities are used to solve various trigonometric situations in which one trigonometric ratio is given and the remaining ratios must be determined. Pythagorean identities are useful in simplifying trigonometric expressions, especially in writing expressions as a function of either sin or cos, as in statements of the double angle formulas. In the differentiation of trigonometric functions, they are sometimes used to further simplify the derivatives of particular functions.
EXAMPLE 1. FINDING THE DERIVATIVE OF TRIGONOMETRIC FUNCTIONS
This example was taken from my Quiz 2. From the given y = 6sec(3x), it can be inferred that it is a product of two functions (constant function and trigonometric function), 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 first function. For instance, 6 times the derivative of sec(3x) plus sec(3x) times the derivative of 6. To find the derivative of sec(3x), we will follow the secant differentiation rule which says that sec(x) is equal to sec(x)tan(x). Therefore, the derivative of sec(3x) is sec(3x)tan(3x) to be multiplied to the derivative of 3x which is 3. The derivative of the constant function 6 is 0, thus the second term will be 0. We will be left with the first term in which we need to combine 6 and the derivative of 3x which is 3, and 6 times 3 is 18. Therefore, the final answer would be 18sec(3x)tan(3x).
This example was taken from my Quiz 2. From the given y = 6sec(3x), it can be inferred that it is a product of two functions (constant function and trigonometric function), 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 first function. For instance, 6 times the derivative of sec(3x) plus sec(3x) times the derivative of 6. To find the derivative of sec(3x), we will follow the secant differentiation rule which says that sec(x) is equal to sec(x)tan(x). Therefore, the derivative of sec(3x) is sec(3x)tan(3x) to be multiplied to the derivative of 3x which is 3. The derivative of the constant function 6 is 0, thus the second term will be 0. We will be left with the first term in which we need to combine 6 and the derivative of 3x which is 3, and 6 times 3 is 18. Therefore, the final answer would be 18sec(3x)tan(3x).
EXAMPLE 2. FINDING THE DERIVATIVE OF TRIGONOMETRIC FUNCTIONS
This example was again taken from my Quiz 2. The given is a difference of two trigonometric functions, so it would be easy since the difference rule states that the derivative of the difference of two functions is as simple as getting the derivative of each term. As can be observed, the first thing I did is rewrite the function from this: y = sec4(x) - tan4(x) to this: y = (secx)4 - (tanx)4 so that I can begin with the chain rule. I applied the chain rule on both terms by bringing down the exponent to be multiplied to the inside function with minus 1 exponent, and then to be multiplied to the derivative of the inside function. From there, I need to get the derivative of the two inside functions which happened to be trigonometric functions, thus the need to apply the rules or formulas shown above. The derivative of sec(x) [by secant differentiation rule] is sec(x)tan(x) multiplied to the derivative of x which is 1; the derivative of tan(x) [by tangent differentiation rule] is sec2(x) multiplied to the derivative of x which is 1. Then, by combing like terms and factoring, I arrived to the answer 4sec2(x)tan(x) [sec2(x) - tan2(x)]. Here comes the application of the Pythagorean identities. 1 + tan2(x) = sec2(x), therefore, 1 = sec2(x) - tan2(x). Hence, what is inside the bracket of my answer is just equal to 1. By simplifying, my final answer would be 4sec2(x)tan(x).
This example was again taken from my Quiz 2. The given is a difference of two trigonometric functions, so it would be easy since the difference rule states that the derivative of the difference of two functions is as simple as getting the derivative of each term. As can be observed, the first thing I did is rewrite the function from this: y = sec4(x) - tan4(x) to this: y = (secx)4 - (tanx)4 so that I can begin with the chain rule. I applied the chain rule on both terms by bringing down the exponent to be multiplied to the inside function with minus 1 exponent, and then to be multiplied to the derivative of the inside function. From there, I need to get the derivative of the two inside functions which happened to be trigonometric functions, thus the need to apply the rules or formulas shown above. The derivative of sec(x) [by secant differentiation rule] is sec(x)tan(x) multiplied to the derivative of x which is 1; the derivative of tan(x) [by tangent differentiation rule] is sec2(x) multiplied to the derivative of x which is 1. Then, by combing like terms and factoring, I arrived to the answer 4sec2(x)tan(x) [sec2(x) - tan2(x)]. Here comes the application of the Pythagorean identities. 1 + tan2(x) = sec2(x), therefore, 1 = sec2(x) - tan2(x). Hence, what is inside the bracket of my answer is just equal to 1. By simplifying, my final answer would be 4sec2(x)tan(x).
MODULE 6 - DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS
The above-mentioned six basic trigonometric functions have their corresponding inverse functions (also called arcus functions) namely arcsine (arcsin x), arccosine (arccos x), arctangent (arctan x), arcsecant (arcsec x), arccosecant (arccsc x), and arccotangent (arccot x). Note that these inverse functions exist when appropriate restrictions are placed on the domain of the original functions. Furthermore, inverse trigonometric functions are used to calculate the angle of a given trigonometric value. They are also used in variety of fields such as engineering, geometry, and navigation among others. Shown on the right are the formulas of the derivatives of inverse trigonometric functions with u as a function of x. Note that u′ denotes du/dx.
EXAMPLE 3. FINDING THE DERIVATIVE OF INVERSE TRIGONOMETRIC FUNCTIONS
This example was taken from my Quiz 2. From the given, y = arccsc(x) - 4arccot(x), it can be inferred that it is a difference of two inverse trigonometric functions. Thus, it would be easy finding its derivative since the difference rule is as simple as getting the derivative of each term. Now, in finding the derivative of each term, we need to apply the derivative rules of differentiation of inverse trigonometric functions. For instance, to determine the derivative of the first term arccsc(x), we need to apply the arccosecant differentiation rule shown above. Note that the "u" in the formula is the function x. Meaning, the numerator -u′ denotes the negative of the derivative of x. The derivative of x is 1, and its negative is -1. For the denominator, no need to find the derivative, but just replace u with x. Thus, it would be the absolute value of x multiplied to the square root of x2 - 1. Furthermore, I applied the arccotangent differentiation rule in determining the derivative of the second term 4arccot(x). The presence of the function 4 implies the need for product rule. However, the derivative of a constant is 0, therefore, I just focused on the primary statement "the first function times the derivative of the second." Hence, 4 times the derivative of arccot(x). From the formula, the numerator is again -u′ which denotes the negative of the derivative of x, which is -1. For the denominator, just replace the u with x, thus 1 + x2. Negative 1 times negative 4 is positive 4. Thus the answer shown on the right is correct.
This example was taken from my Quiz 2. From the given, y = arccsc(x) - 4arccot(x), it can be inferred that it is a difference of two inverse trigonometric functions. Thus, it would be easy finding its derivative since the difference rule is as simple as getting the derivative of each term. Now, in finding the derivative of each term, we need to apply the derivative rules of differentiation of inverse trigonometric functions. For instance, to determine the derivative of the first term arccsc(x), we need to apply the arccosecant differentiation rule shown above. Note that the "u" in the formula is the function x. Meaning, the numerator -u′ denotes the negative of the derivative of x. The derivative of x is 1, and its negative is -1. For the denominator, no need to find the derivative, but just replace u with x. Thus, it would be the absolute value of x multiplied to the square root of x2 - 1. Furthermore, I applied the arccotangent differentiation rule in determining the derivative of the second term 4arccot(x). The presence of the function 4 implies the need for product rule. However, the derivative of a constant is 0, therefore, I just focused on the primary statement "the first function times the derivative of the second." Hence, 4 times the derivative of arccot(x). From the formula, the numerator is again -u′ which denotes the negative of the derivative of x, which is -1. For the denominator, just replace the u with x, thus 1 + x2. Negative 1 times negative 4 is positive 4. Thus the answer shown on the right is correct.
EXAMPLE 4. FINDING THE DERIVATIVE OF INVERSE TRIGONOMETRIC FUNCTIONS
This time, this example was taken from my Problem Set 3. Observing the given on the right, we can already say the need to apply the chain rule because of how it is written. By following the chain rule, I brought down the exponent 2 to be multiplied to the inside function with minus 1 exponent (which is arccos(4x2)) and which will be multiplied to the derivative of the inside function arccos(4x2). To determine the derivative of this function, we need to apply the arccosine differentiation rule. From the formula, the -u′ is the negative derivative of the function. The negative derivative of 4x2 is -8x (by constant multiple rule). For the denominator, I just replaced u of the u2 with 4x2. The square of 4x2 is 16x4. To simplify, I just multiplied the -8x to 2arccos(4x2) which is equal to -16xarccos(4x2). Kindly see my solution on the right for clarification.
This time, this example was taken from my Problem Set 3. Observing the given on the right, we can already say the need to apply the chain rule because of how it is written. By following the chain rule, I brought down the exponent 2 to be multiplied to the inside function with minus 1 exponent (which is arccos(4x2)) and which will be multiplied to the derivative of the inside function arccos(4x2). To determine the derivative of this function, we need to apply the arccosine differentiation rule. From the formula, the -u′ is the negative derivative of the function. The negative derivative of 4x2 is -8x (by constant multiple rule). For the denominator, I just replaced u of the u2 with 4x2. The square of 4x2 is 16x4. To simplify, I just multiplied the -8x to 2arccos(4x2) which is equal to -16xarccos(4x2). Kindly see my solution on the right for clarification.
MODULE 7 - DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
DERIVATIVES OF EXPONENTIAL FUNCTIONS
An exponential function is a mathematical function in the form f (x) = ax, where “x” is a variable and “a” is a constant which is the base of the function and must be positive and not equal to 1. The Euler's number or letter "e" is the most commonly used function base, and ex is the most commonly used exponential function which is also called the natural exponential function. In differentiating exponential functions, the power rule is not applicable since the variable of an exponential function is in the exponent. Shown on the left is the basic derivatives of exponential functions. Note that the exponential function ex has the property that it is its own derivative. It can be proved by using the basic exponential rule where the derivative of au is equal to au In(a) du/dx. Observe that the e is the constant a and x is the exponent u. In(e) is equal to 1 and the derivative of x is 1. Thus ex times 1 times 1 would still be ex.
An exponential function is a mathematical function in the form f (x) = ax, where “x” is a variable and “a” is a constant which is the base of the function and must be positive and not equal to 1. The Euler's number or letter "e" is the most commonly used function base, and ex is the most commonly used exponential function which is also called the natural exponential function. In differentiating exponential functions, the power rule is not applicable since the variable of an exponential function is in the exponent. Shown on the left is the basic derivatives of exponential functions. Note that the exponential function ex has the property that it is its own derivative. It can be proved by using the basic exponential rule where the derivative of au is equal to au In(a) du/dx. Observe that the e is the constant a and x is the exponent u. In(e) is equal to 1 and the derivative of x is 1. Thus ex times 1 times 1 would still be ex.
EXAMPLE 5. FINDING THE DERIVATIVE OF EXPONENTIAL FUNCTIONS
I got this example from cuemath.com. The link is on the references below. To start, the given f(x) = 3x + 3x2 is a sum of two functions, so we will just get the derivative of each term according to the sum rule. To get the derivative of 3x, we will use the exponential rule au = au In(a) du/dx. Take note that a or the constant is 3 and u or the exponent is x. Therefore, the derivative would be 3x times In(3) times the derivative of x which is 1. So it would look like this: 3x In(3)(1) or 3x In(3). The last term 3x2 would apply the constant multiple rule discussed in the previous modules. Factor out 3. Get the derivative of x2 by power rule: bring down the exponent 2 and multiply it to the variable x with minus 1 exponent, thus 2x. 2x times 3 is 6x. Therefore, the derivative of 3x + 3x2 is 3x In(3) + 6x.
I got this example from cuemath.com. The link is on the references below. To start, the given f(x) = 3x + 3x2 is a sum of two functions, so we will just get the derivative of each term according to the sum rule. To get the derivative of 3x, we will use the exponential rule au = au In(a) du/dx. Take note that a or the constant is 3 and u or the exponent is x. Therefore, the derivative would be 3x times In(3) times the derivative of x which is 1. So it would look like this: 3x In(3)(1) or 3x In(3). The last term 3x2 would apply the constant multiple rule discussed in the previous modules. Factor out 3. Get the derivative of x2 by power rule: bring down the exponent 2 and multiply it to the variable x with minus 1 exponent, thus 2x. 2x times 3 is 6x. Therefore, the derivative of 3x + 3x2 is 3x In(3) + 6x.
EXAMPLE 6. FINDING THE DERIVATIVE OF EXPONENTIAL FUNCTIONS
Again, this example was taken from cuemath.com. The given f(x) = ex / (1 + x) implies the need for quotient rule in order to be differentiated as stated on the left because technically it is a quotient of two functions. The quotient rule states that the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the square of the denominator. The ex has the property that it is its own derivative. I have discussed above the proof about this property. The derivative of (1 + x) is 1 by constant rule and power rule. Therefore the numerator would look like this: (1 + x) (ex) - (ex)(1), while the denominator would just be the square of the denominator of the function: (1 + x)2. To simplify the numerator, distribute the xex.o (1 + x). Hence, ex times 1 is ex and ex times x is xex. The simplified form would be like this: ex + xex - ex. ex minus ex is 0, therefore, what will be left in the numerator is the term xex. Finally the derivative of ex / (1 + x) is equal to xex / (1 + x)2.
Again, this example was taken from cuemath.com. The given f(x) = ex / (1 + x) implies the need for quotient rule in order to be differentiated as stated on the left because technically it is a quotient of two functions. The quotient rule states that the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the square of the denominator. The ex has the property that it is its own derivative. I have discussed above the proof about this property. The derivative of (1 + x) is 1 by constant rule and power rule. Therefore the numerator would look like this: (1 + x) (ex) - (ex)(1), while the denominator would just be the square of the denominator of the function: (1 + x)2. To simplify the numerator, distribute the xex.o (1 + x). Hence, ex times 1 is ex and ex times x is xex. The simplified form would be like this: ex + xex - ex. ex minus ex is 0, therefore, what will be left in the numerator is the term xex. Finally the derivative of ex / (1 + x) is equal to xex / (1 + x)2.
DERIVATIVES OF LOGARITHMIC FUNCTIONS
Logarithmic function is an inverse function of exponential function. Thus, any exponential function can be expressed into logarithmic form and any logarithmic function can be expressed into exponential form. For instance, the logarithm of any number N if interpreted as an exponential form, is the exponent to which the base of the logarithm should be raised, to obtain the number N. Note that logarithms are particularly important because they allow us to work with very huge quantities while manipulating smaller numbers. Logarithms have numerous applications in astronomical and scientific calculations involving huge numbers. Furthermore, there are some special logarithms that occur on a very regular basis. These are the common logarithm and the natural logarithm. The common logarithm is the log base 10 but can be written without the base 10 (log x = log10 x). The natural logarithm is the log base "e" but can be written with a different notation In x (In x = loge x). Finding the derivative of logarithmic functions, whether special logarithms or not, requires formula. Shown on the right are the derived formulas in differentiating logarithmic functions.
Logarithmic function is an inverse function of exponential function. Thus, any exponential function can be expressed into logarithmic form and any logarithmic function can be expressed into exponential form. For instance, the logarithm of any number N if interpreted as an exponential form, is the exponent to which the base of the logarithm should be raised, to obtain the number N. Note that logarithms are particularly important because they allow us to work with very huge quantities while manipulating smaller numbers. Logarithms have numerous applications in astronomical and scientific calculations involving huge numbers. Furthermore, there are some special logarithms that occur on a very regular basis. These are the common logarithm and the natural logarithm. The common logarithm is the log base 10 but can be written without the base 10 (log x = log10 x). The natural logarithm is the log base "e" but can be written with a different notation In x (In x = loge x). Finding the derivative of logarithmic functions, whether special logarithms or not, requires formula. Shown on the right are the derived formulas in differentiating logarithmic functions.
EXAMPLE 7. FINDING THE DERIVATIVE OF LOGARITHMIC FUNCTIONS
I got this example from math24.net. The link is on the references below. To find the derivative of the given function y = In (x2 - 2x), we need to apply the second formula above on natural logarithm. As the formula states, we need to multiply 1 / f(x) to the derivative of f(x) or simply the derivative of f(x) over f(x). Note that the f(x) is the x2 - 2x. To find its derivative, we will apply the power rule and constant rule to get 2x - 2 as the derivative. Thereby, the derivative of In (x2 - 2x) is (2x -2) / (x2 - 2x).
I got this example from math24.net. The link is on the references below. To find the derivative of the given function y = In (x2 - 2x), we need to apply the second formula above on natural logarithm. As the formula states, we need to multiply 1 / f(x) to the derivative of f(x) or simply the derivative of f(x) over f(x). Note that the f(x) is the x2 - 2x. To find its derivative, we will apply the power rule and constant rule to get 2x - 2 as the derivative. Thereby, the derivative of In (x2 - 2x) is (2x -2) / (x2 - 2x).
EXAMPLE 8. FINDING THE DERIVATIVE OF LOGARITHMIC FUNCTIONS
I also got this example from math24.net. With the given, y = In (sin x), we will be dealing with the derivative of the natural logarithm of a trigonometric function. Same in the previous example, we will use the second formula about natural logarithm. As the formula states, we need to multiply 1 / f(x) to the derivative of f(x) or simply the derivative of f(x) over f(x). Note that the f(x) in this example is the sin x. As discussed in Module 5, the derivative of sin x is cos x. Therefore we need to multiply 1 / sin x by cos x which will give us cos x / sin x. Note that according to the quotient identities of the fundamental trigonometric identities, cos x / sin x is equal to cot x. Thus, the derivative of In (sin x) is cot x.
I also got this example from math24.net. With the given, y = In (sin x), we will be dealing with the derivative of the natural logarithm of a trigonometric function. Same in the previous example, we will use the second formula about natural logarithm. As the formula states, we need to multiply 1 / f(x) to the derivative of f(x) or simply the derivative of f(x) over f(x). Note that the f(x) in this example is the sin x. As discussed in Module 5, the derivative of sin x is cos x. Therefore we need to multiply 1 / sin x by cos x which will give us cos x / sin x. Note that according to the quotient identities of the fundamental trigonometric identities, cos x / sin x is equal to cot x. Thus, the derivative of In (sin x) is cot x.
MODULE 8 - POLYNOMIAL CURVES
OVERVIEW
This module focuses on tangents and normals, extrema and first derivative test,concavity and second derivative test, and sketching polynomial curves.
This module focuses on tangents and normals, extrema and first derivative test,concavity and second derivative test, and sketching polynomial curves.
TANGENTS AND NORMALS
The tangent is a straight line which just touches the curve at one point. Note that the derivative of the function at this point is the slope of the tangent line. Meanwhile, the normal is a straight line which is perpendicular to the tangent line. In order to calculate the equations of these lines, it should be noted that the equation of a straight line passing through the point (x1 , y1) which has a slope or gradient "m" is given by m = (y - y1) / (x - x1). Also, it should be remembered that the slope of perpendicular lines are negative reciprocals of one another.
The tangent is a straight line which just touches the curve at one point. Note that the derivative of the function at this point is the slope of the tangent line. Meanwhile, the normal is a straight line which is perpendicular to the tangent line. In order to calculate the equations of these lines, it should be noted that the equation of a straight line passing through the point (x1 , y1) which has a slope or gradient "m" is given by m = (y - y1) / (x - x1). Also, it should be remembered that the slope of perpendicular lines are negative reciprocals of one another.
EXAMPLE 9. FINDING THE EQUATION OF THE LINE TANGENT TO A FUNCTION
I got this example from my Problem Set 4. Finding the equation of the line with a given point and a function is an easy work because it would be more on substitution of the values. It was stated above that "the derivative of the function at a given point is the slope of the tangent line." Meaning, we need to find the derivative of the given function and substitute the given point to get the slope of the equation that we are tasked to find. Note that a slope and a point can produce an equation through the slope-intercept form, y - y1 = m(x - x1). Through power rule, constant multiple rule, and constant rule, I was able to determine the derivative of x3 - 3x2 + 2 which is 3x2 - 6x. Then, substitute the x-value in this derivative and solve. It would be like this: 3(3)2 - 6(3). 32 is 9. 3 times 9 is 27 and 6 times 3 is 18. 27 minus 18 is 9. Thus, 9 is the slope of the tangent line. Now, substitute these values: m = 9, x1 = 3, y1 = 2 to the slope-intercept form: y - y1 = m(x - x1). It would be like this: y - 2 = 9(x - 3). Distribute 9 inside the parenthesis to become 9x - 27. Transpose -2 from the other side to become 9x - 27 + 2. Therefore the equation is y = 9x -25.
I got this example from my Problem Set 4. Finding the equation of the line with a given point and a function is an easy work because it would be more on substitution of the values. It was stated above that "the derivative of the function at a given point is the slope of the tangent line." Meaning, we need to find the derivative of the given function and substitute the given point to get the slope of the equation that we are tasked to find. Note that a slope and a point can produce an equation through the slope-intercept form, y - y1 = m(x - x1). Through power rule, constant multiple rule, and constant rule, I was able to determine the derivative of x3 - 3x2 + 2 which is 3x2 - 6x. Then, substitute the x-value in this derivative and solve. It would be like this: 3(3)2 - 6(3). 32 is 9. 3 times 9 is 27 and 6 times 3 is 18. 27 minus 18 is 9. Thus, 9 is the slope of the tangent line. Now, substitute these values: m = 9, x1 = 3, y1 = 2 to the slope-intercept form: y - y1 = m(x - x1). It would be like this: y - 2 = 9(x - 3). Distribute 9 inside the parenthesis to become 9x - 27. Transpose -2 from the other side to become 9x - 27 + 2. Therefore the equation is y = 9x -25.
INSTRUCTION: Find the equation of the line tangent to the function at the given point. Write your answers in slope-intercept form.
EXAMPLE 10. FINDING THE EQUATION OF THE LINE NORMAL TO THE FUNCTION
Supposedly, I was tasked to find the equation of the normal line from the example above taken from my Problem Set 4. Since we already computed the slope of the tangent line, we can already determine the slope of the normal line knowing that the normal line is perpendicular to the tangent line. Meaning, their slopes are negative reciprocal of one another. Therefore, to determine the slope of the normal line, get the negative reciprocal of 9. Reciprocal in math is simply the inverse of a number. The inverse of 9 is 1/9. Since we are asked the "negative" reciprocal, we will use -1/9 as the slope of the normal line. Same in the example above, we need to substitute these values: m = -1/9, x1 = 3, y1 = 2 to the slope-intercept form: y - y1 = m(x - x1). Through substitution, we get: y - 2 = -1/9(x - 3). Distribute -1/9 inside the parenthesis by multiplying to become -1/9x + 1/3. Transpose the -2 from the other side to become -1/9x + 1/3 +2. By combining like terms, we get the equation of the normal line; y = -1/9x + 7/3.
Supposedly, I was tasked to find the equation of the normal line from the example above taken from my Problem Set 4. Since we already computed the slope of the tangent line, we can already determine the slope of the normal line knowing that the normal line is perpendicular to the tangent line. Meaning, their slopes are negative reciprocal of one another. Therefore, to determine the slope of the normal line, get the negative reciprocal of 9. Reciprocal in math is simply the inverse of a number. The inverse of 9 is 1/9. Since we are asked the "negative" reciprocal, we will use -1/9 as the slope of the normal line. Same in the example above, we need to substitute these values: m = -1/9, x1 = 3, y1 = 2 to the slope-intercept form: y - y1 = m(x - x1). Through substitution, we get: y - 2 = -1/9(x - 3). Distribute -1/9 inside the parenthesis by multiplying to become -1/9x + 1/3. Transpose the -2 from the other side to become -1/9x + 1/3 +2. By combining like terms, we get the equation of the normal line; y = -1/9x + 7/3.
EXTREMA AND THE FIRST DERIVATIVE TEST (FDT)
According to cliffsnotes.com, "if the derivative of a function changes sign around a critical point, the function is said to have a local (relative) extremum at that point. If the derivative changes from positive (increasing function) to negative (decreasing function), the function has a local (relative) maximum at the critical point. If, however, the derivative changes from negative (decreasing function) to positive (increasing function), the function has a local (relative) minimum at the critical point. When this technique is used to determine local maximum or minimum function values, it is called the First Derivative Test for Local Extrema. Note that there is no guarantee that the derivative will change signs, and therefore, it is essential to test each interval around a critical point." The picture on the right shows the summary of how the first derivative test works to determine the relative extrema also known as local extrema. The first thing you need to do is to find the first derivative of the given function. Then, set the derivative equal to zero and solve for x. The value/s of x will be the critical number/s to be used in the number line. Then, from the number line, pick a number in each region or interval to be plugged into the first derivative and notice whether the result is positive or negative. Then, as stated above, if the derivative changes from positive to negative, there exists a local maximum at the critical point and if otherwise, there exists a local minimum at the critical point. Furthermore, to find the points of the local extrema, just plug in the x-value to the original function to get the y-value, thus the point.
According to cliffsnotes.com, "if the derivative of a function changes sign around a critical point, the function is said to have a local (relative) extremum at that point. If the derivative changes from positive (increasing function) to negative (decreasing function), the function has a local (relative) maximum at the critical point. If, however, the derivative changes from negative (decreasing function) to positive (increasing function), the function has a local (relative) minimum at the critical point. When this technique is used to determine local maximum or minimum function values, it is called the First Derivative Test for Local Extrema. Note that there is no guarantee that the derivative will change signs, and therefore, it is essential to test each interval around a critical point." The picture on the right shows the summary of how the first derivative test works to determine the relative extrema also known as local extrema. The first thing you need to do is to find the first derivative of the given function. Then, set the derivative equal to zero and solve for x. The value/s of x will be the critical number/s to be used in the number line. Then, from the number line, pick a number in each region or interval to be plugged into the first derivative and notice whether the result is positive or negative. Then, as stated above, if the derivative changes from positive to negative, there exists a local maximum at the critical point and if otherwise, there exists a local minimum at the critical point. Furthermore, to find the points of the local extrema, just plug in the x-value to the original function to get the y-value, thus the point.
FINDING ABSOLUTE EXTREMA
How is local extrema different from absolute extrema? A local maximum or relative maximum is a point a within the given interval of a function that satisfies the condition of being the largest value within the given interval while an absolute maximum or global maximum is the largest value of f(x) across its entire domain. Furthermore, a local minimum or relative minimum is a point a within a given interval of a function that satisfies the condition of being the smallest value within the given interval while an absolute minimum or global minimum is the smallest value of f(x) across its entire domain. On the above example, we are shown how to find the local extrema using the first derivative test. This time, we will use the first derivative test to find the absolute extrema. First, we need to find the first derivative of the given function. Set it equal to zero and solve for x. That x-value will be used as one of the test values as long as it is included in the domain. In the example on the right, the domain is [1,4] and obviously 2 as the x-value is included in it. These three x-values will be plugged into the original function to determine their corresponding y-values. From this, the ordered pair with the lowest y-value is the absolute minimum, and the one with highest y-value is the absolute maximum.
How is local extrema different from absolute extrema? A local maximum or relative maximum is a point a within the given interval of a function that satisfies the condition of being the largest value within the given interval while an absolute maximum or global maximum is the largest value of f(x) across its entire domain. Furthermore, a local minimum or relative minimum is a point a within a given interval of a function that satisfies the condition of being the smallest value within the given interval while an absolute minimum or global minimum is the smallest value of f(x) across its entire domain. On the above example, we are shown how to find the local extrema using the first derivative test. This time, we will use the first derivative test to find the absolute extrema. First, we need to find the first derivative of the given function. Set it equal to zero and solve for x. That x-value will be used as one of the test values as long as it is included in the domain. In the example on the right, the domain is [1,4] and obviously 2 as the x-value is included in it. These three x-values will be plugged into the original function to determine their corresponding y-values. From this, the ordered pair with the lowest y-value is the absolute minimum, and the one with highest y-value is the absolute maximum.
CONCAVITY AND THE SECOND DERIVATIVE TEST (SDT)
If the first derivative describes the direction of the function, the second derivative describes the concavity of the function. The Second Derivative Test will tell us which region a function is concave upward or concave downward. Here's how it works. The first step is to find the second derivative of the function by simply getting the derivative of the first derivative of the function. Set the second derivative equal to zero and solve for x. The x-value/s will serve as the critical point/s of your number line. From the number line, pick a number in each region or interval to be plugged in the second derivative and observe the sign. As shown on the right, if f"(x) > 0, or simply positive, then the graph of the function is concave upward on that particular interval. If f"(x) < 0, or simply negative, then the graph of the function is concave downward on that particular interval.
If the first derivative describes the direction of the function, the second derivative describes the concavity of the function. The Second Derivative Test will tell us which region a function is concave upward or concave downward. Here's how it works. The first step is to find the second derivative of the function by simply getting the derivative of the first derivative of the function. Set the second derivative equal to zero and solve for x. The x-value/s will serve as the critical point/s of your number line. From the number line, pick a number in each region or interval to be plugged in the second derivative and observe the sign. As shown on the right, if f"(x) > 0, or simply positive, then the graph of the function is concave upward on that particular interval. If f"(x) < 0, or simply negative, then the graph of the function is concave downward on that particular interval.
POINTS OF INFLECTION
Note that if the graph of the function changes from concave upward to concave downward, or from concave downward to concave upward, there will be a point of inflection (though not all the time). The point of inflection or inflection point is a point in which the concavity of the function changes. In other words, if the function changes from positive to negative, or from negative to positive, at a specific point x = c, then that point is known as the point of inflection on a graph. Most of the time, that x-value is one of the critical points you have solved from the second derivative. Note that you can find the corresponding y-value by substituting the x-value to the original function. "There are times that a graph has no point of inflection and that time would be if the function is undefined at some value of x." Note that if a function is undefined to a value of x, there will be a vertical asymptote that the graph approaches but never meets.
Note that if the graph of the function changes from concave upward to concave downward, or from concave downward to concave upward, there will be a point of inflection (though not all the time). The point of inflection or inflection point is a point in which the concavity of the function changes. In other words, if the function changes from positive to negative, or from negative to positive, at a specific point x = c, then that point is known as the point of inflection on a graph. Most of the time, that x-value is one of the critical points you have solved from the second derivative. Note that you can find the corresponding y-value by substituting the x-value to the original function. "There are times that a graph has no point of inflection and that time would be if the function is undefined at some value of x." Note that if a function is undefined to a value of x, there will be a vertical asymptote that the graph approaches but never meets.
SKETCHING POLYNOMIAL CURVES
Sketching polynomial curves is a bit challenging. Yet, once you familiarized the guidelines shown on the right, it would be as easy as 1, 2, 3. Aside from these three guidelines, I will provide below a step-by-step process of graphing polynomial curves by giving an example taken from my Problem Set 5.
Sketching polynomial curves is a bit challenging. Yet, once you familiarized the guidelines shown on the right, it would be as easy as 1, 2, 3. Aside from these three guidelines, I will provide below a step-by-step process of graphing polynomial curves by giving an example taken from my Problem Set 5.
A. DOMAIN
Note that the domain of a function is the set of numbers to which the function can be applied, that is, the complete set of possible values of the independent variable. In other words, it is the set of all possible inputs for the function. In the example, we are tasked to find the domain of the given rational function. In order to find the domain of a rational function, we need to set the denominator equal to zero. Then, find the zeros of the denominator by solving the equation we created. The domain of the rational function would be all real numbers except the zeros of the denominator since they will make the function undefined. For instance, the domain of the example is all real numbers except +2 and -2.
Note that the domain of a function is the set of numbers to which the function can be applied, that is, the complete set of possible values of the independent variable. In other words, it is the set of all possible inputs for the function. In the example, we are tasked to find the domain of the given rational function. In order to find the domain of a rational function, we need to set the denominator equal to zero. Then, find the zeros of the denominator by solving the equation we created. The domain of the rational function would be all real numbers except the zeros of the denominator since they will make the function undefined. For instance, the domain of the example is all real numbers except +2 and -2.
B. INTERCEPTS
The point at which a line or curve crosses the axis of the graph is known as the intercept. The x-intercept is defined as a point that crosses the x-axis. The y-intercept is defined as a point that crosses the y-axis. Note that if the axis isn't given, the y-axis is normally used on which the letter 'b' is usually used to represent it. However, in the example, I decided to determine both the x-intercept and y-intercept because they can help us in graphing the curve. For instance, to find the x-intercept, you need to let y be equal to zero and solve for x. By algebraic manipulation, I was able to get two values of x which are -3 and +3. Thus, the x-intercepts are (-3,0) and (3,0). For the y-intercept, set x be equal to zero and solve for the value of y. As shown on my solution, I got y = 9/2. Therefore, the y-intercept is (0,9/2).
The point at which a line or curve crosses the axis of the graph is known as the intercept. The x-intercept is defined as a point that crosses the x-axis. The y-intercept is defined as a point that crosses the y-axis. Note that if the axis isn't given, the y-axis is normally used on which the letter 'b' is usually used to represent it. However, in the example, I decided to determine both the x-intercept and y-intercept because they can help us in graphing the curve. For instance, to find the x-intercept, you need to let y be equal to zero and solve for x. By algebraic manipulation, I was able to get two values of x which are -3 and +3. Thus, the x-intercepts are (-3,0) and (3,0). For the y-intercept, set x be equal to zero and solve for the value of y. As shown on my solution, I got y = 9/2. Therefore, the y-intercept is (0,9/2).
C. SYMMETRY
If a graph does not change when reflected over a line or rotated around a point, the graph is symmetric with respect to that line or point. To find the symmetry of the given function f(x), you need to solve the f(-x) and -f(x). Please note that if f(-x) = f(x), then the function is an even function and the curve is symmetric about the y-axis. If f(-x) = -f(x), then the function is an odd function and the curve is symmetric about the origin. Upon computation as shown on the left, I found out that f(-x) is equal to f(x). Thus, the rational function is an even function and symmetric about the y-axis.
If a graph does not change when reflected over a line or rotated around a point, the graph is symmetric with respect to that line or point. To find the symmetry of the given function f(x), you need to solve the f(-x) and -f(x). Please note that if f(-x) = f(x), then the function is an even function and the curve is symmetric about the y-axis. If f(-x) = -f(x), then the function is an odd function and the curve is symmetric about the origin. Upon computation as shown on the left, I found out that f(-x) is equal to f(x). Thus, the rational function is an even function and symmetric about the y-axis.
D. ASYMPTOTES
According tp byjus.com, an asymptote is a straight line that constantly approaches a given curve but does not meet at any infinite distance. In other words, Asymptote is a line that a curve approaches as it moves towards infinity. There are two most used asymptotes and these are the horizontal and vertical asymptotes. In finding the horizontal asymptotes of a rational function, there are three things to remember. If the degree of the polynomials both in numerator and denominator is equal, then divide the coefficients of highest degree terms to get the horizontal asymptotes. If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptotes will be y = 0. If the degree of the numerator is greater than the degree of the denominator, then there are no horizontal asymptotes. When it comes to the vertical asymptotes of a rational function, they are usually the zeros of the denominator. In other words, they will be the exemptions of the domain.
According tp byjus.com, an asymptote is a straight line that constantly approaches a given curve but does not meet at any infinite distance. In other words, Asymptote is a line that a curve approaches as it moves towards infinity. There are two most used asymptotes and these are the horizontal and vertical asymptotes. In finding the horizontal asymptotes of a rational function, there are three things to remember. If the degree of the polynomials both in numerator and denominator is equal, then divide the coefficients of highest degree terms to get the horizontal asymptotes. If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptotes will be y = 0. If the degree of the numerator is greater than the degree of the denominator, then there are no horizontal asymptotes. When it comes to the vertical asymptotes of a rational function, they are usually the zeros of the denominator. In other words, they will be the exemptions of the domain.
E. INTERVALS OF INCREASE OR DECREASE
To find the intervals of increase or decrease, you need to calculate the first derivative and find the critical points of the given function. Note that critical points are the points where the first derivative is zero or does not exist. That is why in the example, I included the points 0, 2, and -2 as the critical points to be considered on the number line and in the intervals. Then, pick a number in each region or interval to be plugged into the first derivative of the function. Notice if the sign is negative or positive. If the sign is positive, the interval is increasing. If the sign is negative, the interval is decreasing. For instance, from the three critical points, I got four intervals namely (-∞, -2), (-2, 0), (0, 2), and (2, ∞). From these intervals, I picked the test values -3, -1, 1, and 3, respectively. Then, I plugged these test values to the first derivative 20x/(x2 - 4)2. Consequently, I noted the signs of the answers after plugging in. Then, I identified which interval in increasing and which interval is decreasing. In conclusion, the increasing intervals are (0, 2) and (2, ∞) and the decreasing intervals are (-∞, -2) and (-2, 0).
To find the intervals of increase or decrease, you need to calculate the first derivative and find the critical points of the given function. Note that critical points are the points where the first derivative is zero or does not exist. That is why in the example, I included the points 0, 2, and -2 as the critical points to be considered on the number line and in the intervals. Then, pick a number in each region or interval to be plugged into the first derivative of the function. Notice if the sign is negative or positive. If the sign is positive, the interval is increasing. If the sign is negative, the interval is decreasing. For instance, from the three critical points, I got four intervals namely (-∞, -2), (-2, 0), (0, 2), and (2, ∞). From these intervals, I picked the test values -3, -1, 1, and 3, respectively. Then, I plugged these test values to the first derivative 20x/(x2 - 4)2. Consequently, I noted the signs of the answers after plugging in. Then, I identified which interval in increasing and which interval is decreasing. In conclusion, the increasing intervals are (0, 2) and (2, ∞) and the decreasing intervals are (-∞, -2) and (-2, 0).
F. LOCAL EXTREMA
It was already discussed above about how to find the local extrema using the First Derivative Test (FDT). In finding the intervals of increase or decrease, we had already used the first derivative test. That is why, looking at my solution, I only mentioned "from E." After all, the same process will be used in finding the local extrema. For instance, when we observe the sign of the first derivative shown on the table above, it can be inferred that change of sign happened. From the interval (-2, 0) to (0, 2), the sign changed from negative to positive. If the sign of the first derivative changes from negative to positive, we can say that there exists a local minimum at x = 0. To find the exact point, we need to determine the corresponding y-value by finding f(0). In other words, substitute 0 to the original function and solve for f(0). By further computation, I got f(0) is equal to 9/2. Thus, the point is (0, 9/2).
It was already discussed above about how to find the local extrema using the First Derivative Test (FDT). In finding the intervals of increase or decrease, we had already used the first derivative test. That is why, looking at my solution, I only mentioned "from E." After all, the same process will be used in finding the local extrema. For instance, when we observe the sign of the first derivative shown on the table above, it can be inferred that change of sign happened. From the interval (-2, 0) to (0, 2), the sign changed from negative to positive. If the sign of the first derivative changes from negative to positive, we can say that there exists a local minimum at x = 0. To find the exact point, we need to determine the corresponding y-value by finding f(0). In other words, substitute 0 to the original function and solve for f(0). By further computation, I got f(0) is equal to 9/2. Thus, the point is (0, 9/2).
G. CONCAVITY AND POINTS OF INFLECTION
Concavity and points of inflection were already discussed above. For the application of the information I provided above, analyze my solution on the left. To find the concavity of the given rational function, I applied the Second Derivative Test (SDT). I solved for the second derivative of the given function by differentiating the first derivative. In the process of differentiation, I followed the quotient rule since I am dealing with a rational function. Of course, one must remember all the other basic derivative rules such as the power rule, constant multiple rule, constant rule, and the sum and difference rule. After getting the second derivative, set it equal to zero and find its critical points. Note that critical points are the points where the second derivative is zero or does not exist. However, in this example, I used the same critical points I got from the first derivative test namely, 0, -2, and 2. Then, I got the intervals (-∞, -2), (-2, 0), (0, 2), and (2, ∞). From these intervals, I picked random number that is included in each interval such as -3, -1, 1, and 3, respectively. Then, I plugged in these values to the second derivative and observed the sign if positive or negative. If f" > 0 or positive, the interval is concave upward. If f" < 0 or negative, the interval is concave downward. In conclusion, I found out that the graph is concave upward at the intervals (-2, 0) and (0, 2) and is concave downward at the intervals (-∞, -2) and (2, ∞). Meanwhile, despite the change of concavity, I found out that the graph has no point of inflection. And it is because the original rational function is undefined to some values of x specifically 2 and -2. Note that if a function is undefined to a value of x, there will be a vertical asymptote that the graph approaches but never meets. Thus, there will be no point of inflection.
Concavity and points of inflection were already discussed above. For the application of the information I provided above, analyze my solution on the left. To find the concavity of the given rational function, I applied the Second Derivative Test (SDT). I solved for the second derivative of the given function by differentiating the first derivative. In the process of differentiation, I followed the quotient rule since I am dealing with a rational function. Of course, one must remember all the other basic derivative rules such as the power rule, constant multiple rule, constant rule, and the sum and difference rule. After getting the second derivative, set it equal to zero and find its critical points. Note that critical points are the points where the second derivative is zero or does not exist. However, in this example, I used the same critical points I got from the first derivative test namely, 0, -2, and 2. Then, I got the intervals (-∞, -2), (-2, 0), (0, 2), and (2, ∞). From these intervals, I picked random number that is included in each interval such as -3, -1, 1, and 3, respectively. Then, I plugged in these values to the second derivative and observed the sign if positive or negative. If f" > 0 or positive, the interval is concave upward. If f" < 0 or negative, the interval is concave downward. In conclusion, I found out that the graph is concave upward at the intervals (-2, 0) and (0, 2) and is concave downward at the intervals (-∞, -2) and (2, ∞). Meanwhile, despite the change of concavity, I found out that the graph has no point of inflection. And it is because the original rational function is undefined to some values of x specifically 2 and -2. Note that if a function is undefined to a value of x, there will be a vertical asymptote that the graph approaches but never meets. Thus, there will be no point of inflection.
H. GRAPH
After determining the domain, intercepts such as x-intercept and y-intercept, symmetry, asymptotes such as horizontal and vertical asymptotes, intervals of increase or decrease, local extrema, and concavity and points of inflection, you may now start graphing the rational function. My own strategy was to start plotting the points I found on my previous computations such as the x-intercept and y-intercept. Also, I plot the asymptotes, horizontal and vertical asymptotes in broken lines. Note that the graph will never meet these asymptotes. I also bear in mind the concavity and intervals of increase or decrease before tracing the graph. Finally, I started sketching the graph by considering all the information I gathered above. For checking, one can use graphing software such as desmos.com to verify the sketched graph. Shown on the right is the graph I have made for the rational function f(x) = [2(x2 - 9)] / (x2 - 4).
After determining the domain, intercepts such as x-intercept and y-intercept, symmetry, asymptotes such as horizontal and vertical asymptotes, intervals of increase or decrease, local extrema, and concavity and points of inflection, you may now start graphing the rational function. My own strategy was to start plotting the points I found on my previous computations such as the x-intercept and y-intercept. Also, I plot the asymptotes, horizontal and vertical asymptotes in broken lines. Note that the graph will never meet these asymptotes. I also bear in mind the concavity and intervals of increase or decrease before tracing the graph. Finally, I started sketching the graph by considering all the information I gathered above. For checking, one can use graphing software such as desmos.com to verify the sketched graph. Shown on the right is the graph I have made for the rational function f(x) = [2(x2 - 9)] / (x2 - 4).
Salazar (BSCE 1B) - Problem Set 5.pdfPROBLEM SET 3 (25/25)
Salazar (BSCE 1B) - Problem Set 4.pdfQUIZ 2 SOLUTIONS (40/40)
Quiz 3 - Salazar (BSCE 1B).pdfPROBLEM SET 4 (25/25)
Salazar (BSCE1B) Midterm Exam.pdfQUIZ 3 SOLUTIONS (25/25)
Quiz 2 - Salazar (BSCE1B).pdfPROBLEM SET 5 (45/45)
Salazar (BSCE 1B) - Problem Set 3.pdfMIDTERM EXAM (50/50)
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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