diff cal

MATH 401 – DIFFERENTIAL CALCULUS Batangas State University PB Main I Rizal Ave., Batangas City MODULE in MATH 401 DIFFERENTIAL CALCULUS EMIL C. ALCANTARA, Ph.D. RONIE A. MENDOZA, M.Sc. RENSON A. ROBLES, Ph.D. i

MATH 401 – DIFFERENTIAL CALCULUS Table of Contents CHAPTER I Functions, Limits and Continuity 1 Definition and Notation of Functions 4 Domain and Range 7 Graph of Functions 12 Types of Functions 24 Operations on Functions 27 Limit of a Function 32 One – sided Limits 41 Continuity 45 Chapter Test CHAPTER II Derivatives of Algebraic Functions 46 Differentition 49 The Increment Method 52 Theories on Differentiation of Algebraic Functions 63 Higher order Derivatives 65 Implicit Differentiation 68 Chapter Test CHAPTER III Derivatives of Transcendental Functions 69 Trigonometric Functions 76 Chain Rule of Derivatives of Trigonometric Functions 85 Implicit Differentiation of Trigonometric Functions 88 Derivatives of Inverse Trigonometric Functions 91 Chain Rule of Inverse Trigonometric Functions 92 Higher – order Derivative of Inverse Trigonometric Functions 93 Implicit Differentiation of Inverse Trigonometric Functions 96 Logarithmic Functions 98 Higher – order Derivatives of Logarithmic Functions 99 Implicit Differentiation of Lograithmic Functions 101 Logarithmic Differentiation 104 Exponential Functions 105 Chain Rule of the Derivatives of Exponential Functions 107 Higher – order Derivaives of Exponential Functions 107 Implicit Differentiation of Exponential Functions 112 Hyperbolic Functions 115 Chain Rule of the Derivatives of Hyperbolic Functions 118 Higher – order Derivatives of Hyperbolic Functions 122 Implicit Differentiation of Hyperbolic Functions 123 Chapter Test ii

MATH 401 – DIFFERENTIAL CALCULUS CHAPTER IV Applications of Derivatives of Algebraic and Transcendental Functions The Differential 127 Application of the Differential 130 Approximation Formulas 132 Error Propagation 135 Tangent Line and Normal Line to a given Curve 138 Relative Exrema 147 Increasing and Decresing Functions and the 150 First Derivative Test Concavity, Points of Inflection and the Second 159 Derivative Test Curve Tracing 169 Optimization Problems 174 Optimization Problems Involving Algebraic Functions 174 Optimization Problems Involving Transcedental Functions189 Number Problems 195 Related Rates 199 Related Rates ProblemsInvolving Algebraic Functions 200 Related Rates Problems Involving Transcendental 206 Functions Motion Problems 211 Chapter Test 218 CHAPTER V Partial Differentiation 220 Definition of Parital Derivatives of a Functions 220 Partial Derivatives by Formulas of Differentiation 230 Higher – order Partial Derivatives 233 Total Derivatives 236 Chain Rule of Partial Differentiation 239 Implicit Partial Derivatives 241 Chapter Test REFERENCES iii

MATH 401 – DIFFERENTIAL CALCULUS CHAPTER I FUNCTIONS, LIMITS AND CONTINUITY One of the most useful tools in modeling real – life problems and situations is the concept of function. For instance, if we want to determine the dimension of a rectangular field enclosed by 400m of fence, then we can express the area A as a function of its length l or width w. Similarly, the concept of function is used to model the total cost of a product Cx as a function of the amount of product ordered x. This function is used as a rule that describes the relationship between the dependent and independent variables. In the study of Differential Calculus, functions are used to present the concept of differentiation which is based on the notion of limits. In this chapter, we shall learn functions and their graphs, operations on functions, definition of limits and how limit theorems are used to evaluate limits of a function. Also, this chapter presents one–sided limits, limits at infinity and infinite limits, and continuity of functions on an open or closed intervals. Definitions and theorems presented in this chapter are taken from [1] Larson, R. (2018), [2] Leithold, L. (2002) and [3] Stewart, J. (2016) At the end of this chapter, the learners might be able to: 1. Define and sketch the different types of algebraic and transcendental functions; 2. Classify functions and recognize combinations and operations of functions; 3. Evaluate limits using properties of limits; 4. Determine continuity at a point and continuity on an open interval; and 5. Apply different theorems to evaluate one - sided limits and continuity on a closed interval. 1.1. FUNCTIONS Function assigns rule to describe how a certain quantity depends on the values of other variables. The equation of a line, y  mx  b , for example sets the relationship between the variables x and y . This relationship can be thought of as a correspondence from a set X of real numbers x to a set Y of real numbers y . Below is the formal definition of a function and is due to Leithold, L. (2002) Definition1.1. A function is a set of ordered pairs x, y in which no two distinct ordered pairs have the same first number. 1

MATH 401 – DIFFERENTIAL CALCULUS Symbols such as f , g , and h are used to denote functions unless stated otherwise. If the function is expressed in terms of the variable x , f x , gx , and hx are used to denote this function. For instance, in the finding the area of circle, Ar  r 2 is used to describe the relationship between the area and the radius of the given circle. Here, it can be observed that the area, A is expressed as a function of r and the value of A depends on the value of r . So, we say that A is the dependent variable and r is the independent variable. Example1.1. Let f be defined by f x  2x 1. It can be observed that f is function since when x is replaced by any real numbers there is exactly one value of y obtained. x 3  2 1 0 1 2 3 f x  7  5  3 1 1 3 5 Example1.2. The circle x2  y2  9 cannot be a function since when x is 0, y assumes two values such as 3 and –3. Thus, we have the ordered pairs 0,3 and    0,3. Similarly, the ordered pairs 3,0,  3,0, 1,2 2 and 1,2 2 can also be obtained from the given circle. Observe that 1 is assigned to two values of y , that is  2 2 . Figure 1.1. The circle x2  y 2  9 Example1.3. Let g be defined by gx  x2 1. Refer to the next table for the values of x and its corresponding values of gx. Although, several values of gx appear similar in the table, g can still be thought as a function since no two or more values of x are repeated. x 3  2 1 0 1 2 3 gx 10 5 2 1 2 5 10 The relations presented in Example 1.1, 1.2 and 1.3 can be illustrated by mapping. Here, the set of real numbers x is called the domain while the set of real numbers f x, y and gx are the range. 2

MATH 401 – DIFFERENTIAL CALCULUS Figure 1.2 shows the mapping of the domain to the range of the given relations in Example 1.1, 1.2 and 1.3. Figure 1.2 (a) shows a one–to–one mapping since there is exactly one value of x mapped to exactly one value of y . In (b), a mapping of one–to–many is observed since there is exactly one value of x , say 0 and 1 mapped to two values of y , i.e. -3 and 3; and  2 2 and 2 2 , respectively. In (c), a mapping of many–to–one can be seen as there are two values of x mapped to one value of y . The ordered pairs  3,10 and 3,10;  2,5 and 2,5 ; and 1,2 and 1,2 show this relation. Domain Range Domain Range Domain Range y x f x x x gx -3 -3 -7 -3 -3 1 -2 -5 0 2 2 -2 2 -1 -3 1 0 -1 5 0 -1 3 0 10 1 1 22 1 2 3 3 2 3 5 3 a b c Figure 1.2. Mapping of a f x  2x 1, b x2  y2  9 and c gx  x2 1 Based on the examples presented in Figure 1.2 and Definition 1.1, a function possesses a one–to–one correspondence (bijection) and many–to–one correspondence (surjection) but not one–to–many or many–to–many. Another way to determine whether a relation is function or not is by using the vertical line test. Given the graph of a function, draw vertical lines overlaying the graph. If the vertical lines pass through exactly one point on the graph, then it is a function. If it passes through two or more points, then it is not a function. To evaluate function, a straight forward substitution is used. For instance, given a function, f x  x2  2 , if we wish to find f  3 we shall replace x by -3 and perform the operation leading us with f  3  11 . Example1.4. Let f be a function defined by f x  x2  2x  3, find a. f 2; b. f  1  ; 2 c. f 1; d. f 2x and e. f x  h. Solution: a. f x  x2  2x  3 d. f x  x2  2x  3 3

MATH 401 – DIFFERENTIAL CALCULUS f 2  22  22 3 f 2x  2x2  22x 3 f 2  3 f 2x  4x2  4x  3 b. f x  x2  2x  3 e. f x  x2  2x  3 f x  h  x  h2  2x  h 3 f  1    1 2  2 1   3 f x  h  x2  2xh  h2  2x  2h  3 2 2 2 f  1   1 1 3 2 4 f  1   9 2 4 c. f x  x2  2x  3 f 1  12  21  3 f 1  6 1.1.1. Doman and Range Domain and range as described on the first section helps characterized a function. Domain is also classified as the set of the independent variables while the range is the set of the dependent variables. Definition1.2. [2] The set of all admissible values of x is called the domain, and the set of all the resulting values of y is called the range of the function. Example1.5. Let f be a function defined by f x  2x  3 . Observed that x can be replaced by any real numbers that will give a defined value for f x.  Thus, the domain of f is the set of real numbers, D  x x  . In interval notation, it is given by  ,. Similarly, x is replaced by any number on this interval, the range is also a number on this interval. So the range is also an element of real numbers  ,. Example1.6. Let g be a function defined by gx  x .Since g involves a square root, then x cannot be replaced by any negative numbers. Thus, the domain of g is the set of non–negative real numbers. In interval notation, it is given by 0,. Also, the range of this function are the numbers on the same interval, 0,. 4

MATH 401 – DIFFERENTIAL CALCULUS Example1.7. Let f be a function defined by f x  x  3 . Here, x  3 cannot be a negative number, so we can let x  3  0. Solving this inequality, we have x  3 . So, the domain of this function is the set of all numbers greater than or equal to -3 or  3,. The range of this function is 0,. Example1.8. [2] Let f be a function defined by f x  x2  9 . Since f involves square root, then x2  9  0. Here, we shall think of any number that when replaced to x will give x2  9  0. Solving this inequality, we have x  3 or x  3 . So the domain of f is given by  ,33, and the range is 0,. Example1.9. Find the domain and range of the function y  3 x . Recall that the cube root of any negative number is defined on the set of negative real number, the cube root of 0 is 0 and the cube root of a positive real number is still defined on the same set of positive real numbers. Therefore, the domain of y  3 x is  , and the range is  ,. From this example, we can conclude that for any function f defined by f x  n x where n is any odd positive integer, then the domain and range of f is  ,. Example1.10. Find the domain and range of the following functions: a. f x  2 c. f x  1 x x2 b. gx  x d. f x  1 x2 x2 16 Solution: a. Since x can be found on the denominator, then x cannot be replaced by 0. So the domain of f x  1 is the set of real numbers except 0 or x interval notation we have  ,0 0, and the range is also the set of real numbers except 0,  ,0 0, since the numerator is constant. b. For gx  x , the denominator x  2 should be equal to 0, i.e. x  2 x2 . The domain of g is the set of real numbers except -2 or  ,2  2,. For the range of g , observed that the numerator is no longer constant so gx assumes any number except 1. Thus, the range of g is  ,1 1,. 5

MATH 401 – DIFFERENTIAL CALCULUS c. Similar with a, x appears on the denominator. So we have x2  0 giving us with the domain  ,0 0,. For the range, take note that the numerator is constant, 1 and the denominator is x2 . Given this facts, f x do not assumes values such as 0 and any negative numbers, thus the range of this function is 0,. d. The domain of f x  1 is the set of real numbers except  4 x2 16 since when x is 4 or -4, the denominator becomes 0. The range is   , 1   0,.  16  Example 1.11. Let y be function defined by y  2x . The function y  2x is an exponential function whose domain is the set of real numbers,  , and the range is is the set of all positive real numbers, 0, . Example 1.12. The domain of the function f x  sin 2x is the set of real numbers  , while the range is any numbers on the interval 1,1. Exercise 1.1. Find the domain and range of the following functions. 1. y  2 x  2 3 2. y  x2 3. y  x2  2x  4 4. y  x3  2 5. f x  3 2x  3 6. f x  9  x2 7. gx  x2  4 x2 8. gx  x 2x 9. f x  2x2 25  x2 10. y   1 x 2 6

MATH 401 – DIFFERENTIAL CALCULUS 1.1.2. Graph of Functions Function is described as a set of ordered pairs x, y. This set of ordered pairs or points leads to the graphical representation of a function in R2 plane. Below is the formal definition of the graph of a function. Definition 1.3. [2] If f is a function, then the graph of f is the set of all points x, y in the plane R2 for which x, y is an ordered pair in f. The graph of a function f can also be thought as a graph of the equation y  f x. This determines the behavior of the graph as the variable x and y change. For instance, in the graph of the parabola y  x2 , as the value of x increases, y also increases approaching   . See Figure 1.3. As mentioned in section 1.1, one way to determine whether a relation is a function or not when its graph is given is through the vertical line test (VLT). Referring to the graph of y  x2 on the left, when you draw vertical lines, you will notice that the lines pass through exactly one point on the graph or curve of the parabola. Also as observed on its graph, y  x2 follows many–to–one correspondence as illustrated in Section Figure1.3. Graph of y  x2 1.2. Another way to sketch the graph of a function is by determining its properties. This applies for the graphs of some special curves like the conic sections. Knowledge on the standard forms of these curves will help us trace their graphs. As example, it would be very easy to sketch the circle x2  y2  9 if the center–radius form of this equation is obtained. Particularly, x2  y2  9 is a circle with center at the origin 0,0 and radius equal to 3 by following the center-radius form of a circle. To sketch this curve, one might start at the center 0,0 and locate points 3 units away to the left and right and above and below the point 0,0. Doing it so, we obtained the points 3,0,  3,0, 0,3and 0,3. See Figure 1.1. Functions other than these curves can be graphed by plotting points on the plane and smoothly tracing those points. 7

MATH 401 – DIFFERENTIAL CALCULUS To facilitate the graphing of a function, the following steps are suggested: 1. Identify the domain and range and the properties of the function. 2. Choose suitable values of x from the domain of a function and solve for its corresponding value of y . 3. Determine the behaviour of x and y . 4. Plot the points x, y on the plane. 5. Smoothly trace the curve. Example 1.13. Sketch the graph of the following functions: a. f x  2x  3 e. f x  2 b. f x  x  3 x c. f x  25  x2 f. f x  1 d. f x  x2  4 x2 g. y  2x Solution: a. b. Figure1.4. Graph of f x  2x  3 Figure1.5. Graph of f x  x  3 From Figure 1.4 it can be observed that the graph of f x  2x  3 is a line that passes through the points 0,3 and   3 ,0 . The point 0,3 is the 2 x  intercept and the point   3 ,0 is the y  intercept of f x  2x  3 . We 2 can also verify that the domain and range of this function is any number on the interval  ,. From Example 1.7 the domain of f x  x  3 is  3, and the range is 0,. Also, it can be seen from Figure1.5 that as x increases, y also increases. 8

MATH 401 – DIFFERENTIAL CALCULUS c. d. Figure1.6. Graph of f x  25  x2 Figure1.7. Graph of f x  x2  4 The graph of f x  25  x2 (see Figure 1.6) shows a semi – circle with 0,3 as y  intercept and  3,0 and 3,0 as x  intercepts. The graph of f x  x2  4 in Figure 1.7 confirms that the domain of this function is  ,22, and the range is 0,. e. f. Figure1.8. Graph of f x  2 Figure1.9. Graph of f x  1 x2 x g. Figure1.10. Graph of f x  2x 9

MATH 401 – DIFFERENTIAL CALCULUS The graphs of the functions f x  2 (Figure 1.8), f x  1 (Figure x x2 1.9) and y  2x (Figure 1.10) confirm the result of Example 1.10 (a) and (c) and Example 1.11. In Figure 1.8 we can see that as x increases approaching   , y decreases approaching 0; as x approaches 0 from the right, y increases approaching   ; as x decreases approaching   , y approaches 0 and finally as x approaches 0 from the left, y approaches . Example1.14. Sketch the graph of the following functions: a. f x  2x 2 b. f x  1 c. f x  x 2  9 25  x 2 x2 16 x3 Solution: Figure1.11. Graph of f x  2x 2 Figure1.12. Graph of f x  x2 1  16 25  x 2 The graph of f x  2x2 in Figure c. 25  x2 Figure1.12. Graph of f x  x2  9 1.11 shows that the domain of this function is the set of real numbers x3 except  5 and the range is the set of real numbers except (0,–2]. Figure 1.12 shows that the domain of f x  1 is the set of real x2 16 numbers except  4 and the range is 10

MATH 401 – DIFFERENTIAL CALCULUS   , 1   0,. Finally, it can be seen from Figure 1.12 that domain of  16  f x  x 2  9 is  ,3  3, and the range is  ,6  6,. x3 The lines x  5 and x  5 in Figure 1.11 and the lines x  4 and x  4 are called vertical asymptotes; while the lines y  2 in Figure 1.11 and y  0 in Figure 1.12 are called horizontal asymptotes. The asymptotes of functions determine its discontinuity. For instance in the function f x  1 , when x is replaced by 4 or -4, you’ll have f 4  1 and x2 16 0 f  4  1 . We can also say that when x approaches 4 from the right, f x 0 gets arbitrarily large or approaching   . Similarly, as x approaches 4 from the left f x approaches   . This idea gives the formal definition of asymptotes. Definition 1.4. Rees, R., (2003). Suppose that f x  px is a rational qx function in lowest terms and a is some real number where qx  0 i. The vertical line x  a is a vertical asymptote of the graph of f if as x  a , then f x   . ii. The horizontal line y  a is a horizontal asymptote of the graph of f if as x   , then f x  a . From this definition, we set the rules to determine the asymptote of a function. Suppose that the rational function f x  px  an xn  a x n1  ...  a1 x  a0 where qx  0 qx bm x m n1  b x m1  ...  b1 x  b0 m1 is in lowest terms. If qa  0 , then x  a is a vertical asymptote . If n  m , then the x  axis is a horizontal asymptote. If n  m , then the horizontal asymptote is the line y  an . bm If n  m , then the graph of f has no horizontal asymptote. 11

MATH 401 – DIFFERENTIAL CALCULUS For the function f x  x 2  9 in Figure 1.12, observe that the f is x3 undefined at x  3 , i.e., when x is replaced by -3, the denominator becomes 0. However, based on Definition 1.4, the rational function should be in lowest term leading us with f x  x  3. So now, f  3  6 . The point  3,6 is called point discontinuity. 1.1.3. Types of Functions In the study of Calculus we might encounter different forms of functions. Generally speaking functions can be classified into two such as algebraic functions and transcendental functions. Some forms of algebraic functions are linear function, f x  ax  b ; quadratic function, f x  ax2  bx  c , for a,b,c  and a  0 ; and polynomial functions,  fx  an xn  a xn1  ...  a1 x  a0 . We may also include the rational n1 functions of the forms f x  px where qx  0 qx (a) (b) (c) (d) Figure 1.13. Some examples of (a) linear function; (b) quadratic function; (c) polynomial function, and (d) rational function Transcendental functions are functions that transcend algebraic functions. For instance, 1 x  x2  x3  ... transcends ex ; sin x  x  x3  x5  x7  ... 1! 2! 3! 3! 5! 7! and cosx  1 x2  x4  x6  ... . Some commonly used transcendental 2! 4! 6! functions are: 1. Trigonometric functions (Six Circular Functions) 2. Inverse trigonometric functions 3. Exponential functions 12

MATH 401 – DIFFERENTIAL CALCULUS 4. Logarithmic functions 5. Hyperbolic and inverse hyperbolic functions Trigonometric Functions and Inverse Trigonometric Functions One of the most commonly used transcendental functions is the six circular functions or the trigonometric functions. These functions were taken from the unit circle in Figure 1.14. The cosine, since, tangent, cotangent, cosecant and secant are the six circular functions. Px, y y Definition 1.5. [Stewart (2012)]. Let  be an r angle in standard position and let Px, y be a  point on the terminal side. If r  x2  y2 is the 0 1 x distance from the origin to the point Px, y , then sin   y cos   x tan   y , x  0 r r x Figure 1.14. Unit Circle csc  r , y  0 sec  r , x  0 cot   x , y  0 yx y The graphs of the six circular functions are presented on Figure 1.15. s a function defined by x  n where n  x and nZ . Figure 1.15. Graphs of the six unit circle Reference: https://sites.google.com/site/reimerprecalculus/home/unit-4b/4-6-graphs-other-trig-functions 13

MATH 401 – DIFFERENTIAL CALCULUS The domain and range of the trigonometric functions are: Table1 Domain and Range of Trigonometric Functions Domain Range sin   , 1,1 cos   , 1,1 tan     x    n where n  Z  , x 2   ,  cot  x  x  n where n  Z sec     x    n where n  Z  ,11, x 2   x  x  n where n  Z csc   ,11, Some trigonometric identities are as follows: cos x  cos x cosx  y  cos x cos y  sin x sin y sin x  sin x cosx  y  cos x cos y  sin x sin y sinx  y  sin x cos y  cos x sin y sin x csc x  1 sinx  y  sin x cos y  cos x sin y cos x sec x  1 tan x cot x  1 tanx  y  tan x  tan y sin 2 x  cos 2 x  1 1 tan x tan y 1  tan2 x  sec2 x tanx  y  tan x  tan y 1  cot 2 x  csc2 x 1 tan x tan y cos 2 x  sin 2 x sin x sin y  1 cosx  y 1 cosx  y cos 2x  1  2sin 2 x 22 2 cos 2 x 1 sin 2x  2sin x cos x sin x cos y  1 sinx  y 1 sinx  y tan 2x  2 tan x 22 1  tan2 x cos x cos y  1 cosx  y 1 cosx  y sin 2 x  1  cos 2x 2 22 cos 2 x  1  cos 2x sin x  sin y  2sin 1 x  ycos 1 x  y 2 22 tan2 x  1  cos 2x 1  cos 2x sin x  sin y  2cos 1 x  ysin 1 x  y 22 cos x  cos y  2 cos 1 x  ycos 1 x  y 22 cos x  cos y  2sin 1 x  ysin 1 x  y 22 These identities will help students to simplify both trigonometric expressions and equations. 14

MATH 401 – DIFFERENTIAL CALCULUS Let us consider the sine function sin   y . If       , we see from r 22 Figure 1.14 that the sine function attains the value on the interval 1,1 exactly once and so is one–to–one. For the cosine function, if we restrict the value of  inclusively between 0     and       for tangent 22 function, these gives cosine and tangent a one–t –one correspondence. On these intervals, we obtain their inverse functions as follows: sin 1 x  y  sin y  x cos 1 x  y  cos y  x tan1 x  y  tan y  x Table2 Domain and Range of the Inverse Trigonometric Functions Domain Range sin 1 x 1,1   ,  cos1 x 1,1 2 2  tan1 x all real numbers 0,     ,    2 2 (a) y  sin 1 x (b) y  cos1 x (c) y  tan1 x Figure 1.16. Graphs of (a) y  sin 1 x , (b) y  cos1 x and (c) y  tan1 x Reference: https://www.onlinemathlearning.com/inverse-sine-cosine-tangent.html Exponential and Logarithmic Functions A function y defined by the relation, y  a x where a is a positive number except 1 is called an exponential function of x. Example 1.15. Sketch the graph of (a) f x  2x and (b) f x   1 x 2 Solution: The graphs of f x  2x and f x   1 x are shown in Figure 1.18. 2 15

MATH 401 – DIFFERENTIAL CALCULUS x f x f x  2x f x   1 x x f x -3 1/8 2 -3 8 -2 1/4 -2 4 -1 1/2 -1 2 01 01 12 1 1/2 24 2 1/4 38 3 1/8 Figure 1.18. Graphs of (a) f x  2x and (b) f x   1 x 2 Properties of Exponential Function For f x  a x where a  1 1. Domain :  , Range : 0, 2. As x  , f x increases and gets steeper 3. As x  , f x decreases and flattens 4. Asymptotic with respect to negative x – axis Figure 1.19. Graph of f x  a x where a  1 For f x  a x where 0  a  1 1. Domain :  , Range: 0, 2. As x  , f x decreases and flattens 3. As x  , f x increases and gets steeper 4. Asymptotic with respect to positive x – axis Figure 1.20. Graph of f x  a x where 0  a  1 Some laws on exponential functions are listed below. a0 1 a x  a xy e0 1 ex  exy a1  a ay e1  e ey a xa y  a xy exe y  exy  a x y  a xy  e x y  e xy 16

MATH 401 – DIFFERENTIAL CALCULUS Logarithmic Functions By definition of logarithm, the relation y  a x , can be written as x  loga y . Thus a x and log a x are inverse functions of one another, and it follows that aloga x  x and log a a x  x . Example 1.16. Convert the following exponential to logarithmic functions 1. 23  8 Ans. log2 8  3 2.  1 1  1 log 1 1 1 5 5 5 3. 100  1 5 4. 33  1 log101  0 27 log 3 1  3 5.  1 3  1 27  3  27 log 1 1  3 27 27 The logarithmic function takes a special when a  e , that is called natural logarithm, given by loge x  ln x and when a  10 , called common logarithm and is written as log10 x  log x . Properties of Logarithmic Function For f x  loga x where a  1 1. Domain : 0, Range :  , 2. As x  , f x increases and flattens 3. As x  0 from the right, f x decreases and gets steeper 4. Asymptotic with respect to negative y – axis Figure 1.21. Graph of f x  loga x where a  1 For f x  loga x where 0  a  1 1. Domain : 0, Range :  , 2. As x  , f x decreases and flattens 3. As x  0 from the right, f x increases and gets steeper 4. Asymptotic with respect to positive y – axis Figure 1.22. Graph of f x  loga x where 0  a  1 17

MATH 401 – DIFFERENTIAL CALCULUS Some laws on logarithmic functions are as follows: ln 1  0 log a 1  0 ln e  1 ln xy  ln x  ln y log a a  1 ln x  ln x  ln y log a xy  log a x  log a y y r ln x  ln x r log a x  log a x  log a y y r log a x  log a x r Hyperbolic Functions Hyperbolic functions are functions possessing similar characteristics with that of the six trigonometric functions derived from its relation to the equilateral hyperbola. The hyperbolic functions are defined as follows: sinh x  ex  ex csc hx  2 tanh x  e x  ex 2 ex  ex ex  ex cosh x  e x  ex sec hx  2 coth x  e x  ex 2 ex  ex ex  ex Some properties and identities of hyperbolic functions and their domain and range are presented in Table 3 while their graphs are on Figure 1.23. cosh 2 x  sinh 2 x  1 sinh(x  y)  sinh x cosh y  cosh xsinh y tanh2 x  sec h2 x  1 cosh(x  y)  cosh x cosh y  sinh xsinh y coth2 x  csc h2 x  1 sinh 2x  2sinh x cosh x cosh x  sinh x  ex cosh 2x  cosh 2 x  sinh 2 x cosh x  sinh x  ex Table 3 Domain and Range of Hyperbolic Functions Domain Range cosh x  , 1, sinh x  ,  , tanh x  , sec hx  , 1,1 csc hx  ,00, 0,1 coth x  ,00,  ,00,  ,1 1, 18

MATH 401 – DIFFERENTIAL CALCULUS Figure 1.23. Graph of Hyperbolic Functions Reference: https://www.jobilize.com/calculus/test/graphs-of-hyperbolic-functions-by-openstax Absolute Function The next example shows an absolute function. Example 1.17. Let f be a function defined by (a) f x  x and (b) f x  x 1 . Find the domain and range of f and sketch its graph. Solution: Observed from Figure 1.24 and Figure 1.25 that the domain of Figure 1.24. The absolute function f x  x f is the set of real numbers  , and the range is the set of non negative real numbers, 0,. Recall that x  x if x0  x x0 19

MATH 401 – DIFFERENTIAL CALCULUS It can also be observed from Figure 1.25, that when the graph of f x  x is shifted one unit to the left, the graph of f x  x 1 is obtained. Figure 1.25. The absolute function f x  x 1 This observations lead to the following generalization: To sketch or trace the graph of i. f x  x  a shift the graph of f x  x , a units to the left; ii. f x  x  a shift the graph of f x  x , a units to the right; iii. f x  x  a shift the graph of f x  x , a units to downward; and iv. f x  x  a shift the graph of f x  x , a units to upward. this is true for all a  Another functions that you might want to study is the greatest integer functions. This function is an example of step functions. To investigate the behavior of this function, you may use the concept of one – sided limits which will be introduced on the next section of this chapter. Definition 1.5. [1]. The greatest integer function (gif) is a function defined by x  n where n  x and n  Z . For instance, 2.4  2 ,  2.2  3,  5  5 and 1  0 2 Example 1.18. Find the domain and range of the function f x  x and sketch its graph. Solution: The domain of this function is the set of real numbers  , and the range is the set of all integers, Z . The graph is shown in Figure 1.26. 20

MATH 401 – DIFFERENTIAL CALCULUS Figure 1.26. The greatest integer function f x  x Odd and Even Function If the function f satisfies f  x  f x for all number on its domain, then the function is said to be an even function. The graph of f is symmetric with respect to the y  axis. If the graph of f is symmetric with respect to the origin then f is an odd function. This function satisfies f  x   f x. Example 1.19. Determine whether f defined by (a) f x  4x4  3x2  3; (b) f x  x  3x3  x5 and (c) f x  2x3  3x2  x  3 is even or odd function. Solution: To do this, we replace x by  x and solve for f  x a. f x  4x4  3x2  3 b. f x  x  3x3  x5 f  x  4 x4  3 x2  3 f  x   x 3 x3   x5 f  x  4x4  3x2  3 f  x  x  3x3  x5 f x  4x4  3x2  3 is an even function f  x  x  3x3  x5  f x  x  3x3  x5 is an odd function c. f x  2x3  3x2  x  3 f  x  2 x3  3 x2   x 3 f  x  2x3  3x2  x  3 f  x  2x3  3x2  x  3 Observe that f  x  f x and f  x   f x therefore f is neither even nor odd function. 21

MATH 401 – DIFFERENTIAL CALCULUS Piecewise Functions Piecewise – defined functions will be useful in the study of limits and continuity which will be presented in Section 1.2 of this chapter. For instance the graph of f x  x 2  9 in Figure 1.12 has a break point at x3 x  3. This point is called point discontinuity. This means that f x do not exist at this point of f. Let us consider the piecewise – defined function f x  22x 1 if x 1 x 1 x  2 1 x Here if we replace x by any number less than 1, we shall use f x  2x 1 and if x  1, use f x  x  2 . Obviously when x  1, f x  2 . The graph of this Figure 1.27. The graph of f x  22x 1 if x 1 piecewise – defined function is x 1 shown on the next figure. The domain of this function is the set x  2 1 x of real numbers  , and the range is  ,12 3,. Example 1.20. Sketch the piecewise – defined function hx  13x x 2 if x0  1 x 1 . Solution: Observe that the function h1x  1 x2 is a parabola that opens downward with vertex at 0,1 while h2 x  3x 1 is a line that will pass through 0,1 and   1 ,0 . 3 Taking this into consideration, the graph of h is shown below in Figure 1.28. Figure 1.28. The graph of h 22

MATH 401 – DIFFERENTIAL CALCULUS The domain of h is  ,01, and the range is  ,14,. Example 1.21. Sketch the piecewise function g defined by x 1 x  2  2 x2. gx   4 x2 if 1 x 2 x Solution: The domain of g is  , and the range is  ,1 0,2. Figure 1.29. The graph of g Exercise 1.2. Sketch the graph of the following functions. Also, find its domain and range. 1. f x  x  2 2. gx  2x 1 x 3. y  4  x2 x2 4. y  x2  25 5. y  x3  2x2 x2 6. f x  x  2 7. f x  2  x 8. hx   3x  2 if x 1  x 2 1 x 9. f x  5x 1 x3 if x  3 2x 1 3  x 10. gx   3 if x  2 1 2 x2 2 x 2 Reference: [2] Leithold, L (2002) 23

MATH 401 – DIFFERENTIAL CALCULUS 1.1.4. Operations on Functions A new function can be obtained by combining two or more functions. These functions may undergo operations such as addition, subtraction, multiplication, division and composition. Definition 1.6. [2] Given two functions f and g : i. their sum, denoted by f  g is a function defined by  f  gx  f x gx ii. their difference, denoted by f  g is a function defined by  f  gx  f x gx iii. their product, denoted by f  g is a function defined by  f  gx  f x gx iv. their quotient, denoted by f is a function defined by  f x  f x g g gx where gx  0. The domain of these resulting functions consists of those values of x common to the set of domains of f and g . We only need to exclude those values of x in Case iv where gx  0. Example 1.22. Let f and g be functions defined by f x  2x 1 and gx  x2  2x 1, find (a) f  g ; (b) f  g ; (c) f  g and(d) f . g Solution: b.  f  gx  f x gx a.  f  gx  f x gx  f  gx  2x 1 x2  2x 1  f  gx  2x 1 x2  2x 1  f  gx  x2  4x  f  gx  x2  2 c.  f  gx  f x gx d.  f x  f x g gx  f  gx  2x 1x2  2x 1  f x  2x 1  f  gx  2x3  3x2 1 g x2  2x 1 Another operation involving two or more functions is called composition. Below is the formal definition of the composite function. 24

MATH 401 – DIFFERENTIAL CALCULUS Definition 1.7. [2] Given two functions f and g , the composite function of f and g denoted by f  g is defined by  f  gx  f gx. The domain of f  g is the set of all numbers x in the domain of g such that gx is in the domain of f. Example 1.23. Let f and g be functions defined by f x  x 1 and gx  x2  2 find (a) f gx; (b) g f x Solution:  b. g f x  2 a. f gx  x2  2 1 f gx  x2 1 x 1 2 g f x  x 1 2  x 1 Observed that the domain of f is 1, and the domain of g is  , , therefore the domain of f gx is  ,11, and this set is in the domain of g . Example 1.24. Let f and g be functions defined by f x  2 and x 1 gx  2x 1find (a) f g2; (b) g f  3; (c) f  g  1   .  2  Solution: b. g f  3 c. f  g  1    2  a. f g2 f x  2 gx  2x 1 Since g  1   2 , then g2  221 x 1  2 g2  3 f g2  f 3 f  3  2   1 f  g  1    f  2  2  f g2  2 31 2 f  g  1    2 3 1 g f  3  g 1   2  1 1   f g2  1  2 g f  3  g 1  2  2  f  g  1     4  2 3 g f  3  2 1  1  2  2 g f  3  2 g f  3  2 1  1  2  2 g f  3  2 25

MATH 401 – DIFFERENTIAL CALCULUS Exercise 1.3. Apply the concepts of operations on functions to solve the following. Let f and g be functions defined by in each of the following numbers. Find f  g ; f  g ; f  g ; f ; f  g and g  f . Also, find the domain and range of g the resulting functions. 1. f x  x3 1; gx  x  2 2. f x  4  x2 ; gx  1 x 3. f x  x  2 ; gx  2x 1 x 1 4. f x  1 ; gx  1 x2 x 5. f x  e2x ; gx  ln x Let f, g and h be functions defined by f x  3x 1; gx  1 and x2 hx  x2 1, find the following: 6.  f  g2 7.  f  h 1  2 8.  f  1 g 9.  f  h1 10. f gh2 References: [1] Larson, R. (2018) and [2] Leithold, L (2002) 26

MATH 401 – DIFFERENTIAL CALCULUS 1.2. LIMIT OF FUNTION Limits of a function play a very important role in finding the derivatives of the given functions. For instance, given a function f x , if the limit of this function exists at a certain number a , then f x is said to be differentiable on that value of a. 1.2.1. Definition of Limit and Theorems on Limits The formal definition of the limit of a function is given on Definition 1.8. Here, we consider a function f . Definition 1.8. [2] Let f be a function at every number in some open interval containing a , except possibly at the number a itself. The limit of a f x as x approaches a is L , written as lim f x  L xa if the following statement is true: Given any   0 , however small, there exists a   0 such that if 0  x  a   , then f x L   . Geometric interpretation of Definition 1.8 is presented in Figure 1.30. L  1 L L  1 a a  1 a  1 Figure 1.30. Geometric representation of lim f x  L xa It can be seen from the figure on the left that as the value of x approaches a from the right, f x approaches L . This is also true when you approach the value of a from the left. Theorem 1.1 shows the basic theorems on limits. 27

MATH 401 – DIFFERENTIAL CALCULUS Theorem 1.1. [2] Let f and g be functions and let a  1.1.1. If c is constant, then lim c  c . xa 1.1.2. If c is constant, then lim cx  clim x  ca . xa xa 1.1.3. limmx  b  ma  b (limit of a linear function) xa 1.1.4. If lim f x  L and lim gx  M , then lim f x gx  L  M . xa xa xa 1.1.5. If lim f x  L and lim gx  M , then lim f x gx  L  M . xa xa xa 1.1.6. If lim f1x  L1 , lim f 2 x  L2 ,…, lim fn x  Ln then xa xa xa lim f1 x   f2 x fn1x  L1  L2  Ln . xa 1.1.7. If lim f x  L and n is any positive integer, then lim f xn  Ln . xa xa 1.1.8. If lim f x  L and lim gx  M , then lim  f x  L .  gx M xa xa xa 1.1.9. If lim f x  L and n is any positive integer, then lim n f x  n L xa xa 1.1.10. If a is any real number except 0, then lim 1  1 . 1.1.11. xa x a If a  0 and n is positive integer, or if a  0 and n is an odd positive integer, then lim n x  n a . xa 1.1.12. lim c  0 if n  Z  and c any constant xx n One easy way to evaluate the limits of a function is by straight forward substitution. Consider the following illustrations. Illustration 1.1. Find lim 4 . x2 Solution: By Theorem 1.1.1 it is easy to show that lim 4  4 . x2 Illustration 1.2. Evaluate lim  1  . x2 2  Solution: By Theorem 1.1.1, lim  1    1 . x2 2  2 Illustration 1.3. Evaluate lim x2  2x  3. x2 Solution: lim x2  2x  3  lim x2  2 lim x  lim 3 by Theorem 1.1.4 x2 x2 x2 x2 28

MATH 401 – DIFFERENTIAL CALCULUS   22  2 2 3 by Theorem 1.1.7, 1.1.2 and 1.1.1 lim x2  2x  3  11 x2 Illustration 1.4. Evaluate lim 2x  33 . x 1 2 Solution: 2  1   3 2  3 lim 2x  33    23 by Theorem 1.1.7, 1.1.4, 1.1.2 and x 1 2 1.1.1. lim 2x  33  8 x 1 2 Illustration 1.5. Evaluate lim x2  2x 1 . x0 x  2 Solution: lim x2  2x 1  02  201 x2 0 2 x0 lim x2  2x 1   1 x0 x  2 2 Illustration 1.6. Evaluate lim 2x 1 . x4 x2 Solution: lim 2x 1  241 x4 x  2 42  9  3 3 2 2 22 lim 2x 1  3 2 x4 x  2 2 Illustration 1.7. Evaluate lim 3  x2 . xx1 Solution: lim 3  x2  3  12  4  2 xx1 1 1 Illustration 1.8. Evaluate lim x2 16 . x4 x  4 Solution: lim x2 16  42 16  0 x4 x  4 44 0 29

MATH 401 – DIFFERENTIAL CALCULUS By direction substitution we noticed that the limit is 0 . The 0 expression 0 is called indeterminate form. To evaluate lim x2 16 we 0 x4 x  4 first simplify this algebraically. lim x2 16  lim x  4x  4  lim x  4  4  4  8 . So, lim x2 16  8 x4 x  4 x4 x4 x4 x4 x  4 Illustration 1.9. Evaluate lim x3 1 . x1 x 1 Solution: By straight forward substitution, we have lim x3 1  13 1  0 . x1 x 1 1 1 0 Simplifying x3 1 we have x 1 lim x3 1  lim x 1x2  x 1  limx2  x 1  12 11  3 . x1 x 1 x1 x 1 x1 Therefore, lim x3 1  3. x1 x 1 Illustration 1.10. Evaluate lim 1 x . x1 1  x Solution: It is also an obvious observation that when x is replaced by 1, we’ll have 0 . To evaluate lim 1 x , we first rationalize the numerator. 0 x1 1  x So, we have  lim 1 x  lim 1 x  1 x  lim 1  1 x  lim 1  1  1 . x x x1 1 x 1 1 2 x1 1  x x1 1  x 1  x1 x1 lim 1 x  1 x1 1 x 2 Illustration 1.11. Evaluate lim 3  h  3 . h0 h Solution: Replacing h by 0, we’ll have lim 3  h  3  3  0  3  0 . h0 h 00 Rationalizing the numerator,       32 2    h 3  lim     h0 3 h3 3 h  3  lim h  3 . Simplifying, h 3 h  3 h0 h  3  30

MATH 401 – DIFFERENTIAL CALCULUS 3 h3  3  h  3  3h3  h  lim  lim 3 h3    lim h0   h0       h0 h h 3 h3     lim  h  lim   1  .   h3    h  h0 h 3 h0 3  3 Substituting 0 to h , lim 3  h  3  1  1 h0 h 3 03 2 3 lim 3  h  3   3 h0 h 6 Indeterminate form 0 is introduced in Illustration 1.8–1.11. This 0 expression should be avoided when evaluating limits of function. One way to do that is to simplify the expression algebraically. However, in a more advanced study of limits, we can employ the rule called L’Hospitals Rule. But this rule needs the concept of differentiation. Other indeterminate forms are  ,    , 0   , 00 ,1 and 0 .  Illustration 1.12. Evaluate lim 2 . xx 3 Solution: By theorem 1.1.12 we have lim 2  2  0 . xx 3  Illustration 1.13. Evaluate lim 3x3  2x  4 . x 2  3x2  2x3 Solution: If we substitute  to x we obtain an indeterminate form. To evaluate the limit, we first divide every terms of the numerator and denominator by x raised to the highest power, in this case by x3 . So we’ll have 3x3  2x  4  lim 3x3  2x  4 . Simplifying, lim x x3 x3 x3 x 2  3x2  2x3 2  3x2  2x3 x3 x3 x3 lim 3x3  2x  4 3 2 4 . Applying Theorem 1.1.12  lim x2 x3 x 2  3x2  2x3 x 2  3  2 x3 x 31

MATH 401 – DIFFERENTIAL CALCULUS lim 3x3  2x  4 3 2 4  300  3  2 3 x 2  3x2  2x3 2  3  2 0  0  2 2 3  lim 3x3  2x  4   3 x 2  3x2  2x3 2 Illustration 1.14. Evaluate limsin 2x  2  . x cos x  Solution: lim sin 2x  2   sin 2  2  0  2  2 x cos x  cos  1 Two important theorems in finding the limits of trigonometric functions are lim sin x 1 and lim 1  cos x  0 . The proof of these theorems x0 x x0 x are found in [2] Leithold, L. (2002). These theorems proved that the sine and cosine functions are continuous at 0. The tangent, secant, cosecant and cotangent functions are continuous on their domains. 1.2.2. One – sided Limits The concept of one – sided limits lies when we approach the value of a either from the left or from the right, or a number less than r greater than a , that is when we choose a number on the open interval containing a but not a itself. Figure1.31. Graph of f x  x  3 For instance, the function f defined by f x  x  3 does not exists for all x  3. This means that the lim x  3 x3 has no meaning. However, if x  3, it can be observed that that value of f x gets closer and closer to 0. In this case, we let x approach  3 from the right and this is called the right– hand limit or the one–sided limit from the right. So we, write lim x  3  0 (read as limit of x  3 as x x3 approaches  3 from the right). Below is the definition of right–hand limit. 32

MATH 401 – DIFFERENTIAL CALCULUS Definition 1.9. [2] Let f be a function at every number in some open interval a,c. Then the limit of a f x as x approaches a from the right is L , written as lim f x  L if any   0 , however small, there exists a xa   0 such that if 0  x  a  , then f x L   . When a number less than a is taken into consideration, we say x approaches a from the left. This limit is called left–hand limit or the one– sided limit from the left. Definition 1.10. [2] Let f be a function at every number in some open interval d, a . Then the limit of a f x as x approaches a from the left is L , written as lim f x  L if any   0 , however small, there exists a xa   0 such that if 0  a  x  , then f x L   . Example 1.25. Let f be a function defined by f x  xx  4 if x  4 , find  4  4 x lim f x and lim f x x4 x4 Solution: Since the limit is approach from the right of  4 , then we might choice x  4 . So we have lim x  4  4  4  0 , therefore x4 lim f x  0 . x4 When x is approach from the left of  4 , we shall use x  4 , then lim f x  lim x  4  4  4  8. x4 x4 From Example 1.25 we can say that the lim f x does not exists since x4 lim f x  lim f x. This leads to the following theorem. x4 x4 Theorem 1.2. [2] lim f x exists and is equal to L if and only if lim f x xa xa and lim f x exist and both are equal to L. xa Illustration 1.15. Let g be a function defined by x  11 x 2 if x  2 , find 3  x2 2 x g  lim gx and lim gx and show that lim gx exists. x2 x2 x2 Solution: 33

MATH 401 – DIFFERENTIAL CALCULUS i. lim gx  lim 3  x2   3  22  7 x2 x2 ii. lim gx  lim 11 x2   11 22  7 x2 x2 Since lim gxand lim gx exist and are both equal to 7, then x2 x2 lim gx exists and lim gx  7 x2 x2 x  5 if x  3  Illustration 1.16. Let f be a function defined by f (x)   9 x2 if  3  x  3 if 3  x 3  x Determine whether the lim f x and lim f x exist. x3 x3 Solution: i. To show that lim f x exists we have x3 to show that lim f x and lim f x x3 x3 both exist and are equal. lim f x  lim 3  x  3  3  0 x3 x3 lim f x  lim 9  x2  9  32  0 x3 x3 Since lim f x and lim f x both exists x3 x3 Figure1.32. Graph of f and both are equal to 0, then lim f x x3 exists and lim f x  0 . x3 ii. lim f x  lim 9  x2  9   32  0 x3 x3 lim f x  lim x  5  3  5  2 x3 x3 Since lim f x and lim f x both exists however lim f x  lim f x, x3 x3 x3 x3 then lim f x does not exists. x3 1.2.3. Infinite Limits Let us consider the function f defined by f x  1 . Observed that as x x approaches 0 from the right, the value of f x tends to increase approaching   , so we say that lim 1   . Also, we see that lim 1   . xx0 xx0 34

MATH 401 – DIFFERENTIAL CALCULUS This case of limits is called infinite limits. It also confirms that the line x  0 is the vertical asymptote of f x  1 . x Figure1.33. Graph of f x  1 x Definition 1.11. [2] The line x  a is a vertical asymptote of the graph of the function f if at least one of the following statements is true: i. lim f x   xa ii. lim f x   xa iii. lim f x   xa iv. lim f x   xa Below are the theorems for infinite limits. Theorem 1.3. [2] If r is any positive integer, then i. lim 1   xx0 r ii. lim 1    if r is even xr   if r is odd x0 Illustration 1.17. Use Theorem 1.13 to evaluate lim 1 . xx0 5 Solution: By Theorem 1.3 (ii), it is an obvious observation that r  5 which is an odd positive integer and the limit is approaching 0 from the left. So we have lim 3   . xx0 5 Illustration 1.18. Evaluate lim x  3 . x0 x 4  3x3 35

MATH 401 – DIFFERENTIAL CALCULUS Solution: lim x  3 can be simplified into x0 x 4  3x3  lim x  3  lim x  3  lim 1 x0 x 4  3x3 x0 x3 x  3 xx0 3 Since the limit is approaching 0 from the right, it follows that lim x  3   . x0 x 4  3x3 Theorem 1.4. [2] If a is any real number and if lim f x  0 and xa lim gx  c , where c is any constant not equal to 0, then xa i. If c  0 and if f x  0 through positive values of f x , lim gx   f x xa ii. If c  0 and if f x  0 through negative values of f x , lim gx   f x xa iii. If c  0 and if f x  0 through positive values of f x , lim gx   f x xa iv. If c  0 and if f x  0 through negative values of f x , lim gx   f x xa The theorem is also valid if “ x  a ” is replaced by “ x  a ” or “ x  a ” Illustration 1.19. Apply Theorem 1.4 to evaluate the following limits: (a) lim x  2 (b) lim 3  x2 x1 x  1 x0 x (c) lim x2  x  2 (d) lim x2  4 x3 x 2  2x  3 x2 x  2 Solution: (a) lim x  2 x1 x  1 Let gx  x  2 and f x  x 1, then lim gx  lim x  2  1 2  3 x1 x1 36

MATH 401 – DIFFERENTIAL CALCULUS lim f x  lim x 1  11  0 . By applying Theorem 1.4 (iii) x1 x1  lim x  2   3   x1 x 1 0 (b) lim 3  x2 x0 x lim 3  x2  3  02  3 . The numerator 3  0 and the limit is x0 x 00 approached through the negative values, by applying Theorem 1.4 (ii)  lim 3  x2  3   x0 x 0 (c) lim x2  x  2 x3 x 2  2x  3 lim x2  x  2  32  3  2  14 By Theorem 1.4 (i) x2  2x 3 32  23 3 0 x3  lim x2  x  2  14   x3 x2  2x  3 0 (d) lim x2  4 x2 x  2 lim x2  4  22  4  0 . It is noted from the theorem that the x2 x  2 22 0 numerator should not be 0. So we need to simply x2  4 before x2 applying the theorem. That is   lim x2  4  lim x  2 x  2  lim x  2  2  2  4  2   x2 x  2 x2 x  2 x  2 x2 x  2 22 0 0 By Theorem 1.4 (i) lim x2  4  2   x2 x  2 0 37

MATH 401 – DIFFERENTIAL CALCULUS Theorem 1.5. [2] i. If lim f x  , and lim gx  c , where c is any constant not xa xa equal to 0, then lim f x gx   xa ii. If lim f x  , and lim gx  c , where c is any constant not xa xa equal to 0, then lim f x gx   xa The theorem holds if “ x  a ” is replaced by “ x  a ” or “ x  a ” Illustration 1.20. Apply Theorem 1.5 to evaluate lim  1  1  . x3 x  3 x  3  Solution: Since lim  1   1  1   and lim  1   1  1 , then x3 x  3  3  3 0 x3 x  3  3  3 6 lim  1  1     1   . x3 x  3 x  3  6 Illustration 1.21. Evaluate lim  x2  2x  x 2 1 4  . x2  4x  2   Solution: lim  x2  2x 2     2 2   4   2 x2  4x   22  4 2  6 3  2 lim  1    1   1   x2 x2     0 4   22  4  Therefore, lim  x2  2x 2  1 4    2      2     . x2  4x  x2   3 3 Theorem 1.6. [2] If lim f x   , and lim gx  c , where c is any xa xa constant not equal to 0, then i. if c  0 , lim f x gx   xa ii. if c  0 , lim f x gx   xa The theorem holds if “ x  a ” is replaced by “ x  a ” or “ x  a ” 38

MATH 401 – DIFFERENTIAL CALCULUS Theorem 1.7. [2] If lim f x   , and lim gx  c , where c is any xa xa constant not equal, then i. If c  0 , lim f x gx   xa ii. If c  0 , lim f x gx   xa The theorem holds if “ x  a ” is replaced by “ x  a ” or “ x  a ” Illustration 1.22. Apply Theorem 1.6 and 1.7 to evaluate (a) lim  2  2x 1 and (b) lim  9 x2  x  2  x4 x  4 x  2   3 x x  2  x3 Solution: (a) lim  2  2x 1 x4 x  4 x  2  lim  2   2  2   By Theorem 1.4 (ii) x4 x  4  4  4 0 lim  2x 1  241  9 x4 x  2  4  2 2 By Theorem 1.7 (i) lim  2  2x 1    9    x4 x  4 x  2  2 (b) lim  9 x2  x  2   3 x x  2  x3 lim  9 x2   lim  3  x3  x   lim  3  x3  x   3 x   3  x2  3  x2  x3 x3   x3 lim  3  x   3  3  6   By Theorem 1.4 (i) 3  x x3 33 0 lim  x  2   3  2  5 x3 x  2  3  2 By Theorem 1.6 (i) lim  9 x2  x  2    5    3 x x  2  x3 39

MATH 401 – DIFFERENTIAL CALCULUS Exercise 1.4. Evaluate the following by applying the different theorems on limits. 1. x2  x  2 lim x3 x 2  2x  3 2. lim x2  4 x2 x  2 3. x 4 lim x4 x  4 4. lim 1 x5 x  5 5. lim x  2 x1 x  1 6. lim x  2 x2 x2  4 7. lim 3  x2 x0 x 8. lim x2  9 x3 x  3 9. lim 4  x2 x2 x  2 10. lim 5  x x2 x 2  25 11. lim x  2 x4 4  x 12. lim  1  1  x0 x x2  13. lim  1  3  x2 x  2 x2  4  14. lim x 1 x1 2x  x 2  1 15. lim x  2 x2 2  4x  x2 References: [1] Larson, R. (2018) and [2] Leithold, L (2002) 40

MATH 401 – DIFFERENTIAL CALCULUS 1.3. CONTINUITY Another concept of a function that needs to be Figure1.34. Graph of f x  2x  3 understood is its continuity. For example the function f defined by f x  2x  3 is continuous at every number on the set of real numbers, i.e.  , . Same is also true for polynomial function (see Figure 1.13 (c)) which is continuous at every number. One obvious observation to identify the continuity of a function is the behavior of its domain. That is if the limit of a function f exits at a number a, then f is continuous at a number a. Figure1.35. Graph of f x  x2  9 Figure1.36. Graph of f x   13x x2 if x  0 1 if 1 x x3 Observe from Figure 1.35 that when f x  x2  9 is simplified we have x3 f x  x  3. When x is replaced by -3, f x  6 . However, we cannot substitute - 3 to the function f x  x2  9 since the function is discontinuous at x  3. The x3 point  3,6 is an example of point discontinuity. Another example is the function f on Figure 1.36. Noticed that f is discontinuous on the interval 0,1 . This type of discontinuity is called jump discontinuity. For the functions f x  1 and x f x  1 , the line x  0 is the vertical asymptote. This type of discontinuity is x2 called infinite discontinuity. 41

MATH 401 – DIFFERENTIAL CALCULUS Definition 1.11. [2] The function f is said to be continuous at the number a if and only if the following conditions are satisfied: i. f aexists; ii. lim f (x) exists, and xa iii. lim f x  f a xa If one or more of these conditions fails to hold at a, the function f is said to be discontinuous at a. Example 1.26. Let f be a function defined by f x  1 . Is the function f x3 continuous at x  3 ? Solution: Graphing this function, we’ll observe that the graph of f x  1 x3 has a break at x  3 . This line is the asymptote or the infinite discontinuity of the function. Following Definition 1.11, we have f 3  1 , therefore f 3 0 does not exist. This violates the first condition of the definition of continuity. Therefore, f x  1 is discontinuous at x  3 . x3 Example 1.27. Determine whether the function gx  62  x if x  2 is  x 2 x continuous at x  2 Solution: Find g2. (i) gx  6  x  g2  6  2  4 Since g2  4 , then g2 exists. (ii) Does lim gx exist? x2 By Theorem 1.2 lim gx if lim gx and lim gx exist and are x2 x2 x2 equal. a. lim gx  lim 6  x  6  2  4 x2 x2 lim gx exist. x2 b. lim gx  lim 2  x  2  2  4 x2 x2 42

MATH 401 – DIFFERENTIAL CALCULUS lim gx exist. x2 Therefore, lim gx since lim gx  lim gx x2 x2 x2 (iii) Does lim gx  f 2 ? x2 In (i)and (ii) we see that lim gx  f 2 x2 Since the three conditions hold for g then the g is continuous at x  2 . x  5 if x  3  Example 1.28. Is the function f defined by f (x)   9 x2 if  3  x  3 if 3  x 3  x continuous at (a) x  3 and (b) x  3 ? Solution: a. at x  3 (i) f  3 f x  9  x2  f  3  9   32  0 (ii) Does lim f x exist? x3 lim f x lim x  5  3  5  2 , lim f x exists. x3 x3 x3 lim f x lim 9  x2  9   32  0 , lim f x exists. x3 x3 x3 Since lim f x lim f x, therefore lim f x does not exist. x3 x3 x3 Therefore f x is NOT continuous at x  3. b. at x  3 (i) f 3 f x  9  x2  f 3  9  32  0 (ii) Does lim f x exist? x3 lim f x lim 9  x2  9  32  0 , lim f x exists. x3 x3 x3 lim f x lim 3  x  3  3  0 , lim f x exists. x3 x3 x3 43

MATH 401 – DIFFERENTIAL CALCULUS Since lim f x lim f x, therefore lim f xexists and x3 x3 x3 lim f x  0 x3 (iii) lim f x  f 3  0 x3 Therefore f x is continuous at x  3 . Exercise 1.5. Determine whether the following functions are continuous at the indicated number. 1. f (x)  0x2 if x  2 at x  2 x2 2.  1 if x2 at x  2   f ( x)  x 2 3 x  2 x2 1, 1 x  0 2x 3. f (x)   2x  4 if 0 x 1 at x  0 , x  1 and x  2 1 x  2 0 2  x  3 4. x 2 x  3 x if x 1 at x  1  1 x g  x x 1 x3  if at x  3 5. hx   x  3 x 2 9  x Reference: [2] Leithold, L. (2002) 44

MATH 401 – DIFFERENTIAL CALCULUS CHAPTER TEST (Problem Set 1) I. Sketch the graph of the following function and determine its domain and range. 1. f (x)  x2  81 x9 1 x  1 f (x)  1 x if 2. 1 x  0 x0  x 0  x  1 1 x  1 II. Let f and g be functions defined by f (x)  x  2; g(x)  x2  4 . Find the following: 3. f  g 6. f 4. f  g g 5. f  g 7. f  g 8. g  f Also, find the domain and range of the each resulting function. III. Find the limits of the following functions. 12. lim 2x 9. lim 1 x x4 16  x 2 x1 1  x 13. lim x 1 10. lim (x  h)2  x2 x2 x 2  4 h0 h 11. lim x4  x3 x 12x3 128 IV. Sketch the graph of the function. Determine whether the function is continuous at the indicated number. 14. f (x)  x2 1 if x 3 at x  3 x  5 3 x x  4 x  4  4  x  4 at x  4 and x  4 15. f (x)   16  x2 if 4  x 4 x References: [1] Larson, R. (2018) and [2] Leithold, L. (2002) 45

MATH 401 – DIFFERENTIAL CALCULUS CHAPTER II THE DERIVATIVES AND DIFFERENTIATION OF ALGEBRAIC FUNCTIONS The concepts of limit of a function and its continuity play a significant role in finding the derivative of a function. If the function f is continuous on an open interval (a,b) and the limit also exists on this interval, then we say that the function f is differentiable on that interval. In this chapter, we shall introduce the concepts of derivatives first by geometric interpretation as the slope of a tangent line to the graph of the given function. This gives the formal definition of the derivatives of a function. Second, this chapter also includes how theorems and formulas are used to find the derivative of a function and the high order derivatives of a function. At the end of this chapter, the student might be able to: 1. Define the process of differentiation and determine the derivative of a function by increments. 2. Define derivative as slope of tangent line to the curve and solve problems related to it. 3. Apply the different theorems on differentiation of algebraic functions 4. Apply the concept of chain rule and the general power rule on algebraic functions. 5. Determine higher order derivatives and derivatives of implicit functions. 2.1 Differentiation The process of finding the derivative of a function is called differentiation and the branch of calculus that deals with this process is called differential calculus. Differentiation is an important mathematical tool in physics, mechanics, economics and many other disciplines that involve change and motion. Consider a continuous function y  f (x) . Let P and Q be any distinct points of the curve which determine secant PQ . Representing the point P by (x, f (x)) and Q by (x  x, f (x  x)) we know that the slope of PQ will be f (x  x)  f (x) x 46

MATH 401 – DIFFERENTIAL CALCULUS We can see that as x approaches zero, the point Q becomes nearer and nearer to point P . At the same time that point Q approaches P , the secant PQ rotates about the point P . Intuitively, the limiting position of the secant PQ as Q approaches P is that of the position of the tangent line to the curve y  f (x) at point P . In symbols, lim y  lim f x  x  f x xx0 x0 x is equal to the slope of the tangent line at P . This shows that the derivative of y  f (x) at the point P is equal to the slope of the tangent line at the same point. Using the functional notation y  f (x) , the following are the usual symbols used to mean the first derivative: y'; f '(x) ; dy ; d ( y) ; d ; Dx y ; Dx f (x) ; etc. dx dx [ f (x)] dx Considering a continuous function y  f (x) , we define y' f 'x  dy  lim f x  x f x the first derivative of the function f . dx x0 x Note:  The symbol x , read as “delta x ,” is a single entity which means increment or change in x .  To find the slope of the tangent line to the curve at point P means that we are to find the value of the derivative at that point P. There are two ways of finding the derivative of a function: 1. By using the increment method 2. By using the differentiation formulas Definition2.1. Suppose that x1 is in the domain of the function f, the tangent line to the curve y=f(x) at the point P(x1,f(x1)) is the line with equation, y  f (x1)  m(x  x1) where m  lim f (x1  x)  f (x1) provided the limit exists, and P(x1, f (x1)) x0 x is the point of tangency. 47