Preface This book is intencled primarily to cater for the needs of mathematics 1 Technology {VUT}, Areas covered include: % Unit 1—Basic algebra, functions and trigonometry students at Vaal University of 2 Unit 2 Differentiation © Unit 3~ integration © Unit 4— Matrices and vectors * Unit S— Complex numbers © Unit 6~ Statistics Special care has been taken in the presentation of steps in problem solving for easy understanding. The mathematical steps are so simple that an average student may follow it independently. Worked examples and revision exercises are incorporated in each unit. The exercises could be done in class or as homework, Even though the learning material is presented in such a way that it can easily be followed, siudents are encouraged to consult their lecturers and tutorial assistants should they encounter any problems from their studies. It is therefore expected that when this book is used alongside classroom tuition it will train the student to systematically attempt any question, however complicated, with the utmost confidence. - Although every care has been taken to check mistakes and misprints, some errors may be expected, Please draw the editors’ attention to any error that you may come across. All such errors when brought to the attention of the editors will be corrected and incorporated in the next edition. Several lecturers in the department of mathematics have contributed to this edition, Unit 1 - Binomial theorem Functions and Trigonometry, was written by Dr R. Mahlobo. Unit 2 Differentiation, was written by Or J. Owusu-Mensah Unit 3— Integration, was compiled by Mrs C. Makhaiemele Unit 4— Matrices and Vectors was written by Dr J. Owusu-Mensah Unit S— Complex numbers was compiled by Dr J, Owusu-Mensah Unit 6— Statistics was compiled by Mr M. Mukamuri Special thanks (o Mrs. Mulder whose contribution cannot be overlooked. . Editors: Dr J. Owusu-Mensah and Mrs C.R. Makhalemele February, 2016Table of contents 1. Basic algebra, functions and trigonometry 1.1 Introduction 1.2 Revision: exponents, logarithms and manipulation of formula 1.3 Functions 1.4 Trigonometry 1.5 Binomial Expansion 6 Radian measure 1.7 Revision exercises 2. Differentiation 2.21 introduction 2.2 Limit of @ function 2.3 Differentiation from first principles 2.4 Basic differentiation rules 2.4.1 Derivatives of polynomials 2.4.2 Derivative of a sum or difference 243° The product rule 24.4 The quotient rule 24,5 The chain rule 2.5, Derivatives of trigonometric functions 2.6 Derivatives of exponential functions 2.7. Derivatives of logarithmic functions 2.8, Higher order derivatives 2.9. Derivatives of implicit functions 2.10. Logarithmic differentiation 2.11. Applications of differentiation 2.11.1 Curve sketching 2.11.2 Velocity and acceleration 2.11.3 Newton-Raphson method 2.10 Revision exercise 3, Integration 3.1 Introduction 3.2 Indefinite integrals 3.2.1 Integration by substitution 23 23 7 29 29 29 30 an 32 33 35 35 36 7 39 40 42 45 46 49 49 503.2.2 Integration involving trigonometric functions 3.2.3 Integration involving exponential and logarithmic functions 3.3 Definite integrals 3.4 Applications of integration 3.4.1 Area and the definite integral 3.4.2 Area between curves 3.4.3 Numerical integration (Simpson's rule) Revision exercise 4, Matrices and Vectors 4.1 Introduction 4.2 Matrices 4.2.1 Elementary operations of 2x2 matrices 4.2.2 Determinant of a 2x2 matrix 4.3 Vectors 43.1 Notation 4.3.2 Representation of vectors 4.3.3 Magnitude/ Length of a vector 43.4 Resolution of a vector 435. Unitvectors 43.6 Direction vector 4.3.7 Negative or inverse of a vector 4.3.8 Multiplication of a vector by a scalar 43.9 Parallel vectors 4.3.10 Addition and subtraction of vectors 43.11. Scalar or dot product 43.12 Angle between two vectors 4.3.13, Work done 4.3.14 Vector or cross product 4.3.15 Moment ofa vector 4.4 Revision exercise 5. Complex numbers 5.1 Introduction 5.2 Representation of a complex number 5.3 Addition and subtraction of complex numbers 5.4 Multiplication of complex numbers 5.5 Division of complex numbers 5.6 Equality of complex numbers ii 82 53 4 55. 55 56 58 60 62 62 63 agee 65. 65 67 o7 68 68 68 69 69 70 71 7 72 3 74 74 74 75 18 16 765.7 Atgand diagram and polar form of complex numbers 5.8 De Moivre’s theorem 5.9 Calculating roots 5.10. Revision exercise 6. Statistics 6.1 Introduction 6.2 Data collection 6.3 Tabulation 6.4 Data representation 6.5 Measures of central tendency 6.6 Measures of dispersion 6.7 Correlation and regression 6.8 Revision exercise ‘Answers to exercises and Revision exercises Test and examination questions 7 73 80 81 83 383 83 83 84 86 93, 96 105 aut a7UNIT2 BASIC ALGEBRA, FUNCTIONS AND TRIGONOMETRY 4.1. INTRODUCTION In Unit 1 the following topics are fevised: exponents, logarithms, manipulation of formulae, functions (domain, range and inverse), trigonometric functions and inverse irigonometric functions, These are followed by binomial expansion, radian measure and its applications. By the end of this Unit the learner should be able to: Apply the laws of exponent and logarithm. ¢ Manipulate formulae to solve equations © Expand (a + b)" where n is a rational number, using binomial theorem Convert from degree to radian measure of angles. To determine the arc-length, area of a sector and area of a segment of a circle. 4.2, REVISION: EXPONENTS, LOGARITHMS, FRACTIONS AND MANIPULATION OF FORMULA, In this section exponents, logarithms, fractions and manipulation of formula are briefly introduced as part of the revision process. The following are the exponential laws: 1.2.4. Exponents If a and b are bases, m and n exponents, then the following exponential laws apply: amin b)t=a™™ at0 a) a™ xa" < o)(ax by” = a™5™ d)a=1,a%0 e)ar= 4.2.2. Logarithms A logarithm can be deduced from an exponential equation. The components of an exponential equation are indicated below: exponent/index number hase ie. a =ce>log, c=. For example, logs = —3 because 3? = =, log,625 = 4 because 5+ = 625, log0,01 = ~2 because 10°? = 0,01Note that in the last example base 10 is not indicated. Whenever the base is not indicated, the assumption is that it is 10. A logarithm of a number with 10 as the base is called a common logarithm. The table below shows the laws of logarithm. Table 1: Laws of Logarithm Logarithm with base a,a> | Common logarithm, base | Natural logarithm, base e 0 10 logaMN = log,M+log,N | logMN = logM + logN | InMN = InM + InN M we Lega = l0gaM ~ loge | log a= logM —togn | Iny = nt — Inn logM” = nlogit InM” = nin Tog togea "| Change of base:log,M Exercise 1.2, 1. Express the following equations in logarithmic form: ee (a) a) 25° c) 3? d) 814 = 27 @) 139 =1 2. Express the following equations in exponential form: a) tog, 2 b)In (&) = ~2x c)logyy33 = 0.2 d) Ine -1 ¢) logs 36 = 3. Determine x in each case: b)Inx=4 cyeMe = d) logy(x — 2) = a e) alBa 43 = flog, 2 = log, x 4. Solve for the indicated variable: a) log =—kt, Nyt) C=—“e, Ry )W=klogsZ, Vy No loge) % Ay =x? 42x41, x e)N = Sate 6 chtdf’ 5. Express y in terms of x ; a) logy y = logy 2 + log, xb) log, y = —2log,(x + 1) + log,7 1 o) ex loge = 108Y—logx d) (logy)? = 2logx -1 e) wlog,x + logy =FUNCTIONS Definition A function is a rule which gives only one output for every given input. The rule is often expressed in the form of an equation y = f(x). In terms of mapping of elements, a function is either a one-to-one or a many-to-one relation. A function can also be described in terms of a formula or algorithm that tells how to compute the output for a given input. An example is the function that relates each real number x to its square x*. This is written as f(x) = x* The output of a function f corresponding to an input x is denoted by f(x) (read "fof x’). f£3) =9. This is a many-to-one relation, and thus a function. Not all formulae define a function. For instance, if f(x) = +x, two outputs, f(x) = 2 or—2 correspond to the same input x.=4. This makes the relation a one-to-many relation, and thus the relation is not a function. 4.3.2, Domain and range of a function The domain of the function is the set of elements of the inputs for which the function is defined. The range is the set of outputs corresponding to the inputs. Notation: Domf means the domain of the function f Rangef means the range of the function f Examples. 16) = This is a parabola with axis of symmetry x = + and turing point (3; 2) as the following diagram shows:The domain is the set of all real numbers, while the range = fvve3ye x} 2. fe) sx*-x41 isxs4 The graph is actually part of f(x -xt1 Dom f =(:13} Exercise 1.3.2. 1. Determine the domain and range of the functions defined by the following equations: a) Y= -V=2e x*4+4 c)y=Vx-5 d)y=x242 4f)y=In@—t) g)y=sine 1.3.3. Inverse Functions a) Definition Let be a one-to-one function with domain A and range B. Then the inverse function “thas the domain B and range A and is defined by f'Q)=xe2f)=y for any y inB Examples 1. Determine the inverse of the function y = 3x — 2 Solution Method 1: interchange the variables and make y the subject y=3x-2 =3y-253y-x42a y= eft) = Method 2: First make the other variable the subject of the formula, then interchange y=3x-2 3x = y +2 x = 2% interchanging the variables, we have -1(y) = fA == 2. Determine the inverses of the following functions: a)y=2* b)y=xt-ax49x22 af =H5 Solutions a) y=2* x= 2% slog,x=ylog,2 [log.2 = 1] cy = logy x hence f~?(x) = log, x b) yx? 3x 49,025 x=y?—3y+9 xan(r-J-}aly x-9_ oy CE dare S25 04 99= 9 -)=3e 9=> f(x) =The graphs of the function and its inverse are found below: Exercise 1.3.3 “ Determine the inverses of the following: 3x-2 2x-3 ay = by =log(2x-3) oy = log(x? — 3x + 5) dy =2x-3 1.4 TRIGONOMETRY Consider the right-angled triangle shown below:The following are the trigonometric functions defined from the triangle: opposite sing = hypotenuse adjacent cos@ = — hypotenuse opposite an adjacent cosec@ = —— seco = —— cot@ = —— tan TRIGONOMETRIC IDENTITIES sin? 0+ cos? @ (etan?@=sec@ i+ cot? @ = cos ec*8 sin(A EB) = sin Acos B+ cos Asin B sin2A=2sinacosd cos AB) = cos Acos BF sin Asin B cos 24 = 2eos? A =2sin? A sin? A= zl - cos 24] cos? A= Li 400824) 1.4.1, Inverse trigonometric functions Sometimes we want 0 such that sin @ = +. To do that we use the inverse sine (sin™?) or arcsine ofS, defined as follows: sin@ => @ = sin (2) or8 =arcsin @) i In essence, sin (2) is an angle whose sine value is }- This should not be confused with the cosecant function, a ratio that is a reciprocal of the sine function. Generally we use @ between and including —90° and 90° in order for the inverse to be a function. This leads to the following definition of the arcsine. 9 = sin x @ sind = x,-90° < 9 < 90° We know that the angles for which sine isSare 30° and 150°. However, because of the restriction —90° < 8 < 90° we accept the solution 1 sind > sin“? (2) = 30°Similarly, we define the inverse cosine and inverse tangent as follows: @ =cos*x + cos@ x, 0° <6 < 180° ee @ =tan tx o tang =x, —90°< 8 < 90° ees Examples 1 sing=8 oo=sint(S 2 z 2. tan@ = 1,810 @ = tan’ 60° 1(—1,810) = —61,08° Suppose we want to determine cosec™ty = 1, a # -1(2 Let x = cosec"ty @ cosecx =yor=sinx ax= sin? (> Hence cosec™ Similarly, sec~? y = cos 6) and cot~? y = tan“ (6) Example 1. cosee x = 1,245 x = cosec”1(1,245) = sin? &) =,sin71(0,80321) = 53,4377°~53,44° Exercise 1.4.1 Determine the value of the unknown: a) x = sec"*(2) b) t = cosec (cosec™*2) ¢) Ift = sin~'(@), determine sin 2t in terms of a d) x = sec75° — cosec 60° + cot115°1.5. BINOMIAL EXPANSION Binomial expansion involves the algebraic expansion of powers of a binomial, like (” + y)". When nis large and/or rational, then the binomial theorem is applied. 1.5.1. Binomial Theorem/Formula The binomial theorem makes it possible to expand the power (x + y)" into a sum involving terms of the form ax®y° 21) gage 4 M-D@-2) cy 3! (a+b)” + nah + ar-3p3 .,, Factorial notation The factorial expression can be generalised to the factorial of any arbitrary natural number n as: nl =nx(n—1)x (n-2) x 0K 2X1 Thus 3!1=3x2x1=6 S!=5x4x3x2x1=120 1.5.2. Some properties of the Binomial Formula For n a positive integer, the series is finite with n + 1 terms. For 7a fraction or a negative integer, the series is an infinite series. The sum of the powers of a and b in each term is equal to n, and the powers of a decrease while the powers of b increase to n. a) For 7a positive integer (at by" =a" + nah + MAD gray? ¢ MEDOED gaps yn Examples (ty) = xt axdy + So xty? 4 BM ytys + yt =xt+ 4x’ Stee: +4xy3 + y* (Gx -2y)* = Gx)* + 4(5x)*(—2y) + 2 Gx)*(-2y)? + 2S (5x (-2y)? + -2y)* 625x* — 1000x*y + 600x7y? — 160xy3 + Loy* 10(x? + 3y2)3 = x9 + Ixy? + 27x38 y4 + 27y6 b) For negative integers or fractional n (a+ bx)" = a" + na by + MOY gr2p2y? + MEE gn 3pins The series is valid for |bx| ) ta coo 11x10%9 ays s egret (-5) = 48850 Hence the required coefficient is = —4455p°Example 3 Determine the coefficient of p® in the exeansion of (p + Solution: using the k term The = “12-k+1-2k+2=6 12x11 T= 10 the required coefficient is 1650 Now using the method of counting terms, check the table below: Term Resulting power p 7 pe ft y Pp a r, & Exercise 1.5 1. Expand the following, using the binomial theorem ay(2 +298 b) (2x ~ 3y)* ‘yaa -208 6) Bx + Sy)? 2. Write down the term indicated in the binomial expansions of the following: a) (1 4x)?“ 34 term b)(2—x)5 2x11 9 2 =)" = 1650p® =. the required coefficient is 1650 J Gxt — y)P Write down only the first four terms. Write down only the first four terms.xe Va th (1 +2) 6° term ny (3-3) 6! term e) (12x) 4" term 10 # (-27 +2) Yeni term 3. Expand (1 + x)7? in ascending powers of x up to the fourth term, using the binomial theorem. ‘ : av 4. Determine the coefficient of p° in (p + =) 7 5, Write down the first four terms of the binomial expansion of (2 -2) simplifying all the coefficients and state the limits of x for which the series is valid. wi 1.6 RADIAN MEASURE OF AN ANGLE Consider a positive angle 8. There are two ways by which the angle can be measured, as the following diagrams show: Using a Protractor Using arclength and radius Terminal side sesmall arc resmall radius @ Initial side The first way of measuring the angle is quite familiar, while the second one uses the principle that: _ arc length (s) radius(r) radius r. where @ is the angle subtended by an arc (s) of a circle with If the length of the arc is the same as the radius, angle @ is said to measure 1 radian. Note that, strictly speaking, a radian is not a unit of measurement of angle 0, 15but an indication of how the angle is measured, por t When using a scientific calculator, a radian measure is a measure of an angle with the calculator in the ‘radian mode’. If, for instance, we want sin 1 it is understood to mean we determine sin 1 with the calculator in the radian mode. If the calculator is in the degree mode, it will give us the answer to sin 1°. Note that: sin 1 = 0,8414709 and sin 1° = 0,017452406 1.6.1 Converting angles from degrees to radian measure Consider the following equation (stated earlier): — arc length radius 7 2er Now for one revolution, 6 ===" = 2 But to make one revolution we sweep an angle of 360° Hence we have 360° = 27 rn : 180° == x x radians and yrad =——xy 80 Examples 1. Convert the following angles from degrees to radians. a 90° b. 75° ¢. 43° Solutions 90° = 90 x 2 2% ” 180 2 ase 6) 75° =75x——=-> Hate 3 180 12 c) B°=43x— 180 , 750491578 2. Convert the following angles from radians to degrees a= p32 634 3 4 Solutions 16180° 7 34 3.4 = 184.8057 © Exercises 1.6 1.Express the given angles in radian measure as a multiple of 1 a)35° © b) 135° ©) 225° d)-240° —e) 120° f) 300° 2. Express the given angles in radians 2x x 2x ayy b)-5 OF 1.6.2 Applications of radian measure Note: angles used in this section must be in radians. a) Arc length arc length (s) radius(r) Example Given the radius of a circle as 3,3 cm. Determine the length of the arc which subtends an angle @ = 7 Solution s 6r = x 3,3 cm = 345575 om b) Area ofa sector (Ar) The ratio of the area of a sector to the circle area is the same of the ratio of the angle to 2m, ie An 8 1,2 a2 1 Ay =4r°0 mr? an Ast 2Ast 9 mr? 2x a) Area of a segment (4,,,) 1 o Ase = 5778 Given: Circle cantre 0 Radius 7° Central angle @ Segment ABC Ag = Ast — Area of ad0B = 3720 — 31? sing =372(6 — sind) Examples 1. A lawn sprinkler can water up to a distance of 25m. It turns through an angle of 115° as shown below. What area can it water? eo ae 1S? ra250 Solution Age = 3726 = 3 (25m)? (115 x 5) = 627,2277 m? 182. What is the floor area of the hallway shown below? The outside and inside of the hallway are circular arcs. Solution We first determine the angle 9 0,908926417 9 = arelength _7835m _ = Fadius ~ 290m Note that the radius of inner circle is 8,290 m and radius of outer circle is 12,045 m. « Floor area = 5 [(12,045m)?(0,908926417) — (8,290m)?(0,908926417)] = 34,701 m® 3. Determine the shaded area: ° 20 eo, © € e Aree ee DC=20em, AB=12em AOB=0 Solution Considering ABOC OC = (20-7) and CB =6 Using the Pythagoras theorem, we have (20 —r)? +6? This simplifies to r = 10,9 19x ope ML + cos = 5 = 25 = 0,83486 9 = 2cos™*(0,83486) = 1,16583 + the shaded area = + (10,9)[1,16583 ~ sin(1,16583)] = 14,6561cm? Exercise 1.7 Refer to the diagram below for questions 1 and 2 Ez ———i —+ sngth of chord AB -ngth of minor height CO gth of major height EC Ss archlength AES S=archlength ADB AOB \dius of the cicle i 1. For an arc length S, area of sector As, and central angle @ of a circle of radius r, determine the indicated quantity for the given values: a)s = 1010 mm, @ = 136,0° r=? b)s = 0,3913 km r = 0,9449 km Ag. =? dr=49cm 6 =2,443 Ase =? d)@=300° Ag, = 0,0119 m2 r=? e) Age = 16,5 m2, r= 4,02m s=? 2. For an arc length S, area of sector Age, and central angle 8 of a circle of radius r, determine the area of the segment (Agn) in each case. a) s=100m, r=50m b)d=60m 0=120° Ase = 10 m2, r=3m d)r=10m, h=2m 20REVISION EXERCISE: UNIT 1 1.Expand completely using the binomial theorem: 4 a) (@x-y)? ») (3 +2) <) (a—3b)5 2. Determine the first four terms in each of the following: a) (12x) b)(3-+ x)? °) (1 2) 3. a) Use the binomial theorem to determine the first four terms in. /(1 +2) . Simplify all the terms and hence. estimate the value of 1,005 ,giving your answer in 4 decimal places. b) Use the binomial theorem to determine the first four terms in (1 — x)? . Simplify all the terms and hence estimate the value of 4/(0,75)? , giving your answer in 4 decimal places. 4. Write down the term indicated in each of the following series: a) (2x + 3y)7 4!" term ne b) @ - 2) middle term c) (13x)? 5" term 5. Determine the coefficient of p? in each of the following: ai(p+5) ales) 6. Determine the area of the shaded regions 14 a) 2b) Given that triangle ABC is equilateral and the radius is 50m <—" ¢) Given CD = 60 cm and radius= 40 cm 2UNIT 2 DIFFERENTIATION 2.1 INTRODUCTION This unit on differential calculus deals with finding the rate of change of one quantity with respect to the other, for example the rate of change of displacement with respect to time or the rate of change of electric charge with respect to time. After completing this unit the learner should be able to: ‘+ Evaluate limits of functions * Determine the derivative of a polynomial function from first principles * Determine the derivatives of a function using basic differentiation rules * Determine higher order derivatives * Determine the derivatives of implicit functions © Differentiate products, Aotients and powers of functions using logarithm properties + Apply differentiation techniques to sketch the graph of a function by determining maxima and minima © Use differentiation to solve velocity and acceleration problems for linear and circular motions * Approximate roots of equations using the Newton-Raphson method. 2.2. LIMIT OF A FUNCTION There are situations where a function is not defined at a particular value of x, say x = Xo, however it may be defined for values of x that are very close to xo. Let us look at the following functions. @r@ ==) Fo) In (@), the function is not defined at x = 1, because at that point the denominator is xia _ ert) ol x1 1). Which means we still cannot permit x = 1, but we can say that as the value of x approaches 1, the value of f(x) approaches 2. Clearly the value of f(x) never actually attains the value 2 but it does get as close to it as we select a value of x which zero. However we can see that f(x) = = X41 (provided x # iat is sufficiently close to 1. We then say that the limit of ~—* as x approaches 1 is 2, xtad and written as lim =2 xol 23Similarly, in (b), x = 0 renders the function undefined. However if we select a value of x which is sufficiently close to zero, then the function approaches 1 This is demonstrated in the table below. x z sin x ¥ j 1 7 0,8414709 I ot IE 0,9983341 0,01 0,9999833 0,001 0,9999908 | 0,0007 q 0,00007 1 0,0000007 i As x > 0 (i.e. as n becomes exceedingly small), “"* -, 1. We write tim, , “2% x Note that "is not defined at x=0 a) Evaluating limits using algebra lif is a real-valued (or complex-valued) function, then taking the limit is compatible with the algebraic operations, provided the limits on the right sides of the equations below exist (the last identity only holds if the denominator is non-zero). This fact is often called the algebraic limit theorem. lin (£60) +a(«)) = tim f(2) + lim g(2) Had (Fe) —o(e)) = tim Fe) — lim ot2) Him (Y(2)-9(2)) = a fe) inte) fim (2)fa(2)) = tim Je) im ole) In each case above, when the limits on the right do not exist, or, in the last case, when the limits in both the numerator and the denominator are zero, nonetheless the limit on the left, called an indeterminate form, may stil exist—this depends on the functions f and g. 24The following are the steps followed in evaluating limits of functions: Step 1: If direct substitution of by the input value p leads to zero in the denominator, simplify the function if possible. If the function cannot be simplified, then the limit does not exist, unless the numerator is also zero, in which case the limit may exist, and if it does, can be evaluated using L’Hopital's Rule, which is not going to be covered in this unit. Step 2: Substitute the input value into the simplified form of the function. Examples Determine the following limits: Lim.) Solution The function f(x) = x - is in its simplest form. So go straight to step 2 Tim, .@¢-1)=@2-) Solution Step 1 lim, lim, aig 3 (x+3) ry x3 Step 2: lim, (+3) =G+3)=6 3.lim, , V1=2x Solution Step 1: The function f(x) = WT=2xeannot be simplified further 25,Step 2: Direct substitution leads to a square root of a negative number which is undefined for real numbers. Therefore lim, ,. /I—2x does not exist (for a real-valued function). tim, x Solution Direct substitution does not lead to indeterminate form. So simplifying is not necessary. Step 2 leads to the following solution. sinx sine iM, g(———) = Him, y ~*~ lim, x=1-0=1 x x sin 2x . cos xsin x sinx IM yo = Lor lim, yp =>" * = Him, ,, cos. xX lim , y --* = cos 0x1 =1 2x x x b) Limits to infinity The rule for limits to infinity is based on observations demonstrated in the following table: 1 50,1 ——-=0,0001 10 10000 i 10000000 ,0000001 As the denominator increases, the fraction becomes smaller and smaller. Here we have tim, To determine the limit of a rational polynomial at infinity, divide both the numerator and denominator by the highest power in the denominator and then apply the limit algebra. 26Examples Exercises 2.1 1. Determine the following limits if they exist: 2x? 3 z ae Aim, +! Bxt+x45 a) tim b) lim... o) Tim hot ‘xv4-3 olim,.,, 2.3. DIFFERENTIATION FROM FIRST PRINCIPLES Remember, the gradient of a line joining two points A(x, 74) and B(x, yz) on a curve y= f(@) isgivenas Mag = peal 7If we allow the point B to be closer to A, the slope of AB will be more closely approximate the slope of a tangent drawn to the curve at A. The slope of the tangent, often referred to as the slope of the curve, is the limiting value of the slope of the secant line AB as B approaches A. Hence we could find the slope of a tangent to a curve at a point (x,, f(%,)) by calculating the limit (provided it exists) of the difference f(x, + h) — f (x1) divided fGi+h)-f(a) by has h= 0. We can write this as. Mean = lim, 7 The limit on the right is defined as the derivative of f(x) at x. From the above we have the derivative of the function f(x) as: f°) = lim LEH ~FE) provided the limit exists, This is the formula for determining the derivative of a function from first principles. Example 1. Determine the derivative of each of the following functions from first principles: a) f(x) =-3x? —b)_ f(x) = vx Solution: “oe) a Vir FPDAEOD ej AEM? _ | -3x? ext? 32? ©) FG) = fg SLE yy EON = yy toe = Yim MEO) gy mh jm LEROPO) pe VERVE VERE (VER BE = EE = jig I Se) ti (e+h)—x ik 1 1 = lim = im ————. = — non(vx+h+vx) MeOVx+h+Vx 2x Exercise 2.2 Determine the derivatives of the following functions from first principles: 1 a) f(x) =—5x3b) FG) ==) f(x) = 2x d) fx) =x? +x e)f@=2 28Notation: The derivative of a function y = f(x) can be represented as y’, f'(x), 2, Dy or Df(x) 2.4, BASIC DIFFERENTIATION RULES 2.4.1. Derivatives of polynomials ify = k, where kis any constant, then = =0 Ify = kx, where kis a constant, then 2 = ke For example ify = 4, then © = 0 also if y = —9x, then & = a Ify =x” where n is any real number, positive, negative or fraction, Then @ = nx" Thus the general rule for differentiating x" is to multiply the term by the index or exponent and decrease the index or exponent by one, 2.4.2, Derivative of a sum or difference The derivative of a sum or difference of functions is the sum or difference of the derivatives of functions. The sum or difference is thus differentiated term by term. i.e. £0 @) £9@) =4f@1 tl =3x2-1 Be 43y2) 2-1) 4 4 = Thatis, ify = 3x° ~~ + 4x, then = = 5 (3x7) — 2 (e ) + 3G, (4x) 6x+ 1 até Example Differentiate with respect to x 1) y= 4x3 3x45 Solution 1) y= 4x3 = 3x45 Da 1ox*-3 Qy= 29& ayy t ae = 2X ta S)y =—axt— 2 gy Ba -12845 dx 3 Exercise 2.3 3. Determine the derivatives of each of the following a) y= 3x4 7 ) f(@Q) sax? tx" — 04% b) yaRoxt +2x-5 — e) Dl -x)?] © y=2Vxtx?—3 HFG) =v +2 2.4.3. Product rule ify = u(x) v(x) where both u and v are differentiable, then by the product rule, ay du, ee dx” dx - dx The derivatives of functions such as 2x*(x — 2) can be obtained using the product rule. Examples 1. Determine 2 in each of the following a)y = (x? +.4)(2x — 3) by = (@?-1)Gx-2) Solution a) y = (x? +.4)(2x —3) Letu=x?+4 ul’ =2x v=2x-3 v=2 B= (2x -3)(2x) + (x? +.4)(2) 4x? — 6x + 2x7 +8 6x? — 6x +8 b) y = (x? — 1)(3x — 2) Letu=x?-1 w=2x v=3x-2 v'=3 302 = Gx -2)(2x) + @?-@) = 6x? — 4x + 3x? -3 = 9x? - 44-3 Exercise 2.4 4. Determine the derivative of each of the following: a) y = (x —2)(1 — 2x?) d) f(x) = (2 + 2Vx)(2x? - 1) b) y=x7(x? -1) e) f(x) = (3 - 2vx)(3 + 2vx) co) y =x*(1 — 2x) 2.4.4. Quotient rule » The quotient rule is applied to functions of the type y = _ where both functions are differentiable and v(x) # 0 [Quotient Rule] : x=5 The derivatives of functions such as => or > = can be obtained by using the quotient rule. Examples 1. Determine the derivatives of. _ Asx? _ tx ax OY 0a Oe Solution a)y=* Letu=1—x?, then “ = -2x andv=1tx?then 2 = 2x +x? dx dx dy _ (4x2)(-2x)-(1-x2)(Qx) te dx G+?) ~ Ga? gaz = a =1-2x? ee by=—S Tow Let=1+x then 7 = Landlet v = 1 — 2x’ then AX = 12x20) CD40) _ axed ao (1-202)? (nax2? xo du a Oya, betu=x—t then “= Land let y= 1 +x then >= 1 31dy _ G40)-(e-9 ax (+1)? Exercise 2.5 1. Find the derivative of each of the following: _ 3xt1 x21 ays dy=e by=i> re 242VE oye 9 fe) = 228 2.4.5. The chain rule The chain rule states that if y is a function of u and u is a function of x then dy _dy | du dx du“ dx The chain rule is also known as the function of a function rule. Example 1. Use the chain rule to differentiate with respect to x a)y=(Sx+2)> b)y=(3-5x)3 oc) y = (4x3 - 5x) d) Vx — 4x? Solution a)y = (5x +2)? Letu=5x+2 =>y= a ® = ay? wm and qu 7 OU ay dy du yee 2 Oa O&M = 15u = 15(5x + 2) b)y=(3-5x)* Letu=3-5xtheny =u} du dy _ ae 8 ond Bu Oe oa —5x)* ae age 15 15(3 — 5x) oy = (4x3 —5x)5 Letu=4x3—5x then y 32du 4 = 12x7-5 and 2 = 5ut ax au @ = @ B= 5(12x? — Sut = 5(12x? — 5)(4x9 - Sx) a) f(x) = Vesa? Letusx—4x? theny =ur Exercise 2.6 1. Differentiate each of the following: a) y=(3x*-5)" d) y= V3x7—-1 b) y=G-2x?)% o) y= (\e+1) y= Sato 2.5. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS NOTE: lim Ty and fim St = 0 if f(x) = sinx “oe) ex Yj in GetnDesinx Then f (x) = ara sinxcosh+cosxsinhsinx Fe) = memes (cosh-1)+cosxsinh = lim no h : a sinh = lim [-sinz =" + cosy ho h 7” = (—sinx)(0) + (cosx)(1) = cosx Similarly, using cos(x + A) = cosxcosh — sinxsinh, we may find the derivative of y =cosx. ify = cosx, then & = —sinx 33You will not be required to find the derivatives of sinx and cos.x from first principles. ; e a a ify = sinax, then = acos dx,and if y = cos ax, then = —asingx In general, if y = sin f(x), then £ Se fx) cos f (x), using the chain rule. For example, y = sin(3x - 1) Letu=3x-1 — theny = sinw ay _ ay du dx du dx au ay 3 ond = 3 and qu = COSU ay _ _ then 7 = 3 cos(3x — 1) Example 1. Determine the derivatives of the following: a)y =sin3x+cos3x — b) y = 2cos(3x? — 1) ©) y = sin(vx) — 2 cos(x? — 3) Solution a) y= sin3x+cos3x 2 = 3cos3x —3sin3x dx b) y = 2cos(3x? — 1) Oy i 2 & = -12xsin(3x? — 1) ©) y=sin(Vz) ~ 2.cos(x? - 3) & = cos( VE) + 4x sin(x? ~ 3) Exercise 2.7 ww : du # Determine the derivative of each of the following. cosx a) y= x%cos2x b)y =sin?x c) y = sinx? —cos(x2 + 2) d)y= i-sinx sinx e)y =tanx = 34