SEMESTER PELANGI BESTSELLER PRE-U 2 Mathematics (T) Lee Yoon Woh Chin Siew Wui Tan Ah Geok 1 COPIES SOLD MORE THAN STPMText
Mathematics (T) Lee Yoon Woh Chin Siew Wui Tan Ah Geok PRE-U SEMESTER 2 © Penerbitan Pelangi Sdn. Bhd. 2022 All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, photocopying, mechanical, recording or otherwise, without the prior permission of Penerbitan Pelangi Sdn. Bhd. ISBN: 978-967-2779-96-4 eISBN: 978-967-2779-97-1 (eBook) First Published 2022 Tel: 03-8922 3993 Fax: 03-8926 1223 / 8920 2366 E-mail: [email protected] Enquiry: [email protected] Printed in Malaysia by Herald Printers Sdn. Bhd. Lot 508, Jalan Perusahaan 3, Bandar Baru Sungai Buloh, 47000 Selangor Darul Ehsan. Please log on to https://plus.pelangibooks.com/errata/ for up-to-date adjustments to the contents of the book (where applicable). STPMText
ii PREFACE This book ‘Pre-U STPM Text Mathematics (T) - Semester 2’ is one in a series of three books, specially written to meet the requirements of the new revised Mathematics (T) syllabus in the STPM Examination, which will take effect from 2012. Under the new system introduced by the Malaysian Examinations Council (MEC), the new Form Six curriculum will be spread over three semesters, with candidates sitting for an examination at the end of each semester, and also coursework. This is to enhance the teaching and learning orientation of Form Six, so as to be in line with the orientation of teaching and learning in colleges and universities. The new syllabus fulfils the requirements of this new system, and this book covers all the topics specified in the syllabus of Mathematics (T) for Semester 2. It is also suitable for use by students pursuing a foundation course in science or matriculation programmes at local universities. This book seeks to fulfil the aims and objectives as set out in the new mathematics syllabus, i.e. to develop the understanding of mathematical concepts and mathematical thinking, and acquire skills in problem solving and the applications of mathematics related to science and technology. This will prepare the students with an adequate foundation, before they proceed to programmes in the field of science and technology at institutions of higher learning. The contents of this book, in six chapters based on the syllabus, are organised and planned in a systematic manner so as to make learning more effective. Each new topic in a chapter is clearly presented via a simple and practical approach, which leads to a better overall picture and understanding of the topic concerned. The concepts presented in each subtopic include relevant explanations in detail, followed by the all-important worked examples, presented clearly in a step by step manner. This approach presumes that the student has only prior basic mathematical knowledge and skills up to the SPM level. The questions at the end of each subtopic are planned in such a way that the student will be able to test his understanding of, and apply, the concepts learned to solve basic problems. They are also suitable for the coursework. A summary of the concepts and other important formulae is given at the end of each chapter. A revision exercise covering all the subtopics in the chapter is also included. The questions chosen are all planned such that they are of equivalent standard as those in the STPM examination. To help the student in his revision and to assess his preparedness in facing the examination, a set of sample STPM examination paper is included at the end of the book, together with complete worked solutions. This book may be used as a textbook in the classroom, for coursework or by the student studying on his own. It is hoped that both teachers and students alike will benefit from using this book, and find the learning process both effective and enjoyable.
iii CONTENTS • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 1 LIMITS AND CONTINUITY 1 1.1 Limits 2 1.2 Continuity 6 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 2 DIFFERENTIATION 16 2.1 Derivatives 17 2.2 Applications of Differentiation 39 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 3 INTEGRATION 80 3.1 Indefinite Integrals 81 3.2 Definite Integrals 104 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 4 DIFFERENTIAL EQUATIONS 126 4.1 Differential Equations 127 4.2 First Order Differential Equations with Separable Variables 130 4.3 First Order Linear Differential Equations 134 4.4 Transformations of Differential Equations 136 4.5 Problems Modelled by Differential Equations 140 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 5 MACLAURIN SERIES 153 5.1 Maclaurin Series 154 5.2 Applications of Maclaurin Series 167 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 6 NUMERICAL METHODS 172 6.1 Numerical Solution of Equations 173 6.2 Numerical Integration 186 STPM Model Paper (954/2) 193 Answers 195
iv Symbols = is equal to ≠ is not equal to is identical to or is congruent to ≈ is approximately equal to , is less than < is less than or equal to . is greater than > is greater than or equal to ∞ infinity Set notation N set of natural numbers, {0, 1, 2, 3, …} Z set of integers Z+ set of positive integers Q set of rational numbers R set of real numbers [a, b] closed interval {x | x R, a < x < b} (a, b) open interval {x | x R, a , x , b} [a, b) interval {x | x R, a < x , b} (a, b] interval [x | x R, a , x < b} Functions f a function f f(x) value of a function f at x ex exponential function of x loga x logarithm to base a of x ln x natural logarithm of x, loge x sin, cos, tan, trigonometric functions csc, sec, cot sin–1, cos–1, tan–1, inverse trigonometric csc–1, sec–1, cot–1 functions Derivatives and integrals lim x → af(x) limit of f(x) as x tends to a dy dx first derivative of y with respect to x f(x) first derivative of f(x) with respect to x d2 y dx2 second derivative of y with respect to x f(x) second derivative of f(x) with respect to x dn y dxn nth derivative of y with respect to x f (n) (x) nth derivative of f(x) with respect to x ∫ y dx indefinite integral of y with respect to x ∫ a b y dx definite integral of y with respect to x for values of x between a and b Mathematical Notation
1 CHAPTER LIMITS AND CONTINUITY 1 Subtopic Learning Outcome 1.1 Limits (a) Determine the existence and values of the left-hand limit, right-hand limit and limit of a function. (b) Use the properties of limits. 1.2 Continuity (a) Determine the continuity of a function at a point and on an interval. (b) Use the intermediate value theorem. continuity – keselanjaran continuous – selanjar function – fungsi interval – selang left-hand limit – had kiri limit – had right-hand limit – had kanan stationary point – titik pegun Bilingual Keywords
2 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 1.1 Limits Limits Let f be a function defined on an open interval containing a, possibly not defined at a. The limit (or limiting value) of f(x) as x approaches a, written as lim x → a f(x), is the value that f(x) approaches as x approaches a. If the limit of f(x) as x approaches a is l, then we write lim x → a f(x) = l or f(x) → l as x → a 1. If f(x) → l as x → a from the left (i.e. values of x less than a), we write lim x → a– f(x) = l and this is known as the left-hand limit. 2. If f(x) → l as x → a from the right (i.e. values of x more than a), we write lim x → a+ f(x) = l and this is known as the right-hand limit. 3. Take note that lim x → a– f(x) may not be the same as lim x → a+ f(x). Hence, lim x → a f(x) = l if and only if lim x → a+ f(x) = lim x → a– f(x) = l. Properties of limits Let a be any real number and k any constant. 1. lim x → a c = c, c is a constant 2. lim x → a k f(x) = k lim x → a f(x) 3. The limit of a sum (or difference) is the sum (or difference) of the limits. lim x → a [f(x) ± g(x)] = lim x → a f(x) ± lim x → a g(x) 4. The limit of a product is the product of the limits. lim x → a [f(x) . g(x)] = lim x → a f(x) . lim x → a g(x) 5. The limit of a quotient is the quotient of the limits lim x → a f(x) g(x) = lim x → a f(x) lim x → a g(x) , provided lim x → a g(x) ≠ 0 Limit INFO Introduction to Limit VIDEO
3 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 Example 1 Evaluate (a) lim x → 1 (x2 – x – 2) (b) lim x → 2 x2 + 2x – 3 x2 + x + 1 (c) lim x → 6 36 – x2 6 – x (d) lim x → 0 x3 – 2x2 + 3x x2 – 2x Solution: (a) lim x → 1 (x2 – x – 2) = lim x → 1 x2 – lim x → 1x – lim x → 1 2 = 12 – 1 – 2 = –2 Alternative Method Substitute x = 1, lim x → 1 (x2 – x – 2) = 12 – 1 – 2 = –2 (b) lim x → 2 x2 + 2x – 3 x2 + x + 1 = 22 + 2(2) – 3 22 + 2 + 1 = 5 7 (c) lim x → 6 1 36 – x2 6 – x 2 The function f(x) = 36 – x2 6 – x is not defined when x = 6 If x ≠ 6, then lim x → 6 36 – x2 6 – x = lim x → 6 (6 – x)(6 + x) 6 – x = lim x → 6 (6 + x) = 6 + 6 = 12 Note : We can take values of x as near as possible to 6, but not equal to 6. (d) lim x → 0 1 x3 – 2x2 + 3x x2 – 2x 2 = lim x → 0 x(x2 – 2x + 3) x(x – 2) = lim x → 0 1 x2 – 2x + 3 x – 2 2, x ≠ 0 = 0 – 0 + 3 0 – 2 = – 3 2 Example 2 Determine whether the limit exists in each of the following cases. (a) lim x → 1 f(x) where f(x) = 2 + ex , x , 1 1 + e – x , x . 1 (b) lim x → 2 f(x) where f(x) = (x – 2)2 , x < 2 1 – 2 x , x . 2
4 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 Solution: (a) lim x → 1– f(x) = lim x → 1– (2 + ex ) = 2 + e lim x → 1+ f(x) = lim x → 1_ (1 + e – x) = 1 + e – 1 = e Since lim x → 1– f(x) ≠ lim x → 1+ f(x), lim x → 1 f(x) does not exist. (b) lim x → 2– f(x) = lim x → 2– (x – 2)2 = (2 – 2)2 = 0 lim x → 2+ f(x) = lim x → 2+ 11 – 2 x 2 = 1 – 2 2 = 0 Since lim x → 2– f(x) = lim x → 2+ f(x) = 0, lim x → 2 f(x) exist. Limits at infinity For n Z+ , lim x → ∞ 1 xn = 0. Example 3 Evaluate (a) lim x → ∞ 2 + x 3x + 1 (b) lim n → ∞ 3n2 – n + 1 n2 + 1 Solution: We cannot substitute x = ∞ as ∞ ∞ is undefined. Instead, both the numerator and the denominator are divided by the highest power of x. (a) lim x → ∞ 2 + x 3x + 1 = lim x → ∞ 1 2 x + 1 3 + 1 x 2 = 1 3 (b) lim n → ∞ 1 3n2 – n + 1 n2 + 1 2 = lim n → ∞ 1 3 – 1 n + 1 n2 1 + 1 n2 2 = 3
5 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 Exercise 1.1 1. Find the limit of f(x) as x → 2 if f(x) is given by (a) 3x + 2 (b) 2 + x (c) x2 – 1 (d) 1 + x2 – x3 2. Find the limit of f(x) as x → 3 if f(x) is given by (a) 1 x + 1 (b) 2 + 1 x (c) x – 5 x + 1 (d) x2 – 9 x – 3 3. Find the limit of f(x) as x → ∞ if f(x) is given by (a) 1 + 2 x – 3 x2 (b) 3x + 1 x (c) x2 + 2x – 1 2x2 + 3x + 1 (d) x3 3x2 (x – 1) 4. Evaluate (a) lim x → 0 1 3x + x2 x 2 (b) lim x → 1 1 x2 + 1 x + 1 2 (c) lim x → 2 1 x2 – 4 x – 2 2 (d) lim x → ∞ 1 x2 – 4x + 3 5x + 2 2 (e) lim x → 2 11 + 1 x 212 + 1 x 2 (f) lim x → ∞ 1 2x + 1 5x + 2 2 5. Evaluate (a) lim x → 1 1 3x2 + x – 2 2x3 + 5x + 3 2 (b) lim x → –1 1 x3 – 1 2x3 – 3x + 1 2 6. A function f is defined by f(x) = | x – 3| x – 3 , x ≠ 3 0 , x = 3 Find (a) lim x → 3– f(x), (b) lim x → 3+ f(x). Hence, determine whether lim x → 3 f(x) exists. 7. The function is defined by 3x – 1, x , 0, f(x) = 0, x = 0, 2x + 5, x . 0. Evaluate (a) lim x → 2 f(x), (b) lim x → 3– f(x), (c) lim x → 0– f(x), (d) lim x → 0+ f(x), (e) lim x → 0 f(x).
6 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 8. A function f is defined by 2x, x < 1, f(x) = x + 1, 1 , x < 2, –x – 1, x . 2. Evaluate (a) lim x → 1– f(x), (b) lim x → 1+ f(x), (c) lim x → 1 f(x), (d) lim x → 2– f(x), (e) lim x → 2+ f(x), (f) lim x → 2 f(x). 1.2 Continuity Continuity at a number A function f is said to be continuous at a number a if the graph of f is unbroken at the point (a, f(a)). The definition implicitly requires that the following three conditions hold: (a) f(a) is defined, (b) lim x → a f(x) exists, (c) lim x → a f(x) = f(a) By definition, function f is continuous at a if lim x → a f(x) = f(a) 0 x y a f(a) y = (x) x a y 0 Figure 1.1 Figure 1.2 Function f is continuous at x = a Function f is discontinuous at x = a Example 4 Sketch the graphs of each of the following functions and state whether the function is a continuous function. (a) f(x) = 4 + x2 , x , 0 (b) f(x) = x – 1, x < 0 4 – x , x > 0 x2 , x . 0
7 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 Solution: (a) x y 0 1 2 2 –2 –2 –1 3 4 4 5 6 6 8 7 f is a continuous function since the graph is unbroken in its domain. (b) x y 0 2 –1 – 3 –1 – 4 4 f is not a continuous function. Example 5 The function f is defined by ex + 3 , x , 1, f(x) = 4 , x = 1, e – x + 4 , x . 1. (a) Find lim x → 1– f(x) and lim x → 1+ f(x). Hence, determine whether f is continuous at x = 1. (b) Sketch the graph of f. Solution: (a) lim x → 1– f(x) = lim x → 1– (ex + 3) = e + 3 lim x → 1+ f(x) = lim x → 1+ (e – x + 4) = e + 3 lim x → 1 f(x) = e + 3 But f(1) = 4 Since lim x → 1 f(x) ≠ f(1) \ f is not continuous at x = 1.
8 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 (b) x y 0 1 4 + e 4 3 f(x) = e f(x) = e – x + 4 x + 3 Continuity at an endpoint A function f is continuous from the left at a if lim x → a– f(x) = f(a); a function is continuous from the right at a if lim x → a+ f(x) = f(a). x a y 0 x a y 0 f is continuous from the left at a f is continuous from the right at a. Example 6 The function f is defined by 0, x , 0, f(x) = 1, x > 0 Determine whether f is continuous from the left at 0 and continuous from the right at 0. Solution: lim x → 0– f(x) = lim x → 0– (0) = 0 f(0) = 1 Since lim x → 0– f(x) ≠ f(0), f is not continuous from the left at 0. \ lim x → 0+ f(x) = lim x → 0+ (1) = 1 Since lim x → 0+ f(x) = f(0) = 1, f is continuous from the right at 0.
9 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 Continuity on an interval A function f is continuous on an open interval (a, b) if it is continuous at every number in the interval; a function f is continuous on a closed interval [a, b] if it is continuous on (a, b) and is also continuous from the right at a and from the left at b. Example 7 Show that the function f defined by f(x) = 1 – x2 is continuous on the closed interval [–1, 1]. Solution: We first show that f is continuous on (–1, 1). Let –1 , a , 1. Then lim x → a f(x) = lim x → a 1 – x2 = lim x → a (1 – x2 ) = 1 – a2 = f(a) Next, we show that f is continuous from the right at –1 and from the left at x = 1. lim x → –1+ f(x) = lim x → –1+ 1 – x2 = lim x→–1+ (1 – x2 ) = 0 = f(–1) lim x → 1– f(x) = lim x → 1– 1 – x2 = lim x→1– (1 – x2 ) = 0 = f(1) Notice that a polynomial function is continuous on (–∞, ∞). A rational function is continuous on its domain. The functions ex , ln x, sin x, cos x, tan x, sin–1 x, cos–1 x and tan–1 x are also continuous on their respective domains. Exercise 1.2 1. The function f is defined by f(x) = sin x, x < 0, x, x . 0. Sketch the graph of f for – π 2 , x , π 2 , and state whether f is continuous at 0. 2. The function f is defined by f(x) = x(x – 1), 0 < x , 2, 2(3 – x), 2 < x < 3. Sketch the graph of f, and state whether it is continuous on its domain. 3. Sketch the graph of the function f in each of the following cases, and state whether f is a continuous function. (a) y = cos x, 0 < x < 2π (b) y = tan x, 0 < x < π (c) y = e–x , x R (d) y = (x + 1)2 , x R (e) y = ln(1 + x), x . –1 (f) y = 2 x – 1 , x R, x ≠ 1
10 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 4. The function f is defined by |5 – x| 5 – x , x ≠ 5, f(x) = 0, x = 5. (a) Find lim x → 5– f(x) and lim x → 5+ f(x). Hence, determine whether f continuous at 5. (b) Sketch the graph of f. 5. The function f(x) is defined as follows: 3x – 1, x , 0, f(x) = 0, x = 0, 2x + 5, x . 0. (a) Determine whether lim x → 0 f(x) exist. Hence, determine if f is continuous at 0. (b) Sketch the graph of f. 6. The function f is defined by x(x – 2), 0 < x , 2, f(x) = 1 – 1 2 x, 2 < x , 3, 1 2 x – 2, 3 < x , 5. Determine whether f is continuous at 2 and 3. Sketch the graph of f. 7. A function f is defined by f(x) = 2 – x2 , –3 < x , 0, 2 – (x – 2)2 , 0 < x < 5. (a) Show that f is discontinuous at 0. (b) Determine whether f is continuous from the left, from the right, or neither at 0. (c) Sketch the graph of f. 8. The function f is defined by 2 – x 3 + x , 0 < x , 1, f(x) = 1 + kx2 , x > 1 where k R. Given that lim x → 1 f(x) exists, find the value of k. With the value of k, determine whether f is continuous at 1. 9. The function f is defined by x – 2 x2 – 3x + 2 , x ≠ 1, x ≠ 2, f(x) = c, x = 2. If f is continuous at x = 2, find the value of c.
11 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 10. The function f is defined as follows: x, 0 < x < 2, f(x) = C, 2 , x < 4, x2 – 4x + 2, 4 , x < 5. (a) Determine the value of C which makes f continuous on the interval 0 < x < 5. (b) Sketch the graph of f. Intermediate value theorem Let f be a function which is continuous on the closed interval [a, b]. Suppose that d is a real number between f(a) and f(b), then there exists a number c in (a, b) such that f(c) = d. x a f(a) f(b) c d b f(x) 0 Figure 1.3 Corollary: Let f be a function which is continuous on the closed interval [a, b]. If f(a) · f(b) , 0, then there exists c in (a, b) such that f(c) = 0. In other words f has at least one zero in the interval (a, b). An application of the intermediate value theorem is to prove the existence of roots of equations. Example 8 Show that f(x) = 2x3 – 5x2 – 10x + 5 has a zero in the interval (–1, 2). Solution: We have to show that there is a number c such that –1 , c , 2 and f(c) = 0. Applying the intermediate value theorem, we need to show that f is continuous and that d = 0 is between f(–1) and f(2), i.e. f(–1) , 0 , f(2) or f(2) , 0 , f(–1). f(–1) = 8 and f(2) = –19 we have f(2) , 0 , f(–1) \ d = 0 is between f(–1) and f(2). Since f(x) is a polynomial, it is a continuous function. So by the intermediate value theorem, there must be a number –1 , c , 2 such that f(c) = 0. \ The function does have a zero between –1 and 2.
12 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 Example 9 Use the intermediate value theorem to show that the equation x5 – x + 2 = 0 has at least one real root. Solution: Let f(x) = x5 – x + 2 f(x) is continuous since it is a polynomial f(–1) = 2 f(–2) = –28 Since f(–2) , 0 and f(–1) . 0, the intermediate value theorem tells us that f(c) = 0 for some c in the interval (–2, –1). Hence, the equation x5 – x + 2 = 0 has at least one real root. Graph of f in the interval [–2, –1]. Example 10 Given f(x) = x3 – x2 + x. Show that there is a number c R such that f(c) = 10. Solution: f(0) = 0 and f(3) = 21. We have f(0) , 10 , f(3) Since f is continuous, there must be a number c R such that f(c) = 10. Example 11 Show that there is a positive real root of the equation x7 = x + 1. Solution: Let f(x) = x7 – x – 1 f(0) = –1 f(2) = 125 Since f(0) , 0 and f(2) . 0, the intermediate value theorem tells us that f(c) = 0 for some c in the interval (0, 2), all of whose elements are positive numbers. Hence, there is a positive real root of the equation x7 = x + 1. x 0 –1 –28 2 –2 c f(x)
13 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 Exercise 1.3 1. Use the intermediate value theorem to show that there is a root of the given equation in the given interval. (a) x3 – 4x + 1 = 0 ; (0, 1) (b) x5 – 2x4 – x – 4 = 0 ; (2, 3) (c) x3 + 3x = x2 + 1 ; (0, 1) (d) x2 = x + 2 ; (1, 2) 2. Show that the equation x3 + x – 5 = 0 has at least one root a where 1 , a , 2. 3. Show that the equation e–x + 2 = x has at least one real root. 4. Use the intermediate value theorem to show that the equation x4 = 2x has at least one root. 5. Use the intermediate value theorem to show that the equation x3 + x + 1 = 0 has a root in the interval (–2, 0). Summary 1. lim x → a f(x) exists if and only if lim x → a– f(x) = lim x → a+ f(x). 2. Properties of limits Let a be any real number and k any constant. (a) lim x → a c = c, c is a constant (b) lim x → a k f(x) = k lim x → a f(x) (c) The limit of a sum (or difference) is the sum (or difference) of the limits. lim x → a [f(x) ± g(x)] = lim x → a f(x) ± lim x → a g(x) (d) The limit of a product is the product of the limits. lim x → a [f(x) . g(x)] = lim x → a f(x) . lim x → a g(x) (e) The limit of a quotient is the quotient of the limits lim x → a f(x) g(x) = lim x → a f(x) lim x → a g(x) , provided lim x → a g(x) ≠ 0 3. A function f is continuous at a if lim x → a f(x) = f(a). 4. A function f is continuous on an open interval (a, b) if it is continuous at every number in the interval; a function f is continuous on a closed interval [a, b] if it is continuous on (a, b) and is also continuous from the right at a and from the left at b.
14 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 STPM PRACTICE 1 1. Evaluate (a) lim x → –3 √x + 4 – 1 x + 3 (b) lim x → 1 4e 21 x – 1 x + 1 2 – 1 2e 1 x – 1 x + 1 2 – 1 2. (a) State the three conditions in which function f, f(x) is not continuous at point x = c. (b) A function g is defined by 8 – x 7 – x , x , 6 g(x) = x2 – 2x – 6 kx , x > 6 where k ≠ 0 Determine the set of values of k such that g is not continuous at point x = 6. 3. The function f is defined by x – a, x < –3, f(x) = bx2 + 2, –3 , x < 1, 3x + 2, x . 1. Determine the value of a and the value of b if lim x → –3 f(x) and lim x → 1 f(x) exist. 4. The function f is defined by f(x) = x + 2 , –2 < x , 2, |x| – 2, otherwise. (a) Find lim x → –2– f(x), lim x → –2+ f(x), lim x → 2– f(x) and lim x → 2+ f(x). (b) Determine whether f is continuous at –2 and 2. 5. Evaluate (a) lim x → 3 8 (x – 3) x3 – 27 (b) lim x → 10 x – 10 √7 – √x – 3 6. The function f is defined by x2 – 4 x – 2 , x , 2, f(x) = a2 – 2, x = 2 |2 – x| x – 2 + b , x . 2 Where a and b are constants. (a) If lim x → 2 f(x) exists, find the value of b. (b) If f(x) is continuous at x = 2, determine the values of a.
15 Mathematics Semester 2 STPM Chapter 1 Limits and Continuity 1 7. The function f is defined by 3 – 2ex , x , 0 f(x) = 2, x = 0 3ex – 2, x . 0 (a) Determine the existence of limit f(x) as x approaches 0. (b) State, with a reason whether f(x) is continuous at x = 0. Hence, determine the intervals on which f is continuous. 8. The function f is defined by f(x) = ln x, 0 , x , 1, ax2 + b, 1 < x , ∞. Given f(2) = 3, determine the values of a and b for which f is continuous on (–∞, ∞). 9. The function f is defined by f(x) = x2 – 9 |x – 3| , x ≠ 3 6, x = 3 Determine whether f is continuous at x = 3. 10. Evaluate (a) lim h → 0 e 4h – 1 e 2h – 1 (b) lim h → –∞ x – 5 4x2 + 7 11. A function f is defined by f(x) = ln (x + 2), –2 , x , ∞. (a) Show that f is continuous on its domain. (b) Sketch the graph of f. 12. Use the intermediate value theorem to show that there exists a solution to the equation cos x = x in the interval [0, π 2 ]. 13. Evaluate (a) lim x → 0 3x2 + 16 – 4 x2 (b) lim x → ∞ 16x2 + 1 2x – 1 14. Show that the graph of y = x2 – 4x + 1 intersects the x-axis in the interval [0, 2]. Can the same be said for the graph of y = 2x – 3 x – 1 ? 15. The function f is defined by f(x) = x2 + 2x + 8, x , 0 2ex + c, x > 0 (a) Find lim x → 0– f(x) and lim x → 0+ f(x). Hence, determine the value of c such that function f is continuous at x = 0. (b) Describe the continuity of the function f for (i) x = 0 (ii) x , 0 (iii) x . 0
CHAPTER Subtopic Learning Outcome 2.1 Derivatives (a) Identify the derivative of a function as a limit. (b) Find the derivatives of xn (n Q), ex , In x, sin x, cos x, tan x, sin–1x, cos–1 x, tan–1 x, with constant multiples, sums, differences, products, quotients and composite. (c) Perform implicit differentiation. (d) Find the first derivatives of functions defined parametrically. 2.2 Applications of differentiation (a) Determine where a function is increasing, decreasing, concave upward and concave downward. (b) Determine the stationary points, extremum points and points of inflexion. (c) Sketch the graphs of functions, including asymptotes parallel to the coordinate axes. (d) Find the equations of tangents and normals to curves, including parametric curves. (e) Solve problems concerning rates of change, including related rates. (f) Solve optimisation problems. asymptote – asimptot concave downward – cekung ke bawah curve – lengkung decreasing – menyusut derivative – terbitan differentiation – pembezaan extremum point – titik ekstremum implicit function – fungsi tersirat increasing – menokok limit – had point of inflexion – titik lengkok balas stationary point – titik pegun DIFFERENTIATION 2 Bilingual Keywords
17 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 2.1 Derivatives Derivative of a function Let A(x, y) be a fixed point on the curve y = f(x) and B be a neighbouring point with coordinates (x + x, y + y), where x (read as ‘delta-ex’) denotes a small increase in x and y denotes a corresponding small increase in y. In Figure 2.1, AC = x, BC = y. Gradient of chord AB is BC AC = dy x . As the point B is moved along the curve towards the fixed point A, x → 0 and the direction of the chord AB approaches closer and closer to the direction of AT, the tangent to the curve at A. Thus, gradient of curve at A = lim δx → 0 dy x (read as ‘the limit as x tends to 0’) = dy dx (read as ‘dee’ y by ‘dee’ x) If y = f(x), then dy dx = f(x) defined by f(x) = lim δx → 0 f(x + x) – f(x) x This formal definition of a derivative can be used to differentiate any function. This process is known as differentiation with respect to x (abbreviated to w.r.t.) from first principles. dy dx or f(x) is the ‘derivative of f(x)’, the ‘differential coefficient’ or ‘derived function’ of f(x) w.r.t. x. Note: dy and dx do not have any meaning in themselves; in particular we cannot think of dy dx as dy ÷ dx. Example 1 Find the derivatives of the following functions with respect to x, from the first principles. (a) f(x) = x2 (b) f(x) = 1 x Solution: (a) Using the definition d dx f(x) = lim δx → 0 f(x + x) – f(x) δx 4 d dx (x2 ) = lim δx → 0 (x + x) 2 – x2 x 4 = lim δx → 0 x2 + 2x x + (x) 2 – x2 x 4 y C T x 0 B(x + δx, y + δy) A(x, y) y = f(x) Figure 2.1 Differentiation INFO Feynman's Differentiation VIDEO
18 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 = lim δx → 0 2xx + (x) 2 x 4 = lim δx → 0 (2x + x) = 2x (b) d dx 1 1 x 2 = lim δx → 0 1 x + x _ 1 x x 4 = lim δx → 0 x – (x + x) (x + x)x (x)4 = lim δx → 0 –1 x(x + x)4 = – 1 x2 Example 2 If y = 3x2 + x + 1, find dy dx from the first principles. Solution: dy dx = lim δx → 0 3(x + x) 2 + (x + x) + 1 – (3x2 + x + 1) x 4 = lim δx → 0 3x2 + 6xx + 3(x) 2 + x + x + 1 – 3x2 – x – 1 x 4 = lim δx → 0 6x x + 3(x) 2 + x x 4 = lim δx → 0 [6x + 3 x + 1] = 6x + 1 Exercise 2.1 Find the derivatives of the following functions from the first principles. 1. y = x3 2. y = x4 3. y = 5x2 4. y = 1 x2 5. y = x2 + 5x 6. y = x2 – x + 3 7. y = 4x3 8. y = 1 x3 9. y = 2x2 + 3 10. y = 2x2 – 3x + 1
19 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Differentiation of standard functions Derivative of a constant Consider the function y = c, where c is a constant. From the derived definition dy dx = lim δx → 0 f(x + x) – f(x) δx As f(x + x) = c = f(x), ∴ f(x + x) – f(x) δx = c – c δx = 0 Hence, dy dx = lim δx → 0 0 = 0 Geometrically, y = c represents a straight line parallel to the x-axis. Therefore, its gradient is zero. d dx (c) = 0, where c is a constant Derivative of af(x) where a is a constant From the derived definition, d dx [a f(x)] = lim δx → 0 a f(x + x) – a f(x) δx = lim δx → 0 a[f(x + x) – f(x)] x 4 = lim δx → 0a lim δx → 0 f(x + x) – f(x)] x 4 = a d dx [f(x)] Hence, d dx [a f(x)] = a d dx [f(x)] Derivative of xn Consider the following differentiation from first principles. d dx x 2 = lim δx → 0 (x + x) 2 – x2 δx = lim δx → 0 (2x + x) = 2x = 2x2 – 1 c y x y = c 0 Figure 2.2
20 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 d dx x 3 = lim δx → 0 (x + x) 3 – x3 δx = lim δx → 0 [3x2 + 3x x + (x) 2 ] = 3x2 = 3x3 – 1 d dx x –1 = lim δx → 0 (x + x) –1 – x–1 δx = lim δx → 0 – 1 x(x + x) = – 1 x2 = –x–2 = –1(x–1 – 1) Hence, d dx xn = nxn – 1, n R. Example 3 Differentiate with respect to x. (a) x–3 (b) x (c) –5x4 (d) 3 x6 Solution: (a) Let y = x–3 dy dx = –3x–3 – 1 = –3x–4 = – 3 x4 (b) Let y = x = x —1 2 dy dx = 1 2 x —1 2 – 1 = 1 2 x –—1 2 = 1 2x (c) Let y = –5x4 dy dx = –5(4x4 – 1) = –20x3 (d) Let y = 3 x6 = 3x–6 dy dx = 3[–6x–6 – 1] = –18x–7 = – 18 x7 Exercise 2.2 Find the derivatives of the following functions with respect to x. 1. 5 2. 3x 3. –4x 4. x5 5. x–4 6. x —1 3 7. x–1 8. x3 9. 1 x 10. 3 x2 11. 5x–3 12. 7x10 13. 5 4x4 14. 7 8 x3 15. –2 x8
21 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Derivative of ex Consider the exponential function f(x) = ax in Figure 2.3 where a is a positive constant. For any value of a, a0 = 1. Hence, the curve passes through the point A(0, 1). The gradient at point A is given by f(0) = lim δx → 0 f(0 + x) – f(0) x = lim δx → 0 f(x) – f(0) δx = lim δx → 0 ax – a0 δx = lim δx → 0 ax – 1 δx For f(x) = ax d dx (ax ) = lim δx → 0 ax + x – ax δx = lim δx → 0 ax 1 ax – 1 x 2 = ax lim δx → 0 ax – 1 x = ax f(0) f(0) = lim δx → 0 aδx – 1 δx Hence, d dx (ax ) = ax f(0) Now there must be a value of a for which f(0) = 1, i.e. the gradient of the graph at (0, 1) is 1. We call this value e. Thus d dx (ex ) = ex e is an irrational number. e ≈ 2.718. ex is known as the exponential function. Derivative of ln x Let y = f(x), where f(x) is any function of x, hence dy dx = lim δx → 0 1 δy δx 2 = lim δx → 0 1 1 δx δy 2 Figure 2.3 y = ax 0 y x A(0, 1)
22 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 But y → 0 when x → 0, dy dx = 1 lim δx → 0 1 δx δy 2 dy dx = 1 dx dy When ln x = y, From the definition of logarithm i.e. if loga y = x, then y = ax . x = ey Differentiate both sides w.r.t. y, dx dy = d dy (ey ) = ey = x dy dx = 1 dx dy = 1 x Hence, d dx (ln x) = 1 x Derivative of ax Let y = ax Taking log of both sides to base e, ln y = ln ax = x ln a Differentiate both sides w.r.t. y, d dy (ln y) = dx dy ln a 1 y = dx dy ln a dx dy = 1 y ln a dy dx = y ln a = ax ln a Hence, d dx (ax ) = ax ln a Example 4 Differentiate each of the following with respect to x. (a) 3x (b) 10x Solution: (a) Let y = 3x (b) y = 10x Hence, dy dx = 3x ln 3 Hence, dy dx = 10x ln 10
23 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Differentiation of trigonometric functions Derivative of sin x Let y = sin x dy dx = lim δx → 0 sin (x + dx) – sin x dx (x + dx) + x (x + dx) – x 2 cos sin 2 2 = lim δx → 0 4 Using the formula sin A – sin B A + B A – B = 2 cos ––––– sin ––––– 2 2 dx dx dx 2 cos 1x + 2 sin 2 2 = lim δx → 0 dx dx sin dx 2 = lim δx → 0 cos 1x + 2 2 dx 2 dx sin dx 2 In the limit, as dx → 0, cos 1x + 2 → cos x and → 1 2 dx 2 Hence d dx (sin x) = cos x, where x is in radians. Derivative of cos x Let y = cos x dy dx = lim δx → 0 cos (x + dx) – cos x dx (x + dx) + x (x + dx) – x –2 sin sin 2 2 = lim δx → 0 4 Using the formula cos A – cos B A + B A – B = –2 sin ––––– sin ––––– 2 2 dx –2 sin 1x + dx 2 2 sin dx 2 = lim δx → 0 dx sin dx 2 = lim δx → 0 –sin 1x + dx 2 2 dx 2 sin dx 2 In the limit, as dx → 0, sin 1x + dx 2 2 → sin x and → 1 dx 2 Hence d dx (cos x) = –sin x, where x is in radians.
24 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Derivative of tan x Let y = tan x dy dx = lim δx → 0 tan (x + dx) – tan x dx 4 = lim δx → 0 sin (x + dx) cos (x + dx) – sin x cos x 4 1 dx = lim δx → 0 sin(x + dx) cos x – cos(x + dx) sin x cos x cos (x + dx) · dx 4 = lim δx → 0 sin [(x + dx) – x] cos x cos (x + dx) · dx Using the formula sin A cos B – cos A sin B = sin (A – B) = lim δx → 0 sin dx cos x cos (x + dx) · dx = lim δx → 0 1 cos x cos (x + dx) · sin dx dx In the limit, as dx → 0, cos (x + dx) → cos x and sin dx dx → 1. Hence, d dx (tan x) = 1 cos2 x = sec2 x Derivative of arcsin x (sin–1 x) Let y = sin–1 x x = sin y Differentiating w.r.t. y, dx dy = cos y \ dy dx = 1 cos y but cos2 y = 1 – sin2 y = 1 – x2 Hence, dy dx = 1 (1 – x2 ) i.e. d dx (sin–1 x) = 1 (1 – x2 ) Note: y = sin–1 x ⇒ – π 2 < y < π 2 , since y is a principal value angle. For this range of values of y, cos x > 0 \ cos x = (1 – x2 ) [not – (1 – x2 )]
25 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Derivative of arc cos x (cos–1 x) Let y = cos–1 x x = cos y dx dy = –sin y dy dx = – 1 sin y but sin2 y = 1 – cos2 y = 1 – x2 \ dy dx = – 1 (1 – x2 ) Hence, d dx (cos–1 x) = – 1 (1 – x2 ) Derivative of arc tan x (tan–1 x) Let y = tan–1 x x = tan y dx dy = sec2 y dy dx = 1 sec2 y = 1 1 + tan2 y = 1 1 + x2 Hence, d dx (tan–1 x) = 1 1 + x2 Rules of differentiation Differentiation of sums and differences of functions Consider two functions of x, p(x) and q(x), and let f(x) = p(x) + q(x). From the derived definition, d dx [f(x)] = lim δx → 0 [p(x + x) + q(x + x)] – [p(x) + q(x)] δx = lim δx → 0 [p(x + x) – p(x)] + [q(x + x) – q(x)] δx = lim δx → 0 p(x + x) – p(x) δx + lim δx → 0 q(x + x) – q(x) δx = d dx [p(x)] + d dx [q(x)] Hence, d dx [f(x)] = d dx [p(x)] + d dx [q(x)]
26 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Example 5 Differentiate each of the following w.r.t. x. (a) 2x4 + 3 x2 (b) x2 – ln 2x3 (c) ex + cos x Solution: (a) Let y = 2x4 + 3 x2 dy dx = d dx (2x4 ) + d dx 1 3 x2 2 = d dx (2x4 ) + d dx (3x–2) = 2(4x4 – 1) + 3(–2x–2 – 1) = 8x3 – 6 x3 (b) Let y = x2 – ln 2x3 = x2 – (ln 2 + ln x3 ) = x2 – ln 2 – 3 ln x dy dx = d dx (x2 ) – d dx (ln 2) – d dx (3 ln x) = 2x2 – 1 – 0 – 3 1 1 x 2 = 2x – 3 x (c) Let y = ex + cos x dy dx = d dx (ex ) + d dx (cos x) = ex – sin x Exercise 2.3 Differentiate each of the following w.r.t. x. 1. 2x 2. 1 1 5 2 x 3. 5x2 + 3x 4. x4 – 6x3 + 5 5. (x – 2)(3x + 1) 6. x3 + 4 x 7. x–2 (1 + x) 8. ex + sin x 9. ln (3x2 ) 10. 4x2 – 2 x – 3 x3 11. ln (x –—2 3 ) – x2 12. 1 + ln 1 3 x 2 13. 5x2 + 3 x – 4 x2 14. (2x + 1)(3x – 2) x2 15. ln 12 x3 2 16. y = 2 sin x + 3 cos x Using the law of logarithm loga xy = loga x + loga y
27 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 17. Given y = 12x + 3 x 2 2 , find dy dx when x = 2. 18. If y = (x + 3)(2x – 1), find dy dx at the point (2, 15). Differentiation of products of functions Consider y = uv ……………, where u and v are functions of x. If x is a small increase in x, and y, u, v are corresponding increases in y, u and v, then y + y = (u + u)(v + v) = uv + uv + v u + uv …………… – : y = uv + v u + uv Dividing both sides by x, δy δx = u δv δx + v δu δx + u δv δx In the limit, as x → 0, u → 0 and v → 0 and δy δx → dy dx , δv δx → dv dx , δu δx → du dx and u δv δx → 0 Hence, dy dx = u dv dx + v du dx This rule is known as the product rule. Example 6 If y = (x2 – 4x + 6)(1 – 3x3 ), find dy dx using the product rule. Solution: Let u = x2 – 4x + 6 and v = 1 – 3x3 Then du dx = 2x – 4 and dv dx = –9x2 Using the product rule, dy dx = u dv dx + v du dx = (x2 – 4x + 6)(–9x2 ) + (1 – 3x3 ) (2x – 4) = –9x4 + 36x3 – 54x2 + 2x – 4 – 6x4 + 12x3 = –15x4 + 48x3 – 54x2 + 2x – 4
28 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Example 7 Differentiate y = ex sin x w.r.t. x. Solution: u = ex and v = sin x du dx = ex and dv dx = cos x dy dx = u dv dx + v du dx = ex cos x + (sin x) ex = ex (cos x + sin x) Exercise 2.4 Using product rule, differentiate the following functions w.r.t. x. 1. x2 (1 + x2 ) 2. (3x – 2)(4x + 1) 3. (2x3 + 1) (x – 5) 4. 1 4 x – x21x – 1 x2 2 5. (x2 – 2x) 1x – 1 x 2 6. 5x cos x 7. 2x2 sin x 8. sin x ln x 9. x ln x 10. ex (x2 – 1) 11. ex (tan x – cos x) 12. sin x tan x 13. x2 cos x 14. x2 (sin x + cos x) 15. sin x cos x Differentiation of quotients of functions Consider y = u v ……………, where u and v are functions of x, thus y + y = u + u v + v …………… – : y = u + u v + v – u v = v(u + u) – u(v + v) v(v + v) = v u – uv v(v + v) Dividing both sides by x δy δx = v δu δx – u δv δx v(v + v)
29 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 In the limit, as x → 0, u → 0 and v → 0. δy δx → dy dx , δv δx → dv dx , δu δx → du dx , v + v → v Hence, dy dx = v du dx – u dv dx v2 This is known as the quotient rule. Example 8 If y = x3 1 + x2 , find dy dx . Solution: Let u = x3 and v = 1 + x2 du dx = 3x2 and dv dx = 2x Hence dy dx = (1 + x2 )3x2 – x3 · (2x) (1 + x2 ) 2 = x2 (3 + 3x2 – 2x2 ) (1 + x2 ) 2 = x2 (3 + x2 ) (1 + x2 ) 2 Example 9 Differentiate ex sin x w.r.t. x. Solution: Let y = ex sin x u = ex and v = sin x du dx = ex and dv dx = cos x Hence dy dx = (sin x) ex – ex cos x sin2 x = ex (sin x – cos x) sin2 x
30 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Exercise 2.5 Using quotient rule, differentiate the following functions w.r.t. x. 1. x2 2x2 – 3 2. x – 1 x + 1 3. 2x (x + 2)2 4. 3x2 – 2 x3 + 2x2 5. sin x x 6. x ln x 7. sin x 1 + cos x 8. ex x – 1 9. tan x (hint: tan x = sin x cos x ) 10. cot x (hint: cot x = cos x sin x ) 11. sec x (hint: sec x = 1 cos x ) 12. cosec x (hint: cosec x = 1 sin x ) 13. cos x + sin x cos x – sin x 14. 1 + ex 2 + ex 15. 1 + x 1 – x Differentiation of composite functions Consider the function y = g(u), where u is a function of x. If x is a small increase in x while u and y are corresponding increases in u and y, then dy dx = lim δx → 0 1 δy δx 2 = lim δx → 0 1 δy δu × δu δx 2 = lim δx → 0 1 δy δu2 × lim δx → 0 1 δu δx 2 In the limit, as x → 0, u → 0 and y → 0. Hence, dy dx = dy du × du dx This is known as the chain rule. Example 10 Using the chain rule, differentiate the following w.r.t. x. (a) (2x – 3)5 (b) (5 + x2 ) 10 Solution: (a) Let y = (2x – 3)5 and u = 2x – 3 y = u5 dy du = 5u4 and du dx = 2
31 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 dy dx = dy du × du dx = 5u4 × 2 = 5 × 2(2x – 3)4 = 10(2x – 3)4 (b) Let y = (5 + x2 ) 10 and u = 5 + x2 y = u10 dy du = 10u9 and du dx = 2x dy dx = dy du × du dx = 10u9 × 2x = 10 × 2x(5 + x2 ) 9 = 20x(5 + x2 ) 9 In general, dy dx (axn + b) k = k(axn + b) k – 1 · d dx (axn + b) , where a, b, n and = kan xn – 1 (axn + b) k – 1 k are constants. Example 11 Differentiate the following functions w.r.t. x. (a) ln (x3 + 1) (b) ln (2x – x5 ) Solution: (a) Let y = ln (x3 + 1) and u = x3 + 1 y = ln u dy du = 1 u and du dx = 3x2 dy dx = dy du × du dx = 1 u × 3x2 = 3x2 x3 + 1 (b) Let y = ln (2x – x5 ) and u = 2x – x5 y = ln u dy du = 1 u and du dx = 2 – 5x4 dy dx = dy du × du dx = 1 u × (2 – 5x4 ) = (2 – 5x4 ) (2x – x5 )
32 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 In general, d dx ln f(x) = f(x) f(x) Example 12 Find the derivative of each of the following. (a) e3x2 (b) e(2x3 + x) Solution: (a) Let y = e3x2 and u = 3x2 Then y = eu dy du = e u and du dx = 6x dy dx = dy du × du dx = eu × 6x = 6x e3x2 (b) Let y = e(2x3 + x) and u = 2x3 + x Then y = eu dy du = eu and du dx = 6x2 + 1 dy dx = dy du × du dx = eu × (6x2 + 1) = (6x2 + 1) e(2x3 + x) In general, d dx ef(x) = f(x)ef(x) Example 13 Differentiate each of the following w.r.t. x. (a) sin 3x (b) cos (1 – 7x5 ) Solution: (a) Let y = sin 3x and u = 3x Then y = sin u dy du = cos u and du dx = 3 dy dx = dy du × du dx = cos u × 3 = 3 cos 3x
33 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 (b) Let y = cos (1 – 7x5 ) and u = 1 – 7x5 Then y = cos u dy du = –sin u and du dx = –35x4 dy dx = dy du × du dx = –sin u × (–35 x4 ) = 35x4 sin (1 – 7x5 ) In general, d dx sin f(x) = f’(x) cos f(x) d dx cos f(x) = –f’(x) sin f(x) Example 14 Differentiate each of the following w.r.t. x. (a) cos3 x (b) sin4 x Solution: (a) Let y = cos3 x and u = cos x = (cos x) 3 Then y = u3 dy du = 3u2 and du dx = –sin x dy dx = dy du × du dx = 3u2 × (–sin x) = –3 cos2 x sin x (b) Let y = sin4 x and u = sin x = (sin x) 4 Then y = u4 dy du = 4u3 and du dx = cos x dy dx = dy du × du dx = 4u3 × cos x = 4 sin3 x cos x In general, if y = cosn x, then dy dx = –n cosn – 1 x sin x and if y = sinn x, then dy dx = n sinn – 1 x cos x
34 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Exercise 2.6 Differentiate each of the following functions with respect to x. 1. e3x 2. ex2 3. ln (x – 1)2 4. (x + 1)6 5. (2x3 + 1)6 6. 4 ln (x2 + 1) 7. sin2 x 8. (x + 1)–1 9. e(x2 + 3) 10. sin (3x + π 4 ) 11. sin2 (x2 + 1) 12. ln (sin x cos x) 13. (ex – e–x ) –1 14. ln [ sin x 1 – cos x ] 15. 2 cos2 (x2 + 1) 16. (2x – 3)2 (3x + 4)3 17. ln (1 + cos x) 18. ln (1 + x2 ) —1 2 19. ln (tan 2x) 20. (x2 + 5x + 3)—1 2 21. Differentiate x – tan x + 1 3 tan3 x with respect to x and express your answer in terms of tan x. 22. Given f(x) = ln (1 + x) – x + 1 2 x2 , prove that f(x) > 0 for all values of x greater than –1. 23. If f(x) = x – ln (1 + x2 ), show that f(x) > 0 for all values of x. Deduce that x ln (1 + x2 ) for all values of x greater than 0. 24. If y = 1 (x – 1) sin x , find dy dx when x = π 2 . 25. Find dy dx for the curve y = ln 1 + sin 2x at the point where x = π 2 . Differentiation of implicit functions and parametric equations Implicit functions All the functions you have differentiated so far are of the form of y = f(x). However, many functions cannot be arranged in this way at all, for example x3 + y3 = xy and others can look clumsy when you try to make y the subject. When a function is specified by an equation connecting x and y and is not easily transposed to the form of y = f(x), it is called an implicit function. Differentiation of implicit functions The chain rule dy dx = dy du × du dx and the product rule d dx (uv) = u dv dx + v du dx are used extensively to help in the differentiation of implicit functions. Consider the implicit function y3 + xy + y = 3.
35 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Differentiate term by term, we have (a) d dx (y3 ) = d dy (y3 ) · dy dx (using chain rule) = 3y2 dy dx (b) xy is a product. Therefore using the product rule, d dx (xy) = x d dx (y) + y d dx (x) = x dy dx + y (c) d dx (y) = dy dx (d) d dx (3) = 0 Therefore, differentiate y3 + xy + y = 3 w.r.t. x, we get 3y2 dy dx + x dy dx + y + dy dx = 0 or (3y2 + x + 1) dy dx + y = 0. Note: 1. Every term in the equation is differentiated w.r.t. x. 2. If g(y) is any function of y where y = f(x), then g(y) is a function of a function of x. Hence, the derivative of g(y) w.r.t. x is d dx g(y) = d dy g(y) · dy dx (using chain rule) = g(y) dy dx Example 15 If 2x2 – 3y2 = 2xy, find dy dx . Solution: 2x2 – 3y2 = 2xy Differentiate w.r.t. x, d dx (2x2 ) – d dx (3y2 ) = d dx (2xy) 4x – 6y dy dx = 2y + 2x dy dx dy dx = 4x – 2y 2x + 6y = 2x – y x + 3y
36 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Example 16 Find dy dx in terms of x and y if x2 sin y + 2x = y. Solution: x2 sin y + 2x = y Differentiate w.r.t. x, d dx (x2 sin y) + d dx (2x) = d dx (y) x2 d dx (sin y) + sin y d dx (x2 ) + 2 = dy dx x2 cos y dy dx + 2x sin y + 2 = dy dx dy dx (1 – x2 cos y) = 2(1 + x sin y) dy dx = 2(1 + x sin y) 1 – x2 cos y Example 17 Find the value of dy dx at the point (4, 2) which lies on the curve x2 – xy – y2 – 2y = 0. Solution: x2 – xy – y2 – 2y = 0. Differentiate w.r.t. x, 2x – 1y + x dy dx 2 – 2y dy dx – 2 dy dx = 0 x dy dx + 2y dy dx + 2 dy dx = 2x – y dy dx (x + 2y + 2) = 2x – y dy dx = 2x – y x + 2y + 2 At the point (4, 2), dy dx = 2(4) – 2 4 + 2(2) + 2 = 3 5 Exercise 2.7 1. Differentiate each of the following implicit functions w.r.t. x. (a) x2 + y2 = 2 (b) 4x2 + 9y2 = 13x (c) x2 + xy + y2 = 0 (d) x3 + y3 = 9y (e) x3 + xy2 = 5xy 2. Find dy dx for each of the following functions in terms of x and y. (a) x2 y3 – 2xy2 = 3y (b) sin x cos y = 2 (c) x2 – y2 = a2 (d) x2 + 5xy – y2 + 2x – 3y + 7 = 0 (e) 2ye3x + 1 x2 sin 2x = 0
37 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 3. If x2 + y2 = y, show that dy dx = 2x 1 – 2y . 4. If 2x2 y – 2cx2 + 4x + 3 = 0, where c is a constant, show that x3 dy dx = 2x + 3. 5. Given (1 + x2 )(1 + y2 ) – cx2 = 0, where c is a constant, show that xy(1 + x2 ) dy dx = 1 + y2 . 6. For the curve x(y2 + 1) – y(x2 + 1) + 4 = 0, find dy dx at the point (–1, 1) on the curve. 7. Given x2 + xy – y2 = 1. Find dy dx at the point (1, 1) on the curve. 8. Find the value of dy dx at the point (1, –4) on the curve y2 + 2xy + 3x2 = 11. Parametric equations In some cases, it is more convenient to represent a function by expressing x and y separately in terms of a third variable, called a parameter. The two expressions for x and y are known as parametric equations. For example x = t 3 y = t 2 – t, where t is a parameter. Parametric differentiation To differentiate a function which is defined in terms of a parameter t, we need to use the chain rule: — dy dy dt = — × — . dx dt dx dt 1 Since — = –––– , dx dx — dt It follows that dy — — dy dt dx = –––– provided that — ≠ 0. dx dx dt — dt dy We can find — in terms of the parameter t, as shown below. dx y = t2 – t and x = t 3 — dy dx = 2t – 1 and — = 3t 2 dt dt —dt 1 = — dx 3t 2 — dy dy dt = — × — dx dt dx 2t – 1 = ––––– 3t 2
38 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Example 18 The parametric equations of a curve are x = t + 3 t and y = t 2 – 4 t . Find dy dx in terms of t. Solution: Given y = t 2 – 4 t and x = t + 3 t = t – 4t –1 = 1 + 3t –1 dy dt = 1 + 4t –2 and dx dt = –3t –2 = – 3 t 2 = 1 + 4 t 2 dt dx = – t 2 3 dy dx = dy dt × dt dx = (1 + 4 t 2 )(– t 2 3 ) = – t 2 3 – 4 3 = – 1 3 (t 2 + 4) Example 19 A curve has parametric equations x = et and y = sin t. Find dy dx in terms of t. Solution: Given y = sin t and x = et dy dx = cos t and dx dt = et dt dx = 1 et dy dx = dy dt × dt dx = cos t × 1 et = e–t cos t Exercise 2.8 1. Find dy dx in terms of t for the curves with the following parametric equations. (a) x = t 2 , y = 2t (b) x = t 2 + 3, y = t(t 2 + 3) (c) x = 1 t – 1 , y = t 3 (d) x = t(t 2 + 1), y = t 2 + 1 2. Find dy dx in terms of q for the curves with parametric equations. (a) x = cos q, y = sin q (b) x = a cos3 q, y = a sin3 q 3. If x = 1 + t 1 – 2t and y = 1 + 2t 1 – t , find dy dx when t = 0.
39 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 4. A curve has parametric equations x = et and y = sin t. Find dy dx in terms of t. 5. A curve is given by the parametric equations x = t – 1 t and y = t + 1 t , t ≠ 0. Find the coordinates of the point where dy dx = 0. 6. If x = a(q – sin q) and y = a(1 – cos q), show that dy dx = cot 1 2 q. 7. Given x = a[ln (cot 1 2 q) – cos q] and y = a sin q where a is a positive constant and 0 q < 1 2 π, show that if q ≠ 1 2 π, dy dx = –tan q. 8. A curve has parametric equations x = 2 sin t and y = 3 cos t. Show that dy dx = – 3 tan t 2 . 9. The parametric equations of a curve are x = 3 – t 2 , y = 3t – t 3 . Find the coordinates of the point where dy dx = 0. 2.2 Applications of Differentiation Gradient of a curve The gradient of a curve, y = f(x), at a point P on the curve is given by the gradient of the tangent at P. It is also given by the value of dy dx at the point P, which can be calculated using the equation of the curve. Thus we can calculate the gradient of the tangent to the curve at any point P. Figure 2.4 Example 20 Find the gradient of the curve y = x2 – 6x + 2 at the point where x = 0. Solution: y = x2 – 6x + 2 Differentiate w.r.t. x, dy dx = 2x – 6 When x = 0 dy dx = 2(0) – 6 = –6 Hence, the gradient of the curve y = x2 – 6x + 2 at x = 0 is –6. 0 y x A P B T
40 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Exercise 2.9 1. Find the gradient of each of the following curves at the point with the given x-coordinates. (a) y = x2 – 3, x = 1 (b) y = x3 + 7x – 4, x = –3 (c) y = 4x2 + x – 2 2x , x = 2 (d) y = (x – 3)(x2 + 2), x = 1 2. Find the coordinates of the point on the curve y = 2 x2 at which the gradient of the curve is 1 2 . 3. Find the gradient of each of the following curves at the given point. (a) x2 + 3xy + 2y2 = 5, (1, –4) (b) x3 + 3y3 = 3, (3, –2) 4. Find the coordinates of the point on the curve y = x2 – 2x – 8 at which the gradient is (a) 4, (b) 2, (c) –2. 5. The parametric equations of a curve are x = q – cos q and y = sin q. Find the coordinates of the point on the curve at which the gradient is zero. 6. Show that the gradient of the curve x3 + 3xy + y3 = 5 at the point (1, 1) is –1. Increasing and decreasing functions 1. y 0 x f(x2) f(x1) x1 x2 Increasing Increasing Figure 2.5 (a) A function f is an increasing function if f(x) increases as the values of x increase i.e. if x1 , x2 , f(x1 ) , f(x2 ). (b) If y = f(x) is differentiable in the interval (x1 , x2 ) and dy dx is positive (or f(x) . 0), then f(x) is an increasing function.
41 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 2. y 0 x f(x1) f(x2) x1 x2 Decreasing Increasing Figure 2.6 (a) A function f is a decreasing function if f(x) decreases as the values of x increases i.e. if x1 , x2 , f(x1 ) . f(x2 ). (b) If y = f(x) is differentiable in the interval (x1 , x2 ) and dy dx is negative (or f(x) , 0), then f(x) is a decreasing function. Hence, f(x) increases as x increases if f(x) . 0 f(x) decreases as x increases if f(x) , 0 3. Consider the function f(x) = x2 y 0 x f(x) = x2 Figure 2.7 for x . 0, f(x) increases as x increases and f(x) . 0 \ f(x) is an increasing function for x , 0, f(x) decreases as x increases and f(x) , 0 \ f(x) is a decreasing function Hence, a function may increase on an interval and decrease on another. Example 21 Show that (a) f(x) = x3 + 2 is an increasing function for all real values of x, (b) f(x) = 1 x is a decreasing function for x . 0. Solution: (a) f(x) = x3 + 2 f(x) = 3x2 . 0 for all real values of x \ f(x) = x3 is an increasing function
42 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 (b) f(x) = 1 x = x–1 f(x) = – 1 x2 , 0 for x . 0 \ f(x) = 1 x , x . 0 is a decreasing function Example 22 Given f(x) = x4 – 8x2 + 2, determine the intervals of increase and decrease. Solution: f(x) = x4 – 8x2 + 2 f(x) = 4x3 – 16x = 4x(x2 – 4) = 4x(x – 2)(x + 2) When f(x) = 0, x = –2, 0, 2 –2 0 2 – + – + Sign of f(x) Hence, f(x) is decreasing in the interval (–∞, –2) and (0, 2) and increasing in the interval (–2, 0) and (2, ∞). Exercise 2.10 Determine the intervals of increase and decrease for each of the following functions. 1. f(x) = x2 – 6x 2. f(x) = x2 – 8x + 3 3. f(x) = 7 – 2x – x2 4. f(x) = x + 4 x Concave upwards and concave downwards y 0 x Concave upwards Concave downwards Figure 2.8 A curve y = f(x) is (a) concave upwards if f(x) increases as x increases or f(x) . 0, (b) concave downwards if f(x) decreases as x increases or f(x) , 0.
43 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Stationary points A stationary point of the curve y = f(x) is a point at which the tangent to the curve is parallel to the x-axis, i.e. the gradient of the curve is zero. In Figure 2.9, A, B and C are stationary points of the curve y = f(x). 0 y = f(x) y A B C x x1 x2 x3 Figure 2.9 To find the coordinates of the stationary point on the curve y = f(x), solve the equation dy dx = 0. For each solution of x = , the point P[, f()] is a stationary point and each value of f() is known as the stationary value. Example 23 Find the coordinates of the stationary points on the curve y = x3 – 3x2 + 6. Solution: y = x3 – 3x2 + 6 dy dx = 3x2 – 6x = 3x(x – 2) At the stationary point, dy dx = 0 i.e. 3x(x – 2) = 0 3x = 0 or x – 2 = 0 x = 0 or x = 2 When x = 0, y = 6 When x = 2, y = 2 Hence, the stationary points are (0, 6) and (2, 2).
44 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 Exercise 2.11 Find the coordinates of the stationary points of each of the following curves. 1. y = (x – 2)(x + 3) 2. y = x3 – 5x2 + 3x – 1 3. y = (x – 1)3 4. y = 5 – 4x – x2 5. y = x + 1 x 6. y = x2 (5 – x) 3 7. y = 4e4x + 9e–4x 8. y = x + 2 cos x, 0 < x < 2π 9. y = ln x x2 , (x 0) 10. y = ex + 2e–x Extremum points and point of inflexion Consider a function y = f(x) with the graph shown below. 0 y = f(x) y (max) A (min) B Point of inflexion C x x1 x2 x3 Figure 2.10 At the point A, i.e. at x = x1 , a maximum value of y occurs since the y-value at A is greater than the y-values on either side of it. Similarly, at B, y is a minimum value since the y-value at the point B is less than the y-values on either side of it and close to it. The point C looks like ‘half a maximum and half a minimum’. The point flattens out at C but instead of dipping down, it has an increasingly positive gradient. Such a point is an example of a point of inflexion. If we consider the gradient of the graph as we travel from left to right, we can draw a graph to show how this gradient varies. We have no actual values for the gradient but we can see whether it is positive or negative, more or less steep. The graph we obtained is the first derived curve of the function and we are really plotting the values of dy dx against the values of x.
45 Mathematics Semester 2 STPM Chapter 2 Differentiation 2 y = f(x) A B C 0 y x x1 x2 x3 + + + – 0 x x1 x2 x3 dy dx– Figure 2.11 At x = x1 , x2 , x3 (corresponding to the three stationary points), the graph of dy dx is at the x-axis. Therefore, for stationary points, A, B and C, dy dx = 0. (a) At A (a local maximum), (i) the y-value at A is greater than the y-values on either side of it, (ii) the gradient of the curve dy dx 0 if x x1 dy dx = 0 if x = x1 dy dx 0 if x x1 (b) At B (a local minimum), (i) the y-value at B is smaller than the y-values on either side of it, (ii) the gradient of the curve dy dx 0 if x x2 dy dx = 0 if x = x2 dy dx 0 if x x2