e
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ℝ
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Mathematical Differentiation for College Students Abdul Latiff Md Ahood Mohd Amirul Mahamud
ℝ y α
x
cos
f π
δ
sin
tan
β
e
μ
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e
tan
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ALL ABOUT
ℝ
δ μ
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Mathematical Differentiation for College Students
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Abdul Latiff Md Ahood Mohd Amirul Mahamud
cos
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ℝ y α
x
f π
δ
sin
tan
β
e
μ
Copyright Š 2018 by Sunway University Sdn Bhd Published by Sunway University Press An imprint of Sunway University Sdn Bhd
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No. 5, Jalan Universiti Sunway City 47500 Selangor Darul Ehsan Malaysia
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press.sunway.edu.my
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All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, now known or hereafter invented, without permission in writing from the publisher.
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Perpustakaan Negara Malaysia
Cataloguing-in-Publication Data
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Abdul Latiff Md. Ahood ALL ABOUT DERIVATIVES / ABDUL LATIFF BIN MD AHOOD, MOHD AMIRUL BIN MAHAMUD. ISBN 978-967-13697-3-9 1. Derivatives (Mathematics). 2. Mathematics--Study and teaching (Higher). I. Mohd. Amirul Mahamud. II. Title. 515.33
Printed by Vivar Printing Sdn Bhd, Selangor
Cover image: Studiojumpee/Shutterstock.com Image used under license from Shutterstock.com
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This book is dedicated to my wife, Nasrah Hanum, and daughter, Fairuz Hanim, who have always been a source of encouragement and inspiration to me.
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Abdul Latiff Md Ahood
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List of Graphs List of Acronyms and Abbreviations About the Authors Preface Foreword Acknowledgements
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CONTENTS
vi vi vii viii ix xi
DIFFERENTIATION OF POLYNOMIALS
CHAPTER 2
DIFFERENTIATION OF TRIGONOMETRIC FUNCTIONS
29
CHAPTER 3
DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
47
IMPLICIT DIFFERENTIATION
71
PARTIAL DERIVATIVES
87
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CHAPTER 4
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CHAPTER 1
CHAPTER 5
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Answers Appendix Further Reading
1
109 121 123
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Figure 1.1 Figure 1.2 Figure 1.3 Figure 1.4
Gradient of a straight line Gradient of a curve Tangent line Secant line (cos(x) − 1) Graph of y = x sin(x) Graph of y = x
Figure 2.1
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Figure 2.2
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LIST OF GRAPHS
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Negative Positive Left Hand Side Right Hand Side
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–ve +ve LHS RHS
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LIST OF ACRONYMS AND ABBREVIATIONS
1 2 3 4 31 32
vii
ABOUT THE AUTHORS
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Abdul Latiff Md Ahood is an accomplished lecturer whose teaching career in tertiary education spans over more than 40 years. A respected educator in Electrical Engineering and Mathematics, he has authored several books including Mesin Elektrik and Let’s Talk Mathematics. His knowledge and experience landed him prominent positions in higher education institutions, such as Deputy Dean of the School of Applied Sciences at Universiti Sains Malaysia and Associate Professor Universiti Teknikal TeknikalMalaysia. MalaysiaHeMelaka. He teaching is currently teaching at University is currently Mathematics Mathematics Unit 1 in the Monash University Foundation Unit 1 in the Monash University Foundation Year programmeYear at programme at Sunway College in capacity of Senior Lecturer. Sunway College in the capacity of the Senior Lecturer.
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Mohd Amirul Mahamud is the author of several journal publications and the co-author of Let’s Talk Mathematics. He has accumulated almost 10 years of experience in education as Lecturer, teaching Algebra and Calculus in various post-secondary courses, and played a crucial role in restructuring the Mathematics syllabus at Albukhary International University. His areas of expertise include, but are not limited to, Pure Mathematics, Mathematical Analysis, Algebraic Geometry and Differential Algebra.
viii
PREFACE
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All About Derivatives explains the techniques of differentiating all types of mathematical functions, ranging from polynomials to exponential and logarithmic functions. The basic principles are carefully explained and comprehensively illustrated with numerous worked examples taken from lectures presented over the years.
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This book has been written specially to cater to students studying Foundation or Matriculation Mathematics at tertiary level. Some of the advanced problems posed in the book, such as those concerned with partial derivatives and implicit differentiation, might even require students to carry out research beyond the scope of the syllabus.
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However, every effort has been made to account for all the differentiation steps without skipping the fundamental concepts. The rigorous worked examples and problem sets found in every chapter will further help test and develop students’ skills and understanding. We genuinely hope that our book will help students grasp the fundamental concepts of Calculus and improve their analytical and critical-thinking skills. Abdul Latiff Md Ahood Mohd Amirul Mahamud
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FOREWORD
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All About Derivatives discusses the concepts and techniques of differentiation of all kinds of functions, ranging from polynomials to exponential and logarithmic functions. In this practical handbook, all the essential principles are explained and demonstrated with complete worked examples. It has been written with the intention to provide study materials and guidelines in such a way that readers would feel guided by a teacher while exploring the book themselves.
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The worked examples provided, along with the problem sets at the end of each chapter, run the gamut from fundamental questions to more advanced ones. These have been meticulously prepared to enhance problem-solving skills and encourage independent learning. Students will tackle problems that are related to applications in real life instead of deliberating about limits and continuity, and this makes for a practical guide for students, especially those majoring in Engineering.
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Efforts made by the writers to account for all the steps in the worked examples are well-intended, as this will ensure that readers are not only able to understand and work out all the problem sets provided in the book, but are also able to solve problems that they will encounter in the future. The book is suitable for students who are taking Mathematics at
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ALL ABOUT DERIVATIVES
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Dr Hailiza Kamarulhaili Dean and Professor School of Mathematics Universiti Sains Malaysia (USM)
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diploma level, as well as for foundation students preparing to undergo higher level of studies in Sciences, Technology, Engineering, and Mathematics (STEM)-related fields.
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ACKNOWLEDGEMENTS
This book would not have been possible without the support of many individuals.
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First, we would like to recognise the numerous students in our courses whom we have enjoyed teaching over the years. They have inspired us to produce simplified books to help students learn, understand and appreciate Mathematics.
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The support given by the Director of the Monash University Foundation Year programme, Mr Lee Thye Cheong, is most encouraging and has spurred us on to get this book published as soon as possible. Ms Carol Wong, the Head of Sunway University Press, has also played an active role in supporting and facilitating all our efforts.
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Most importantly, we would also like to thank our families who constantly encouraged us as we worked on completing this book. Abdul Latiff Md Ahood Mohd Amirul Mahamud
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xii
CHAPTER 1
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DIFFERENTIATION OF POLYNOMIALS
INTRODUCTION
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By definition, a straight line is a graph with a constant gradient. For example, the gradient of the straight line f (x) = 2x, shown in Figure 1.1, is always the same whether it is measured at a or b. y
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f (x) = 2x
10 8
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6
−10
−8
−6
a
4 2
−4
−2
−2
2
4
6
−4
b
−6 −8 −10
Figure 1.1 Gradient of a straight line
8
10
x
2
ALL ABOUT DERIVATIVES
On the other hand, the gradient of the curve f(x) = x2– 2 changes all the time as you move along the curve. Its gradient varies from point to point as shown in Figure 1.2. y
f (x) = x 2 − 2
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10 8
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6 4
Negative gradient −3
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−4
−2
−1
−2
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−5
Positive gradient
2
1
2
3
4
5
x
Gradient = 0
−4
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−6
Figure 1.2 Gradient of a curve
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The gradient of a polynomial function at any point can be determined by drawing a tangent line at the point. While it is not too difficult to draw the tangent line, a computational method for determining the gradient would, of course, be more convenient and efficient. N OTE A tangent line is not just a straight line that touches the graph at one point, it also acts as the straight line that best approximates the graph at that point.
CHAPTER 1 DIFFERENTIATION OF POLYNOMIALS
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Let us say that you have a polynomial equation y = x2 and the tangent line drawn at point P1 as shown in Figure 1.3. y
Tangent line x
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y = x2
Figure 1.3 Tangent line
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From Figure 1.3, the gradient line or tangent line can be moved and adjusted to produce a secant line intersecting the curve at two neighbouring points as shown in Figure 1.4. The transition from the tangent line to the secant line has produced two points, P1(x, y) and P2(x + dx, y + dy), which can be used to help you determine the gradient of the tangent line. N OTE The coordinates of P2 are written as (x + dx, y + dy) where dx represents a small increment in the value of x and dy is the corresponding increment in the value of y.
ALL ABOUT DERIVATIVES
y
y = x2
Secant line
P₂
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Tangent line
P₁
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δy
Q
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δx
x
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Figure 1.4 Secant line
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How is the secant line useful when its gradient is obviously different from the gradient of the tangent line? This is where the idea of ‘limits’ in differentiation come into the picture. Notice that as you move the secant line back to the position of the original tangent line, point P2 gets closer and closer to point P1, which in turn means that both dx and dy approach zero.
Let us consider the polynomial function y = x2. Then, at point P1, you have y + dy = (x + dx)2 y + dy = (x + dx)(x + dx)
Apply the expansion rule
CHAPTER 1 DIFFERENTIATION OF POLYNOMIALS
y = x2 Bring x2 to the RHS
[Equation 1]
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y + dy = x2 + x(dx) + x(dx) + (dx)2 y + dy = x2 + 2x(dx) + (dx)2 x2 + dy = x2 + 2x(dx) + (dx)2 dy = x2 – x2 + 2x(dx) + (dx)2 dy = 2x(dx) + (dx)2
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dy 2x(dx) + (dx)2 = dx dx
Divide by dx on both sides
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dy dx[2x + dx] = dx dx
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The gradient of the line P1P2 can be derived by dividing the change in y with the change in x. Thus, you have
Factorising
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dy dx[2x + dx] = = 2x + dx dx dx
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[Equation 2]
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If you look at Figure 1.4, as point P2 moves closer and closer to point P1, the value of dx gets smaller and smaller, approaching zero. You say that dx tends to zero, or dx→0. Therefore, the gradient of the line P1P2 approaches the gradient of the tangent. This can be expressed as follows: Gradient of tangent = lim
dy
dx → 0 dx
= lim (2x + dx) = 2x dx → 0
where ‘lim’ stands for ‘limit’ and you can say that as dx tends to zero, 2x + dx approaches 2x.
6 ALL ABOUT DERIVATIVES
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You can take the gradient of P1P2 as an approximation to the gradient of the tangent at P1, and consequently you have the rate of change of y with respect to dy x at the point P1. The expression lim can be applied dx → 0 dx dy to other curves, and in calculus, it is denoted by dx (read as ‘the derivative of y with respect to x). It is known as the gradient function of the curve, or the derivative of the function f(x).
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Now, if you look at Figure 1.4, the point P1 is (x, f(x)) and point P2 is (x + dx, f(x + dx)). Therefore, taking the change in y divided by the change in x, you have
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dy f(x + dx) – f(x) f(x + dx) – f(x) = = dx (x + dx) – (x) dx
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So, the derivative of the function f(x) is dy = lim dy dx dx → 0 dx dy = lim f(x + dx) – f(x) dx dx → 0 dx dy , also denoted by f (x), is called dx differentiation of the function y = f(x) with respect to x. In
The process of finding
CHAPTER 1 DIFFERENTIATION OF POLYNOMIALS
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f (x) = lim
h→0
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other words, when you differentiate y = f(x) with respect to x, you are determining the gradient of the function at any point on the curve or at a specific point on the curve. This can be done based on the concept f(x + h) – f(x) h
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Example 1
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where dx has been replaced by h. This method of finding the gradient of a function at any point on the curve is called differentiation from first principles.
Differentiate from first principles: 1 x2
(b) f (x) =
(c) f (x) = x
(d) f (x) = (x – 3)2
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(a) f (x) = x2 – 2x
Solution:
(a) f (x) = x2 – 2x Let h be a small increment in x. Then, f (x + h) = (x + h)2 – 2(x + h) = x2 + 2xh + h2 – 2x – 2h
8 ALL ABOUT DERIVATIVES
f (x) = lim f(x + h) – f(x) h→0 h (x2 + 2xh + h2 – 2x – 2h) – (x2 – 2x) h h→0
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= lim
x2 + 2xh + h2 – 2x – 2h – x2 + 2x h→0 h
= lim
2xh + h2 – 2h h→0 h
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= lim
h(2x + h – 2) h→0 h
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= lim
= lim (2x + h – 2) h→0
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= (2x – 2) when h → 0
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= (2x – 2)
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(b) f (x) =
1 x2
f (x + h) =
1 (x + h)2
f (x) = lim
h→0
f (x + h) – f (x) h
1 1 – 2 2 (x + h) x = lim h h→0
CHAPTER 1 DIFFERENTIATION OF POLYNOMIALS
x2 – (x2 + 2xh + h2) h→0 h x2 (x + h)2
= lim
h→0
= lim
h(–2x – h) h x2 (x + h)2 (–2x – h) (x + h)2
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= lim
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–2xh – h2 = lim h → 0 h x2 (x + h)2
h → 0 x2
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–2x when h → 0 x4
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=
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x2 – (x + h)2 x2 (x + h)2 = lim h→0 h
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= –23 x
(c) f (x) = x f (x + h) = x + h f(x + h) – f(x) h→0 h
f (x) = lim
= lim x + h – x h→0 h
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10 ALL ABOUT DERIVATIVES
= lim
h→0
x + hh – x ( xx ++ hh ++ xx )
x+h–x h → 0 h( x + h + x )
Rationalising the numerator
h h → 0 h( x + h + x )
= lim
when h → 0
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1 x+ x 1 = 2 x =
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1 h→0 ( x + h + x )
= lim
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= lim
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(d) f(x) = (x – 3)2
f(x + h) = (x + h – 3)2 f (x + h) – f (x) h→0 h
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f (x) = lim
(x + h – 3)2 – (x – 3)2 h→0 h
= lim
(x2 + xh – 3x + xh + h2 – 3h – 3x – 3h + 9) – (x2 – 3x – 3x + 9) = lim h h→0 h2 + 2xh – 6h h→0 h
= lim
CHAPTER 1 DIFFERENTIATION OF POLYNOMIALS
= lim
h→0
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h(h + 2x – 6) h
= lim (h + 2x – 6) h→0
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= 2x – 6 when h → 0 = 2x – 6
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Example 2
1 x
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Solution: f (x) =
1 , find f (x) from first principles. x
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Given that f (x) =
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Let h be a small increment in the value of x. Then, 1 x+h 1 1 – f (x + h) – f (x) x+h x = h h
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f (x + h) =
=
1 h
x – x+h x+h x
=
1 h
x – x+h x+h x
xx ++
Rationalising the numerator
x+h x+h
ALL ABOUT DERIVATIVES
=
x – (x + h) 1 h x x + h + (x + h) x
=
–h 1 h x x + h + (x + h) x
=
–1 x x + h + (x + h) x
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f (x) = lim
h→0
f (x + h) – f (x) h
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–1 h → 0 x x + h + (x + h) x
–1 when h → 0 x x +x x
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=
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= lim
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=– =–
1 2x x 1
3
2x2
Fortunately, you do not have to go through this tedious affair of differentiating from first principles all the time. There are some useful formulas and rules which can be utilised to simplify and hasten the process of differentiation. These are explained in the following sections.
CHAPTER 1 DIFFERENTIATION OF POLYNOMIALS
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THE POWER RULE
Example 3
p with respect to x. 2
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Find the derivative of 2x3 – 4x2 + Solution:
[Result (1)]
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f (x) = nxn–1
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The Power Rule enables you to find the derivatives of 1 1 functions involving powers such as x5, 3, (4x) 2 and so on. x If f (x) = xn (where n ∈ ),
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Let y = 2x3 – 4x2 +
p 2
Then, the derivative is given by
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dy = 6x2 – 8x + 0 dx = 6x2 – 8x
p
is a constant, and differentiating 2 a constant gives zero
N OTE The Sum and Difference Rule: To differentiate a function consisting of a sum (or difference) of several terms, as in Example 3, you find the derivative of each term, then add (or subtract) the derivatives. dy If y = f(x) ± g(x), then = f (x) ± g(x). dx
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ALL ABOUT DERIVATIVES
Example 4
Solution: 1
1
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If f (x) = x + 2x + xe (where e = 2.718), find f (x).
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f (x) = x + 2 x + xe = x 2 + 2x 2 + xe 1 –1 1 –1 f (x) = x 2 + 2 x 2 + exe–1 2 2 1 2 + + exe–1 2 x 2 x
=
1+ 2 + exe–1 2 x
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=
e is the constant 2.718
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Example 5
Differentiate the following with respect to x. (a)
(b)
5x4 + 3x3 – x2 + 2x x (3 – 4x)2 3
x2
CHAPTER 1 DIFFERENTIATION OF POLYNOMIALS
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Solution:
=
5x4 + 3x3 – x2 + 2x x
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5x4 + 3x3 – x2 + 2x
1 = x–a xa 2. xa xb = xa +b (Multiplication) xa 3. b = x a–b (Division) x where a and b are constants. 1.
1
x2
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4 3 2 = 5x1 + 3x1 – x1 + 2x1 x2 x2 x2 x2
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(a) Let y =
5
3
1
–1 dy 35 52 15 32 3 12 = x + x – x +x 2 2 2 dx 2
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Therefore,
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= 5x 2 + 3x 2 – x 2 + 2x 2
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(b) Let y =
= =
=
35 52 15 32 3 1 x + x – x + 2 2 2 x
(3 – 4x)2 3
x2 9 – 24x + 16x2 3
x2 9
3 x2
16x2 + – 24x 3 3 x2 x2
–3 2
= 9x
–1 2
– 24x
1
+ 16x 2