1 | P a g e
2 | P a g e Writers : Nor Aidawati binti Nor Khalim, Zahratul Laily binti Edaris, Junaidah binti Jaafar Editor : Zahratul Laily binti Edaris Content reviewer : Asmarini binti Mohamed Proof reader : Zakiah binti Adzmi Graphic designer : Nor Aidawati binti Nor Khalim Copyright ©2023 All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the publisher. Published By : Division of Instructional and Digital Learning Department of Polytechnic and Community College Ministry of Higher Education Level 4, Galeria PjH, Jalan P4W, Persiaran Perdana Presint 4, 62100 Putrajaya Cataloguing-in-Publication Data Perpustakaan Negara Malaysia A catalogue record for this book is available from the National Library of Malaysia eISBN 978-629-7514-16-1
3 | P a g e PREFACE Alhamdulillah, puji dan syukur kehadrat Ilahi, kerana dengan izin dan limpah kurnia-Nya, eBook ini telah berjaya dihasilkan dan diterbitkan seperti yang telah dirancang. Ebook ENGINEERING MATHEMATICS 2 : DIFFERENTIATION ini dihasilkan bagi memudahkan pensyarah serta pelajar yang mengambil kursus Engineering Mathematics 2 untuk mengikuti kursus ini secara berstruktur dan sistematik. Melalui eBook yang dihasilkan ini juga diharapkan supaya dapat menjadi sumber penyebaran ilmu berdasarkan bidang yang dipilih di samping meningkatkan kemantapan intelek penulis sendiri. Akhir kata, kami memohon kemaafan di atas kekurangan yang terdapat dalam penghasilan eBook ini. Sekalung penghargaan kepada pihak Pengurusan Politeknik Tuanku Syed Sirajuddin kerana telah memberi ruang dan peluang dalam penghasilan eBook ini. Harapan kami agar pensyarah dan pelajar mendapat ilmu yang bermanfaat dan dapat dijadikan sebagai rujukan.
4 | P a g e CONTENTS PREFACE iii 1.0 Describe rules of differentiation 5 2.0 Use trigonometric, logarithmic and exponential functions 12 3.0 Apply second order differentiation 16 4.0 Demonstrate the application of differentiation 17 5.0 Solve parametric equation 25 6.0 Apply implicit differentiation 27 7.0 Construct partial differentiation 28 8.0 Apply the technique of total differentiation 32 REFERENCES 34
5 | P a g e DIFFERENTIATION 1.0 DESCRIBE RULES OF DIFFERENTIATION 1.1 IDENTIFY BASIC RULES OF DIFFERENTIATION Differentiate the following functions with respect to : 1. = 5 = 0 Constant Rule 2. = 5 = 5 5−1 = 5 4 Power Rule 3. = 3 = 3 1−1 = 3 0 0 = 1 = 3 Constant Multiple Rule 4. = −3 7 = (7)(−3) 7−1 = −21 6 Constant Multiple Rule 5. = (2 + 4)(−3 − 2) = −6 2 − 4 − 12 − 8 = −6 2 − 16 − 8 = (2)(−6) 2−1 − 16 − 0 = −12 − 16 6. = − + √ = [−3 −2 ] + [ −4 3 ⁄ ] Sum /Difference Rule = [(−2)(−3) −2−1 ] + [(− 4 3 ) − 4 3 −1 ] = 6 −3 − 4 3 − 7 3 7. = 5−3 4 3 = 5 3 − 3 4 3 = 5 2 − 3 = 5 −2 − 3 = (−2)(5) −2−1 − 3 = −10 −3 − 3 8. = ( 3 + 7) 5 = 5( 3 + 7) 4 × (3 2 ) = 15 2 ( 3 + 7) 4 Extended Power Rule 9. = 2 ( 3−5) 3 = 2( 3 − 5) −3 = −3(2)( 3 − 5) −4 × (3 2 ) • = 0, ℎ CONSTANT RULE • = POWER −1 RULE • = −1 CONSTANT MULTIPLE RULE •Differentiate each terms separately SUM AND DIFFERENCE RULE EXAMPLE Differentiate 3 +7 Differentiate 3 −5 Expand and simplify
6 | P a g e = −18 2 ( 3 − 5) −4 = −18 2 ( 3−5) 4 10. Given = 1 3 3 + 2 − 5 −2 , find the value of if = 2. = 2 + 2 + 10 −3 = (2) 2 + 2 + 10(2) −3 = 7.25 11. If () = 4 5 − 10 3 , find the value of ′ (−3) ′ () = 20 4 − 30 2 ′ (−3) = 20(−3) 4 − 30(−3) 2 ′ (−3) = 1350 Differentiate with respect to x : 1. = 4 − 3 3 : = −9 2 2. = 1 − 5 2 : = −10 3. = 1 2 2 : = − 1 3 4. = −2 3 + 3 − 4 : = − 1 3 5. = 3 (7−3) 2 : = 18 (−3+7) 3 6. = (8 + 3)(7 + 5) : = 112 + 61 7. = 5√−3 2 3 : = −2 + 5 6 1 2 ⁄ 8. = (√ − 3)(2 2 + 5) : = −12 + 5 3 2 ⁄ + 5 2 1 2 ⁄ 9. = (5 − ) 2 : = −10 + 2 10. = 3 5 5 − 5 −2 : = 3 4 + 10 −3 1.2 DESCRIBE THE TECHNIQUES OF DIFFERENTIATION IN SOLVING PROBLEMS PRODUCT RULE Find for the following functions. 1. = 2 (10 − 2) = 2 = 10 − 2 = 2 = 10 Applying the Product Rule = + = 2 (10) + (10 − 2)(2) = 10 2 + 20 2 − 4 = 30 2 − 4 2. = ( 5 + 1)( − 3) = 5 + 1 = − 3 = 5 4 = 1 Applying the Product Rule = ( 5 + 1)1 + ( − 3)(5 4 ) = 5 + 1 + 5 5 − 15 4 = 6 5 − 15 4 + 1 3. = (2√ + 1)( 2 − 2) = (2√ + 1) = ( 2 − 2) = − 1 2 = 2 Applying the Product Rule () = + FORMATIVE PRACTICE 1 EXAMPLE Substitute = 2
7 | P a g e = (2√ + 1)2 + ( 2 − 2) − 1 2 = 4 3 2 + 2 + 3 2 − 2 − 1 2 = 5 3 2 + 2 − 2 − 1 2 = 5 3 2 + 2 − 2 √ 4. = (2 − 2 )(5 + 2) = (2 − 2 ) = (5 + 2) = −2 = 5 = (2 − 2 )5 + (5 + 2)(−2) = 10 − 5 2 − 10 2 − 4 = −15 2 − 4 + 10 5. = (4 − 2)(3x 2 + 7) = (4 − 2) = (3x 2 + 7) = 4 = 6 = (4 − 2)6 + (3x 2 + 7)4 = 24 2 − 12 + 12 2 + 28 = 36 2 − 12 + 28 6. = (2 − 5)(3√ 3 − 2) = (2 − 5) = 3√ 3 − 2 = 2 = − 2 3 = (2 − 5) − 2 3 + (3 1 3 − 2) 2 = 2 1 3 − 5 − 2 3 + 6 1 3 − 4 = 8 1 3 − 5 − 2 3 − 4 Differentiate the following functions. 1. = (−4 2 + 3)( − 1) = −12 2 + 8 + 3 2. = 6 −2 (x 2 + 5) = −60 −3 3. = (4 3 + 5)(10x 2 + 3) = 200 4 + 36 2 + 100 4. = (6 4 − 2)(6 4 + 2) = 288 7 5. = (2 − 5)(3√ − 2) = 9 1 2 ⁄ − 4 − 15 2 1 2 ⁄ QUOTIENT RULE ` Find the first derivative of the following equations. 1. = 2 (10−2) = 2 = 10 − 2 = 2 = 10 Applying the Quotient Rule = − 2 = (10−2)(2)− 2 (10) (10−2) 2 = 20 2−4−10 2 (10−2) 2 = 10 2−4 (10−2) 2 2. = ( 5+1) (−3) = 5 + 1 = − 3 = 5 4 = 1 FORMATIVE PRACTICE 2 ( ) = − 2 EXAMPLE 3√ 3 is written as 3 1 3
8 | P a g e Applying the Quotient Rule = (−3)(5 4 )−( 5+1)(1) (−3) 2 = 5 5−15 4− 5−1 (−3) 2 = 4 5−15 4−1 (−3) 2 3. = (2− 2 ) (+3) = (2 − 2 ) = ( + 3) = −2 = 1 = (+3)(−2)−(2− 2 )(1) (+3) 2 = −2 2−6−2+ 2 (+3) 2 = − 2−6−2 (+3) 2 4. = (3+2) (1−2) = (3 + 2) = (1 − 2) = 3 = −2 = (1−2)(3)−(3+2)(−2) (1−2) 2 = 3−6+6+4 (1−2) 2 = 7 (1−2) 2 5. = 3−2 −2 = 3 − 2 = − 2 = −2 = 1 = −2(−2)−(3−2)(1) (−2) 2 = −2+4−3+2 (−2) 2 = 1 (−2) 2 Differentiate the following functions. 1. = −4 2+3 −1 : = − (2−1)(2−3) (−1) 2 2. = 6 x 2+5 : = − 6( 2−5) ( 2+5) 2 3. = 4 3+5 10x 2+3 : = 4(10 3+9−25) (10 2+3) 2 4. = 6 4−2 6 4+2 : = 24 3 (3 4+1) 2 5. = 2−5 3√−2 : = 6−8 1 2 ⁄ +15 2 1 2 ⁄ (3 1 2 ⁄ −2) 2 CHAIN RULE AND EXTENDED POWER RULE Find for the following functions. 1. = (−2 − 7) 4 Method 1 : Chain Rule = (−2 − 7) 4 = −2 − 7 = 4 = −2 = 4 3 = −2 × 4 3 = −8 3 = −8(−2 − 7) 3 Method 2 : Extended Power Rule = (−2 − 7) 4 = 4(−2 − 7) 3 (−2) = −8(−2 − 7) 3 FORMATIVE PRACTICE 3 EXAMPLE = × = [()] −1
9 | P a g e 2. = (10 − 2 2 + 3) 3 Method 1 : Chain Rule = (10 − 2 2 + 3) 3 = 10 − 2 2 + 3 = 3 = −4 + 3 = 3 2 = (−4 + 3) × 3 2 = (−12 + 9)(10 − 2 2 + 3) 2 Method 2 : Extended Power Rule = (10 − 2 2 + 3) 3 = 3(10 − 2 2 + 3) 2 (−4 + 3) = (−12 + 9)(10 − 2 2 + 3) 2 3. = −3 (6 2−3) 3 Method 1 : Chain Rule = −3 (6 2−3) 3 = −3(6 2 − 3) −3 = 6 2 − 3 = −3 −3 = 12 = 9 −4 = 12 × 9 −4 = 108 −4 = 108(6 2 − 3) −4 = 108 (6 2−3) 4 Method 2 : Extended Power Rule = −3 (6 2−3) 3 = −3(6 2 − 3) −3 = 9(6 2 − 3) −4 (12) = 108(6 2 − 3) −4 = 108 (6 2−3) 4 Find the derivative of the following functions by using Chain Rule and Extended Power Rule. 1. = 3(−4 2 + 3) 2 : = 192 3 − 144 2. = √7 2 + 3 : = 7(7 2 + 3) −1 2 ⁄ 3. = −3(2 2 − 10) −3 : = 36(2 2 − 10) −4 4. = 5 (−3 4+3) 4 : = 240 3 (−3 4+3)5 5. = −1 √5 2+3 3 : = 10 3(5 2+3) 1 2 ⁄ = × = [()] −1 = × = [()] −1 FORMATIVE PRACTICE 4
10 | P a g e COMBINATION OF PRODUCT RULE/ QUOTIENT RULE AND EXTENDED POWER RULE Find for the following functions. 1. = (2 − 2 ) 2 (5 + 2) 3 = (2 − 2 ) 2 = (5 + 2) 3 = 2(2 − 2 ) 1 . −2 = 3(5 + 2) 2 . 5 Applying Product Rule = (2 − 2 ) 23(5 + 2) 2 . 5 + (5 + 2) 32(2 − 2 ). −2 = 15(2 − 2 ) 2 (5 + 2) 2 + (−4)(5 + 2) 3 (2 − 2 ) = (2 − 2 ) 1 (5 + 2) 2 [15(2 − 2 ) − 4(5 + 2)] = (2 − 2 )(5 + 2) 2 [30 − 15 2 − 20 2 − 8] = (2 − 2 )(5 + 2) 2 (30 − 35 2 − 8) 2. = (4 − 2) 4 (3x 2 + 7) 2 = (4 − 2) 4 = (3x 2 + 7) 2 = 4(4 − 2) 3 × 4 = 2(3x 2 + 7) × 6 = (4 − 2) 42(3x 2 + 7).6 + (3x 2 + 7) 24(4 − 2) 3 . 4 = 12(4 − 2) 4 (3x 2 + 7) + 16(3x 2 + 7) 2 (4 − 2) 3 = 4(4 − 2) 3 (3x 2 + 7)[3(4 − 2) + 4(3x 2 + 7)] = 4(4 − 2) 3 (3x 2 + 7)(12 2 − 6 + 12 2 + 28) = 4(4 − 2) 3 (3x 2 + 7)(24 2 − 6 + 28) 3. = (2 − 5) 3 (3√ 3 − 2) 4 = (2 − 5) 3 = (3√ 3 − 2) 4 = 3(2 − 5) 2 . 2 = 4(3 1 3 − 2) 3 . − 2 3 = 6(2 − 5) 2 = (2 − 5) 34(3 1 3 − 2) 3 . − 2 3 + (3 1 3 − 2) 46(2 − 5) 2 EXAMPLE √ 3 is written as 1 3
11 | P a g e = 4 − 2 3(2 − 5) 3 (3 1 3 − 2) 3 + 6(3 1 3 − 2) 4 (2 − 5) 2 = 2(2 − 5) 2 (3 1 3 − 2) 3 [2 − 2 3(2 − 5) 1 + 3(3 1 3 − 2) 1 ] = 2(2 − 5) 2 (3 1 3 − 2) 3 (4 1 3 − 10 − 2 3 + 9 1 3 − 6) = 2(2 − 5) 2 (3 1 3 − 2) 3 (13 1 3 − 10 − 2 3 − 6) 4. = (2− 2 ) 3 (+3) = (2 − 2 ) 3 = ( + 3) = 3(2 − 2 ) 2 (−2) = 1 Applying the Quotient Rule 6. = √3−2 (−2) 3 = (3 − 2) 1 2 = ( − 2) 3 = 1 2 (3 − 2) − 1 2. (−2) = 3( − 2) 2 . (1) = (+3)3(2− 2 ) 2 .(−2)−(2− 2 ) 3 (1) (+3) 2 = (−6 2−18)(2− 2 ) 2−(2− 2 ) 3 (+3) 2 = (2− 2 ) 2 [(−6 2−18)−(2− 2 ) 1 ] (+3) 2 = (2− 2 ) 2 (−5 2−18−2) (+3) 2 5. = (3+2) 5 (1−2) 3 = (3 + 2) 5 = (1 − 2) 3 = 5(3 + 2) 4 . (3) = 3(1 − 2) 2 . (−2) = (1−2) 35(3+2) 4 .(3)−(3+2) 53(1−2) 2 .(−2) ((1−2) 3) 2 = 15(1−2) 3 (3+2) 4+6(3+2) 5(1−2) 2 (1−2) 6 = (1−2) 2 (3+2) 4 [15(1−2) 1+6(3+2) 1 ] (1−2) 6 = (3+2) 4[15−30+18+12] (1−2) 4 = (3+2) 4[−12+27] (1−2) 4 = (−2) 31 2 (3−2) − 1 2.(−2)−(3−2) 1 2 3(−2) 2 .(1) ((−2) 3) 2 = −(−2) 3(3−2) − 1 2−3(3−2) 1 2 (−2) 2 (−2) 4 = (−2) 2(3−2) − 1 2[−(−2) 1−3(3−2) 1 ] (−2) 4 = (3−2) − 1 2[−+2−9+6 ] (−2) 2 = 5−7 √3−2(−2) 2 Find the derivative of the following functions. 1. = ( − 4)( − 3) 1 2 2. = (5 − ) 6√1 − 5 2 3. = (3+2) 2 1−3 4. = (5−7) 2 (−2 2+3) 4 5. = 4 ( 2−2) 3 ANSWER 1. = 3−10 2(−3) 1 2 ⁄ 2. = (−6+30 2)(5−) 5−5(5−) 6 √1−5 2 3. = 9 2+12+4 1−3 4. = 25 2−70+49 (−2 2+3) 4 5. = 24−24 ( 2−2) 4 √3 − 2 is written as (3 − 2) 1 2 FORMATIVE PRACTICE 5
12 | P a g e 2.0 USE TRIGONOMETRIC, LOGARITHMIC AND EXPONENTIAL FUNCTIONS 2.1 EXPLAIN DIFFERENCE BETWEEN LAWS OF DIFFERENTIATION 2.2 USE VARIOUS DIFFERENTIATION TECHNIQUES TO SOLVE PROBLEMS Derivative of a Trigonometric Function Differentiate the following functions : 1. = 2 = cos 2 × 2 = 2cos 2 Using Chain Rule = 2 = = 2 = cos = × = cos × 2 = 2cos = 2cos 2 2. = ( 2 3 − 5) = −sin( 2 3 − 5) × 2 3 = − 2 3 sin( 2 3 − 5) Using Chain Rule = 2 3 − 5 = = 2 3 = −sin = × = −sin × 2 3 = − 2 3 sin = − 2 3 sin( 2 3 − 5) 3. = (−3 2 + 2 − 3) = 2 (−3 2 + 2 − 3) × (−6 + 2) = (−6 + 2) 2 (−3 2 + 2 − 3) Using Chain Rule = −3 2 + 2 − 3 = = −6 + 2 = sec2 = × = sec2 × (−6 + 2) = (−6 + 2) sec2 = (−6 + 2) sec2 (−3 2 + 2 − 3) (sin ) = cos ( ) = − ( ) = 2 [sin( + )] = cos( + ) × ( + ) [( +)] = −( + ) × ( + ) [( + )] = 2 ( + ) × (ax+ b) EXAMPLE Using Extended Power Rule Using Extended Power Rule
13 | P a g e Derivative of a Trigonometric Function (Involving Power) Differentiate the following functions: 1. = 3 2 Using Extended Power Rule = 3sin2 2 × cos 2 × 2 = 6sin2 2 cos 2 Using Chain Rule = sin 2 = 3 = 2 cos 2 = 3 2 = × = 3 2 × 2 cos 2 = 32 2 × 2 cos 2 = 62 2 cos 2 2. = 2 (3 2 − 4) Using Extended Power Rule = 2cos(3 2 − 4) × −sin (3 2 − 4)× 6 = −12 (3 2 − 4) sin (3 2 − 4) Using Chain Rule = (3 2 − 4) = 2 = − sin(3 2 − 4) × 6 = 2 = −6 sin(3 2 − 4) = × = 2 × −6 sin(3 2 − 4) = 2 (3 2 − 4) × −6 sin(3 2 − 4) = −12 (3 2 − 4) sin(3 2 − 4) Differentiate the following functions using suitable method: 1. = cos( + ) 2. = sin( 1 2 2 − 3) 3. = 23 (5 2 − 2) 4. = 23 5. = 3 ( 2 − 1) ANSWER 1. = −sin( + ) 2. = cos( 1 2 2 −3) 3. = 6sin(5 2 − 2) 2 (5 2 − 2)(10 −2) 4. = −6 cos 3 3 5. = 6 2 2( 2 − 1) 2( 2 − 1) + 3 ( 2 − 1) = −1 × cos × = −1 × − × = −1 × 2 × cot = −2 = sec = − cot EXAMPLE FORMATIVE PRACTICE 6
14 | P a g e Derivative of Logarithm Function Differentiate the following functions: 1. = ln(2 3 + ) Using Differentiation Rule = 1 2 3+ × (6 2 + 1) = 6 2+1 2 3+ Using Chain Rule = 2 3 + = ln = 6 2 + 1 = 1 = × = 1 × (6 2 + 1) = 1 2 3+ (6 2 + 1) 2. = ln(−3 + 7) 3 = 3ln(−3 + 7) Applying Law of Logarithms = 3. 1 −3+7 × (−3) = −9 −3+7 Using Chain Rule = 3ln(−3 + 7) Applying Law of Logarithms = −3 + 7 = 3 ln = −3 = 3 = × = 3 × (−3) = −9 −3+7 3. = 2 3 ln Applying Product Rule = 2 3 = ln = 6 2 = 1 = + = 2 3 ( 1 ) + ln (6 2 ) = 2 2 + 6 2 ln 4. = 5 ln 3+2 Applying Quotient Rule = 5 ln = 3 + 2 = 5 = 3 2 = − 2 = ( 3+2)( 5 )−5 ln (3 2 ) ( 3+2) 2 = ( 3+2)( 5 )−5 ln (3 2 ) ( 3+2) 2 = 5 2+ 10 −15 2 ln ( 3+2) 2 = 5 3+10−15 3 ln ( 3+2) 2 Differentiate the following functions : 1. = ln() 2 : = 2 2. = ln( 2 ) : = 1 3. = 5−2 ln 2+3 : = −12 2−6+4 2 ln 5+6 3+9 4. = √ − ln(3 3 − 1) : = 1 2√ − 6 3 3−1 5. = 2 ln( 2 ) : = + 2 ( 2 ) (ln||) = 1 (ln| + |) = 1 + × ( + ) EXAMPLE FORMATIVE PRACTICE 7
15 | P a g e Derivative of Exponential Function Differentiate the following functions with respect to : 1. = 2 −3+5 = 2 −3+5 × −3 = −6 −3+5 Using Chain Rule = −3 + 5 = 2 = −3 = 2 = × = 2 × −3 = −6 −3+5 2. = 7 3 Applying Product Rule = 7 = 3 = 7 = 3 × 3 2 = 3 2 3 = + = 73 2 3 + 3 (7) = 21 3 3 + 7 3 = 7 3 (3 3 + 1) 3. = √ +2 Applying Quotient Rule = 1 2 = +2 = 1 2 − 1 2 = +2 × 1 = +2 = − 2 = ( +2 )( 1 2 − 1 2)− 1 2( +2 ) ( +2) 2 = ( +2 )[( 1 2 − 1 2)− 1 2] ( +2) 2 = 1 2 − 1 2− 1 2 +2 = 1−2 2√ +2 = 1−2 2√( +2) Differentiate the following functions: 1. = − 10+5 : = −10 10+5 2. = 3 2 3 3+2 : = 6 3(3 3+2)−27 2 2 (3 3+2) 2 3. = −3 ( 2 + 6) : = − −− 18 −3 4. = (1 − 2 ) 2 : = −4 2 + 4 4 5. = (1 − ) : = − 2 2 ( ) = ( + ) = + × ( + ) EXAMPLE FORMATIVE PRACTICE 8
16 | P a g e 3.0 APPLY SECOND ORDER DIFFERENTIATION 3.1 EXECUTE SECOND ORDER DIFFERENTIATION BY USING VARIOUS DIFFERENTIATION TECHNIQUES FOR VARIOUS FUNCTIONS Find the second derivative for each of the following functions. 1. = −2 3 + 3 2 − 4 = −6 2 + 6 − 4 2 2 = −12 + 6 2. = √3 − 2 = (3 − 2) 1 2 = 1 2 (3 − 2) − 1 2 × (−2) = −1(3 − 2) − 1 2 2 2 = 1 2 (3 − 2) − 3 2 × (−2) 2 2 = −(3 − 2) − 3 2 3. = 3+1 5−3 = 3 + 1 = 5 − 3 = 3 = −3 = (5−3)(3)−(3+1)(−3) (5−3) 2 = 15−9+9+3 (5−3) 2 = 18 (5−3) 2 = 18(5 − 3) −2 2 2 = −36(5 − 3) −3 (−3) 2 2 = 108(5 − 3) −3 2 2 = 108 (5−3) 3 Given () = 4 + 5 3 − 6, find : 1. "(2) () = 4 + 5 3 − 6 ′() = 4 3 + 15 2 − 6 "() = 12 2 + 30 "(2) = 12(2) 2 + 30(2) "(2) = 108 2. "(−2) () = 4 + 5 3 − 6 ′() = 4 3 + 15 2 − 6 "() = 12 2 + 30 "(2) = 12(−2) 2 + 30(−2) "(2) = −12 Find the second derivative for each of the following functions. 1. = sin 3 − 4 2 : 2 2 = −9 sin 3 − 16 2 2. = (4 − 3) 2 : 2 2 = 18 3. = (7 2 − )(3 + 5) 2 : 2 2 = 2100 2 + 1110 + 66 4. = 5 2(6−3) 2 : 2 2 = 135 (6−3)4 Find the second derivative for = (3 + 6) 3 if = 1 : 2 2 = 486 EXAMPLE Second order differentiation is the derivative of It is denoted by 2 2 or "() FORMATIVE PRACTICE 9
17 | P a g e 4.0 DEMONSTRATE THE APPLICATION OF DIFFERENTIATION 4.1 USE DIFFERENTIATION TO FIND THE GRADIENT OF A CURVE The gradient at a point on a curve • ′ () = is the gradient function of = () • To find the gradient at a particular point on the curve = (), substitute the -coordinate of that point into the derivative. 1. Find the gradient of the curve = 2 − 5 − 7 at the point on the curve whose -coordinate is −2. Solution = 2 − 5 − 7 = 2 − 5 Substitute = −2 into = 2(−2) − 5 = −9 At point (−2,7) the gradient is −9 2. Find the gradient of the curve = 3 2 + 8 + 4 at the point (2,32) Solution = 3 2 + 8 + 4 = 6 + 8 Substitute = 2 into = 6(2) + 8 = 20 at (2,32) 1. Find the gradient of the curve = 3 + 5 2 + 5 + 5 at the point on the curve whose - coordinate is −5. 2. Find the gradient of the curve = −2 3 + 2 + 3 at the point (1,2) ANSWER 1. = 30 2. = −4 FORMATIVE PRACTICE 10 = 2 − 5 − 7 = −9 = 3 2 + 8 + 4 = 20 EXAMPLE
18 | P a g e 4.1 EXECUTE TURNING POINTS/STATIONARY POINTS • Turning points / stationary points are the points on a function where its derivative is equal to zero, = 0 (Tangent to the curve is horizontal = no gradient) • Types of stationary points There are three types of stationary points: I. Local maximums II. Local minimums III. Stationary points of inflection. Each of these are illustrated on the graph shown. 1. Find the stationary point of = 2 – 2 + 5. Solution = 2 – 2 + 5 = 2 − 2 = 2 − 2 = 0 2 = 2 = 1 into = (1) 2 – 2(1) + 5 = 4 ∴Stationary point (1,4) 2. Find the stationary points of = 3 + 9 2 + 15 + 3. Solution = 3 + 9 2 + 15 + 3 = 3 2 + 18 + 15 EXAMPLE = 2 – 2 + 5 (1,4) •1. Differentiate the function, •2. Set the derivative equal to zero, = 0 •3. Solve for •4. Substitute back in to find Step to determine the coordinate of stationary point = ()
19 | P a g e = 3 2 + 18 + 15 = 0 ( + 1)( + 5) = 0 1 = −1, 2 = −5 Stationary point 1 1 = −1 into 1 = 3 + 9 2 + 15 + 3 1 = (−1) 3 + 9(−1) 2 + 15(−1) + 3 1 = −4 1 = (−1, −4) Stationary point 2 2 = −5 into 2 = 3 + 9 2 + 15 + 3 2 = (−5) 3 + 9(−5) 2 + 15(−5) + 3 2 = 28 2 = (−5,28) 1. Find the stationary point on the curve, = 4 − ² 2. Find the stationary points of = 3 + 12 2 + 21 − 5 ANSWER 1. (2, 4) 2. (−1, −15) and (−7,93) 4.2 DESCRIBE MAXIMUM, MINIMUM AND POINT OF INFLECTION • The nature of any stationary point can be determined by substituting the coordinate of the stationary point into the second derivative of the function, 2 2 . • Nature of stationary point: Maximum Point (∩-shaped) 2 2 < 0 Minimum Point (∪-shaped) 2 2 > 0 Stationary Inflections Point, (curvature changes) 2 2 = 0 FORMATIVE PRACTICE 11 = 3 + 9 2 + 15 + 3 (−5,28) (−1, −4)
20 | P a g e 1. Find the stationary points on the curve = 3 + 7 2 − 5 − 4 (including point of inflection) and determine their nature. Solution = 3 + 7 2 − 5 − 4 = 3 2 + 14 − 5 3 2 + 14 − 5 = 0 1 = 1 3 , 2 = −5 2 2 = 6 + 14 When 1 = 1 3 2 2 = 6 ( 1 3 ) + 14 2 2 = 16 > 0 (Minimum point) 1 = ( 1 3 ) 3 + 7 ( 1 3 ) 2 − 5 ( 1 3 ) − 4 = −4 23 27 Thus ( 1 3 , −4 23 27 )is a minimum point When 2 = −5 2 2 = 6(−5) + 14 2 2 = −16 < 0 (Maximum point) 2 = (−5) 3 + 7(−5) 2 − 5(−5) − 4 = 71 Thus (−5,71)is a maximum point. When 2 2 = 0, 6 + 14 = 0 = − 7 3 = (− 7 3 ) 3 + 7 (− 7 3 ) 2 − 5 (− 7 3 ) − 4 = 33 2 27 Thus (− 7 3 , 33 2 27 )is a point of inflection. 2. Determine the nature of the stationary points for the function = 6 + 3 2 − 3 Solution = 6 + 3 2 − 3 = 6 − 3 2 6 − 3 2 = 0 3(2 − ) = 0 1 = 0 , 2 = 2 2 2 = 6 − 6 When 1 = 0 2 2 = 6 − 6(0) 2 2 = 6 > 0 (Minimum point) 1 = 6 + 3(0) 2 − (0) 3 1 = 6 Thus (0,6) is a minimum point When 2 = 2 2 2 = 6 − 6(2) 2 2 = −6 < 0 (Maximum point) 2 = 6 + 3(2) 2 − (2) 3 2 =10 Thus (2,10) is a maximum point 1. Find the stationary points on the curve = 4 3 − 3 2 − 6 and determine the nature of the points: 2. Determine the nature of the stationary points for the function = 4 3 + 3 2 − 6 − 1 ANSWER EXAMPLE FORMATIVE PRACTICE 12
21 | P a g e 1. (−0.5, 1.75) is a maximum point (1, −5) is a minimum point 2. ( 1 2 , −2 3 4 ) is a minimum point (−1,4) is a maximum point 4.3 SHOW THE GRAPH OF A CURVE Curve sketching, the steps I. Locate the stationary points using the criterion that at a stationary point II. Determine the character of each stationary point using the criteria III. Sketch the graph by marking on the stationary points first and their character. Between stationary points a function must be either always increasing or always decreasing. 1. A curve whose equation is = 2 3 − 3 2 − 12 + 4 has stationary points at (−1, 11) and (2, −16). Determine the nature of these points. hence sketch the curve. Solution Nature of stationary point; = 2 3 − 3 2 − 12 + 4 = 6 2 − 6 − 12 2 2 = 12 − 6 At = −1 2 2 = 12(−1) − 6 2 2 = −18 < 0 (Maximum point) Thus (−1, 11) is a maximum point At = 2 2 2 = 12(2) − 6 2 2 = 18 > 0 (Minimum point) Thus (2, −16) is a minimum point 2. Given = 3 4 − 4 3 , find if exist the point of inflection, the maximum and the minimum point. Hence sketch the graph. Solution Find the stationary point = 3 4 − 4 3 = 12 3 − 12 2 12 3 − 12 2 = 0 2 ( − 1) = 0 = 0, = 1 2 2 = 36 2 − 24 When = 0 EXAMPLE maximum point (−1, 11) minimum point (2, −16) = 2 3 − 3 2 − 12 + 4
22 | P a g e 2 2 = 36(0) 2 − 24(0) 2 2 = 0 is a point of inflection = 3(0) 4 − 4(0) 3 = 0 Thus (0,0)is a point of inflection. When = 1 2 2 = 36(1) 2 − 24(1) 2 2 = 12 > 0 (Minimum point) = 3(1) 4 − 4(1) 3 = −1 Thus (1, −1) is a minimum point When 2 2 = 0, 36 2 − 24 = 0 = 2 3 = 3( 2 3 ) 4 − 4( 2 3 ) 3 = − 16 27 Thus ( 2 3 , − 16 27 )is a point of inflection 1. Locate the stationary points (−3,54) and (3, −54) on the curve = 3 − 27 and determine the nature of each one. Hence sketch the graph 2. Find the stationary values of = 3 4 − 8 3 + 6 2 − 5. and determine the nature of the points and hence sketch the graph ANSWER 1. FORMATIVE PRACTICE maximum point (−3, 54) minimum point (3, − 54) = 3 − 27 Minimum point ((1.-1) point of inflection ( 2 3 , − 16 27 ) point of inflection (0.0) = 3 4 − 4 3
23 | P a g e 2. 4.4 SOLVE RATE OF CHANGE 1. The surface area of a spherical ball bearing increases at a rate of 15.52 −1 when heated. (a) Find the rate of change in the radius of the ball bearing when its radius is 3.5 cm. = 15.52 −1 = 3.5 =? = 4 2 = 8 = × 15.5 = 8 × = 15.5 8 = 15.5 8(3.5) = 0.176 −1 (b) Find the corresponding rate of change in the volume of the ball bearing. =? = 4 3 3 = 4 2 = × = 4(3.5) 2 × 0.176 = 8.6243 −1 2. The area of a circle is increasing at a uniform rate of 102 −1 . Calculate the rate of increase of the radius when the circumference of the circle is 110. = 102 −1 = 2 = 110 =? 1. Identify the information given. State the symbol to represent the information. 2. Determine the suitable formula or equation that relates all the variables 3. Find the derivative of the equation 4. Solve the problem by substituting the given information (0.-5) minimum point (1.-4) point of inflection = 3 4 − 8 3 + 6 2 − 5 EXAMPLE The rate of change of a quantity refers to the rate of increase or decrease of the quantity with respect to time, t. = × are the rates of change of y and x
24 | P a g e = 2 = 2 2 = 110 = 110 2 = 55 = × 10 = 2 × = 10 2 = 10 2( 55 ) = 1 11 −1 3. A spherical ball is being inflated at a rate of 203 −1 . Find the rate of increase of its radius when its surface area is 1002 . = 203 −1 =? = 1002 = 4 3 3 = 4 2 4 2 = 100 2 = 100 4 = 5 1. A spherical balloon is inflated at a rate of 33 /. Find the increment rate of the radius when the radius is 2cm and 4cm. Answer: = 0.0597 = 0.0149 / 2. The length of sides of a cube increases at the rate of 2 −1 . When the side of the cube is 5 , find the rate of change of its : a) Volume b) Surface area 3. A rectangular water tank (see figure below) is being filled at the constant rate of 20 liters / second. The base of the tank has dimensions w = 1 meter and L = 2 meters. What is the rate of change of the height of water in the tank?(express the answer in cm / sec). FORMATIVE PRACTICE 13
25 | P a g e Extended Power Rule 5.0 SOLVE PARAMETRIC EQUATION 5.1 USE THE CHAIN RULE TO FIND DERIVATIVE OF PARAMETRIC EQUATION FOR ALGEBRAIC PROBLEM • Parametric equation is an equation where the variables (usually and ) are expressed in terms of a third parameter, usually expressed as . For example: = 2 + 5 = 3 2 • is known as the parameter () and () are known as parametric equations • ∴ can be found using the chain rule: Find in terms of the parameter for each of the following: 1. = 4 , = 3 − 2 = 4 3 and = 3 2 ∴ = × = 3 2 × 1 4 3 = 3 4 2. = 4 , = 6 2 = − 4 2 and = 12 ∴ = × = 12 × − 2 4 = −3 3 3. = 4 2 − , = 6 2 + + 3 = 8 − 1 and = 12 + 1 ∴ = × = (12 + 1) × 1 (8 − 1) = 12 + 1 8 − 1 4. = 2√ , = 5√ − 4 = 1 1 2 and = 5−8 1 2 2 1 2 ∴ = × = ( 5 − 8 1 2 2 1 2 ) × ( 1 2) = 5 − 8 1 2 1 2 5. = 6 + 5 , = (2 2 + 3) 3 = 6 and = 3(2 2 + 3) 2 (4) = 12(2 2 + 3) 2 ∴ = × = 12(2 2 + 3) 2 × 1 6 = 2(2 2 + 3) 2 EXAMPLE = ×
26 | P a g e 6. = 1−2 , = 1+3 Applying the Quotient Rule for both and = (1 − 2)(1) − (−2) (1 − 2) 2 = 1 − 2 + 2 (1 − 2) 2 = 1 (1 − 2) 2 And = (1 + 3)(1) − (3) (1 + 3) 2 = 1 + 3 − 3 (1 + 3) 2 = 1 (1 + 3) 2 ∴ = × = 1 (1 + 3) 2 × (1 − 2) 2 1 = (1 − 2) 2 (1 + 3) 2 = ( 1 − 2 1 + 3 ) 2 Find in terms of the parameter for each of the following: 1. = 5 , = 3 − 1 2. = 2 , = 2 + 5 − 3 3. = 7 2 − 1 , = 3 − 2 4. = 6√ − , = 5 2 − 4√ 5. = 2( − 4) , = ( − 3) 4 6. = 5+ 1−2 , = 1+2 1−5 ANSWER 1. = 3 5 2 2. = − 1 2 2 (2 + 5) 3. = 3 2−2 14 4. = 10√ 3−2 3−√ 5. = (−3) 3 (−2) 6. = 7 11 ( 1−2 1−5 ) 2 FORMATIVE PRACTICE 14
27 | P a g e 6.0 APPLY IMPLICIT DIFFERENTIATION 6.1 SOLVE FIRST ORDER IMPLICIT DIFFERENTIATION FOR TWO VARIABLES FUNCTION Implicit differentiation method is used when the relationship between x and y are impossible to express explicitly. Differentiate both sides of the equation with respect to x. Derivative of any term involving y must include the factor. Find for each of the following functions. 1. 3 2 + 2 = 5 6 + 2( ) = 0 2 ( ) = −6 ( ) = − 6 2 = − 3 2. 2 − 2 = 3 2 − 2( ) = 3 −2 ( ) = 3 − 2 ( ) = 3−2 −2 3. 2 = 4 2 is a product of two variables. Apply Product Rules. = = 2 = 1 = 2 ( ) 2 ( ) + 2 = 0 2 ( ) = − 2 ( ) = − 2 2 = − 2 4. 2 2 3 = − + 3 = 2 2 = 3 = 4 = 3 2 ( ) 2 2 3 = − + 3 2 23 2 ( ) + 4 3 = −1 2 23 2 ( ) = −1 − 4 3 ( ) = −1−4 3 2 232 ( ) = −1−4 3 6 22 5. 2 2 + 4 − 2 2 = 9 = 4 = = 4 = 1 ( ) 2 2 + 4 − 2 2 = 9 4 + [4 ( ) + 4] − 4 ( ) = 0 4 + 4 ( ) + 4 − 4 ( ) = 0 4 ( ) − 4 ( ) + 4 + 4 = 0 ( ) (4 − 4) = −4 − 4 ( ) = −4−4 4−4 ( ) = − (+) (−) 6. 2 + sin = 2 2 ( ) + cos ( ) = 2 ( ) (2 + cos ) = 2 ( ) = 2 2+cos EXAMPLE Differentiate every term with respect to x
28 | P a g e Differentiate the following equations. 1. 2 − 3 + 2 2 = 7 2. 2 + 2 = 9 3. 7 2 − cos = 3 4. 2 + 2 = 3 + ln 5. cos sin = 10 6. 6 2 + 4 = 2 ANSWER 1. = 3−2 −3+4 2. = − 2−9 2 3. = 3 2−14 sin 4. = − 2 2−1 5. = sin sin cos cos 6. = − 3 3 7.0 CONSTRUCT PARTIAL DIFFERENTIATION 7.1 DEFINE PARTIAL DIFFERENTIATION Partial differentiation is used when differentiating a function of more than one variable. 7.2 EXECUTE PARTIAL DIFFERENTIATION TO FIND: a) First order partial differentiation assume y as a constant assume x as a constant Determine and of the following functions. 1. = 2 2 + 2 − 3 assume y as a constant = 4 + 2 assume x as a constant = 2 − 3 2 2. = ( 3 − ) 3 assume y as a constant = 3( 3 − ) 2 ( 3 − ) = 3( 3 − ) 23 2 = 9 2 ( 3 − ) 2 assume x as a constant = 3( 3 − ) 2 ( 3 − ) = 3( 3 − ) 2 (−1) = −3( 3 − ) 2 3. = sin(2 2 + ) = cos(2 2 + ) (2 2 + ) = cos(2 2 + )4 = 4 cos(2 2 + ) = cos(2 2 + ) (2 2 + ) FORMATIVE PRACTICE 15 EXAMPLE
29 | P a g e = cos(2 2 + )(1) = cos(2 2 + ) 4. = 6 3 − − 2 = 18 2 − − 2 (− 2 ) = 18 2 − − 2 (− 2 ) = 18 2 + 2 − 2 = 18 2 − − 2 (− 2 ) = 18 2 − − 2 [−(2)] = 18 2 + 2 − 2 5. = cos 2 + sin 3 = 3 cos 3 = −2 sin 2 6. = 2 2 + 2 = 2 2 + 2 (2) = 2 2 + 2 2 = 2 2 (2) + 2 = 2 2 2 + 2 7. = − +2 Apply Quotient Rules to obtain the derivative. = − = + 2 = 1 = 1 = (+2)−(−) (+2) 2 = +2−+ (+2) 2 = 3 (+2) 2 = − = + 2 = −1 = 2 = (−1)(+2)−2(−) (+2) 2 = −−2−2+2 (+2) 2 = −3 (+2) 2 Compute and 1. = 3 3 + 2 2. = ( + )( − ) 3. = (8 + )(2 − 5) 4. = sin 3 + 2 sin ANSWER 1. = 9 2 + sin 2 = 3 3 + 2 2 2. = 2 = −2 3. = 32 − 38 = −38 − 10 4. = 3 cos 3 + 2 cos = 2 FORMATIVE PRACTICE 16
30 | P a g e b) Second order partial differentiation 2 2 = ( ) 2 2 = ( ) 2 = ( ) 2 = ( ) Determine 2 2 , 2 2 , 2 , 2 of the following functions. 1. = 3 − 4 = 3 2 − 4 = −4 2 2 = ( ) 2 2 = (3 2 − 4 ) 2 2 = 6 2 2 = ( ) 2 2 = (−4) 2 2 = 0 2 = ( ) 2 = (−4) 2 = −4 2 = ( ) 2 = (3 2 − 4) 2 = −4 2. = 2 2 + 3 − 2 2 = 4 + 3 = 3 − 4 2 2 = ( ) 2 2 = (4 + 3 ) 2 2 = 4 2 2 = ( ) 2 2 = (3 − 4 ) 2 2 = −4 2 = ( ) 2 = (3 − 4 ) 2 = 3 2 = ( ) 2 = (4 + 3 ) 2 = 3 3. = 3 2 + 3 = 3 2 2 + 3 EXAMPLE
31 | P a g e = 23 + 32 22 = ( ) 22 = (322 + 3 ) 22 = 62 22 = ( ) 22 = (23 + 32) 22 = 23 + 6 2 = ( ) 2 = (23 + 32 ) 2 = 62 + 32 2 = ( ) 2 = (322 + 3) 2 = 62 + 32 4. = 2 sin + = 2 cos + cos = − 22 = ( ) 22 = (2 cos + cos ) 22 = −2 sin 22 = ( ) 22 = (− ) 22 = − 2 = ( ) 2 = (− ) 2 = − sin 2 = ( ) 2 = (2 cos + cos ) 2 = − sin Determine , , 22 , 22 , 2 , 2 of the following functions. 1. = 22 + 2 − 32 2. = 53 − 323 − 2 3. = − sin + 52 4. = 2 cos + 2 sin ANSWER 1. = 4 + 2 22 = 4 2 = 2 = 2 − 6 22 = 2 − 6 2 = 2 2. = 152 −63 22 = 30 − 63 2 = 182 = 922 −2 22 = 182 2 = 182 3. = 10 22 = 10 2 = 0 = 1 −cos 22 = sin 2 = 0 4. = −2 sin + 2 22 = −2 cos + 2 sin = 2 + 2 cos 22 = 2 cos − 2 sin 2 = −2 + 2 2 = −2 + 2 FORMATIVE PRACTICE 17
32 | P a g e 8.0 APPLY THE TECHNIQUE OF TOTAL DIFFERENTIATION 8.1 EXECUTE TOTAL DIFFERENTIATION FOR TWO VARIABLES FUNCTION 1. Determine the total differentiation of the following function: = 6 2 − 4 + 3 = 12 − 4 = −4 + 3 2 Substitute the values in the formula : () = () + () () = (12 − 4)() + (−4 + 3 2 )() 2. If = 3 + 6, find the change of z if (, ) change from (2,3) (2.01, 2.98) = 2.01 − 2 = 0.01 = 2.98 − 3 = −0.02 = 3 2 + 6 = 3 Substitute the values in the formula : () = () + () () = (3 2 + 6 )() + ( 3 )() () = [3(2) 2 (3) + 6](0.01) + (2 3 )(−0.02) () = [3(2) 2 (3) + 6](0.01) + (2 3 )(−0.02) () = 0.26 ∴ increase by 0.26 units 3. A right circular cone radius increases at the rate of 2cm/min. Calculate how fast is the cone’s volume changing when the radius is 15cm and the height is 23cm. = 1 3 2ℎ = 1 3 2ℎ = 2 3 ℎ ℎ = 1 3 2 = × = 3 × 2 3 ℎ = 3 × 2 3 (15)(23) = 690 3 / 4. The height of a cylinder is 5cm and increases at the rate of 0.7 cm/s. The radius of the cylinder is 3cm and decreases at the rate of 0.5 cm/s. Calculate the rate of change for the volume of the cylinder. = 2ℎ = 2ℎ ℎ = 2 = ( ) + ℎ ( ℎ ) = (2ℎ) ( ) + ( 2 ) ( ℎ ) = [2(3)(5)(−0.5)] + [(3) 2 (0.7)] = −15 + 6.3 = −8.7 3 / Formula for total derivative : () = () + () EXAMPLE
33 | P a g e 5. The radius and height of a cylinder is 2.5cm and 6cm respectively with the error of measurement is ±0.02cm. Find the maximum error of its volume. = 2ℎ = 2ℎ ℎ = 2 = ( ) + ℎ ( ℎ ) = (2ℎ) ( ) + ( 2 ) ( ℎ ) = [2(2.5)(6)(0.05)] + [(2.5) 2 (0.05)] = 1.5 + 0.3125 = 1.8125 3 / 1. Given = 3 2 + 2 . Determine the total differential of z. Answer () = (6 + 2 2 )() +3 2 () 2. The height of a cone is 9mm and increases at the rate of 0.4mm/s. The radius of the base is 8.5mm and decreases at the rate of 0.3mm/s. Calculate the rate of change for the volume of the cone where = 1 3 2ℎ. Answer : -5.667mm3 /s FORMATIVE PRACTICE 18
34 | P a g e REFERENCES Chai Mun, Aw, Y.J. & Ting, S.K. (2022). Latest Syllabus Mathematics For Matriculation Science Semester 1. Sap Publications (M) Sdn. Bhd. (ISBN 978-967-321-825-7) Bird, J. (2017). Higher Engineering Mathematics (8th Edition). New York, NY : Routledge. Zuraini Ibrahim, Suria Masnin, Fatin Hamimah, Mohamed Salleh, Myzatul Mansor dan Baharudin Azit (2018). Engineering Mathematics 2. Shah Alam : Oxford Fajar Sdn Bhd. P.M. Cohn (2017). Algebraic Numbers and Algebraic Functions. Taylor & Francis Ltd. Dr. Wong Mee Kiong, Grace Lien Poh Choo, Chang Tze Hin, Moh Sin Yee. Masterclass SPM Form 4&5 KSSM. Sasbadi Sdn. Bhd
35 | P a g e