RI Differentiation C6A Lect Notes

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ________________________________ Chapter 6A: Differentiation Techniques Page 1 of 18 Chapter 6A: Differentiation Techniques SYLLABUS INCLUDES  Differentiation of simple functions defined implicitly or parametrically.  Finding the approximate value of a derivative at a given point using a graphing calculator. CONTENTS 1 Limits and Derivatives 1.1 Limits 1.2 Derivative as a Limit 2 Basic Rules of Differentiation 3 Derivatives of Standard Functions 3.1 Trigonometric Functions 3.2 Inverse Trigonometric Functions 4 Higher Order Derivatives 5 Implicit Differentiation 6 Parametric Differentiation 6.1 Parametric Equations 6.2 Parametric Differentiation Conclusion Appendix: Limits (For Your Information) INTRODUCTION Differentiation deals with the concept of change. It is useful for understanding phenomena in many fields of study, including business, economics and social sciences. This topic reinforces the ideas of derivatives that students learnt in ‘O’-Level Additional Mathematics and further develops the idea of derivatives as a limit. In this chapter, students will learn more techniques of differentiation such as implicit differentiation and parametric differentiation. Students will also learn to use the graphing calculator to calculate derivatives.

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________________________ ________________________________ Chapter 6A: Differentiation Techniques Page 2 of 18 1 Limits and Derivatives 1.1 Limits Consider the expression 2 1x  . Let x approach 2 from the left. Let x approach 2 from the right. We write x 2 . We write x 2 . From the table above, we see that From the table above, we see that as 2x  , 2 1x  approaches 5, as 2x  , 2 1x  approaches 5, i.e. the limiting value of 2 1x  is 5. i.e. the limiting value of 2 1x  is 5. We write  2 2 lim 1 5 x x   . We write  2 2 lim 1 5 x x   . Since   22 22 lim 1 lim 1 5 xx xx    , the limit of 2 1x  as 2x exists and is given by the value 5. We write  2 2 lim 1 5 x x   . If  lim f lim f xa xa xx k   , where k is real, then lim f xa xk   . (Please refer to Appendix for more information and results on Limits.) Key in Y1 = X2 + 1 and go to y- and put TblStart = 1.9989 or 2.0011 and Tbl = 0.0001 or 0.0001 respectively. x y O 22 2

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________________________ ________________________________ Chapter 6A: Differentiation Techniques Page 3 of 18 1.2 Derivative as a Limit Consider a curve with equation f( )yx . Let A  ,fxx and B  ,fxx xx be two points on the curve, where x is a small increment in x . Note: A variable quantity, prefix by  , means a small change in that quantity. Let y be the corresponding small increment in y due to x . Then y  f( )xx  − f( )x and the gradient of the chord AB is given by y x   . Notice that the gradient of the curve at A can be approximated by the gradient of the chord AB . The closer B is to A , the better is the approximation. Thus, d at d y Ax = gradient of the curve at A  gradient of the tangent to the curve at A lim B A  (gradient of chord AB ) 0 lim x y x    (since 0 xa s B A  ) 0 f( ) f( )lim x xx x x    . Note: dd ()dd y yx x is not a fraction. Definition of derivative: If f( )yx , then d d y x  0 f( ) f( )lim x xx x x    . Using the definition of derivative, we can show that  1d ,d nnxn xx  where n   . (Please refer to Appendix for the proof.) Derivatives of Elementary Functions In Secondary School, you learnt the derivatives of the following elementary functions: y d d y x nx for fixed n 1nnx  ex ex ln x 1 x sin x cos x cos x sin x tan x 2sec x 0 x y y = f(x) Chord AB

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________________________ ________________________________ Chapter 6A: Differentiation Techniques Page 4 of 18 2 Basic Rules of Differentiation Let f and g be two functions of x . We denote  dd f and gdd x xx x   by    f and gx x . The following are some basic rules of differentiation: (1)    d fg fgd ax b x a x b xx   , where a and b are constants. (2)    d fg g f + fgd x xx x x xx  (Product Rule) (3)             2 fg f f gd dg g x xx xx xx x     (Quotient Rule) (4)    ddfg f g f g gdd x xx xx x  (Chain Rule) If we write fyu and  g,ux then the chain rule can be written as ddd=ddd yyu x ux . (5) d1 dd d x yy x  Applying Chain Rule, we have the following results: y d d y x f n x   1 ff n nx x   f e x   f ef x x   ln f x ,  f0 x   1 ff xx   sin f ( )x  c o sf () f ()x x  cos f ( )x  s i nf () f ()x x  tan f ( )x   2sec f ( ) f ( )x x

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________________________ ________________________________ Chapter 6A: Differentiation Techniques Page 5 of 18 Example 1 Differentiate the following with respect to x . (a) 53(1 )x  , (b) 32x , (c) 2ln 1x  , (d) sin x . Solution   35d 1d xx (a)      254 245 31 5 15 1 x x xx    3d 2d x x (b)      3ln 2 +3 ln 2 +3 ln 2 3 d e, s e e b e l o w d d ed el n 2 2l n 2 x x x x x x       Note   2d ln 1d xx (c)  2 22 d1 ln 1d2 11 221 1 xx xxx x         d sin d xx (d) 1cos 2 1 cos 2 x x x x   Example 2 Differentiate the following with respect to x . (a) (2 3) 5x x , (b) 2 1 3 x x   , (c) ln ln x , 1x , (d) 2 5log ( 1)x  Solution   d 23 5d xxx (a)   125 2 3 25 45 2 3 62 3 25 25 xx x xx x xx        2 d1 d3 x xx    (b)          2 22 22 2 2222 31 12 3 32 2 2 3 33 xx x x xx x x x xx           d ln lnd xx (c) 11 ln 1 ln x x x x      2 5 d log 1d xx (d)    2 2 2 ln 1d dl n 5 11 2 2ln 5 1 1l n 5 x x xxx x         Note: In Examples 1(b), 1(c) and 2(d) we have used the following properties of logarithm:  ln log=e , 0 log , , , 0, , 1 log yxyy z x z yx xy x y z x z x     ln ln , 0yxy x x

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________________________ ________________________________ Chapter 6A: Differentiation Techniques Page 6 of 18 3 Derivatives of Standard Functions 3.1 Trigonometric Functions y d d y x y d d y x cot x 2cosec x cot f ( )x   2cosec f ( ) f ( )x x sec x sec tanx x sec f ( )x    sec f ( ) tan f ( ) f ( )x xx  cosec x cosec cotx x  cosec f ( )x    cosec f ( ) cot f ( ) f ( )x xx  Note: 1. The above results hold for x measured in radians only. If x is measured in degrees, then convert x to 180 x radians. 2. The derivative of sec x and cosec x can be found in the Formula List. We can verify the formula easily. For example, dd 1secdd c o sxx xx   2 1 sincos 1s i n sec tancos cos xx x x