1. Techniques of Differentiation (For Upload)

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 1 H2 MATHEMATICS TOPIC TECHNIQUES OF DIFFERENTIATION 2022/JC1 1 Limits 1.1 Limit of a Function Consider the behaviour of a function  2f2 xxx  for values of x close to 2. x f x x f x 1.0 2.000000 3.0 8.000000 1.5 2.750000 2.5 5.750000 1.8 3.440000 2.2 4.640000 1.9 3.710000 2.1 4.310000 1.95 3.852500 2.05 4.152500 1.99 3.970100 2.01 4.030100 1.995 3.985025 2.005 4.015025 1.999 3.997001 2.001 4.003001 x values approaching 2 from the left x values approaching 2 from the right The above table gives the corresponding values of f x for values of x close to 2 but not equal to 2. Content Outline:  Differentiation of simple functions defined implicitly.  Finding the approximate value of a derivative at a given point using a graphing calculator.

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 2 From the graph and table of values, we observe that as x approaches 2, f x gets closer and closer to 4. We express this in words as “the limit of the function  2f2 xxx  as x approaches 2 is 4”. Mathematically, we write it as  2 lim f 4 x x   . In general, given a function f , if f x approaches a fixed number L as x approaches a , then L is the limit of f x as x approaches a . We write it as lim f xa xL   __________________________________________________________________________________  Remark: To find the limit of a function f , f x need not be defined at xa . __________________________________________________________________________________ For example, consider the graph of  sinf xx x for π3π 3x  as shown below. We observe that even though f0 does not exist, it is clear that the limit of the function  sinf xx x as x approaches 0 is 1. If f is defined at xa and  lim f f xa xa   , we say f is continuous at xa . If f is continuous at every point on an interval  ,ab , f is a continuous function on the interval  ,ab . Intuitively, continuous functions are the functions whose graphs can be drawn without lifting the pen off the paper. In the above example, f is not continuous at 0x (hence it is excluded graphically with an empty circle). In H2 Maths syllabus, the func tions are all assumed to be c ontinuous on the interval under consideration. y O x 3 3 1 Note that when 0x ,  f x is not defined.

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 3 1.2 Properties of Limits (  Self-Reading) If lim f xa x  and lim g xa x  both exist, then Properties Examples (a)  lim f lim f xa xa kx k x    , where k is a constant    33 44 3 lim 5 5 lim 54 320 xx x x     (b)    lim f g lim f lim g xa xa xa x xx x       22 33 3 2 lim lim lim 33 6 xx x x xxx       (c)   lim f g lim f lim g xa xa xa x xx x                  22 2 lim ln lim ln lim ln 2 2 2ln2 ln 4 xx x x xx x               (d)     lim fflim g lim g xa xa xa xx x x        , where lim g 0 xa x       2 2 1 1 1 2 lim 11lim 23 l i m 23 11 21 3 2 x x x xx xx           Example 1 ( Self-Reading): Evaluate the following limits (a) 2 1 41lim 2x xx x     , (b) 1lim x x   , (c) 31lim 2x x x    . Solution: (a)    2 2 1 1 1 lim 4 141lim 2l i m 2 x x x xxxx xx           2 14 1 1 12 4 3    (b) 1lim 0 x x  

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 4 (c) Method :   31 2 1 2 1 2 31lim lim 2 3lim 1 lim 3 lim 1 30 10 3 x x xxx x x x x xx xx x x                      Method : 31 7lim lim 322 73l i m 2 30 3 xx x x xx x              2 The Derivative as a Limit (First Principles) Consider a continuous function fyx . Let  ,Pxy and  ,Qx xy y be two points on the curve fyx where x is a small change in x and y is a small change in y . Therefore, gradient of     ff xx x yy y yPQ xx x x x         . As Q moves closer to P , 0x  and gradient of PQ  gradient of AB . A B x y P (x, y) Q (x + x, y + y) x + x y y + y x O Divide throughout by the highest power of x Perform long division B A

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 5 Thus, the gradient of tangent to the curve at    00 fflim lim xx xx x yP xx     . We denote 0 lim x y x   as d d y x . This is known as the first derivative of f (with respect to x ) at x and is also denoted by f' x or d fd xx . The process of finding the derivative, f' x , from f x is called differentiation. The first derivative of f x from First Principles is Example 2: Let 2f: ,xx x  . Using the first principles of differentiation, show that f' ( ) 2xx . Solution:      22f and f δδxx xx xx   From first principles,       2 2 222 ff 2 2 xx x xx x xx xx x x x x xx                0 f' l i m 2 2( s h o w n ) x xx x x        Refer to Section 5: Appendix on Page 17 - 20 (Examples A to C ) for finding the derivatives of 1 , x sin x and ex from the first principles.     0 ffd f' l i md x xx xy xxx     , provided the limit exists. Expand and simplify

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 6 3 Rules of Differentiation 3.1 Basic Rules (  Self-Reading) Given that a and n are real constants, d 0d ax   1d d nnxn xx   dd ffdd ax a xxx         dd dfg f gdd d x xxxxx x         Example 3: Differentiate 53 3 1 xxx  with respect to x . Solution:  3 2 1 2 53 3 5 3 44 d1 d dd 335 2 xx x x xxx x xxx          3.2 Chain Rule (  Self-Reading) If y is a function of u , where u is a function of x , then dd d ddd yy u xu x . Example 4: Differentiate  22 1 64 1xx with respect to x . Solution:         22 22 232 32 32 d1 d 64 1dd 64 1 d26 4 1 6 4 1 d 26 4 1 1 2 4 83 1 64 1 xxxx xx xx xx x xx x x xx                      “Outermost to innermost” i.e. differentiate the function with negative power followed by the expression .

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 7 3.3 Product Rule (  Self-Reading) Let yu v where u and v are both functions of x . Then dd d d dd d d yv u uv u vxx x x  . Example 5: Differentiate   1 2 231 2 7xx with respect to x . Solution:       11 1 22 2 22 2 dd d3 1 27 3 1 27 27 3 1dd d xx x x x xxx x                                11 22 11 22 1 2 2 2 1 222 1 22 2 131 2 722 762 31 2 7 6 2 7 27 3 1 6 27 2 7 3 1 12 42 2 7 15 42 1 xx x x xx x x xx x x xx x x xx x               