DIFFERENTIATION IKMJB 31 DUM 20132 UNIT 3: DIFFERENTIATION ______________________________________________ 3.1 INTRODUCTION TO DIFFERENTIATION Differentiation is an operation of obtaining f’(x) using the theory of differentiation and several rules that will be stated in this modules. UNIT LEARNING OUTCOMES After completing the unit, students should be able to: 1. find the differential of a function from first principle. 2. differentiate the algebraic functions or polynomials using the Basic Rule of differentiation or formula. 3. differentiate the product functions using Product Rule. 4. differentiate the quotient of functions using Quotient Rule. 5. differentiate the composite of functions using Chain Rule 6. find higher order derivatives 3.1.1 LIMITS The concept of a limit is a very important one in mathematics, and students need to understand it before they proceed to the study of differentiation. In this section, we consider the word limit with, say, ”the limit of the function f(x) as x approaches a value a”. The notation used for the limit of f(x) as x approaches a is: lim f (x) x→a Where “lim” is the abbreviation for “limit” and “ x → a “ means “x approaches the value of a”. In general, if the function f(x) approaches some value L as x approaches a, we would indicate that with the notation
DIFFERENTIATION IKMJB 32 DUM 20132 f x L x a = → lim ( ) Consider the function f(x) = x2 . What is happening to f(x) as x approaches 2. To get the answer we can calculate some values of f(x) for x near 2 using the table below: x y = x2 2.2 4.84 2.1 4.41 2.01 4.0401 2.001 4.004001 2.000 ? 1.999 3.996001 1.99 3.9601 1.9 3.61 From the information above we can conclude that f(x) approaches 4 as x approaches 2. In mathematical symbol, we write lim 4 2 2 = → x x This is read “the limit as x approaches 2 of x2 is 4 “. Sometimes, finding the limiting value of an expression means simply substituting a number.
DIFFERENTIATION IKMJB 33 DUM 20132 Example 3.1: Find lim(4 5) 3 − → x x Solution: lim(4 5) 3 − → x x = 4(3) – 5 = 7 We write lim(4 5) 3 − → x x = 7 Example 3.2: Find the limit of the following functions: a. lim( 18) 2 5 + → x x b. 9 1 lim 3 x→−2 x − c. lim 10 x→−1 d. x 3 x x 6 lim 2 x 3 − − − → e. 16 4 lim 2 4 − − → x x x Solution: a. lim( 18) 2 5 + → x x = (5)2 + 18 = 43 Substitute x with 3
DIFFERENTIATION IKMJB 34 DUM 20132 b. 9 1 lim 3 x→2 x − = (2) 9 1 3 − = 8 9 1 − = -1 c. lim 10 x→−1 = 10 d. x 3 x x 6 lim 2 x 3 − − − → = 3 3 3 3 6 2 − − − = 0 0 Sometimes in a rational function (the form of b a ), using substitution method can give meaningless form; 0 0 at x = c, so one should look for an algebraic simplification (factorization) of the quotient before trying to take the limit. x 3 x x 6 lim 2 x 3 − − − → = ( )( ) ( 3) 3 2 lim 3 − − + → x x x x = lim ( 2) 3 + → x x = 3 + 2 = 5 This method is also known as the cancellation method. The quotient is undefined Limits of a constant is the constant itself: b b x c = → lim
DIFFERENTIATION IKMJB 35 DUM 20132 e. 16 4 lim 2 2 − − → x x x = ( 4)( 4) ( 4) lim 4 − + − → x x x x = ( 4) 1 lim x→4 x + = (4 4) 1 + = 8 1 Notes: Limit does not exist when the denominator of a rational function is equal zero. EXERCISE 3.1 1. Evaluate the following limits whenever they exists. a. lim (4 8) 2 3 − → x x b. lim (2x 4x 3x 1) 3 2 x 1 − + − →− c. lim ( 3) x 4 − →− d. lim (4x x 1) 2 x 1 − − → e. lim ( 16) 2 4 − →− x x f. → +1 3 lim 2 x x x g. 7 3 2 3 lim 2 2 5 − + → x x x h. − − → x 3 x 9 lim 2 x 3 i. + − →− x 3 x 9 lim 2 x 3 j. + − → − 7 49 lim 2 7 x x x k − − + → 1 6 5 lim 2 1 x x x x .
DIFFERENTIATION IKMJB 36 DUM 20132 3.1.2 DIFFERENTIATION FROM FIRST PRINCIPLE To differentiate a function f with respect to x means to find its derivative at the point (x, f(x) ). The derivative of f at x is given by provided this limit exists. This process is called differentiation from first principles. The first derivative of the function y = f(x) is the first order differentiation and it is denoted by dx dy or f ‘ (x). f ‘ (x) is read as ‘ f prime of x ‘. Example 3.3: Find dx dy of the following functions by using differentiation from first principles. a. y = 5x b. y = 3x + 2 c. y = x2 Solution: a. y = f(x) = 5x f(x + x ) = 5(x + x ) = 5x + 5 x f(x + x ) – f(x) = 5x + 5 x - 5x = 5 x ( ) ( ) x f x x f x dx dy f x x ( ) ' lim 0 + − = = →
DIFFERENTIATION IKMJB 37 DUM 20132 therefore 5 lim 5 5 lim ( ) ( ) lim 0 0 0 = = = + − = → → → x x x x x x f x x f x dx dy b. y = f(x) = 3x + 2 f(x + x ) = 3(x + x ) + 2 = 3x + 3 x +2 f(x + x ) – f(x) = (3x + 3 x + 2) – (3x + 2) = 3 x therefore lim 3 3. 3 lim ( ) ( ) lim 0 0 0 = = = + − = → → → x x x x x x f x x f x dx dy c. y = f(x) = x2 f(x + x ) = (x + x ) 2 = x2 + 2x x +( x ) 2 f(x + x ) – f(x) = [x2 + 2x x +( x ) 2 ] – [x2 ] = 2x x +( x ) 2 therefore
DIFFERENTIATION IKMJB 38 DUM 20132 x x x x x x x x x f x x f x dx dy x x x 2 2 0 lim 2 2 ( ) lim ( ) ( ) lim 0 2 0 0 = = + = + + = + − = → → → EXERCISE 3.2 Find dx dy of the following functions using First Principles. a. y = 8x b. y = -3x c. y = 4x - 5 d. y = -2x +10 e. y = 2x2 f. y = x2 + 1
DIFFERENTIATION IKMJB 39 DUM 20132 3.2 DERIVATIVES OF FUNCTIONS 3.2.1 DIFFERENTIATION OF POLYNOMIAL FUNCTIONS Differentiation is the process of calculating a derivative. The derivative of a function represents an infinitesimal change in the function with respect to whatever parameters it may have. Sometimes the terms differentiation can be replaced by the words ‘differentiate’, ‘derivate’ or ‘differential coefficient’. There are several notation for differentiation. The most common notations for differentiation are : dx dy - this is pronounced dy by dx or dy dx. dx d - this is pronounced d by d x or d dx. D - this is pronounced d Given y = f(x) . To find differential coefficient, dx d operates on y = f(x). The process can be write as i) ( ) f(x) dx d y dx d = ii f (x) dx dy = iii) y = Df(x) f(x) f (x) y Df(x) dx d dx dy = = = =
DIFFERENTIATION IKMJB 40 DUM 20132 i. DERIVATIVES OF A CONSTANT If y = f(x) = k then f '(x) 0 dx dy = = , where k is a constant. Example 3.4: Find the derivative of the following functions: a. y = 2 b. y = π c. y = 100 Solutions : a. y = 2, (2) 0 dx d dx dy = = b. y = , ( ) 0 dx d dx dy = = c. y = 100, (100) 0 dx d dx dy = = ii. DERIVATIVES OF kx (kx) k, where k is a cons tan t. dx d f '(x) dx dy If y = f(x) = kx then = = = Example 3.5 : Find the derivative of the following functions: a. y = 5x b. y = πx c. y = -100x
DIFFERENTIATION IKMJB 41 DUM 20132 Solutions : a. y = 5x , 5 dx dy = b. y = πx , π dx dy = c. y = −100x , 100 dx dy = − iii. DERIVATIVES OF xn n n 1 x nx dx d f '(x) dx dy then n If y f(x) x − = = = = = Example 3.6 : Find the derivative of the following functions: a. y = x6 b. y = x - 3 c. y = 3 1 x d. 5 4 y x − = Solutions : a. y = x6 , 6 6 1 5 (x ) 6x 6x dx d dx dy = = = − b. y = x-3 , 3 3 1 4 x 3x 3x dx d dx dy − − − − = − = − =
DIFFERENTIATION IKMJB 42 DUM 20132 c. y = 3 1 x , 3 2 1 3 1 3 1 x 3 1 x 3 1 x dx d dx dy − − = = = d. 5 4 y x − = , 1 5 4 5 4 x 5 4 x dx d dx dy − − − = − = 5 9 x 5 4 − = − iv. DERIVATIVES OF axn where a is a cons tan t and n is an int eger. ax anx , dx d f '(x) dx dy then n If y f(x) ax n n − 1 = = = = = Example 3.7 : Differentiate each of the following with respect to x: a. y = 5x, b. y = 4x3 c. y = -12x-5 d. y = 2 x 3 Solutions : a. y = 5x, (5x) (5)(x ) dx d dx dy 1−1 = = = 5x0 = 5
DIFFERENTIATION IKMJB 43 DUM 20132 b. y = 4x3 , (4x ) 4(3x ) dx d dx dy 3 3−1 = = = 4(3)x3 – 1 = 12x2 c. y = -12x-5 , 5 5 1 ( 12x ) 12( 5)x dx d dx dy − − − = − = − − 6 60x − = d. y = 2 x 3 , 2 2 1 3 (3x ) 3( 2)x 6x dx d dx dy − − − − = = − = − = 3 x 6 − v. DIFFERENTIATION OF SUM AND DIFFERENCES m n ax bx where a is a cons tan t and n is an int eger. amx bnx dx dy bx dx d ax dx d f '(x) dx dy If y f(x) ax bx then , m 1 n 1 m n m n − − = = = = = Example 3.8 : Find dx dy of the following functions: a. y = x 3 + 4x2 b. y = 4x3 – 5x
DIFFERENTIATION IKMJB 44 DUM 20132 c. x 4 y x 2 = + d. y = (x – 1)(2x + 3) e. y = (x – 5)2 f. y = x 5x 3x 2 − Solution : a. y = x 3 + 4x2 ( ) ( ) 3x 8x 3(x ) 2(4x ) 4x dx d x dx d dx dy 2 3 2 2 1 3 2 = + = + = + − − b. y = 4x3 – 5x 12x 5 3(4x ) 5x (5x) dx d (4x ) dx d dx dy 2 3 1 1 1 3 = − = − = − − − c. 2 2 1 x 4x x 4 y x − = + = + (4x ) dx d (x ) dx d dx dy 2 −1 = + 1 1 2x 4( 1)x − − = + − 2 2x 4x − = −
DIFFERENTIATION IKMJB 45 DUM 20132 d. y = (x – 1)(2x + 3) 2x x 3 2 = + − (3) dx d (x) dx d (2x ) dx d dx dy 2 = + − = 4x + 1 e. y = (x – 5)2 = x2 – 10x + 25 (25) dx d (10x) dx d (x ) dx d dx dy 2 = − + = 2x – 10 f. y = x 5x 3x 2 − = x 3x x 5x 2 − y = 5x – 3 (3) dx d (5x) dx d dx dy = − = 5 Expand first divide first Expand first
DIFFERENTIATION IKMJB 46 DUM 20132 UNIT EXERCISE 3.3 1. Find dx dy of the following functions: a. y = - 9 b. y = sin π c. y = 4x d. y = -x -5 e. y = x3 f. y = - 6x g. x 1 y = h. y = x i. 5 x 1 y = j. y = 8x2 k. 4 5x 2 y = l. 7 x 5 y = − m. 4 y 25x − = − n. 4 3 y = x o. y = 6 x 3 1 p. y = 6 x 3 2. Differentiate the following with respect to x. a. f(x) = 2 3 x + 3x b. f(x) = x 2x 7 2 − + c. f(x) = 3x x x 4 2 − + d. f(x) = 4 2 x − 3x e. f(x) = x 3 3 − f. f(x) = 3x x 1 2 + − g. y = 4x2 + 3x – 2 h. y = 3x6 – 4x5 – x 4 i. y = (x + 3) (x – 4) j. y = (2x – 1)2 k. y 3x 5x x 4 = − + l. y (x 2)(3x 1) 2 = + − m. n. o. p. y =
DIFFERENTIATION IKMJB 47 DUM 20132 FORMULA 3.2.2 DIFFERENTIATION OF TRIGONOMETRIC FUNCTIONS (u) dx d (tan u) sec u. dx d vi) (u) dx d (cos u) sinu. dx d v) (u) dx d (sin u ) cosu . dx d iv) (tan x) sec x dx d iii) (cos x) sin x dx d ii) (sin x) cos x dx d i) 2 2 = = − = = = − = Example 3.9 : Differentiate the following functions with respect to x. a. y = sin 2x b. 2 y = cos 5x c. y = tan 10x d. y = −2cos 3x e. x 1 tan 2x y 2 + =
DIFFERENTIATION IKMJB 48 DUM 20132 Solution : 2 cos 2x cos 2x.(2) = = 2 2 10x sin 5x sin 5x (10x) = − = − c. y = tan 10x let u = 10x10 dx du then = (u) dx d sec u. dx dy 2 = 10 sec 10x sec 10x.(10) 2 2 = = a. y = sin 2x let u = 2x 2 dx du then = (u) dx d cos u. dx dy = d. b. 2 y = cos 5x 2 let u = 5x 10x dx du then = (u) dx d sin u. dx dy = −
DIFFERENTIATION IKMJB 49 DUM 20132 d. y = −2cos 3x let u = 3x 3 dx du then = (u) dx d ( 2sinu). dx dy = − − = 2sin 3x.(3) = 6 sin 3x e. x 1 tan 2x y 2 + = By using Quotient Rule : Let u = tan 2x v = x2 + 1 2sec 2x sec 2x..(2) dx du 2 2 = = 2x dx dv = ( ) 2 2 2 2 2 2 2 2 [(x 1)] 2sec 2x (x 1) (2x)(tan 2x) [(x 1)] (x 1)(2sec 2x) (tan 2x)(2x) dx dy + + − = + + − = Quotient Rule :
DIFFERENTIATION IKMJB 50 DUM 20132 UNIT EXERCISE 3.4 Differentiate the following functions with respect to x. (a) y = cos 3x (b) y = sin 10x (c) sin 5x 3 1 y = (d) y = 3cos ( 2x + 1) (e) tan 2x 4 1 y = − (f) = − − x 5 1 y 5 sin 3 (g) = x − 4 2 1 y 10 tan (h) cos (5 4x) 4 3 y = −
DIFFERENTIATION IKMJB 51 DUM 20132 3.2.3 DIFFERENTIATION OF EXPONENTIAL FUNCTIONS Exponential functions have the following differentiation formulas : Example 3.10: Differentiate the following functions with respect to x. a. b. 2 3x y =e c. d. 4x e 1 y = e. x 2x 1 y e .e − = f. 3x 2 5x 3 e e y + − = g. Solution : a. Let u = 2x , 2 dx du = Hence, (u) dx d e . dx dy u = 2x y e = 3x 4 y e + = 2x y e ln x sin2x = + − 2x y e = ۩ ۩
DIFFERENTIATION IKMJB 52 DUM 20132 2x 2x 2e e .(2) = = b. 2 3x y = e Let u = 2 3x , 6x dx du = Hence, (u) dx d e . dx dy u = 2 3x 2 3x 6xe e .(6x) = = c. Let u = 3x + 4 , 3 dx du = Hence, (u) dx d e . dx dy u = 3x 4 (3x 4) 3e e .(3) + + = = d. 4x e 1 y = 4x e − = Let u = -4x , 4 dx du = − Hence, (u) dx d e . dx dy u = 4x 4x 4e e .( 4) − − = − = − 3x 4 y e + = From the Law of Indices :
DIFFERENTIATION IKMJB 53 DUM 20132 e. x 2x 1 y e .e − = 3x 1 x 2x 1 e e − + − = = Let u = 3x – 1 , 3 dx du = Hence, (u) dx d e . dx dy u = 3x 1 3x 1 3e e .(3) − − = = f. 3x 2 5x 3 e e y + − = 2x 5 5x 3 3x 2 5x 3 (3x 2) e e e − − − − − − + = = = Let u = 2x – 5 , 2 dx du = Hence, (u) dx d e . dx dy u = 2x 5 2x 5 2e e .(2) − − = = From the Law of Indices : From the Law of Indices :
DIFFERENTIATION IKMJB 54 DUM 20132 g. 2cos 2x x 1 2e cos 2x.(2) x 1 e .(2) dx dy 2x 2x = + − = + − UNIT EXERCISE 3.5 Differentiate the following with respect to x. (a) (b) (c) (d) (e) (f) 2x 5 x 3 e e y + + = 2 x 6x e 1 e 1 (g) y = + 4x 1 x 2 2x e e 1 (h) y 5e + = − + (i) 2x y e ln x sin2x = + − 2 x 1 y e = 4x y 6e = 6x 4 y e = − 2x 5x 1 y 3e 5e + = + 3 8x y e 4 = 3x y e sin x =
DIFFERENTIATION IKMJB 55 DUM 20132 FORMULA 3.2.4 DIFFERENTIATION OF LOGARITHMIC FUNCTIONS (u) dx d u 1 (ln u) dx d x 1 (ln x) dx d • = = Example 3.11 : Differentiate the following functions with respect to x. a. y = ln 5x b. y = 3lnx c. y = 4ln 2x d. 3 y = ln x e. y = ln(2x +1) f. y = ln10x −5ln(3 − 2x) g. y = sin x ln x Solution : a. y = ln 5x let u = 5x , 5 dx du = (u) dx d u 1 dx dy = (5) 5x 1 dx dy = 5x 5 = x 1 =
DIFFERENTIATION IKMJB 56 DUM 20132 b. y = 3ln x x 3 = = x 1 3 dx dy c. y = 4ln 2x let u = 2x , 2 dx du = ( ) x 4 2 2x 1 4 (u) dx d u 1 dx dy = = = d. 3 y = ln x Let 3 u = x , 2 3x dx du = ( ) x 3 3x x 1 (u) dx d u 1 dx dy 2 3 = = = e. y = ln(2x +1) Let u = 2x + 1 , 2 dx du = ( ) 2x 1 2 2 2x 1 1 (u) dx d u 1 dx dy + = + = =
DIFFERENTIATION IKMJB 57 DUM 20132 f. y = ln 10x − 5 ln (3 − 2x) let u =10x , 10 dx du = let u = 3 − 2x , 2 dx du = − ( ) ( 2) 3 2x 1 10 5 10x 1 dx dy − − − = 3 2x 10 x 1 − = + g. y = sin x ln x By using Product Rule ( ) ln x (cos x) x sin x ln x cos x x 1 sin x dx dy = + + = Product Rule :
DIFFERENTIATION IKMJB 58 DUM 20132 UNIT EXERCISE 3.6 Differentiate the following functions with respect to x. (a) y = ln 10x x 3 2 (b) y = ln ln 10x 2 1 (c) y = 2 (d) y = ln x ( ) 4 x 4 1 e y = ln 6 (f) y = 4ln x 5 x 5 2 (g) y = 8 ln (h) y = ln (3 − 5x) (i) y = − 5ln (8x +12) j) ln x x (j) y =
DIFFERENTIATION IKMJB 59 DUM 20132 3.3 TECHNIQUES OF DIFFERENTIATION 3.3.1 DIFFERENTIATE PRODUCT OF FUNCTIONS USING PRODUCT RULE To differentiate the product of two functions such as (x3 + 5) (2x2 – 1) we expand the expression before differentiating it. It is tedious and takes time if the functions involved are quite complicated. As an alternative we can use the rule which is also known as the product rule. If p(x)= f(x) g(x) , where f(x) and g(x) are differentiable, then f(x) or dx d g(x) g(x) dx d (f(x)g(x)) f(x) dx d = + If set u = f (x) and v = g(x) then the product rule may also be expressed in the form ( ) dx du v dx dv uv u dx d = + Example 3.12 : Find the derivative of the following function. a. y = (x2 + 2)(x3 – 5) b. y = (x2 – 4x + 6)(1 - x 3 ) Solution : a. Given y = (x2 + 2)(x – 5)
DIFFERENTIATION IKMJB 60 DUM 20132 Let u = x 2 + 2 → dx du = 2x and v = x 3 – 5 → dx dv = 3x2 Use the product rule : dx dy = u dx dv + v dx du = (x 2 + 2)(3x 2 )+ (x3 - 5)(2x) = 3x 4 + 6x 2 + 2x 4 – 10x = 5x 4 + 6x 2 – 10x b. Given y = (x2 – 4x + 6) (1 - x 3 ) Let u = x 2 – 4x + 6 → dx du = 2x - 4 and v = 1 - x 3 → dx dv = -3x2 Use the product rule : dx dy = u dx dv + v dx du = (x 2 – 4x + 6)(-3x 2 ) + (1 - x 3 )(2x - 4) = -3x 4 + 12x 3 – 18x2 + 2x – 4 – 2x4 + 4x3 = -5x 4 + 16x 3 – 18x2 + 2x – 4
DIFFERENTIATION IKMJB 61 DUM 20132 3.3.2 DIFFERENTIATE QUOTIENT OF FUNCTIONS USING QUOTIENT RULE An easier method known as the quotient rule can be used to differentiate a function which involves the quotient of two polynomials. If ( ) ( ) g(x) f x p x = , where f(x) and g(x) are differentiable, then 2 [g(x)] g(x) dx d f(x) f(x) dx d g(x) g(x) f(x) dx d − = or If set u = f (x) and v = g(x) then the quotient rule may also be expressed in the form 2 v dx dv u dx du v v u dx d − = Example 3.13 : Find the derivative of 2 2 3 ) ) 3 x a y x x b y x + = = +
DIFFERENTIATION IKMJB 62 DUM 20132 Solution: ( )( ) ( )( ) 2 2 2 2 2 2 2 2 2 x x 3 x 2x x 3 x x 2x x 3 1 dx dy 1 dx dv 2x dx du Let u x 3 v x x x 3 a) y − = − − = − + = = = = + = + = ( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 x 3 3 x x 3 x 3 2x x 3 x 3 1 x 2x dx dy 2x dx dv 1 dx du Let u x v x 3 x 3 x b) y + − = + + − = + + − = = = = = + + =
DIFFERENTIATION IKMJB 63 DUM 20132 3.3.3 DIFFERENTIATE COMPOSITE FUNCTIONS USING CHAIN RULE Functions such as ( ) ( ) 2 2 5 3 x + 2 , 1+ x and 1+ x − can be stated in the form f g(x) and are called composite functions. If y = f(u) and u = g(x) and both of these functions are differentiable, then the composite function y = f[g(x)] is differentiable and dx du . du dy dx dy = Example 3.14 : Differentiate the following function with respect to x. a. y = (x3 + 2)4 b. y = (2x + 3)4 c. y = 4(x – 1)3 Solutions: a. Given y = (x3 + 2)4 Let u x 2 3 = + 4 y = u ( ) 3 2 3 2 3 3 2 2 3 12x x 2 12x u 4u 3x dx du du dy dx dy 4u du dy 3x and dx du = + = = = = =
DIFFERENTIATION IKMJB 64 DUM 20132 b. Given y = (3x + 3)4 Let u = 3x + 3 4 y = u 3 dx du = and 3 4u du dy = dx du dx dy dx dy = ( ) 3 3 3 12u 12 2x 3 4u 3 = = + = c. Given y = 4(x – 1)3 Let u = x − 1 3 y = 4u 1 dx du = and 2 12u du dy = ( ) 2 2 12 x 1 12u 1 dx du dx dy dx dy = − = = Alternative method n y = f(x) = au dx du nau (au dx d dx dy n 1 n ) = • − =
DIFFERENTIATION IKMJB 65 DUM 20132 Example 3.15 : Find the derivative of y = (x3 + 2)4 Solutions: ( ) 3 4 3 4 3 4 1 3 3 3 2 2 3 3 ( 2) ( 2) 4( 2) . ( 2) 4( 2) (3 ) 12 ( 2) y x dy d x dx dx d x x dx x x x x − = + = + = + + = + = +
DIFFERENTIATION IKMJB 66 DUM 20132 UNIT EXERCISE 3.7 1. Find the derivatives of the following functions using Product Rule. a. y = (3x2 + 7)(x3 – 1) b. y = (x +1)(3x2 – 5) c. y = (x + 7)(1 – x 2 ) d. y = 5x3 (1 – x 5 ) e. y = 2x (4x 1) 6 2 + f. y =(5x2 + 6)( 2x 5) 3 − g. y = (3x 1)(2x 1) 2 − + h. y = ( )( ) 4 x 2 3x 5 + i. y = 3x (x 5) 3 2 − j. y = (3x 2)(2x 1) 2 + +
DIFFERENTIATION IKMJB 67 DUM 20132 2. Differentiate the following with respect to x. a. y = 2x 1 x 2 + b. y = 5x 2 3x 2 3 − c. y = x 1 5x 2 + d. y = 3x 2 2x 1 2 4 + − e. y = 5 2 1 x 3x 5 − + f. y = 1 3x x 1 2 − − g. y = x 2 4x 3 2 + − h. y = 1 x x 3 − + i. y = x 1 x 2 2 + − j. y = 4x 1 2x 3 + k. 3x 1 x 5 y 2 3 − + = l. y = x 1 2x 5 4 + −
DIFFERENTIATION IKMJB 68 DUM 20132 3. Differentiate the following with respect to x using Chain’s Rule. a. y = ( ) 3 x −1 b. y = ( ) 5 2x +1 c. y = ( ) 4 3 2x −1 d. y = x 2 1 − e. y = ( ) 2 2x 1 5 + f. y = ( ) 3 2x 5 3 + − g. y = – 2(5 + 3x) 5 h. y = 3(x + 4)3 4. Find dx dy a. y = x (x 1) 3 + b. y = 2x (x 1) 3 − c. y = (3x −1)(2x +1) d. y = (x + 2)(3x −1) e. y = 3x (x 5) 3 2 − f. y = (3x 2)(2x 1) 2 + +
DIFFERENTIATION IKMJB 69 DUM 20132 3.4 HIGHER DERIVATIVES Let y be a function of x, that is y = f(x). Differentiating y with respect to x, f (x) dx dy = ' is known as the first derivative of f. If we differentiate dx dy or f’(x) with respect to x again, we obtain the second order derivative which is written as f (x) dx d y '' 2 2 = . Hence, if y = f(x) So, f (x) dx dy = ' [ ''( )] 2 2 f x dx d dx d y = = f’’(x) 2 2 2 : dx dy dx d y note This differentiation process can be continued to find the third, fourth, and successive derivatives of f( x), which are called higher order derivatives of f( x). Because the “prime” notation for derivatives would eventually become somewhat messy, it is preferable to use the numerical notation f( n )( x) = y( n ) to denote the nth derivative of f( x). For example fourth order can be denoted by ; 4 4 = (4) () = (4) Example 3.16 Find 2 2 dx d y for each of the following: a. y = 125x – 5 b. y = 18 35 2 1 2 x + x − c. y = 3x3 – x 2 + 10x - 3
DIFFERENTIATION IKMJB 70 DUM 20132 Solution: a. y = 125x – 5 dx dy =125 2 2 dx d y = 0 b. y = 18 35 2 1 2 x + x − dx dy = x + 18 2 2 dx d y = 1 c. y = 3x3 – x 2 + 10x – 3 dx dy = 9x2 – 2x + 10 2 2 dx d y = 18x – 2 Example 3.17 Find the first, second, and third derivatives of () = 5 4 − 3 3 + 7 2 − 9 + 2 Solution: ′ () = 5(4) 4−1 − 3(3) 3−1 + 7(2) 2−1 − 9 = 20 3 − 9 2 + 14 − 9 ′′() = 20(3) 3−1 − 9(2) 2−1 + 14 = 60 2 − 18 + 14 ′′′() = 60(2) 2−1 − 18 = 120 − 18 Differentiate y Differentiate dx dy
DIFFERENTIATION IKMJB 71 DUM 20132 EXERCISE 3.8 1. Find 2 3 2 3 d y d y and dx dx for each of the following equations: a. y = 200x + 30 b. y = x2 – 2x + 16 c. y = 3 2 15 4 3 x − x d. y = x3 + 21x2 + 15x - 5 e. y = x2 - x 2 f. y = 6 8 2 3 2 − x + x g. y = 9x3 - 2x 2 1 + 15x 2. Find the first, second, and third derivatives of y = sin 2 x. 3. Find ′′′(4)if () = √ . :