MAT183/ MAT421 : Calculus I 46 4. Jun 2015/MAT421/Q1c 5. Dec 2015/MAT421/Q1c 6. Jun 2014/MAT421/Q1c 7. Jul 2017/MAT421/Q1c Answer : 7. k = 1 h(x) is continuous at x = 3 Answer : 5. i) = −3 ii) f(x) is discontinuous at x = 1 Answer : 4. = −1, b= 1 Answer : 6. = −1, b = −4
Chapter 2 : Differentiation 47 CHAPTER 2 : DIFFERENTIATION List of Topics 1.0 The First Principle 2.0 Rules of Differentiation 2.1 Basic Rule 2.1.1 Constant Rule 2.1.2 Power Rule 2.1.3 Sum and difference Rules 2.2 Generalized Power Rule 2.3 Chain Rule 2.4 Product Rule 2.5 Quotient Rule 2.6 Trigonometric Rules 2.7 Exponential Rule 2.8 Logarithmic Rule 3.0 Implicit Differentiation 4.0 Higher Order Differentiation 5.0 Linear Approximations and Differentials 1.0 The Definition of Derivative (The First Principle) ( ) ( ) ( ) h f x h f x f' x lim h + − = →0 Simple Examples Find f’(x) using the definition of derivative a) f(x) = 2x b) f(x) = x 2 c) f(x) = x
MAT183/ MAT421 : Calculus I 48 ) () = 2 ( + ℎ) = 2( + ℎ) = 2 + 2ℎ Using the first principle (or the definition of derivative) ′() = ℎ→0 ( + ℎ) −() ℎ = ℎ→0 2 + 2ℎ − 2 ℎ = ℎ→0 2ℎ ℎ = ℎ→0 2 = 2 # ) () = 2 ( + ℎ) = ( + ℎ) 2 = 2 + 2ℎ +ℎ 2 Using the first principle (or the definition of derivative) ′() = ℎ→0 ( + ℎ) −() ℎ = ℎ→0 2 + 2ℎ + ℎ 2 − 2 ℎ = ℎ→0 2ℎ +ℎ 2 ℎ = ℎ→0 ℎ(2 +ℎ) ℎ = ℎ→0 (2 + ℎ) = 2 + 0 = 2 # ) () = √ ( + ℎ) = √ + ℎ Using the first principle (or the definition of derivative) ′() = ℎ→0 ( + ℎ) −() ℎ = ℎ→0 √ + ℎ − √ ℎ = ℎ→0 √ + ℎ − √ ℎ ∙ √ + ℎ + √ √ + ℎ + √ = ℎ→0 (√ + ℎ) 2 − (√) 2 ℎ(√ + ℎ + √) = ℎ→0 + ℎ − ℎ(√ + ℎ + √) = ℎ→0 ℎ ℎ(√ + ℎ + √) = ℎ→0 1 √ + ℎ + √ = 1 √ + 0 + √ = 1 √ + √ = 1 2√ #
Chapter 2 : Differentiation 49 More Examples (From Previous semester papers) Example 1/ MAR 2004/ MAT183/Q4a (ii)/ 5 marks Use the definition of derivatives to find f' (x) for the function f(x) = x − 2x . Solution : ) () = √ − 2 ( + ℎ) = √ + ℎ − 2( + ℎ) = √ + ℎ − 2 − 2ℎ Using the first principle (or the definition of derivative) ′() = ℎ→0 ( + ℎ) − () ℎ = ℎ→0 (√ + ℎ − 2 − 2ℎ) − (√ − 2) ℎ = ℎ→0 √ + ℎ − 2ℎ − √ ℎ = ℎ→0 √ + ℎ − √ ℎ − ℎ→0 2ℎ ℎ = ℎ→0 √ + ℎ − √ ℎ ∙ √ + ℎ + √ √ + ℎ + √ − ℎ→0 2 = ℎ→0 (√ + ℎ) 2 − (√) 2 ℎ(√ + ℎ + √) − 2 = ℎ→0 + ℎ − ℎ(√ + ℎ + √) − 2 = ℎ→0 ℎ ℎ(√ + ℎ + √) − 2 = ℎ→0 1 √ + ℎ + √ − 2 = 1 √ + 0 + √ − 2 = 1 √ + √ − 2 = 1 2√ − 2 # Split using theorem of limits : → [() ± ()] = → () ± → () Note : separate the term containing surd, and the term not containing surd + = 2 √ + √ = 2√
MAT183/ MAT421 : Calculus I 50 Example 2/ OCT 2004/ MAT183/Q2c/ 7 marks Find the derivative of ( ) 2 1 1 + = x f x by using the definition of derivative. Solution : ) () = 1 2√ + 1 ( + ℎ) = 1 2√ + ℎ + 1 Using the first principle (or the definition of derivative) ′() = ℎ→0 ( + ℎ) −() ℎ ′() = ℎ→0 1 ℎ ∙ [( +ℎ) − ()] = ℎ→0 1 ℎ ∙ [ 1 2√ + ℎ + 1 − 1 2√ + 1 ] = ℎ→0 1 ℎ ∙ [ (2√ + 1) − (2√ + ℎ + 1) (2√ + ℎ + 1)(2√ + 1) ] = ℎ→0 1 ℎ ∙ [ 2√ + 1 − 2√ + ℎ −1 (2√ + ℎ + 1)(2√ + 1) ] = ℎ→0 1 ℎ ∙ [ 2√ − 2√ + ℎ (2√ + ℎ + 1)(2√ + 1) ] = 2 ℎ→0 1 ℎ ∙ [ √ − √ + ℎ (2√ + ℎ +1)(2√ +1) ] = 2 ℎ→0 1 ℎ ∙ [ √ − √ + ℎ (2√ + ℎ +1)(2√ +1) ] ∙ √ + ℎ +√ √ + ℎ +√ = 2 ℎ→0 1 ℎ ∙ [ (√) 2 − (√ +ℎ) 2 (2√ +ℎ + 1)(2√ + 1)(√ + ℎ + √) ] = 2 ℎ→0 1 ℎ ∙ [ − ( + ℎ) (2√ + ℎ +1)(2√ +1)(√ +ℎ + √) ] = 2 ℎ→0 1 ℎ ∙ [ −ℎ (2√ + ℎ +1)(2√ +1)(√ +ℎ + √) ] = 2 ℎ→0 [ −1 (2√ + ℎ + 1)(2√ + 1)(√ + ℎ + √) ] = 2 [ −1 (2√ +0 + 1)(2√ + 1)(√ + 0 + √) ] = 2 [ −1 (2√ + 1)(2√ + 1)(√ + √) ] = 2 ( −1 (2√ + 1) 2 (2√) ) = 2 [ −1 2√(2√ + 1) 2 ] = −1 √(2√ + 1) 2 # Answer : In that case, you better write the formula ‘per h’ as multiply 1 ℎ so that it will not appear so messy like this, ′() = ℎ→0 [ 1 2√ + ℎ + 1 − 1 2√ + 1 ℎ ] Question : How to solve limits, if the function is given in fractions? Simplify using ( + )( − ) = − No need to expand! + = 2 √ + √ = 2√
Chapter 2 : Differentiation 51 Example 3/ page 13/ MAR 2005/ MAT183/Q3c/ 6 marks Find the derivative of f(x) = x + 1 from the first principle. Solution : ) () = √ + 1 ( + ℎ) = √ + ℎ + 1 Using the first principle (or the definition of derivative) ′() = ℎ→0 ( + ℎ) − () ℎ = ℎ→0 √ + ℎ + 1 − √ + 1 ℎ = ℎ→0 √ + ℎ + 1 − √ + 1 ℎ ∙ √ + ℎ + 1 + √ + 1 √ + ℎ + 1 + √ + 1 = ℎ→0 (√ + ℎ + 1) 2 − (√ + 1) 2 ℎ(√ + ℎ + 1 + √ + 1) = ℎ→0 ( + ℎ + 1) − ( + 1) ℎ(√ + ℎ + 1 + √ + 1) = ℎ→0 + ℎ + 1 − − 1 ℎ(√ + ℎ + 1 + √ + 1) = ℎ→0 ℎ ℎ(√ + ℎ + 1 + √ + 1) = ℎ→0 1 √ + ℎ + 1 + √ + 1 = 1 √ + 0 + 1 + √ + 1 = 1 √ + 1 + √ + 1 = 1 2√ + 1 # Multiplying with their conjugate, and simplify using : ( + )( − ) = −
MAT183/ MAT421 : Calculus I 52 Example 4/ page 18/ NOV 2005/ MAT183/Q2a/ 5 marks Use the first principle of differentiation to find f' (x) for the function ( ) x f x x 2 2 = − . Solution : ) () = 2 − 2 ( + ℎ) = ( + ℎ) 2 − 2 + ℎ = 2 + 2ℎ + ℎ 2 − 2 + ℎ Using the first principle (or the definition of derivative) ′() = ℎ→0 ( + ℎ) − () ℎ ′() = ℎ→0 1 ℎ ∙ [( + ℎ) − ()] = ℎ→0 1 ℎ ∙ [( 2 + 2ℎ + ℎ 2 − 2 + ℎ ) − ( 2 − 2 )] = ℎ→0 1 ℎ ∙ [ 2 + 2ℎ + ℎ 2 − 2 + ℎ − 2 + 2 ] = ℎ→0 1 ℎ ∙ [2ℎ + ℎ 2 − 2 + ℎ + 2 ] = ℎ→0 1 ℎ ∙ [2ℎ + ℎ 2 + 2 − 2 + ℎ ] = ℎ→0 1 ℎ ∙ [2ℎ + ℎ 2 ] + ℎ→0 1 ℎ ∙ [ 2 − 2 + ℎ ] = ℎ→0 1 ℎ ∙ ℎ(2 + ℎ) + ℎ→0 1 ℎ ∙ [ 2( + ℎ) − 2 ( + ℎ) ] = ℎ→0 (2 + ℎ) + ℎ→0 1 ℎ ∙ [ 2 + 2ℎ − 2 ( + ℎ) ] = ℎ→0 (2 + ℎ) + ℎ→0 1 ℎ ∙ [ 2ℎ ( + ℎ) ] = ℎ→0 (2 + ℎ) + ℎ→0 ( 2 ( + ℎ) ) = (2 + 0) + ( 2 ( + 0) ) = 2 + 2 2 #
Chapter 2 : Differentiation 53 Example 5/ page 36/ APR 2007/ MAT183/Q2a/ 5 marks Use the first principle of differentiation to find f' (x) for the function f(x) = 3 + x . ans : f’(x) = 1 2√3+ Solution : ) () = √3 + ( + ℎ) = √3 + + ℎ Using the first principle (or the definition of derivative) ′() = ℎ→0 ( + ℎ) − () ℎ = ℎ→0 √3 + + ℎ − √3 + ℎ = ℎ→0 √3 + + ℎ − √3 + ℎ ∙ √3 + + ℎ + √3 + √3 + + ℎ + √3 + = ℎ→0 (√3 + + ℎ) 2 − (√3 + ) 2 ℎ(√3 + + ℎ + √3 + ) = ℎ→0 (3 + + ℎ) − (3 + ) ℎ(√3 + + ℎ + √3 + ) = ℎ→0 3 + + ℎ − 3 − ℎ(√3 + + ℎ + √3 + ) = ℎ→0 ℎ ℎ(√3 + + ℎ + √3 + ) = ℎ→0 1 (√3 + + ℎ + √3 + ) = 1 √3 + + 0 + √3 + = 1 √3 + + √3 + = 1 2√3 + # Multiplying with their conjugate, and simplify using : ( + )( − ) = −
MAT183/ MAT421 : Calculus I 54 Example 6/ page 54/ OCT 2008/ MAT183/Q2a/ 5 marks Use the first principle of differentiation to find f' (x) for the function ( ) x x f x − = 3 2 . Ans : 6 (3−) 2 Solution : ) () = 2 3 − ( + ℎ) = 2( +ℎ) 3− ( + ℎ) = 2 + 2ℎ 3− −ℎ Using the first principle (or the definition of derivative) ′() = ℎ→0 ( + ℎ) −() ℎ ′() = ℎ→0 1 ℎ ∙ [( +ℎ) − ()] = ℎ→0 1 ℎ ∙ [ 2 + 2ℎ 3− −ℎ − 2 3 − ] = ℎ→0 1 ℎ ∙ [ (2 + 2ℎ)(3 −)− 2(3 − −ℎ) (3 − − ℎ)(3 −) ] = ℎ→0 1 ℎ ∙ [ 6 −2 2 + 6ℎ − 2ℎ − 6 + 2 2 +2ℎ (3 − − ℎ)(3 − ) ] = ℎ→0 1 ℎ ∙ [ 6ℎ (3 − − ℎ)(3 − ) ] = ℎ→0 [ 6 (3 − − ℎ)(3 − ) ] = 6 (3 − − 0)(3 − ) = 6 (3 − )(3 −) = 6 (3 − ) 2 # Answer : In that case, you better write the formula ‘per h’ as multiply 1 ℎ so that it will not appear so messy like this, ′() = ℎ→0 [ 1 2√ + ℎ + 1 − 1 2√ + 1 ℎ ] Question : How to solve limits, if the function is given in fractions?
Chapter 2 : Differentiation 55 2.0 Rules of Differentiation 2.1 Basic Rule 2.1.1 Constant Rule If y = a, where a is a constant Then = 0 dx dy Or (a) = 0 dx d , where a is a constant 2.1.2 Power Rule If y = axn , where a is a constant Then −1 = n nax dx dy Or ( ) −1 = n n ax nax dx d , where a is a constant 2.1.3 Sum and Difference Rules If y = f(x) g(x) Then f' (x) g' (x) dx dy =
MAT183/ MAT421 : Calculus I 56 Examples Differentiate with respect to x. a) 4 3 2 3 = − + + x y x x b) ( ) 5 2 4 3 2 1 x x x f x = + − + c) ( ) = − + + 3 2 3 1 4 3 2 x x x g x d) ( ) e x h x = x + x − + 3 1 5 2 4 Solution : ) () = 1 4 + 3 − √2 + √ 2 5 () = −4 + 3 − √2 + 2 5 differentiate ′ () = −4 −5 + 3 − 0 + 2 5 2 5 −1 ′ () = −4 −5 + 3 + 2 5 − 3 5 # Before differentiate, move up variable x from denominator using the properties of indices. 1 = − change the surd (@ radical) function √ 2 5 to exponent (power) 2 5 so that we can differentiate using power rule no need to change surd constant √2 to 2 1 2, because differentiate any constant will become zero. ) = 2 3 − √ + 3 √ + 4 = 2 3 − 1 2 + 3 − 1 2 + 4 differentiate = 6 2 − 1 2 − 1 2 − 3 2 − 3 2 + 0 = 6 2 − 1 2√ − 3 2 − 3 2 # #
Chapter 2 : Differentiation 57 ) ℎ() = 5 4 + √2 − 1 √3 + ℎ() = 5 4 + √2√ − 1 √3√ + ℎ() = 5 4 + √2 1 2 − 1 √3 1 2 + ℎ() = 5 4 + √2 1 2 − 1 √3 − 1 2 + differentiate ℎ ′ () = 20 3 + √2 ∙ 1 2 − 1 2 − 1 √3 ∙ − 1 2 − 1 2 −1 + 0 ℎ ′ () = 20 3 + √2 ∙ 1 2 ∙ 1 1 2 + 1 2√3 − 3 2 ℎ ′ () = 20 3 + √2 ∙ 1 2 ∙ 1 √ + 1 2√3 − 3 2 ℎ ′ () = 20 3 + √2 ∙ 1 (√2) 2 ∙ 1 √ + 1 2√3 − 3 2 ℎ ′ () = 20 3 + 1 √2 ∙ 1 √ + 1 2√3 − 3 2 ℎ ′ () = 20 3 + 1 √2 + 1 2√3 − 3 2 # ) () = 2 3 3 − 4 2 + 1 √ 3 + () = 2 3 −3 − 4 2 + − 1 3 + differentiate ′() = 2 3 ∙ −3 −4 − 8 − 1 3 − 4 3 + 0 ′() = −2 −4 − 8 − 1 3 − 4 3 ′() = − 2 4 − 8 − 1 3 − 4 3 #
MAT183/ MAT421 : Calculus I 58 2.2 Generalized Power Rule If = () = • () −1 • ′ Examples Use Generalized Power Rule to find the derivative of ) = ( 3 − 2) 6 ) = 1 2√ + 1 ) = √ + 1 Solution : ) = ( 3 − 2) 6 differentiate = 6( 3 − 2) 5 ∙ (3 2 − 0) = 6( 3 − 2) 5 ∙ 3 2 = 18 2 ( 3 − 2) 5 # ) = 1 2√ + 1 = (2√ + 1) −1 differentiate = −(2√ + 1) −2 ∙ (2 ∙ 1 2√ + 0) = − 1 (2√ + 1) 2 ∙ 1 √ = − 1 √ (2√ + 1) 2 ) = √ + 1 # = ( + 1) 1 2 differentiate = 1 2 ( + 1) − 1 2 ∙ (1 + 0) = 1 2 ∙ 1 √ + 1 = 1 2√ + 1 #
Chapter 2 : Differentiation 59 2.3 Chain Rule = • = • • = • • • Examples Use Chain Rule to find the derivative of ) = ( 3 − 2) 6 ) = 1 2√ + 1 ) = √ + 1 Solution : ) = ( 3 − 2) 6 = 6 Differentiate with respect to u = 6 5 Using Chain Rule, = ∙ = 6 5 ∙ 3 2 = 18 2 5 = 18 2 ( 3 − 2) 5 # ℎ = 3 − 2 = 3 2
MAT183/ MAT421 : Calculus I 60 ) = 1 2√ + 1 = (2√ + 1) −1 = −1 Differentiate with respect to u = − −2 = − 1 2 Using Chain Rule, = ∙ = − 1 2 ∙ 1 √ = − 1 √ 2 = − 1 √ (2√ + 1) 2 # ℎ = 2√ + 1 = 2 1 2 + 1 = 2 ∙ 1 2 − 1 2 + 0 = − 1 2 = 1 √ ) = √ + 1 = ( + 1) 1 2 = 1 2 Differentiate with respect to u = 1 2 − 1 2 = 1 2 ∙ 1 √ = 1 2√ Using Chain Rule, = ∙ = 1 2√ ∙ 1 = 1 2√ + 1 # ℎ = +1 = 1
Chapter 2 : Differentiation 61 2.4 Product Rule 2.5 Quotient Rule Examples Differentiate with respect to x. ) = (3 − 3 )(2 + 5 ) ) = (5 2 + 3) 4 (1 − ) 3 ) = ( 3 + 3) 2 ) = 2 − 4 ( 2 + 2) 3 Solution : ) = (3 − 3 )(2 + 5 ) Differentiate using product rule, = ′ + ′ = (2 + 5 )(3 − 3 2 )+ 5 4 (3 − 3 ) = 6 − 6 2 + 3 5 − 3 2 +15 5 − 5 7 = 6 −9 2 +18 5 − 5 7 # = = ′ + ′ = = ′ − ′ 2 ) = (5 2 +3) 4 (1 −) 3 Differentiate using product rule, = ′ + ′ = 40 (5 2 + 3) 3 (1 − ) 3 − 3(5 2 + 3) 4 (1 − ) 2 = (5 2 + 3) 3 (1 − ) 2 (40(1 −)− 3(5 2 +3)) = (5 2 +3) 3 (1 −) 2 (40 − 40 2 − 15 2 −9) = (5 2 +3) 3 (1 −) 2 (40 − 55 2 − 9) # = 3 − 3 = 2 + 5 ′ = 3− 3 2 ′ = 5 4 = (5 2 + 3) 4 = (1 − ) 3 ′ = 4 (5 2 + 3) 3 ∙ 10 ′ = 3(1 −) 2 ∙−1 ′ = 40 (5 2 + 3) 3 ′ = −3(1 − ) 2
MAT183/ MAT421 : Calculus I 62 ) = ( 3 + 3) 2 Differentiate using quotient rule, = ′ − ′ 2 = ( 3 + 3) 2 − 6 3 ( 3 +3) (( 3 +3) 2) 2 = ( 3 + 3)[( 3 + 3)− 6 3 ] ( 3 + 3) 4 = [( 3 + 3)− 6 3 ] ( 3 + 3) 3 = 3 − 5 3 ( 3 + 3) 3 # ) = 2 − 4 ( 2 + 2) 3 Differentiate using quotient rule, = ′ − ′ 2 = 2 ( 2 +2) 3 − 6( 2 + 2) 2 ( 2 − 4) (( 2 + 2) 3) 2 = 2 ( 2 + 2) 2 [( 2 + 2)− 3( 2 − 4)] ( 2 + 2) 6 = 2 [( 2 + 2)− 3( 2 −4)] ( 2 + 2) 4 = 2 ( 2 + 2 −3 2 + 12) ( 2 +2) 4 = 2 (−2 2 +14) ( 2 + 2) 4 = −4 3 + 28 ( 2 +2) 4 # = = ( 3 + 3) 2 ′ = 1 ′ = 2( 3 + 3) ∙ 3 2 ′ = 6 2 ( 3 + 3) = 2 − 4 = ( 2 + 2) 3 ′ = 2 ′ = 3( 2 +2) 2 ∙ 2 ′ = 6( 2 + 2) 2
Chapter 2 : Differentiation 63 2.6 Trigonometric Rules Examples Differentiate with respect to x. ) = 2 ) = 3 2 ) = √ ) = 5 ) = 3 4 ) = 3 3 Solution : ① ( ) = ② ( ) = − ③ ( ) = 2 ④ ( ) = ⑤ ( ) = − ⑥ ( ) = − 2 ① ( ) = ∙ ′ ② ( ) = − ∙ ′ ③ ( ) = 2 ∙ ′ ④ ( ) = ∙ ′ ⑤ ( ) = − ∙ ′ ⑥ ( ) = − 2 ∙ ′ ) = 2 Differentiate wrt x = 2 ∙ 2 = 2 2 # ) = 3 2 Differentiate wrt x = − 3 2 ∙ 6 = −6 3 2 # = √ = 1 2 ′ = 1 2 − 1 2 ′ = 1 2 ∙ 1 √ ′ = 1 2√ ) = √ Differentiate wrt x = 2 √ ∙ 1 2√ = 2 √ 2√ #
MAT183/ MAT421 : Calculus I 64 for example (f), we can also differentiate in the following way, ) = 5 = 5 5 ∙ 5 = 5 5 5 # ) = 3 4 = ( 4) 3 = 3( 4) 2 ∙ 4 ∙ 4 = 12 4 2 4 # ) = 3 3 = 3 ∙ 3 3 = 3 = − 3 ∙ 3 = −3 3 # ) = 3 3 = = ′ + ′ = ( 3)(3 3)+ ( 3)(−3 2 3) = 3 3 3 − 3 3 2 3 = 3 3 ∙ 3 3 − 3 3 ∙ 1 2 3 = 3 2 3 3 − 3 3 = 3 2 3 − 3 3 = 3(2 3 − 1) 3 = 3 (−2 3) 3 = −3 2 3 3 = −3 3 # = 3 = 3 ′ = 3 ∙ 3 ′ = − 2 3 ∙ 3 ′ = 3 3 ′ = −3 2 3 2 + 2 = 1 2 3 + 2 3 = 1 2 3 − 1 = − 2 3
Chapter 2 : Differentiation 65 2.7 Exponential Rule Examples Differentiate with respect to x. ) = 2 ) = 3 2 ) = √ ) = 4 ) = 3−2 Solution : ) = √ Differentiate wrt x = √ ∙ 1 2√ = √ 2√ # ) = 2 Differentiate wrt x = 2 ∙ 2 = 2 2 # ) = 3 2 Differentiate wrt x = 3 2 ∙ 6 = 6 3 2 # = √ = 1 2 ′ = 1 2 − 1 2 ′ = 1 2 ∙ 1 √ ′ = 1 2√ ( ) = ( ) = • ′ ) = 4 Differentiate wrt x = 4 ∙ − 4 4 ∙ 4 = −4 4 4 4 # ) = 3−2 Differentiate wrt x = 3−2 ∙ (3 2 − 2) = (3 2 − 2) 3−2 #
MAT183/ MAT421 : Calculus I 66 2.8 Logarithmic Rule Examples Differentiate with respect to x. ) = (3) ) = (2 3 ) ) = ( 4) ) = (3 3) ) = ( 2 + 1 2 − 1 ) Solution : ( ) = 1 ( ) = 1 • ′ Hot Tips !! If = () Write = + (using the properties of logarithm) If = ( ) Write = − (using the properties of logarithm) ) = (3) Differentiate wrt x = 1 3 ∙ 3 = 1 # ) = (2 3 ) Differentiate wrt x = 1 2 3 ∙ 6 2 = 3 # ) = ( 4) Differentiate wrt x = 1 4 ∙ 2 4 ∙ 4 = 4 4 2 4
Chapter 2 : Differentiation 67 ) = (3 3) using the properties of logarithms = (3) + ( 3) Differentiate wrt x = 1 3 ∙ 3 + 1 3 ∙ 3 3 ∙ 3 = 1 + 3 3 # ) = (3 3) = () Differentiate wrt x = 1 ∙ ( ′ + ′ ) = 1 3 3 ∙ (3 3 + 9 3 3) = 1 3 3 ∙ 3 3 (1 + 3 3) = 1 (1 + 3 3) = 1 + 3 3 # = 3 = 3 ′ = 3 ′ = 3 3 ∙ 3 ′ = 3 3 3 ) = ( 2 +1 2 −1 ) using the properties of logarithms = ( 2 +1) −( 2 − 1) Differentiate wrt x = 1 2 + 1 ∙ 2 − 1 2 − 1 ∙ 2 = 2 2 + 1 − 2 2 − 1 # Or… = 2( 2 − 1) − 2( 2 + 1) ( 2 + 1)( 2 − 1) = 2 3 − 2 −2 3 − 2 ( 2 + 1)( 2 − 1) = −4 ( 2 + 1)( 2 −1) # ) = ( 2 +1 2 −1 ) = ( ) Differentiate wrt x = 1 ( ) ∙ ( ′ − ′ 2 ) = ( ′ − ′ 2 ) = 2 − 1 2 + 1 ( 2( 2 − 1)− 2( 2 + 1) ( 2 −1) 2 ) = 1 2 + 1 ( 2( 2 − 1)− 2( 2 + 1) ( 2 − 1) ) = 2 3 − 2 − 2 3 − 2 ( 2 + 1)( 2 − 1) = −4 ( 2 + 1)( 2 −1) # = 2 + 1 = 2 − 1 ′ = 2 ′ = 2
MAT183/ MAT421 : Calculus I 68 More Examples (From Previous Semester Paper) Example 1/ MAR 2005/ MAT183/ Q1c (5 marks) Find dx dy if ( ) + + = 3 2 2 2 2 x cos x y ln Solution : Example 2/ MAR 2005/ MAT183/ Q2c (4 marks) Given ( ) 3 1 − = x x f x , find f' (x). Solution : = [ 2 + cos 2 (2 + 2) 3 ] Using the properties of logarithm = (2 + cos 2) − (2 + 2 ) 3 = (2 + cos 2) − 3 (2 + 2 ) Differentiate with respect to x, = 1 2 + cos 2 ∙ − 2 ∙ 2 − 3 2 + 2 ∙ 2 = −2 2 2 + cos 2 − 6 2 + 2 # () = ( 1 − ) 3 Using generalized power rule (combine with quotient rule) ′() = 3 ( 1 − ) 2 ( ′ − ′ 2 ) ′() = 3 ∙ 2 (1 − ) 2 ( (1 − ) − (−1) (1 − ) 2 ) ′() = 3 2 (1 − ) 2 ∙ 1 − + 1 (1 − ) 2 ′() = 3 2 (1 − ) 2 ∙ 2 − (1 − ) 2 ′ () = 3 2 (2 − ) (1 − ) 4 # = = 1 − ′ = 1 ′ = −1
Chapter 2 : Differentiation 69 Example 3/ NOV 2005/ MAT183/ Q1c (6 marks) Differentiate the following functions with respect to x : i) cos x sin x y 1 3 3 + = ii) + + = 3 1 2 2 x x x y ln Solution : ) = 3 1 + 3 = Using quotient rule, = ′ − ′ 2 = (3 3)(1 + 3) − (−3 3)( 3) (1 + 3) 2 = 3 3 + 3 2 3 + 3 2 3 (1 + 3) 2 = 3( 3 + 2 3 + 2 3) (1 + 3) 2 = 3( 3 + 1) (1 + 3) 2 = 3(1 + 3) (1 + 3) 2 = 3 1 + 3 # = 3 = 1 + 3 ′ = 3 3 ′ = −3 3 ) = ( 2 2 + 3 + 1 ) Using the properties of logarithms, = (2) − ( 2 + 3 + 1) Differentiate wrt x, = 1 2 ∙ 2 − 1 2 + 3 + 1 ∙ (2 + 3) = 1 − 2 + 3 2 + 3 + 1 #
MAT183/ MAT421 : Calculus I 70 Example 4/ OCT2006/ MAT183/ Q5a (5 marks) Find dx dy if ( ) 4 3 y = x sin2x . Solution : Example 5/ APR 2009/ MAT183/ Q2c (3 marks) Find dx dy if + − = x x y tan 1 1 . Solution : = 4 ( 2) 3 = Differentiate wrt x (using Product Rule) = ′ + ′ = 4 3 ( 2) 3 + 6 4 2 2 2 = 4 3 3 2 + 6 4 2 2 2 = 2 3 2 2 (2 2 + 3 2) # = 4 = ( 2) 3 ′ = 4 3 ′ = 3( 2) 2 ∙ 2 ∙ 2 ′ = 6 2 2 2 = ( 1 − 1 + ) = ( ) Differentiate wrt x (using Trigo-Rule & Quotient Rule) = 2 ( ) ∙ ( ′ − ′ 2 ) = 2 ( 1 − 1 + ) ∙ ( −(1 + ) − (1 − ) (1 + ) 2 ) = 2 ( 1 − 1 + ) ∙ ( −1 − − 1 + (1 + ) 2 ) = 2 ( 1 − 1 + ) ∙ ( −2 (1 + ) 2 ) = −2 (1 + ) 2 2 ( 1 − 1 + ) # = 1 − = 1 + ′ = −1 ′ = 1
Chapter 2 : Differentiation 71 Example 6 APR 2007/ MAT183/ Q1c (5 marks) Given ( ) 2 1 x e y x + = − , show that (1 ) (1 ) 0 2 2 + + + x y = dx dy x . Solution : = − (1 + 2) = Differentiate wrt x (using Quotient Rule) = ′ − ′ 2 = − − (1 + 2 ) − 2 − (1 + 2) 2 = − −− 2 −− 2 − (1 + 2) 2 = − − (1 + 2 + 2) (1 + 2) 2 = − − ( 2 + 2 + 1) (1 + 2) 2 = − − (1 + ) 2 (1 + 2) 2 = − = 1 + 2 ′ = − − ′ = 2 Hence, (1 + 2 ) + (1 + ) 2 = (1 + 2 ) ∙ − − (1 + ) 2 (1 + 2) 2 + (1 + ) 2 ∙ − (1 + 2) (1 + 2 ) + (1 + ) 2 = − − (1 + ) 2 (1 + 2) + − (1 + ) 2 (1 + 2) (1 + 2 ) + (1 + ) 2 = 0 # (ℎ)
MAT183/ MAT421 : Calculus I 72 Example 7/ APR 2011/ MAT183/ Q2a (4 marks) Find dx dy if ( 1) 2 y = x ln x + . Solution : . Example 8/ OCT2009/ MAT183/ Q1c (5 marks) Find dx dy if − + = 1 4 2 3 x e y ln x . Solution : = 2 ( + 1) = Differentiate wrt x (using Product Rule) = ′ + ′ = 2 ( + 1) + 2 + 1 # = ( 3 + 4 2 − 1 ) = ( ) Differentiate wrt x = 1 ( ) ∙ ( ′ − ′ 2 ) = ( ′ − ′ 2 ) = 2 − 1 3 + 4 ( 3 3 ( 2 − 1) − 2( 3 + 4) ( 2 − 1) 2 ) = 1 3 + 4 ( 3 3 ( 2 − 1) − 2( 3 + 4) ( 2 −1) ) = 3 3 ( 2 − 1)− 2( 3 + 4) ( 3 +4)( 2 − 1) # = 3 3 ( 2 − 1) ( 3 + 4)( 2 −1) − 2( 3 + 4) ( 3 +4)( 2 − 1) = 3 31 3 + 4 − 2 2 − 1 # = 3 + 4 = 2 − 1 ′ = 3 3 ′ = 2 = ( 3 + 4 2 − 1 ) Using the properties of logarithms, = ( 3 + 4) − ( 2 − 1) Differentiate wrt x = 1 3 + 4 ∙ 3 3− 1 2 − 1 ∙ 2 = 3 31 3 + 4 − 2 2 − 1 # = 2 = ( + 1) ′ = 2 ′ = 1 + 1
Chapter 2 : Differentiation 73 Example 9/ OCT 2008/ MAT183/ Q2b (6 marks) Differentiate the following functions with respect to x : i) ( ) ( ) 4 3 f x = 7 − x ii) ( ) − = cos x x f x ln 3 2 5 2 Solution : Example 10/ MAR 2012/ MAT183/ Q2a (6 marks) Find dx dy for the following : i) 2 4 3 1 x x x y = + − (2 marks) ii) 2 2 y = x x − (4 marks) Solution : ) = 2 √ − 2 = Differentiate wrt x (using Product Rule) = ′ + ′ = 2√ − 2+ 1 2 2 ( − 2) − 1 2 = 2√ − 2+ 2 2√ −2 # ) () = √(7 − 3) 4 () = (7 − 3 ) 1 4 Using generalized power rule, ′() = 1 4 (7 − 3 ) − 3 4 ∙−3 2 ′() = − 3 4 2 (7 − 3 ) − 3 4 # ) () = [ 2 − 5 2 3 ] Using the properties of logarithms, () = (2 − 5)− (2 3) () = (2 − 5)− ( 3) 2 () = (2 − 5)− 2 ( 3) Differentiate wrt x, ′() = 1 2 −5 ∙ 2 − 2 ∙ 1 3 ∙− 3 ∙ 3 ′() = 2 2 −5 + 6 3 3 ′() = 2 2 −5 + 6 3 # ) = 1 +3 2 − √ 4 = −1 + 3 2 − 1 4 Differentiate wrt x, = − −2 + 6 − 1 4 − 3 4 # = 2 = √ − 2 = ( − 2) 1 2 ′ = 2 ′ = 1 2 ( −2) − 1 2
MAT183/ MAT421 : Calculus I 74 3.0 Implicit Differentiation Simple Examples a) if A = y A’ = dx dy b) if B = 2y B’ = 2 dx dy c) if C = 3y 2 C’ = 6y dx dy d) If D = y = 2 1 y D’ = dx dy • y • − 2 1 2 1 = dx dy 2 y 1 e) If A = xy u = x v = y u’ = 1 v’ = dx dy Using product rule A’ = vu' + uv' A’ = y + x dx dy Procedure to solve implicit differentiation Differentiate each expression one by one (when we differentiate the terms containing y, it must has ) Separating the terms containing and not containing Factorise Solve for (write as a subject)
Chapter 2 : Differentiation 75 f) If B = y x u = x v = y u’ = 1 v’ = dx dy Using quotient rule A’ = 2 v vu' − uv' A’ = 2 y dx dy y − x g) If B = sin3y B’ = 3 cos3y dx dy h) If C = cosec 3y 2 C’ = − cosec 3y 2 cot 3y 2 • 6y• dx dy C’ = −6y cosec 3y 2 cot 3y 2 dx dy i) If A = e 4y A’ = e 4y • 4 dx dy A’ = 4 e4y dx dy j) If B = e tany B’ = e tany •sec2y• dx dy k) If C = ln 2y C’ = dx dy y 2 2 1 • = dx dy y 1
MAT183/ MAT421 : Calculus I 76 More Examples (From Previous Semester Papers) Example 1/ MAR 2004/ MAT183/ Q4a (i) (5 marks) Given y = sin 2x + tan(xy ) 2 3 , find dx dy by implicit differentiation. Solution : 2 = 3 (2) + Let = + Differentiate : ′ = ′ + ′ − − − − − Where, By substituting A’, B’ and C’ into 2 = 6 3 (2) (2) + 2 + 2 Separating the term containing and not containing 2 − 2 = 6 3 (2) (2) + 2 Factorize (2 − 2 ) = 6 3 (2) (2) + 2 Write as a subject = 6 3 (2) (2) + 2 2 − 2 # = 2 Using generalized power rule, ′ = 2 = 3 (2) = ((2)) 3 Using generalized power rule, ′ = 3((2)) 2 ∙ (2) ∙ 2 ′ = 6 3 (2) (2) = ′ = 2 [ ′ + ′] ′ = 2 [ + ] ′ = 2 + 2 = = ′ = 1 ′ =
Chapter 2 : Differentiation 77 Example 2/ OCT 2004/ MAT183/ Q2b (7 marks) Let y be a function of x. If ( ) ( ) y x y + 1 = 3sec 2x − e 2 3 , find dx dy by using implicit differentiation. Solution : ( + 1) 2 = 33 (2) − Let = + Differentiate : ′ = ′ + ′ − − − − − Where, By substituting A’, B’ and C’ into 2( + 1) = 18 3 (2) (2) + − + 2 Multiplying by 2 2 2 ( + 1) = 18 2 3 (2) (2) + 2 ( − + 2 ) 2 2 ( + 1) = 18 2 3 (2) (2) − + 2 2 ( + 1) − = 18 2 3 (2) (2) − (2 2 ( + 1) − ) = 18 2 3 (2) (2) − = 18 2 3 (2) (2) − 2 2( + 1) − # = ( + 1) 2 Using generalized power rule, ′ = 2( + 1) = 33 (2) = 3((2)) 3 Using generalized power rule, ′ = 9((2)) 2 ∙ (2) (2) ∙ 2 ′ = 18 3 (2) (2) = − ′ = − [ ′ − ′ 2 ] ′ = − − 2 ′ = − + 2 = = ′ = 1 ′ =
MAT183/ MAT421 : Calculus I 78 Example 3/ MAR 2005/ MAT183/ Q4a (5 marks) Let y be a function of x. If 2 1 2 e − x + y = xy , find dx dy by using implicit differentiation. Ans : = 2− 2+ Solution : Try This !
Chapter 2 : Differentiation 79 Example 4/ NOV 2005/ MAT183/ Q2b (6 marks) Let y be a function of x. If ( ) xy y = tan x + y + e 4 2 2 , find dx dy by using implicit differentiation. Ans : = 2 2 ( 2+ 2 )+ 43−2 2( 2+2)− Solution : Try This !
MAT183/ MAT421 : Calculus I 80 Example 5/ APR 2006/ MAT183/ Q2b (7 marks) Find dx dy for ( ) y xy x cos x y e 2 3 + 3 = 2 + − by using implicit differentiation. Solution : ( ) y xy x x y e 2 3 + 3 = 2cos + − Let A + B = C + D Differentiate : A' + B' = C' + D' −−−−−−−−− Where, By substituting A’, B’, C’ and D’ into , ( ) ( ) dx dy e dx dy x x y x y dx dy y x 3y + + 6 = − 2sin + − 2sin + − 3 Separating the terms containing dx dy not containing dx dy ( ) x y (x y ) dx dy x y dx dy e dx dy x y + 3 + 2sin + = − 6 − − 2sin + 3 Factorize dx dy ( x e (x y ) ) x y (x y ) dx dy y + 3 + 2sin + = − 6 − − 2sin + 3 Thus, ( ) x e (x y ) x y x y dx dy y + + + − − − + = 3 2sin 6 2sin 3 # = Using product rule, ′ = + = 3 2 Using Power rule, ′ = 6 = 2 ( + ) Using tigonometric rule, ′ = 2 • − ( + ) • (1 + ) ′ = −2 ( + ) (1 + ) Expand, ′ = −2 ( + ) − 2 ( + ) = − 3 Using exponential rule, ′ = − 3 • 3 ′ = −3 3
Chapter 2 : Differentiation 81 Example 6/ OCT 2006/ MAT183/ Q5b (6 marks) Find dx dy for + = y x x y y tan 2 3 3 4 by using implicit differentiation. Solution : 3 2 + 4 3 = ( ) Let + = Differentiate : ′ + ′ = ′ − − − − − Where, By substituting A’, B’ and C’ into 6 + 3 2 + 12 2 = 2 ( ) − 2 ( ) 2 Multiplying by 2 6( 2 ) + 3 2 ( 2 ) + 12 2 ( 2 ) = ( 2 ) ( 2 ( ) − 2 ( ) 2 ) 6 3 + 3 2 2 + 12 4 = 2 ( ) − 2 ( ) 3 2 2 + 12 4 + 2 ( ) = 2 ( ) − 6 3 (3 2 2 + 12 4 + 2 ( )) = 2 ( ) − 6 3 = 2 ( ) − 6 3 3 2 2 + 12 4 + 2 ( ) # = 3 2 Using product rule, ′ = ′ + ′ ′ = 6 + 3 2 = 4 3 ′ = 12 2 = ( ) ′ = 2 ( ) [ ′ − ′ 2 ] ′ = 2 ( ) − 2 ′ = 2 ( ) − 2 ( ) 2 = 3 2 = ′ = 6 ′ = = = ′ = 1 ′ =
MAT183/ MAT421 : Calculus I 82 Example 7/ OCT 2007/ MAT183/ Q2b (6 marks) Find dx dy for 3 ( ) 2 2 2 2 ln x y − x + y = by using implicit differentiation. Solution : 3ln( ) 2 2 2 2 x y − x + y = Let A − B = 2 Differentiate : A' − B' = 0 A' = B' −−−−−−−−− Where, By substituting A’ and B’ into , 2 2 1 6 3 x y dx dy y dx dy x y + + + = 2 2 6 3 1 x y dx dy y xy dx dy y x + + = + Cross multiply = + + + dx dy xy y dx dy 2x y 6y 3x 1 2 = 3 ( 2) Using the properties of log, = 3( 2 + ) = 3(2 + ) = 6 + 3 Differentiate ′ = 6 • 1 + 3 • 1 ′ = 6 + 3 = √2 + 2 = (2 + 2 ) 1 2 ⁄ Using Generalized Power rule, ′ = 1 2 (2 + 2 ) −1 2 ⁄ (2 + 2 ) ′ = 1 2 • 1 √2 + 2 • 2 (1 + ) ′ = 1 + √2 + 2 Expand : 6√2 + 2 + 3√2 + 2 = + 2 Separating the terms containing not containing 3√2 + 2 − 2 = − 6√2 + 2 Factorize (3√2 + 2 − 2 ) = − 6√2 + 2 Thus, = −6√2+2 3√2+2−2 #
Chapter 2 : Differentiation 83 Exercise (Implicit Differentiation) 1. Find dx dy for 4tan x xy sin2y 2 2 − = by using implicit differentiation. 2. Find dx dy for xy = ln(cos x + sin y ) by using implicit differentiation. 3. Find dx dy for ( ) ( ) 2 4 2 2 x e sin 3x ln y x y = + + by using implicit differentiation. 4. Using the implicit differentiation, find dx dy for 2 ( 1) 2 2 2 2 x y + e = sec x + x y . 5. Find dx dy for xe y sin x xy y + = − 2 3 by using implicit differentiation. 6. Find dx dy for ( ) 2 5cos x y 3e 4x y + − = by using implicit differentiation. 7. Let y be a function of x. If ( ) 2 3 4 e ln xy x y = − , find dx dy by using implicit differentiation. 8. Find dx dy for sin(x y ) x 3xy 6 + = + by using implicit differentiation. = 8 2 − 2 2 2 + 2 = − − − + − = 6 3 3 − 4 2 4+ − 2 4+ 2 4+ − 2 = ( + 1) ( + 1) − 4 2 − 2 2 2 2 4 2 + 2 2 22 = − 3 − − 2 −3 − + 2 − = 8 + 5 ( + ) −5 ( + ) − 3 = 2 2−3 − 4 + 3 2−3 = 6 5 + 3 − ( + ) ( + ) − 3
MAT183/ MAT421 : Calculus I 84 4.0 Higher Order Differentiation Given y = f(x) = 2 2 3 2 5 6 4 3 2 x − x + x + x − x + (first derivate) = f' (x) = dx dy (second derivate) = f' ' (x) = dx d y 2 2 (third derivate) = f' ' ' (x) = dx d y 3 3 (fourth derivate) = f' ' ' ' (x) = dx d y 4 4 (fifth derivate) = f' ' ' ' ' (x) = dx d y 5 5 (sixth derivate) = f' ' ' ' ' ' (x) = dx d y 6 6 Example 1/ APR 2010/ MAT183/ Q2a (3 marks) Find the second derivative of y = ln(cos 3x) Solution : = ( 3) = 1 3 ∙ − 3 ∙ 3 = − 3 3 3 = −3 3 2 2 = −3 2 3 ∙ 3 = −9 2 3 #
Chapter 2 : Differentiation 85 Example 2/ SEP 2011/ MAT183/ Q2a (4 marks) Find 2 2 dx d y if ( 1) 2 y = ln x + . Solution : = ( 2 + 1) = 1 2 + 1 ∙ 2 = 2 2 + 1 Differentiate (again) wrt x (using Quotient Rule) 2 2 = ′ − ′ 2 2 2 = 2( 2 + 1) − 2(2) ( 2 + 1) 2 2 2 = 2 2 + 2 − 4 2 ( 2 + 1) 2 2 2 = 2 − 2 2 ( 2 + 1) 2 2 2 = 2(1 − 2 ) ( 2 + 1) 2 # = 2 = 2 + 1 ′ = 2 ′ = 2
MAT183/ MAT421 : Calculus I 86 Example 3/ OCT 2007/ MAT183/ Q1c (5 marks) If x y xe 2 3 − = , prove that 4 4 0 2 2 + + y = dx dy dx d y . Solution : = 3 −2 Using product rule, = ′ + ′ = 3 −2− 6 −2 Using product rule, 2 2 = −6 −2 + ′ + ′ 2 2 = −6 −2− 6 −2 + 12 −2 2 2 = −12 −2 + 12 −2 Therefore, 2 2 + 4 + 4 = −12 −2 + 12 −2 + 4(3 −2− 6 −2 ) + 4(3 −2 ) 2 2 + 4 + 4 = −12 −2 + 12 −2 +12 −2− 24 −2 +12 −2 2 2 + 4 + 4 = 0 # ( ) = 3 = −2 = 3 = −2 −2 = −6 = −2 = −6 = −2 −2
Chapter 2 : Differentiation 87 Example 4/ MAR 2004/ MAT183/ Q5a(i) (5 marks) Given 1 1 2 + = x y , show that ( ) ( ) 3 2 5 2 2 3 2 2 1 1 1 3 + − + + + = x x x xy dx dy dx d y . Solution : = 1 √ 2 + 1 = ( 2 + 1) − 1 2 → 3 = (( 2 + 1) − 1 2) 3 = ( 2 + 1) − 3 2 Using generalized power rule, = − 1 2 ( 2 + 1) − 3 2 ∙ 2 = −( 2 + 1) − 3 2 Using product rule, 2 2 = ′ + ′ 2 2 = −( 2 + 1) − 3 2 + 3 2 ( 2 + 1) − 5 2 Therefore, 2 2 + + 3 = −( 2 + 1) − 3 2 + 3 2 ( 2 + 1) − 5 2 − ( 2 + 1) − 3 2 + ( 2 + 1) − 3 2 2 2 + + 3 = 3 2 ( 2 + 1) − 5 2 − ( 2 + 1) − 3 2 2 2 + + 3 = 3 2 √( 2 + 1) 5 − 1 √( 2 + 1) 3 # ( ) = − = ( 2 + 1) − 3 2 = −1 = − 3 2 ( 2 + 1) − 5 2 ∙ 2 = −3( 2 + 1) − 5 2
MAT183/ MAT421 : Calculus I 88 5.0 Linear Approximations and Differentials Example 1/ APR 2011/ MAT183/ Q3b (5 marks) Use differential to estimate the value of √36.02 and give your answer correct to four decimal places. Solution : Hot Tips!! x0 = nearest whole number that gives the answer as a whole number when f(x0) is solved x = small changes (less than 1) f(x0) should be the whole number f ’(x0) should be the whole number or fraction (never write in decimals) √36.02 = √ = √0 + ∆ = √36 + 0.02 find , ∆ = 36.02, 0 = 36, ∆ = +0.02 find () ( ) () = √ (0 ) = (36) = √36 = 6 find ′() ′( ) () = √ = 1 2 ′ () = 1 2 − 1 2 = 1 2√ ′ (0 ) = ′ (36) = 1 2√36 = 1 2(6) = 1 12 Estimate/ Approximate √36.02 ≈ (0 ) + ′(0 ) ∙ ∆ ≈ 6 + ( 1 12) (0.02) ≈ 6.0017 # (4 . ) Procedure to estimate/approximate using differentials : Identify the suitable values for 0 and Identify () and find the value of (0 ) Find ′() and the value of ′(0 ) Use (0 + ) ≈ (0 ) +′(0 ) to estimate/approximate the given question () ≈ ( + ∆) ≈ ( ) + ′( ) ∙ ∆
Chapter 2 : Differentiation 89 Example 2/ APR 2009/ MAT183/ Q3a (5 marks) Given ( ) 2 2 3 x f x + = . If x changes from 1 to 1.05, estimate the value of f(1.05) using differentials. Give your answer correct to THREE (3) decimal places. [Hint : (0 + ) ≈ (0 ) + ′(0 )] Solution : 1.05 = = 0 + ∆ = 1 + 0.05 find , ∆ = 1.05, 0 = 1, ∆ = +0.05 find () ( ) () = 3 2 + 2 (0 ) = (1) = 3 2 + (1) 2 = 1 find ′() ′( ) () = 3 2 + 2 = 3(2 + 2 ) −1 ′ () = −3(2 + 2 ) −2 ∙ 2 = −6 (2 + 2) 2 ′ (0 ) = ′ (1) = −6(1) (2 + (1) 2) 2 = − 6 9 = − 2 3 Estimate/ Approximate (1.05) ≈ (0 ) + ′(0 ) ∙ ∆ ≈ 1 + (− 2 3 ) (0.05) ≈ 0.967 # (3 . )
MAT183/ MAT421 : Calculus I 90 Example 3/ OCT 2010/ MAT183/ Q3a (5 marks) Use differential to estimate + 0 01 5 sin . correct to four (4) decimal places. Solution : ( 5 + 0.01) = = 0 + ∆ = 5 + 0.01 find , ∆ = 5 + 0.01, 0 = 5 , ∆ = +0.01 find () ( ) () = (0 ) = ( 5 ) = ( 5 ) find ′() ′( ) () = ′ () = ′ (0 ) = ′ ( 5 ) = ( 5 ) Estimate/ Approximate ( 5 + 0.01) ≈ (0 ) + ′(0 ) ∙ ∆ ≈ ( 5 ) + ( 5 ) (0.01) ≈ 0.5959# (3 . )
Chapter 2 : Differentiation 91 Example 4/ APR 2008/ MAT183/ Q3a (5 marks) Use differential to approximate √1 + (1.1) correct to two (2) decimal places. Solution : (1.1) = = 0 + ∆ = 1 + 0.1 find , ∆ = 1.1, 0 = 1, ∆ = +0.1 find () ( ) () = √1 + (0 ) = (1) = √1 + (1) = √1 + 0 = 1 find ′() ′( ) () = √1 + = (1 + ) 1 2 ′ () = 1 2 (1 + ) − 1 2 ∙ 1 = 1 2√1 + ′ (0 ) = ′ (1) = 1 2(1)√1 + (1) = 1 2 Estimate/ Approximate √1 + (1.1) ≈ (0 ) + ′(0 ) ∙ ∆ ≈ 1 + ( 1 2 ) (0.1) ≈ 1.05 # (2 . ) The best 0 is 1 because it gives (0 ) a whole number, 1 x = the given decimal number. 0 = the nearest whole number ∆ = small changes and between (−0.9999 to 0 .9999)
MAT183/ MAT421 : Calculus I 92 Example 5/ SEP 2011/ MAT183/ Q3b (5 marks) Given () = √20 − 2, use differential to estimate the value of √20 − (1.99) 2. Give your answer correct to three (3) decimal places. Solution : (1.99) = = 0 + ∆ = 2 − 0.01 = 2 + (−0.01) find , ∆ = 1.99, 0 = 2, ∆ = −0.01 find () ( ) () = √20 − 2 (0 ) = (2) = √20 − (2) 2 = √16 = 4 find ′() ′( ) () = √20 − 2 = (20 − 2 ) 1 2 ′ () = 1 2 (20 − 2 ) − 1 2 ∙ −2 = − √20 − 2 ′ (0 ) = ′ (2) = −2 √20 − 2 2 = − 1 2 Estimate/ Approximate √20 − (1.99) 2 ≈ (0 ) + ′(0 ) ∙ ∆ ≈ 4 + (− 1 2 ) (−0.01) ≈ 4.005 # (3 . ) The best 0 is 2 because it gives (0 ) a whole number, 4 x = the given decimal number. 0 = the nearest whole number ∆ = small changes and between (−0.9999 to 0 .9999)
Chapter 2 : Differentiation 93 Example 6/ MAR 2012/ MAT183/ Q3a (5 marks) Use differential to estimate the value of √7.96 3 and give your answer correct to four (4) decimal places. Solution : Try This !
MAT183/ MAT421 : Calculus I 94 END OF CHAPTER 2 Example 7/ OCT 2012/ MAT183/ Q3b (5 marks) Given () = √4 2 − 7. Estimate the value of (2.01) using differential. Give your answer correct to three (3) decimal places. Solution : Try This !
Chapter 3 Applications of Differentiation 95 CHAPTER 3 APPLICATIONS OF DIFFERENTIATION List of Topics 3.1 dx dy as a slope/gradient of the tangent line to the curve y = f(x). 3.2 The graph of polynomial Functions • x-intercept and y-intercept • critical/stationary/turning points • the interval where f(x) is increasing and decreasing and the extremum point(s) • the interval where f(x) is concave up or concave down and the inflection points • based on the above answer sketch the graph of polynomial Functions 3.3 The graph of rational functions (linear/linear) • x-intercept and y-intercept • vertical and horizontal asymptotes • the interval where f(x) is increasing and decreasing and the extremum point(s) • the interval where f(x) is concave up or concave down and the inflection point(s) • based on the above answer sketch the graph of rational Functions 3.4 Maximum and Minimum Problems 3.5 Related Rates (Rate of Change) 3.6 Mean-value Theorem of Differentiation