Lecture 5 - 1 CC2053: Calculus 2 Introduction to Calculus...

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Unformatted text preview: 1 CC2053: Calculus 2 Introduction to Calculus and Linear Algebra 2011/2012 Semester 1 CALCULUS: Techniques of Differentiation 2 CC2053: Calculus 2 Techniques of Differentiation 1. The Constant Rule For any constant c , . 2. The Power Rule For any real number n , . 3. The Constant Multiple Rule If c is a constant and f ( x ) is differentiable, then so is cf ( x ) and 4. The Sum Rule If f ( x ) and g ( x ) are differentiable, then so is the sum S ( x ) = f ( x ) + g ( x ) and S' ( x ) = f ' ( x ) + g' ( x ) ; that is, ) ( = c dx d ) ( = c dx d 1 ) (- = n n nx x dx d 1 ) (- = n n nx x dx d dx df c x cf dx d = )) ( ( dx df c x cf dx d = )) ( ( dx dg dx df x g x f dx d + = + )] ( ) ( [ dx dg dx df x g x f dx d + = + )] ( ) ( [ 3 CC2053: Calculus 2 Techniques of Differentiation – Example 1 Evaluate the followings: a. b. c. d. e. f. ) ( 3 2 x dx d ) 15 (- dx d ) ( 7 x dx d 5 1 x dx d ) 3 ( 4 x dx d ) 7 ( x dx d- 4 CC2053: Calculus 2 Techniques of Differentiation – Example 1 (solution) a. b. c. d. e. f. 3 1 3 2 3 2 ) ( ) ( 1 3 2 3 2 3 2-- = = = x x x dx d x dx d ) 15 ( =- dx d 6 1 7 7 7 7 ) ( x x x dx d = =- ( ) 6 1 5 5 5 5 5 1----- =- = = x x x dx d x dx d 3 3 4 4 12 ) 4 ( 3 ) ( 3 ) 3 ( x x x dx d x dx d = = = 2 3 2 3 2 1 2 7 ) 2 1 ( 7 ) 7 ( ) 7 (--- =-- =- =- x x x dx d x dx d 5 CC2053: Calculus 2 Techniques of Differentiation - Example 2 Differentiate a. b. c. 7 2 + =- x y 8 12 4 5 2 3- +- = x x x y 7 5 3 2-- = x x y 6 CC2053: Calculus 2 Techniques of Differentiation – Example 2 (solution) 3 3 2 2 2 2 ) 7 ( ) ( ) 7 (----- = +- = + = + = x x dx d x dx d x dx d dx dy 12 8 15 ) 8 12 4 5 ( 2 2 3 +- =- +- = x x x x x dx d dx dy 8 4 8 4 7 5 7 5 21 10 ) 7 ( 3 ) 5 ( 2 ) ( 3 ) ( 2 ) 3 2 (---- + =-- =- =- = x x x x x dx d x dx d x x dx d dx dy a) b) c) 7 CC2053: Calculus 2 Techniques of Differentiation – Example 3 It is estimated that x months from now, the population of a certain community will be (a) At what rate will the population be changing with respect to time 15 months from now? (b) By how much will the population actually change during the 16 th month? 000 , 8 20 ) ( 2 + + = x x x P 8 CC2053: Calculus 2 Techniques of Differentiation – Example 3 (solution) a. The rate of change of the population 15 months from now will be 50 people per month b. The actual population change during the 16 th month is people 20 2 ) ( + = ′ x x P 50 20 ) 15 ( 2 ) 15 ( = + = ′ P 51 525 , 8 576 , 8 ) 15 ( ) 16 ( =- =- P P 525 , 8 000 , 8 ) 15 ( 20 ) 15 ( ) 15 ( 2 = + + = P 576 , 8 000 , 8 ) 16 ( 20 ) 16 ( ) 16 ( 2 = + + = P 9 CC2053: Calculus 2 Techniques of Differentiation – Percentage Rate of Change ) ( ) ( 100 of size of change of rate 100 of change of rate percentage x Q x Q Q Q Q ′ = = ) ( ) ( 100 of size of change of rate 100 of change of rate percentage x Q x Q Q Q Q ′ = = 10 CC2053: Calculus 2 Techniques of Differentiation – Example 4 The gross domestic product (GDP) of a certain country...
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Lecture 5 - 1 CC2053: Calculus 2 Introduction to Calculus...

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