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assiganment1solution - MAT 1339 C Assignment 1(Due Wed Sept...

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MAT 1339 C Assignment 1 (Due Wed. Sept. 29th, 5:30 pm) Student Number : Name: Problem 1: [1 mark] Using the definition of a derivative find f 0 ( x ) if f ( x ) = x 3 - 2 x + 2010. Solution: f 0 ( x ) = lim h 0 f ( x + h ) - f ( x ) h = lim h 0 ( x + h ) 3 - 2( x + h ) + 2010 - ( x 3 - 2 x + 2010) h = lim h 0 (3 xh + 3 x 2 + h 2 - 2) h h = lim h 0 3 xh + 3 x 2 + h 2 - 2 = 3 x 2 - 2 . Problem 2: [1 mark] Using the rules of differentiation find the derivative of g ( x ) = 2 x 2010 - 1 2 x 2000 + 10 x 6 . Solution: f 0 ( x ) = (2 x 2010 ) 0 - ( 1 2 x 2000 ) 0 + ( 10 x 6 ) 0 = 2 · 2010 · x 2010 - 1 - 1 2 · 2000 · x 2000 - 1 + 10 · ( - 6) · x - 6 - 1 = 4020 x 2009 - 1000 x 1999 - 60 x 7 .
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Problem 3: [1 mark] If g ( x ) = 2 x 6 - 12 x 3 and f ( x ) = 12 x 3 + 4 x 4 find the derivative of f ( x ) g ( x ) . Solution: f ( x ) g ( x ) · 0 = 12 x 3 + 4 x 4 2 x 6 - 12 x 3 · 0 = (12 + 4 x ) x 3 (2 x 3 - 12) x 3 · 0 = 12 + 4 x 2 x 3 - 12 · 0 = (12 + 4 x ) 0 (2 x 3 - 12) - (12 + 4 x )(2 x 3 - 12) 0 (2 x 3 - 12) 2 = 4(2 x 3 - 12) - (12 + 4 x ) · 6 x 2 (2 x 3 - 12) 2 = - 4 x 3 - 18 x 2 - 12 ( x 3 - 6) 2 . Problem 4: [1 mark] If f ( x ) = 2 x - x 2010 and g ( x ) = x 23 - 2010 + x 22 find the derivative of f ( x ) g ( x ). Solution: ( f ( x ) g ( x )) 0 = ( (2 x - x 2010 )( x 23 - 2010 + x 22 ) ) 0 = (2 x - x 2010 ) 0 ( x 23 - 2010 + x 22 ) + (2 x - x 2010 )( x 23 - 2010 + x 22 ) 0 = (2 - 2010 x 2009 )( x 23 - 2010 + x 22 ) + (2 x - x 2010 )(23 x 22 + 22 x 21 ) .
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Problem 5: [2 marks] If f ( x ) =
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