practiceExam1Solutions

# practiceExam1Solutions - MATH 2200 Fall 2009 Exam 1(1 Find...

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Unformatted text preview: MATH 2200, Fall 2009 Exam 1 (1) Find the derivatives of the following functions: (a) f ( x ) = (2 x 4- x 3 + 7 x ) 7 Solution f ( x ) = 7(2 x 4- x 3 + 7 x ) 6 d dx (2 x 4- x 3 + 7 x ) = 7(2 x 4- x 3 + 7 x ) 6 (8 x 3- 3 x 2 + 7) (b) f ( x ) = (3 x 3- 8 x 2 )(1 + x + x 2 ) 4 x- 1 Solution f ( x ) = (4 x- 1) d dx ( (3 x 3- 8 x 2 )(1 + x + x 2 ) )- (3 x 3- 8 x 2 )(1 + x + x 2 ) d dx ( 4 x- 1 ) (4 x- 1) 2 = (4 x- 1) ( (3 x 3- 8 x 2 ) d dx (1 + x + x 2 ) + (1 + x + x 2 ) d dx (3 x 3- 8 x 2 ) )- (3 x 3- 8 x 2 )(1 + x + x 2 )(4) (4 x- 1) 2 = (4 x- 1) ( (3 x 3- 8 x 2 )(1 + 2 x ) + (1 + x + x 2 )(9 x 2- 16 x ) )- (3 x 3- 8 x 2 )(1 + x + x 2 )(4) (4 x- 1) 2 (c) f ( x ) = 3 x- 3 √ x 2 + 1 Solution f ( x ) = ( √ x 2 + 1 ) d dx (3 x- 3)- (3 x- 3) d dx ( √ x 2 + 1 ) x 2 + 1 = ( √ x 2 + 1 ) (3)- (3 x- 3) 1 2 ( x 2 + 1)- 1 2 d dx ( x 2 + 1) x 2 + 1 = ( √ x 2 + 1 ) (3)- (3 x- 3) 1 2 ( x 2 + 1)- 1 2 (2 x ) x 2 + 1 (d) f ( x ) = q x 2 + √ 1- x Solution f ( x ) = 1 2 ( x 2 + √ 1- x )- 1 2 d dx ( x 2 + √ 1- x ) = 1 2 ( x 2 + √ 1- x )- 1 2 2 x + 1 2 (1- x )- 1 2 d dx (1- x ) = 1 2 ( x 2 + √ 1- x )- 1 2 2 x + 1 2 (1- x )- 1 2 (- 1) (2) Use the definition of the derivative to find the derivatives of the following functions: (a) f ( x ) = 2 x + 4 Solution f ( x ) = lim h → f ( x + h )- f ( x ) h = lim h → ( 2( x + h ) + 4 )- (2 x + 4) h = lim h → 2 x + 2 h + 4- 2 x- 4 h = lim h → 2 h h = lim h → 2 = 2 So,...
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practiceExam1Solutions - MATH 2200 Fall 2009 Exam 1(1 Find...

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