practice1-solutions

practice1-solutions - Math 2250, Spring 2011, Practice Exam...

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Math 2250, Spring 2011, Practice Exam 1 Computation Find the derivatives of the following functions. (1) f ( x ) = 3 x 12 - 4 x 5 - 3 x + 1 solution. Using the power rule: d dx ( 3 x 12 - 4 x 5 - 3 x + 1 ) = 3(12) x 11 - 4(5) x 4 - 3 = 36 x 11 - 20 x 4 - 3 ± (2) f ( x ) = sec( e x + 3 x 2 ) solution. Using sec( x ) = 1 cos ( x ) = ( cos ( x )) - 1 : d dx ( sec( e x + 3 x 2 ) ) = d dx ( cos( e x + 3 x 2 ) ) - 1 = - 1 ( cos( e x + 3 x 2 ) ) - 2 d dx ( e x + 3 x 2 ) = - 1 ( cos( e x + 3 x 2 ) ) - 2 ( e x + 6 x ) or, rewriting a little (optional on this exam) = - sec 2 ( e x + 3 x 2 ) ( e x + 6 x ) ± (3) f ( x ) = sin ( ( x + 1 + x )(3 x 3 - 2 x ) 8 ) solution. d dx sin ( ( x + 1 + x )(3 x 3 - 2 x ) 8 ) = cos ( ( x + 1 + x )(3 x 3 - 2 x ) 8 ) d dx h ( x + 1 + x )(3 x 3 - 2 x ) 8 i = cos ( ( x + 1 + x )(3 x 3 - 2 x ) 8 ) h ( x + 1 + x ) d dx (3 x 3 - 2 x ) 8 + ± d dx ( x + 1 + x ) ² (3 x 3 - 2 x ) 8 i = cos ( ( x + 1 + x )(3 x 3 - 2 x ) 8 ) h ( x + 1 + x )8(3 x 3 - 2 x ) 7 d dx (3 x 3 - 2
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This note was uploaded on 12/10/2011 for the course MATH 2250 taught by Professor Chestkofsky during the Spring '08 term at UGA.

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practice1-solutions - Math 2250, Spring 2011, Practice Exam...

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