Exam 2 - Jolley, Garrett Exam 2 Due: Oct 31 2007, 1:00 am...

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Unformatted text preview: Jolley, Garrett Exam 2 Due: Oct 31 2007, 1:00 am Inst: R Heitmann 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find f ( x ) when f ( x ) = x- 1 x + 1 2 . 1. f ( x ) = 6( x + 1) ( x- 1) 3 2. f ( x ) =- 4( x- 2) ( x + 1) 3 3. f ( x ) = 6( x- 2) ( x + 1) 3 4. f ( x ) =- 4( x + 1) ( x- 1) 3 5. f ( x ) = 4( x- 1) ( x + 1) 3 correct 6. f ( x ) =- 6( x + 2) ( x- 1) 3 Explanation: By the Chain and Quotient Rules, f ( x ) = 2 x- 1 x + 1 ( x + 1)- ( x- 1) ( x + 1) 2 . Consequently, f ( x ) = 4( x- 1) ( x + 1) 3 . keywords: derivative, quotient rule, chain rule 002 (part 1 of 1) 10 points Determine f ( x ) when f ( x ) = 2 x sin x . 1. f ( x ) = 2 n cos x x- sin x o 2. f ( x ) = 1 2 n cos x x- sin x o 3. f ( x ) = cos x x- sin x 4. f ( x ) = 1 2 n sin x x + cos x o 5. f ( x ) = 2 n sin x x + cos x o 6. f ( x ) = sin x x + cos x correct Explanation: By the Chain and Product Rules, f ( x ) = 2 2 x sin x + 2 x 2 x cos x . Consequently, f ( x ) = sin x x + cos x . keywords: chain rule, product rule, trig func- tion 003 (part 1 of 1) 10 points Find the second derivative of f when f ( x ) = cos2 x + 3sin 2 x . 1. f 00 ( x ) = cos2 x 2. f 00 ( x ) =- cos2 x 3. f 00 ( x ) =- 2cos2 x 4. f 00 ( x ) = sin2 x 5. f 00 ( x ) = 2cos2 x correct Jolley, Garrett Exam 2 Due: Oct 31 2007, 1:00 am Inst: R Heitmann 2 6. f 00 ( x ) = 2sin2 x Explanation: Differentiating once we see that f ( x ) =- 2sin2 x + 6sin x cos x . Now 2sin x cos x = sin2 x , so f ( x ) = sin2 x . Consequently, by differentiating once again we obtain f 00 ( x ) = 2cos2 x . keywords: second derivative, trig function 004 (part 1 of 1) 10 points Determine the third derivative, f 000 ( x ), of f when f ( x ) = 2 x + 1 . 1. f 000 ( x ) = (2 x + 1)- 3 / 2 2. f 000 ( x ) =- 3(2 x + 1)- 5 / 2 3. f 000 ( x ) = (2 x + 1)- 5 / 2 4. f 000 ( x ) = 3(2 x + 1)- 5 / 2 correct 5. f 000 ( x ) =- 3(2 x + 1)- 3 / 2 6. f 000 ( x ) =- (2 x + 1)- 3 / 2 Explanation: To usetheChainRulesuccessively itsmore convenient to write f ( x ) = 2 x + 1 = (2 x + 1) 1 / 2 . For then f ( x ) = 1 2 2 (2 x + 1)- 1 / 2 = (2 x + 1)- 1 / 2 , while f 00 ( x ) =- 1 2 2 (2 x + 1)- 3 / 2 =- (2 x + 1)- 3 / 2 , and f 000 ( x ) = 3 2 2 (2 x + 1)- 5 / 2 . Consequently, f 000 ( x ) = 3(2 x + 1)- 5 / 2 . keywords: third derivative, chain rule 005 (part 1 of 1) 10 points Find dy/dx when 5 x 2 + 3 y 2 = 9 . 1. dy dx = 5 x 3 y 2. dy dx = 3 xy 3....
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This note was uploaded on 10/29/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.

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Exam 2 - Jolley, Garrett Exam 2 Due: Oct 31 2007, 1:00 am...

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