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Exam 2 - Jolley Garrett – Exam 2 – Due 1:00 am – Inst...

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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 it’smore 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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Exam 2 - Jolley Garrett – Exam 2 – Due 1:00 am – Inst...

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