# Exam2_2 1 - 5. f ( x ) = 1 2 n cos x x-sin x o 6. f ( x ) =...

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Taylor, Douglas – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: JEGilbert 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points ±ind f 0 ( x ) when f ( x ) = x - 1 x + 1 · 2 . 1. f 0 ( x ) = 4( x - 1) ( x + 1) 3 correct 2. f 0 ( x ) = - 4( x - 2) ( x + 1) 3 3. f 0 ( x ) = 6( x - 2) ( x + 1) 3 4. f 0 ( x ) = - 6( x + 2) ( x - 1) 3 5. f 0 ( x ) = 6( x + 1) ( x - 1) 3 6. f 0 ( x ) = - 4( x + 1) ( x - 1) 3 Explanation: By the Chain and Quotient Rules, f 0 ( x ) = 2 x - 1 x + 1 · ( x + 1) - ( x - 1) ( x + 1) 2 . Consequently, f 0 ( x ) = 4( x - 1) ( x + 1) 3 . keywords: derivative, quotient rule, chain rule 002 (part 1 oF 1) 10 points Determine f 0 ( x ) when f ( x ) = x sin x . 1. f 0 ( x ) = 2 n sin x x + cos x o 2. f 0 ( x ) = cos x x - sin x 3. f 0 ( x ) = sin x x + cos x 4. f 0 ( x ) = 1 2 n sin x x + cos x o correct
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Unformatted text preview: 5. f ( x ) = 1 2 n cos x x-sin x o 6. f ( x ) = 2 n cos x x-sin x o Explanation: By the Chain and Product Rules, f ( x ) = 1 2 x sin x + x 2 x cos x . Consequently, f ( x ) = 1 2 n sin x x + cos x o . keywords: chain rule, product rule, trig Func-tion 003 (part 1 oF 1) 10 points ind the second derivative oF f when f ( x ) = 3cos2 x-sin 2 x . 1. f 00 ( x ) =-7cos2 x 2. f 00 ( x ) =-14sin2 x 3. f 00 ( x ) =-7sin2 x 4. f 00 ( x ) =-14cos2 x correct 5. f 00 ( x ) = 7cos2 x...
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