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Unformatted text preview: roberts (eer474) – Quest HW 5 – seckin – (56425) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the derivative of f when f ( x ) = x 2 sin x + 2 x cos x . 1. f ′ ( x ) = ( x 2 + 2) sin x 2. f ′ ( x ) = ( x 2 + 2) sin x 3. f ′ ( x ) = ( x 2 + 2) cos x 4. f ′ ( x ) = ( x 2 + 2) cos x correct 5. f ′ ( x ) = ( x 2 2) cos x 6. f ′ ( x ) = ( x 2 2) sin x Explanation: By the Product Rule, f ′ ( x ) = 2 x sin x + x 2 cos x + 2 cos x 2 x sin x . Consequently, f ′ ( x ) = ( x 2 + 2) cos x . 002 10.0 points Find the value of f ′ (0) when f ( x ) = (1 3 x ) − 4 . Correct answer: 12. Explanation: Using the chain rule and the fact that ( x α ) ′ = α x α − 1 , we obtain f ′ ( x ) = 12 (1 3 x ) 5 . At x = 0, therefore, f ′ (0) = 12 . 003 10.0 points Determine f ′ ( x ) when f ( x ) = x 2 / 3 ( x 2) 1 / 3 . 1. f ′ ( x ) = 3 x 4 3 x 2 / 3 ( x 2) 1 / 3 2. f ′ ( x ) = 3 x 4 3 x 1 / 3 ( x 2) 2 / 3 correct 3. f ′ ( x ) = x 2 3 x 1 / 3 ( x 2) 2 / 3 4. f ′ ( x ) = 3 x + 2 3 x 2 / 3 ( x 2) 1 / 3 5. f ′ ( x ) = 3 x 2 3 x 2 / 3 ( x 2) 1 / 3 6. f ′ ( x ) = x + 4 3 x 1 / 3 ( x 2) 2 / 3 Explanation: By the Chain and Product Rules, f ′ ( x ) = 2 3 x − 1 / 3 ( x 2) 1 / 3 + 1 3 x 2 / 3 ( x 2) − 2 / 3 = 1 3 braceleftBigg 2( x 2) 1 / 3 x 1 / 3 + x 2 / 3 ( x 2) 2 / 3 bracerightBigg . Bringing the right hand side to a common denominator we thus see that f ′ ( x ) = 2( x 2)+ x 3 x 1 / 3 ( x 2) 2 / 3 . roberts (eer474) – Quest HW 5 – seckin – (56425) 2 Consequently, f ′ ( x ) = 3 x 4 3 x 1 / 3 ( x 2) 2 / 3 ....
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This note was uploaded on 11/08/2010 for the course MATH 408K taught by Professor Gualdani during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Gualdani
 Differential Calculus

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