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# HW10S - Gauthier Joseph Homework 10 Due Nov 1 2007 3:00 am...

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Gauthier, Joseph – Homework 10 – Due: Nov 1 2007, 3:00 am – Inst: JEGilbert 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 Determine f x - f y when f ( x, y ) = 2 x 2 + 2 xy - 4 y 2 - 3 x - 2 y . 1. f x - f y = 6 x + 10 y - 5 2. f x - f y = 6 x - 6 y - 1 3. f x - f y = 6 x - 6 y - 5 4. f x - f y = 2 x - 6 y - 1 5. f x - f y = 2 x + 10 y - 5 6. f x - f y = 2 x + 10 y - 1 correct Explanation: After differentiation we see that f x = 4 x + 2 y - 3 , f y = 2 x - 8 y - 2 . Consequently, f x - f y = 2 x + 10 y - 1 . keywords: partial derivative, first order par- tial derivative, polynomial 002 (part 1 of 1) 10 points Determine f x when f ( x, y ) = cos(2 y - x ) - x sin(2 y - x ) . 1. f x = - cos(2 y - x ) - x sin(2 y - x ) 2. f x = - 2 sin(2 y - x ) - x cos(2 y - x ) 3. f x = x sin(2 y - x ) 4. f x = - x cos(2 y - x ) 5. f x = x cos(2 y - x ) - sin(2 y - x ) 6. f x = 2 sin(2 y - x ) - x cos(2 y - x ) 7. f x = - x sin(2 y - x ) 8. f x = x cos(2 y - x ) correct Explanation: From the Product Rule we see that f x = sin(2 y - x ) - sin(2 y - x )+ x cos(2 y - x ) . Consequently, f x = x cos(2 y - x ) . partial derivative, first order partial deriva- tive, trig function, keywords: 003 (part 1 of 1) 10 points Determine f x when f ( x, y ) = ( xy - 3) e - xy . 1. f x = - y (2 - xy ) e - xy 2. f x = - y (2 + xy ) e - xy 3. f x = y ( xy - 4) e - xy 4. f x = x (4 - 3 xy ) e - xy 5. f x = y (4 - xy ) e - xy correct 6. f x = x (3 xy - 4) e - xy 7. f x = - x (2 + 3 xy ) e - xy 8. f x = - x (2 + xy ) e - xy Explanation: From the Product Rule we see that f x = ye - xy - y ( xy - 3) e - xy .

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Gauthier, Joseph – Homework 10 – Due: Nov 1 2007, 3:00 am – Inst: JEGilbert 2 Consequently, f x = y (4 - xy ) e - xy . keywords: partial derivative, first order par- tial derivative, exp function, 004 (part 1 of 1) 10 points Determine h = h ( x, y ) so that ∂f ∂x = h ( x, y ) ( x 2 + 5 y 2 ) 2 when f ( x, y ) = 4 x 2 y x 2 + 5 y 2 . 1. h ( x, y ) = 20 xy 3 2. h ( x, y ) = 40 xy 2 3. h ( x, y ) = 20 x 3 y 4. h ( x, y ) = 40 x 3 y 5. h ( x, y ) = 40 xy 3 correct 6. h ( x, y ) = 20 xy 2 Explanation: Differentiating with respect to x using the quotient rule we obtain ∂f ∂x = 8 xy ( x 2 + 5 y 2 ) - 8 x 3 y ( x 2 + 5 y 2 ) 2 .
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