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M408K Hm. 7 Solutions

# M408K Hm. 7 Solutions - Mahon Kevin Homework 7 Due 3:00 am...

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Mahon, Kevin – Homework 7 – Due: Oct 13 2005, 3:00 am – Inst: Edward Odell 1 This print-out should have 21 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 the derivative of f when f ( x ) = 3 x sin 5 x + 3 5 cos 5 x. 1. f 0 ( x ) = 15 x cos 5 x + 6 sin 5 x 2. f 0 ( x ) = - 15 x cos 5 x 3. f 0 ( x ) = 15 x cos 5 x correct 4. f 0 ( x ) = 15 x cos 5 x - 6 sin 5 x 5. f 0 ( x ) = 15 cos 5 x Explanation: Since d dx sin x = cos x, d dx cos x = - sin x, it follows that f 0 ( x ) = 3 sin 5 x + 15 x cos 5 x - 3 sin 5 x. Consequently, f 0 ( x ) = 15 x cos 5 x . keywords: Stewart5e, 002 (part 1 of 1) 10 points Find f 0 ( x ) when f ( x ) = x + 1 x - 1 · 2 . 1. f 0 ( x ) = - 6( x + 2) ( x - 1) 3 2. f 0 ( x ) = 4( x - 1) ( x + 1) 3 3. f 0 ( x ) = - 4( x + 1) ( x - 1) 3 correct 4. f 0 ( x ) = 6( x - 2) ( x + 1) 3 5. f 0 ( x ) = - 4( x + 2) ( x - 1) 3 6. f 0 ( x ) = 6( 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: Stewart5e, 003 (part 1 of 1) 10 points Determine f 0 ( x ) when f ( x ) = 1 + 2 x 1 - x 2 . 1. f 0 ( x ) = - 2 + x (1 - x 2 ) 3 / 2 2. f 0 ( x ) = - 2 - x (1 - x 2 ) 1 / 2 3. f 0 ( x ) = 2 + x (1 - x 2 ) 3 / 2 correct 4. f 0 ( x ) = - 2 + x (1 - x 2 ) 1 / 2 5. f 0 ( x ) = 2 - x (1 - x 2 ) 3 / 2 6. f 0 ( x ) = 2 + x (1 - x 2 ) 1 / 2

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Mahon, Kevin – Homework 7 – Due: Oct 13 2005, 3:00 am – Inst: Edward Odell 2 Explanation: By the Product and Chain Rules, f 0 ( x ) = 2 (1 - x 2 ) 1 / 2 + 2 x (1 + 2 x ) 2(1 - x 2 ) 3 / 2 = 2(1 - x 2 ) + x (1 + 2 x ) (1 - x 2 ) 3 / 2 . Consequently, f 0 ( x ) = 2 + x (1 - x 2 ) 3 / 2 . (Note: the Quotient Rule could have been used, but it’s simpler to use the Product Rule.) keywords: Stewart5e, 004 (part 1 of 1) 10 points Determine f 0 ( x ) when f ( x ) = 3 cos 2 x - 2 sin 2 x . 1. f 0 ( x ) = 6 cos x + 4 sin x 2. f 0 ( x ) = - 6 sin x + 4 cos x 3. f 0 ( x ) = - 10 sin x cos x correct 4. f 0 ( x ) = 10 sin x cos x 5. f 0 ( x ) = - 6 sin x - 4 cos x 6. f 0 ( x ) = 2 sin x cos x 7. f 0 ( x ) = - 2 sin x cos x Explanation: Since d dx sin x = cos x , d dx cos x = - sin x , the Chain Rule ensures that f 0 ( x ) = - 6 cos x sin x - 4 sin x cos x . Consequently, f 0 ( x ) = - 10 sin x cos x . keywords: Stewart5e, 005 (part 1 of 1) 10 points Find the value of f 0 (1) when f ( x ) = ( x 2 + 3) 1 / 2 + 3 x . 1. f 0 (1) = - 3 2. f 0 (1) = - 4 3. f 0 (1) = - 9 2 4. f 0 (1) = - 7 2 5. f 0 (1) = - 5 2 correct Explanation: Using the Chain Rule and the fact that d dx x r = r x r - 1 holds for all values of r , we see that f 0 ( x ) = x ( x 2 + 3) 1 / 2 - 3 x 2 . At x = 1, therefore, f 0 (1) = - 5 2 . keywords: Stewart5e, 006 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 1 3 (5 - 3 x ) 5 + 6 x .
Mahon, Kevin – Homework 7 – Due: Oct 13 2005, 3:00 am – Inst: Edward Odell 3 1. f 0 ( x ) = - 9 x 5 + 3 x 2. f 0 ( x ) = - 9 x 5 + 6 x correct 3. f 0 ( x ) = - 3 x 5 + 6 x 4. f 0 ( x ) = 3 x 5 + 6 x 5. f 0 ( x ) = 9 x 5 + 6 x Explanation: By the Product and Chain Rules, f 0 ( x ) = - 5 + 6 x + 5 - 3 x 5 + 6 x .

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