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M408K Hm. 14 Solutions

# M408K Hm. 14 Solutions - Mahon Kevin Homework 14 Due Dec 1...

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Mahon, Kevin – Homework 14 – Due: Dec 1 2005, 3:00 am – Inst: Edward Odell 1 This print-out should have 17 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 Differentiate the function f ( x ) = cos(ln 4 x ) . 1. f 0 ( x ) = sin(ln 4 x ) x 2. f 0 ( x ) = - sin(ln 4 x ) 3. f 0 ( x ) = 1 cos(ln 4 x ) 4. f 0 ( x ) = - sin(ln 4 x ) x correct 5. f 0 ( x ) = 4 sin(ln 4 x ) x 6. f 0 ( x ) = - 4 sin(ln 4 x ) x Explanation: By the Chain Rule f 0 ( x ) = - sin(ln 4 x ) x . keywords: Stewart5e, 002 (part 1 of 1) 10 points Differentiate the function f ( t ) = 1 + ln t 3 - ln t . 1. f 0 ( t ) = 4 t (3 - ln t ) 2 correct 2. f 0 ( t ) = - 4 t (3 - ln t ) 2 3. f 0 ( t ) = 3 ln t (1 + ln t ) 2 4. f 0 ( t ) = - 3 ln t t (1 + ln t ) 2 5. f 0 ( t ) = 3 t (1 + ln t ) 2 6. f 0 ( t ) = - 4 (3 - ln t ) 2 Explanation: By the Quotient Rule, f 0 ( t ) = (3 - ln t )(1 /t ) + (1 + ln t )(1 /t ) (3 - ln t ) 2 = (3 - ln t ) + (1 + ln t ) t (3 - ln t ) 2 . Consequently, f 0 ( t ) = 4 t (3 - ln t ) 2 . keywords: Stewart5e, 003 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = ln s 2 x 2 - 1 2 x 2 + 1 . 1. f 0 ( x ) = 2 x 2 x 2 - 1 2. f 0 ( x ) = - 2 x 2 x 2 - 1 3. f 0 ( x ) = - 4 x 4 x 4 - 1 4. f 0 ( x ) = - 4 x 2 x 4 - 1 5. f 0 ( x ) = - 2 x 4 x 2 - 1 6. f 0 ( x ) = 4 x 4 x 4 - 1 correct

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Mahon, Kevin – Homework 14 – Due: Dec 1 2005, 3:00 am – Inst: Edward Odell 2 7. f 0 ( x ) = 2 x 4 x 2 - 1 8. f 0 ( x ) = 4 x 2 x 4 - 1 Explanation: Properties of logs ensure that ln s 2 x 2 - 1 2 x 2 + 1 = 1 2 n ln(2 x 2 - 1) - ln(2 x 2 + 1) o . By the Chain rule, therefore, f 0 ( x ) = 1 2 n 4 x 2 x 2 - 1 - 4 x 2 x 2 + 1 o . Consequently, f 0 ( x ) = 4 x 4 x 4 - 1 . keywords: Stewart5e, 004 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 6 · 5 x - 2 x 5 ln 5 . 1. f 0 ( x ) = ln 5(6 · 5 x - 10 x 4 ) correct 2. f 0 ( x ) = log 2 5(6 · 5 x - 10 x 4 ) 3. f 0 ( x ) = ln 5(6 · 5 x - 10 x 5 ) 4. f 0 ( x ) = ln 5(6 · 5 x - 1 - 10 x 4 ) 5. f 0 ( x ) = ln 5(6 · 5 x - 2 x 4 ) Explanation: Note that 5 x = e x ln 5 . By the chain rule, therefore, f 0 ( x ) = 6 · e x ln 5 ln 5 - 10 x 4 ln 5 . Thus f 0 ( x ) = ln 5(6 · 5 x - 10 x 4 ) .
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