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M408K Hm. 6 Solutions

M408K Hm. 6 Solutions - Mahon Kevin Homework 6 Due Oct 6...

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Mahon, Kevin – Homework 6 – Due: Oct 6 2005, 3:00 am – Inst: Edward Odell 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 Find the derivative of f when f ( x ) = 5 sec x + 4 cos x . 1. f 0 ( x ) = sin x ( 5 sec 2 x + 4 ) 2. f 0 ( x ) = cos x ( 5 csc 2 x - 4 ) 3. f 0 ( x ) = sin x ( 5 sec 2 x - 4 ) correct 4. f 0 ( x ) = cos x ( 5 csc 2 x + 4 ) 5. f 0 ( x ) = sin x ( 4 - 5 sec 2 x ) 6. f 0 ( x ) = cos x ( 4 - 5 csc 2 x ) Explanation: Since d dx (sec x ) = sec x tan x = sin x sec 2 x while d dx (cos x ) = - sin x , we see that f 0 ( x ) = sin x ( 5 sec 2 x - 4 ) . keywords: Stewart5e, 002 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 2 x - 5 sin x . 1. f 0 ( x ) = 5 - cos x 2. f 0 ( x ) = 5 - sin x 3. f 0 ( x ) = 2 - 5 cos x correct 4. f 0 ( x ) = 2 + 5 cos x 5. f 0 ( x ) = 2 - 5 sin x 6. f 0 ( x ) = 2 + 5 sin x Explanation: Since d dx sin x = cos x , we see that f 0 ( x ) = 2 - 5 cos x . keywords: Stewart5e, 003 (part 1 of 1) 10 points Differentiate y = tan x - 6 sec x . 1. y 0 = cos x + sin x 2. y 0 = 6 cos x + sin x 3. y 0 = 6 cos x - sin x 4. y 0 = cos x - sin x 5. y 0 = cos x + 6 sin x correct Explanation: y = tan x - 6 sec x y 0 = sec x sec 2 x - (tan x - 6) sec x tan x sec 2 x = sec x ( sec 2 x - tan 2 x + 6 tan x ) sec 2 x = 1 + 6 tan x sec x
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Mahon, Kevin – Homework 6 – Due: Oct 6 2005, 3:00 am – Inst: Edward Odell 2 Alternate solution: Simplify y first: y = sin x - 6 cos x y 0 = cos x + 6 sin x keywords: Stewart5e, 004 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 1 + sin x cos x . 1. f 0 ( x ) = 1 sin x - 1 2. f 0 ( x ) = 1 1 + sin x 3. f 0 ( x ) = 1 cos x - 1 4. f 0 ( x ) = - 1 1 + cos x 5. f 0 ( x ) = - 1 sin x + 1 6. f 0 ( x ) = 1 1 - sin x correct 7. f 0 ( x ) = 1 1 - cos x 8. f 0 ( x ) = 1 cos x + 1 Explanation: By the quotient rule, f 0 ( x ) = cos 2 x + sin x (1 + sin x ) cos 2 x = ( cos 2 x + sin 2 x ) + sin x cos 2 x . But sin 2 x + cos 2 x = 1, so f 0 ( x ) = 1 + sin x 1 - sin 2 x = 1 + sin x (1 - sin x )(1 + sin x ) . Consequently, f 0 ( x ) = 1 1 - sin x . keywords: Stewart5e, 005 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 3 tan x + 2 cot x . 1. f 0 ( x ) = 3 - 5 cos 2 x sin 2 x cos 2 x correct 2. f 0 ( x ) = 3 + 5 cos x sin x cos x 3. f 0 ( x ) = 3 + 5 cos 2 x sin 2 x cos 2 x 4. f 0 ( x ) = 3 + 5 sin 2 x sin 2 x cos 2 x 5. f 0 ( x ) = 3 - 5 cos x sin x cos x 6. f 0 ( x ) = 3 - 5 sin 2 x sin 2 x cos 2 x Explanation: After differentiation f 0 ( x ) = 3 sec 2 x - 2 csc 2 x = 3 cos 2 x - 2 sin 2 x = 3 sin 2 x - 2 cos 2 x sin 2 x cos 2 x . Now 3 sin 2 x - 2 cos 2 x = 3 ( 1 - cos 2 x ) - 2 cos 2 x = 3 - 5 cos 2 x .
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Mahon, Kevin – Homework 6 – Due: Oct 6 2005, 3:00 am – Inst: Edward Odell 3 Consequently, f 0 ( x ) = 3 - 5 cos 2 x sin 2 x cos 2 x . keywords: Stewart5e, 006 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = tan x (1 + 8 csc x ) . 1. f 0 ( x ) = sec 2 x (1 + 8 sin x ) correct 2. f 0 ( x ) = csc 2 x (1 + 8 sin x ) 3. f 0 ( x ) = csc 2 x (sin x + 8) 4. f 0 ( x ) = sec 2 x (sin x + 8) 5. f 0 ( x ) = csc 2 x (1 - 8 cos x ) 6. f 0 ( x ) = sec 2 x (1 - 8 sin x ) Explanation: Since d dx tan x = sec 2 x , d dx csc x = - csc x cot x , the Product Rule ensures that f 0 ( x ) = sec 2 x (1 + 8 csc x ) - 8 tan x csc x cot x = sec 2 x + 8 csc x (sec 2 x - 1) .
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