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# CalcHW7Answers - Villarreal Natalie Homework 7 Due Oct 9...

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Villarreal, Natalie – Homework 7 – Due: Oct 9 2007, 3:00 am – Inst: Louiza Fouli 1 This print-out should have 18 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 ) = 4 x cos 5 x . 1. f 0 ( x ) = 20 cos 5 x + 5 x sin 5 x 2. f 0 ( x ) = 20 cos 5 x - 4 x sin 5 x 3. f 0 ( x ) = 4 cos 5 x + 20 x sin 4 x 4. f 0 ( x ) = 4 cos 4 x - 4 x sin 5 x 5. f 0 ( x ) = 4 cos 5 x - 20 x sin 5 x correct Explanation: Using the formulas for the derivatives of sine and cosine together with the Chain Rule we see that f 0 ( x ) = (4 x ) 0 cos 5 x + 4 x (cos 5 x ) 0 = 4 cos 5 x - 20 x sin 5 x . keywords: derivative, trig function, chain rule 002 (part 1 of 1) 10 points Find the value of f 0 (0) when f ( x ) = (1 - 2 x ) - 5 . Correct answer: 10 . Explanation: Using the chain rule and the fact that ( x α ) 0 = α x α - 1 , we obtain f 0 ( x ) = 10 (1 - 2 x ) 6 . At x = 0, therefore, f 0 (0) = 10 . keywords: derivative, chain rule 003 (part 1 of 1) 10 points Find f 0 ( x ) when f ( x ) = p x 2 + 2 x . 1. f 0 ( x ) = 1 2 ( x + 1) p x 2 + 2 x 2. f 0 ( x ) = 2( x + 1) x 2 + 2 x 3. f 0 ( x ) = ( x + 1) p x 2 + 2 x 4. f 0 ( x ) = 2( x + 1) p x 2 + 2 x 5. f 0 ( x ) = x + 1 x 2 + 2 x correct 6. f 0 ( x ) = x + 1 2 x 2 + 2 x Explanation: By the Chain Rule, f 0 ( x ) = 1 2 x 2 + 2 x (2 x + 2) . Consequently, f 0 ( x ) = x + 1 x 2 + 2 x . keywords: derivative, square root, chain rule 004 (part 1 of 1) 10 points

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Villarreal, Natalie – Homework 7 – Due: Oct 9 2007, 3:00 am – Inst: Louiza Fouli 2 Find f 0 ( x ) when f ( x ) = 1 x 2 - 4 x . 1. f 0 ( x ) = 2 - x ( x 2 - 4 x ) 3 / 2 correct 2. f 0 ( x ) = 2 - x ( x 2 - 4 x ) 1 / 2 3. f 0 ( x ) = x - 2 (4 x - x 2 ) 3 / 2 4. f 0 ( x ) = x - 2 ( x 2 - 4 x ) 3 / 2 5. f 0 ( x ) = 2 - x (4 x - x 2 ) 3 / 2 6. f 0 ( x ) = x - 2 (4 x - x 2 ) 1 / 2 Explanation: By the Chain Rule, f 0 ( x ) = - 1 2( x 2 - 4 x ) 3 / 2 (2 x - 4) . Consequently, f 0 ( x ) = 2 - x ( x 2 - 4 x ) 3 / 2 . keywords: derivative, square root, chain rule 005 (part 1 of 1) 10 points Determine f 0 ( x ) when f ( x ) = 1 - x 1 - 2 x 2 . 1. f 0 ( x ) = 2 x - 1 (1 - 2 x 2 ) 1 / 2 2. f 0 ( x ) = 1 - 2 x (1 - 2 x 2 ) 3 / 2 3. f 0 ( x ) = 1 + 2 x (1 - 2 x 2 ) 3 / 2 4. f 0 ( x ) = 1 - 2 x (1 - 2 x 2 ) 1 / 2 5. f 0 ( x ) = 2 x - 1 (1 - 2 x 2 ) 3 / 2 correct 6. f 0 ( x ) = 1 + 2 x (1 - 2 x 2 ) 1 / 2 Explanation: By the Product and Chain Rules, f 0 ( x ) = - 1 (1 - 2 x 2 ) 1 / 2 + 4 x (1 - x ) 2(1 - 2 x 2 ) 3 / 2 = - (1 - 2 x 2 ) + 2 x (1 - x ) (1 - 2 x 2 ) 3 / 2 . Consequently, f 0 ( x ) = 2 x - 1 (1 - 2 x 2 ) 3 / 2 . (Note: the Quotient Rule could have been used, but it’s simpler to use the Product Rule.) keywords: derivative, quotient rule, chain rule 006 (part 1 of 1) 10 points Find f 0 ( x ) when f ( x ) = 2 cos 2 x + cos 2 x .
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