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Unformatted text preview: Lightsey, Steven – Exam 2 – Due: Nov 3 2005, 11:00 pm – Inst: Daniel Allcock 1 This printout should have 18 questions. Multiplechoice 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 f ( x ) when f ( x ) = ‡ x 2 x + 1 · 2 . 1. f ( x ) = 6( x + 1) ( x 1) 3 2. f ( x ) = 6( x + 2) ( x 1) 3 3. f ( x ) = 6( x 2) ( x + 1) 3 correct 4. f ( x ) = 4( x 2) ( x + 1) 3 5. f ( x ) = 4( x 1) ( x + 1) 3 6. f ( x ) = 4( x + 1) ( x 1) 3 Explanation: By the Chain and Quotient Rules, f ( x ) = 2 ‡ x 2 x + 1 · ( x + 1) ( x 2) ( x + 1) 2 . Consequently, f ( x ) = 6( x 2) ( x + 1) 3 . keywords: Stewart5e, derivative, quotient rule, chain rule 002 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = tan x 3 + 2 sec x . 1. f ( x ) = sec 2 x (3 + 2 sec x tan x ) 2 2. f ( x ) = 3 sin x + 2 cos x (3 cos x + 2) 2 3. f ( x ) = 3 + 2 cos x (3 cos x + 2) 2 correct 4. f ( x ) = 3 sec x + 2 (3 + 2 sec x ) 2 5. f ( x ) = sec 2 x 3 + 2 sec x tan x 6. f ( x ) = 3 cos x + 2 (3 cos x + 2) 2 Explanation: Now, tan x 3 + 2 sec x = sin x cos x 3 + 2 cos x = sin x 3 cos x + 2 . Thus f ( x ) = cos x 3 cos x + 2 + 3 sin 2 x (3 cos x + 2) 2 = cos x (3 cos x + 2) + 3 sin 2 x (3 cos x + 2) 2 . Consequently, f ( x ) = 3 + 2 cos x (3 cos x + 2) 2 , since cos 2 x + sin 2 x = 1. keywords: Stewart5e, 003 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 4 x sin 5 x + 4 5 cos 5 x. 1. f ( x ) = 20 x cos 5 x correct 2. f ( x ) = 20 x cos 5 x + 8 sin 5 x Lightsey, Steven – Exam 2 – Due: Nov 3 2005, 11:00 pm – Inst: Daniel Allcock 2 3. f ( x ) = 20 x cos 5 x 8 sin 5 x 4. f ( x ) = 20 cos 5 x 5. f ( x ) = 20 x cos 5 x Explanation: Since d dx sin x = cos x, d dx cos x = sin x, it follows that f ( x ) = 4 sin 5 x + 20 x cos 5 x 4 sin 5 x. Consequently, f ( x ) = 20 x cos 5 x . keywords: Stewart5e, derivative, product rule, chain rule, trig functions 004 (part 1 of 1) 10 points Find the equation of the tangent line to the graph of 2 y 2 xy 3 = 0 , at the point P = (5 , 3). 1. 2 y + x = 1 2. 7 y = 3 x + 6 correct 3. 2 y = x + 1 4. 7 y + 3 x = 6 5. 5 y = 3 x Explanation: Differentiating implicitly with respect to x we see that 4 y dy dx y x dy dx = 0 , so dy dx = y 4 y x . At P = (5 , 3), therefore, dy dx fl fl fl P = 3 7 . Thus by the point slope formula, the equation of the tangent line at P is given by y 3 = 3 7 ( x 5) . Consequently, 7 y = 3 x + 6 . keywords: Stewart5e, 005 (part 1 of 1) 10 points The figure below shows the graphs of three functions: t One is the graph of the position function s of a car, one is its velocity v , and one is its acceleration a . Identify which graph goes with which function....
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This note was uploaded on 10/29/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz

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