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Unformatted text preview: Mahon, Kevin Exam 2 Due: Nov 3 2005, 11:00 pm Inst: Edward Odell 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 + 2) ( x 1) 3 2. f ( x ) = 6( x 2) ( x + 1) 3 correct 3. f ( x ) = 6( x + 1) ( x 1) 3 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 + 7 sec x . 1. f ( x ) = sec 2 x 3 + 7 sec x tan x 2. f ( x ) = 3 sec x + 7 (3 + 7 sec x ) 2 3. f ( x ) = 3 + 7 cos x (3 cos x + 7) 2 correct 4. f ( x ) = 3 cos x + 7 (3 cos x + 7) 2 5. f ( x ) = 3 sin x + 7 cos x (3 cos x + 7) 2 6. f ( x ) = sec 2 x (3 + 7 sec x tan x ) 2 Explanation: Now, tan x 3 + 7 sec x = sin x cos x 3 + 7 cos x = sin x 3 cos x + 7 . Thus f ( x ) = cos x 3 cos x + 7 + 3 sin 2 x (3 cos x + 7) 2 = cos x (3 cos x + 7) + 3 sin 2 x (3 cos x + 7) 2 . Consequently, f ( x ) = 3 + 7 cos x (3 cos x + 7) 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 ) = x sin 4 x + 1 4 cos 4 x. 1. f ( x ) = 4 x cos 4 x 2. f ( x ) = 4 x cos 4 x + 2 sin 4 x Mahon, Kevin Exam 2 Due: Nov 3 2005, 11:00 pm Inst: Edward Odell 2 3. f ( x ) = 4 cos 4 x 4. f ( x ) = 4 x cos 4 x correct 5. f ( x ) = 4 x cos 4 x 2 sin 4 x Explanation: Since d dx sin x = cos x, d dx cos x = sin x, it follows that f ( x ) = sin 4 x + 4 x cos 4 x sin 4 x. Consequently, f ( x ) = 4 x cos 4 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 y 2 xy 6 = 0 , at the point P = (1 , 3). 1. 4 y = 3 x + 9 2. 5 y = 3 x + 12 correct 3. 4 y + 3 x = 9 4. 5 y + 3 x = 12 5. y = 3 x Explanation: Differentiating implicitly with respect to x we see that 2 y dy dx y x dy dx = 0 , so dy dx = y 2 y x . At P = (1 , 3), therefore, dy dx fl fl fl P = 3 5 . Thus by the point slope formula, the equation of the tangent line at P is given by y 3 = 3 5 ( x 1) . Consequently, 5 y = 3 x + 12 . 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 04/15/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz
 Differential Calculus

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