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final3 - Granillo Yvette Final 1 Due 10:00 pm Inst E...

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Granillo, Yvette – Final 1 – Due: May 10 2006, 10:00 pm – Inst: E Schultz 1 This print-out should have 25 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 ( t ) = ln t 4 + ln t . 1. f 0 ( t ) = 4 ln t 4 + ln t 2. f 0 ( t ) = 4 t (4 + ln t ) 2 correct 3. f 0 ( t ) = ln t 4 + ln t 4. f 0 ( t ) = t (4 + ln t ) 2 5. f 0 ( t ) = t (4 + ln t ) 2 6. f 0 ( t ) = 4 ln t (4 + ln t ) 2 Explanation: By the Quotient Rule, f 0 ( t ) = (4 + ln t )(1 /t ) - (ln t )(1 /t ) (4 + ln t ) 2 . Consequently, f 0 ( t ) = 4 t (4 + ln t ) 2 . keywords: Stewart5e, derivative, Quotient Rule, log function, 002 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 3 x cos 2 x - 4 sin 2 x . 1. f 0 ( x ) = - 8 cos 2 x + 6 x sin 2 x 2. f 0 ( x ) = - 6 x sin 2 x - 8 cos 2 x 3. f 0 ( x ) = - 6 x sin 2 x - 5 cos 2 x correct 4. f 0 ( x ) = 8 cos 2 x - 5 x sin 2 x 5. f 0 ( x ) = 6 x sin 2 x - 5 cos 2 x Explanation: Using formulas for the derivatives of sine and cosine together with the Product and Chain Rules, we see that f 0 ( x ) = 3 cos 2 x - 6 x sin 2 x - 8 cos 2 x = - 6 x sin 2 x - 5 cos 2 x . keywords: Stewart5e, 003 (part 1 of 1) 10 points Determine f 0 ( x ) when f ( x ) = cos( e - x ) . 1. f 0 ( x ) = - sin( e - x ) 2. f 0 ( x ) = - e - x sin( e - x ) 3. f 0 ( x ) = e - x sin( e - x ) correct 4. f 0 ( x ) = e - x cos( e - x ) 5. f 0 ( x ) = cos( e - x ) + e - x 6. f 0 ( x ) = - cos( e - x ) - e - x Explanation: By the Chain Rule, f 0 ( x ) = sin( e - x ) d dx e - x . Consequently, f 0 ( x ) = e - x sin( e - x ) .
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Granillo, Yvette – Final 1 – Due: May 10 2006, 10:00 pm – Inst: E Schultz 2 keywords: Stewart5e, 004 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 1 2 ln s 2 + x 2 2 - x 2 . 1. f 0 ( x ) = x 2 - x 4 2. f 0 ( x ) = x 4 - x 4 3. f 0 ( x ) = - x 2 - x 4 4. f 0 ( x ) = 2 x 4 - x 4 correct 5. f 0 ( x ) = 2 x 2 - x 4 6. f 0 ( x ) = - 2 x 2 - x 4 7. f 0 ( x ) = - 2 x 4 - x 4 8. f 0 ( x ) = - x 4 - x 4 Explanation: Properties of logs ensure that 1 2 ln s 2 + x 2 2 - x 2 = 1 4 n ln(2 + x 2 ) - ln(2 - x 2 ) o . By the Chain rule, therefore, f 0 ( x ) = 1 4 n 2 x 2 + x 2 + 2 x 2 - x 2 o . Consequently, f 0 ( x ) = 2 x 4 - x 4 . keywords: Stewart5e, 005 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 4 tan - 1 ( e 3 x ) . 1. f 0 ( x ) = 4 9 + e 6 x 2. f 0 ( x ) = 3 e 3 x 9 + e 6 x 3. f 0 ( x ) = 3 e 3 x 1 + e 3 x 4. f 0 ( x ) = 12 e 3 x 9 + e 3 x 5. f 0 ( x ) = 4 1 + e 3 x 6. f 0 ( x ) = 12 e 3 x 1 + e 6 x correct Explanation: By the Chain Rule, f 0 ( x ) = 3 · 4 e 3 x 1 + e 6 x since d dx tan - 1 x = 1 1 + x 2 , ( e 3 x ) 2 = e 6 x . Consequently, f 0 ( x ) = 12 e 3 x 1 + e 6 x . keywords: Stewart5e, 006 (part 1 of 1) 10 points Determine dy/dx when 3 cos x sin y = 5 . 1. dy dx = tan x
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Granillo, Yvette – Final 1 – Due: May 10 2006, 10:00 pm – Inst: E Schultz 3 2. dy dx = tan xy 3. dy dx = cot x cot y 4. dy dx = tan x tan y correct 5. dy dx = cot x tan y Explanation: Differentiating implicitly with respect to x we see that 3 n cos x cos y dy dx - sin y sin x o = 0 . Thus dy dx cos x cos y = sin x sin y . Consequently, dy dx = sin x sin y cos x cos y = tan x tan y . keywords: Stewart5e, implicit differentiation, trig functions, product rule 007 (part 1 of 1) 10 points Determine if the limit lim x → -∞ 3 + x + 2 x 3 2 - 5 x 3 exists, and if it does, find its value.
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