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cal_final

cal_final - Cantu Corina Final 1 Due 10:00 pm Inst...

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Cantu, Corina – Final 1 – Due: May 10 2006, 10:00 pm – Inst: Castravet 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 ) = 2 ln t 3 + ln t . 1. f 0 ( t ) = 2 ln t 3 + ln t 2. f 0 ( t ) = 6 ln t (3 + ln t ) 2 3. f 0 ( t ) = 2 t (3 + ln t ) 2 4. f 0 ( t ) = 2 t (3 + ln t ) 2 5. f 0 ( t ) = 6 t (3 + ln t ) 2 correct 6. f 0 ( t ) = 6 ln t 3 + ln t Explanation: By the Quotient Rule, f 0 ( t ) = 2(3 + ln t )(1 /t ) - (2 ln t )(1 /t ) (3 + ln t ) 2 . Consequently, f 0 ( t ) = 6 t (3 + 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 ) = 5 x cos 3 x - 6 sin 3 x . 1. f 0 ( x ) = - 15 x sin 3 x - 13 cos 3 x correct 2. f 0 ( x ) = 18 cos 3 x - 13 x sin 3 x 3. f 0 ( x ) = - 18 cos 3 x + 15 x sin 3 x 4. f 0 ( x ) = 15 x sin 3 x - 13 cos 3 x 5. f 0 ( x ) = - 15 x sin 3 x - 18 cos 3 x Explanation: Using formulas for the derivatives of sine and cosine together with the Product and Chain Rules, we see that f 0 ( x ) = 5 cos 3 x - 15 x sin 3 x - 18 cos 3 x = - 15 x sin 3 x - 13 cos 3 x . keywords: Stewart5e, 003 (part 1 of 1) 10 points Determine f 0 ( x ) when f ( x ) = sin( e - x ) . 1. f 0 ( x ) = e - x cos( e - x ) 2. f 0 ( x ) = - e - x sin( e - x ) 3. f 0 ( x ) = - sin( e - x ) + e - x 4. f 0 ( x ) = - e - x cos( e - x ) correct 5. f 0 ( x ) = cos( e - x ) 6. f 0 ( x ) = sin( e - x ) - e - x Explanation: By the Chain Rule, f 0 ( x ) = cos( e - x ) d dx e - x . Consequently, f 0 ( x ) = - e - x cos( e - x ) .

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Cantu, Corina – Final 1 – Due: May 10 2006, 10:00 pm – Inst: Castravet 2 keywords: Stewart5e, 004 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 2 ln s 1 - 2 x 2 1 + 2 x 2 . 1. f 0 ( x ) = 4 x 1 - 2 x 4 2. f 0 ( x ) = 8 x 1 - 2 x 4 3. f 0 ( x ) = - 4 x 1 - 4 x 4 4. f 0 ( x ) = - 8 x 1 - 4 x 4 correct 5. f 0 ( x ) = - 8 x 1 - 2 x 4 6. f 0 ( x ) = - 4 x 1 - 2 x 4 7. f 0 ( x ) = 4 x 1 - 4 x 4 8. f 0 ( x ) = 8 x 1 - 4 x 4 Explanation: Properties of logs ensure that 2 ln s 1 - 2 x 2 1 + 2 x 2 = ln(1 - 2 x 2 ) - ln(1 + 2 x 2 ) . By the Chain rule, therefore, f 0 ( x ) = - 4 x 1 - 2 x 2 - 4 x 1 + 2 x 2 . Consequently, f 0 ( x ) = - 8 x 1 - 4 x 4 . keywords: Stewart5e, 005 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 5 tan - 1 ( e 4 x ) . 1. f 0 ( x ) = 20 e 4 x 16 + e 4 x 2. f 0 ( x ) = 5 1 + e 4 x 3. f 0 ( x ) = 4 e 4 x 16 + e 8 x 4. f 0 ( x ) = 5 16 + e 8 x 5. f 0 ( x ) = 4 e 4 x 1 + e 4 x 6. f 0 ( x ) = 20 e 4 x 1 + e 8 x correct Explanation: By the Chain Rule, f 0 ( x ) = 4 · 5 e 4 x 1 + e 8 x since d dx tan - 1 x = 1 1 + x 2 , ( e 4 x ) 2 = e 8 x . Consequently, f 0 ( x ) = 20 e 4 x 1 + e 8 x . keywords: Stewart5e, 006 (part 1 of 1) 10 points Determine dy/dx when 3 cos x sin y = 7 . 1. dy dx = tan x
Cantu, Corina – Final 1 – Due: May 10 2006, 10:00 pm – Inst: Castravet 3 2. dy dx = tan xy 3. dy dx = cot x tan y 4. dy dx = tan x tan y correct 5. dy dx = cot x cot 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 → -∞ 2 + 3 x - 5 x 3 1 + 2 x 3 exists, and if it does, find its value.

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cal_final - Cantu Corina Final 1 Due 10:00 pm Inst...

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