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Unformatted text preview: white (taw933) – HW14 – benzvi – (55600) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the derivative of f when f ( θ ) = ln (cos 3 θ ) . 1. f ′ ( θ ) = 1 sin 3 θ 2. f ′ ( θ ) = 3 cot 3 θ 3. f ′ ( θ ) = 3 cos 3 θ 4. f ′ ( θ ) = cot 3 θ 5. f ′ ( θ ) = 3 tan3 θ correct 6. f ′ ( θ ) = 3 tan3 θ Explanation: By the Chain Rule, f ′ ( θ ) = 1 cos(3 θ ) d dθ (cos 3 θ ) = 3 sin 3 θ cos 3 θ . Consequently, f ′ ( θ ) = 3 tan 3 θ . 002 10.0 points Find the slope of the line tangent to the graph of ln( xy ) 2 x = 0 at the point where x = 1. 1. slope = 1 2 e − 2 2. slope = e 2 correct 3. slope = 1 2 e − 2 4. slope = e − 2 5. slope = e 2 6. slope = 1 2 e 2 Explanation: Differentiating implicitly with respect to x we see that 1 xy parenleftBig y + x dy dx parenrightBig 2 = 0 , in which case dy dx = y (1 2 x ) x = e 2 x (1 2 x ) x 2 because, by exponentiation, y = e 2 x x . Consequently, at x = 1, slope = dy dx vextendsingle vextendsingle vextendsingle x =1 = e 2 . 003 10.0 points Find the derivative of f ( t ) = 1 + ln t 4 ln t . 1. f ′ ( t ) = 5 t (4 ln t ) 2 correct 2. f ′ ( t ) = 4 t (1 + ln t ) 2 3. f ′ ( t ) = 4 ln t t (1 + ln t ) 2 4. f ′ ( t ) = 5 t (4 ln t ) 2 5. f ′ ( t ) = 4 ln t (1 + ln t ) 2 6. f ′ ( t ) = 5 (4 ln t ) 2 white (taw933) – HW14 – benzvi – (55600) 2 Explanation: By the Quotient Rule, f ′ ( t ) = (4 ln t )(1 /t ) + (1 + ln t )(1 /t ) (4 ln t ) 2 = (4 ln t ) + (1 + ln t ) t (4 ln t ) 2 . Consequently, f ′ ( t ) = 5 t (4 ln t ) 2 . 004 10.0 points Find the derivative of f when f ( x ) = 2 ln( x radicalbig x 2 3) , ( x > 3) . 1. f ′ ( x ) = 4 √ x 2 3 2. f ′ ( x ) = 4 √ x 2 3 3. f ′ ( x ) = 2 √ x 2 3 4. f ′ ( x ) = 2 √ x 2 3 correct 5. f ′ ( x ) = 1 √ x 2 3 6. f ′ ( x ) = 1 √ x 2 3 Explanation: By the Chain Rule f ′ ( x ) = 2 x √ x 2 3 parenleftBig 1 x √ x 2 3 parenrightBig = 2 √ x 2 3 . 005 10.0 points Determine f ′ ( x ) when f ( x ) = e (3 ln( x 5 )) . 1. f ′ ( x ) = 1 x e 3 ln( x 5 ) 2. f ′ ( x ) = 15 x 14 correct 3. f ′ ( x ) = 15(ln x ) e 3 ln( x 5 ) 4. f ′ ( x ) = e 15 /x 5. f ′ ( x ) = 14 x 15 6. f ′ ( x ) = 3 x 2 e 3 ln( x 5 ) Explanation: Since r ln x = ln x r , e ln x = x , we see that...
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 Spring '08
 schultz
 Derivative, Differential Calculus

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