# Exam3 6 - 6 f x = sin e-x-e-x Explanation By the Chain Rule...

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Betancourt, Daniel – Exam 3 – Due: May 1 2007, 11:00 pm – Inst: Diane Radin 6 for all a > 0. But then by properties of limits, lim x →-∞ 5 e 2 x + 1 2 e 2 x - 5 = - 1 5 . Consequently, limit = - 1 5 . keywords: exponential function, limit as in- Fnity 008 (part 1 of 1) 10 points ±ind the value of the derivative of f at x = 1 when f ( x ) = 5 x + e 4 x . 1. f 0 (1) = e 4 2. f 0 (1) = 5 + e 4 3. f 0 (1) = 5 - 4 e 4 4. f 0 (1) = 5 + 4 e 4 correct 5. f 0 (1) = 4 e 4 6. f 0 (1) = 5 - e 4 Explanation: After di²erentiation f 0 ( x ) = 5 + 4 e 4 x . At x = 1, therefore, f 0 (1) = 5 + 4 e 4 . keywords: 009 (part 1 of 1) 10 points Determine f 0 ( x ) when f ( x ) = sin( e - x ) . 1. f 0 ( x ) = - e - x sin( e - x ) 2. f 0 ( x ) = cos( e - x ) 3. f 0 ( x ) = - sin( e - x ) + e - x 4. f 0 ( x ) = e - x cos( e - x ) 5. f 0 ( x ) = - e - x cos( e - x ) correct
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Unformatted text preview: 6. f ( x ) = sin( e-x )-e-x Explanation: By the Chain Rule, f ( x ) = cos( e-x ) d dx e-x . Consequently, f ( x ) =-e-x cos( e-x ) . keywords: 010 (part 1 of 1) 10 points ±ind the derivative of f ( t ) = 4-ln t 1 + ln t . 1. f ( t ) = 5 t (1 + ln t ) 2 2. f ( t ) = ln t t (4-ln t ) 2 3. f ( t ) =-ln t (4-ln t ) 2 4. f ( t ) =-5 t (1 + ln t ) 2 correct 5. f ( t ) = 5 (1 + ln t ) 2 6. f ( t ) =-1 t (4-ln t ) 2 Explanation:...
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