Exam3 7 - -1 ( e x ) . 1. f ( x ) = 3 1 + e 2 x 2. f ( x )...

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Betancourt, Daniel – Exam 3 – Due: May 1 2007, 11:00 pm – Inst: Diane Radin 7 By the Quotient Rule, f 0 ( t ) = - (1 + ln t )(1 /t ) + (4 - ln t )(1 /t ) (1 + ln t ) 2 = - (1 + ln t ) + (4 - ln t ) t (1 + ln t ) 2 . Consequently, f 0 ( t ) = - 5 t (1 + ln t ) 2 . keywords: derivative, Quotient Rule, log function, 011 (part 1 of 1) 10 points Determine the derivative of f ( x ) = 5 · 2 x - 2 x 2 ln 2 . 1. f 0 ( x ) = (5 · 2 x - 1 - 4 x 2 ) ln 2 2. f 0 ( x ) = 10 · 2 x - 1 - 4 x ln 2 3. f 0 ( x ) = 5 · 2 x - 1 - 2 x ln 2 4. f 0 ( x ) = (5 · 2 x - 4 x ) ln 2 correct 5. f 0 ( x ) = 5 x · 2 x - 1 - 2 x ln 2 6. f 0 ( x ) = (5 · 2 x - 2 x ) ln 2 Explanation: Note Frst that 2 x = e x ln 2 . By the Chain Rule, therefore, f 0 ( x ) = 5 · e x ln 2 ln 2 - 4 x ln 2 . Consequently, f 0 ( x ) = (5 · 2 x - 4 x ) ln 2 . keywords: derivative, general base Chain Rule, 012 (part 1 of 1) 10 points ±ind the derivative of f ( x ) = 3 sin
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Unformatted text preview: -1 ( e x ) . 1. f ( x ) = 3 1 + e 2 x 2. f ( x ) = e x 1 + e 2 x 3. f ( x ) = e x 1-e 2 x 4. f ( x ) = 3 e x 1 + e 2 x 5. f ( x ) = 3 1-e 2 x 6. f ( x ) = 1 1-e 2 x 7. f ( x ) = 3 e x 1-e 2 x correct 8. f ( x ) = 1 1 + e 2 x Explanation: Since d dx sin-1 x = 1 1-x 2 , d dx e ax = ae ax , the Chain Rule ensures that f ( x ) = 3 e x 1-e 2 x . keywords: 013 (part 1 of 1) 10 points ind the value of lim x 3 tan 3 x + x 4 tan 5 x-5 x . 1. limit = 11 15...
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.

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