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Unformatted text preview: Version 023 – Final Exam – Radin – (56520) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the value of f ′ (4) when f ( x ) = x 3 / 2 + 8 x 1 / 2 . 1. f ′ (4) = 4 2. f ′ (4) = 5 correct 3. f ′ (4) = 9 2 4. f ′ (4) = 6 5. f ′ (4) = 11 2 Explanation: Since d dx x r = rx r − 1 , we see that f ′ ( x ) = 3 2 x 1 / 2 + 4 x − 1 / 2 . At x = 4, therefore, f ′ (4) = 5 . 002 10.0 points Find f ′ ( x ) when f ( x ) = 2 − x 1 + x . 1. f ′ ( x ) = 1 (1 + x ) 2 2. f ′ ( x ) = − 3 (1 + x ) 2 correct 3. f ′ ( x ) = 3 (1 + x ) 2 4. f ′ ( x ) = − 2 (1 + x ) 2 5. f ′ ( x ) = − 1 (1 + x ) 2 6. f ′ ( x ) = 2 (1 + x ) 2 Explanation: By the Quotient Rule f ′ ( x ) = − (1 + x ) − (2 − x ) (1 + x ) 2 . Consequently, f ′ ( x ) = − 3 (1 + x ) 2 . 003 10.0 points Find the derivative of g when g ( x ) = 8 sec x + tan x . 1. g ′ ( x ) = 8 sec x tan x + 1 − tan 2 x 2. g ′ ( x ) = 8 sec x tan x + sec 2 x correct 3. g ′ ( x ) = 8 sin x sec 2 x − cos 2 x 4. g ′ ( x ) = 8 cos x sec 2 x + 1 − sec x 5. g ′ ( x ) = 8 cos x sec 2 x + 1 + tan x Explanation: When g ( x ) = 8 sec x + tan x its derivative is given by g ′ ( x ) = 8 sec x tan x + sec 2 x . 004 10.0 points Version 023 – Final Exam – Radin – (56520) 2 Find the value of f ′ (0) when f ( x ) = 1 2 e 3 x + 1 3 e − 2 x . 1. f ′ (0) = 4 3 2. f ′ (0) = 5 6 correct 3. f ′ (0) = 1 3 4. f ′ (0) = 7 6 5. f ′ (0) = 1 Explanation: By the Chain rule, f ′ ( x ) = 3 2 e 3 x − 2 3 e − 2 x . Consequently, f ′ (0) = 5 6 . 005 10.0 points Calculate f ′ (ln7) when f ( x ) = ln ( √ 7 + e x ) . 1. f ′ (ln7) = 1 8 2. f ′ (ln7) = 1 2 3. f ′ (ln7) = − 1 4 4. f ′ (ln7) = 1 4 correct 5. f ′ (ln7) = 1 12 Explanation: By properties of logarithms, f ( x ) = 1 2 ln parenleftBig 7 + e x parenrightBig . Using the Chain Rule, we now see that df dx = e x 2(7 + e x ) . Thus, at x = ln 7, f ′ (ln7) = 7 2(7 + 7) = 1 4 since e ln x = x . 006 10.0 points Find the derivative of f ( x ) = sin − 1 ( e 3 x ) . 1. f ′ ( x ) = e 3 x 1 + e 6 x 2. f ′ ( x ) = 3 e 3 x 1 + e 6 x 3. f ′ ( x ) = 3 1 + e 6 x 4. f ′ ( x ) = e 3 x √ 1 − e 6 x 5. f ′ ( x ) = 1 1 + e 6 x 6. f ′ ( x ) = 3 √ 1 − e 6 x 7. f ′ ( x ) = 1 √ 1 − e 6 x 8. f ′ ( x ) = 3 e 3 x √ 1 − e 6 x correct Explanation: Since d dx sin − 1 x = 1 √ 1 − x 2 , d dx e ax = ae ax , Version 023 – Final Exam – Radin – (56520) 3 the Chain Rule ensures that f ′ ( x ) = 3 e 3 x √ 1 − e 6 x . 007 10.0 points Determine dy/dx when 4 cos x sin y = 1 ....
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This note was uploaded on 06/05/2010 for the course MATH 57420 taught by Professor Radin during the Fall '09 term at University of Texas.
 Fall '09
 RAdin

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