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Unformatted text preview: MA115 Mathematical Analysis I Test 2 Instructions The test consists of three sections, Part 1: Questions 1 – 6 Part 2: Questions 7 – 10 Part 3: Bonus Question You are to do all of Part 1. For Part 2, select and complete 2 of the 4 problems given. Please clearly indicate which two problems you choose. The two problems you indicate will be the only problems from Part 2 graded. Part 3 consists of one bonus question worth 5 points. Part 3 is optional; if you do not do it it will not count against you. Show your work. This does not mean your answers should be long—we just need you to tell us how you obtained your answer in addition to what you obtained. Any response to a question consisting of just an answer will be marked incorrect. Please type your answers up as you do your homework, and submit your completed test via email by 9PM Wednesday, July 28, to the course instructor and the two teaching assistants. You are allowed to use your textbook and all of the notes from the course. You are not permitted to use a graphing calculator or computer graphing program. Good luck! 1 1. (10 points) The graph of the green graph a is the only one whose tangent lines have both positive and negative slopes. The red graph c must be its derivative since c is the only graph to have both positive and negative values. In turn, all tangent lines to the red graph have positive slopes, so the blue graph b must be the derivative of c since it is always positive. (The green graph a too is always positive, but we have already determined that it is not the derivative of c .) So a = f , c = f , and b = f 00 . 2. (15 points) (a) Using the chain rule with “inside” function ln x 2 and “outside” function sin x we get f ( x ) = cos(ln x 2 ) d dx (ln x 2 ) To find the derivative of ln x 2 we use the chain rule again, this time with “inside” function x 2 and “outside” function ln x , getting a final answer of f ( x ) = cos(ln x 2 ) 1 x 2 d dx ( x 2 ) = cos(ln x 2 ) 1 x 2 (2 x ) = 2 cos(ln x 2 ) x (b) Using the quotient rule with numerator function ln x and denominator function x 2 + 1 we get g ( x ) = ( x 2 + 1) d dx (ln x ) (ln x ) d dx ( x 2 + 1) ( x 2 + 1) 2 = ( x 2 + 1) ( 1 x ) (ln x )(2 x ) ( x 2 + 1) 2 = x 2 + 1 2 x 2 ln x x ( x 2 + 1) 2 (c) The first few derivatives of h ( x ) = xe x can be found quickly using the product rule, and a pattern emerges: h ( x ) = ( x 1) e x h 00 ( x ) = ( x 2) e x h 000 ( x ) = ( x 3) e x h (4) = ( x 4) e x . . ....
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This note was uploaded on 04/07/2008 for the course MA 115 taught by Professor Mahalanobis during the Fall '08 term at Stevens.
 Fall '08
 Mahalanobis
 Math

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