2140_Test_4__Solutions_Fall_07

2140_Test_4__Solutions_Fall_07 - Review for Test 4: 1....

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Review for Test 4: Solution Key 1. Find the slope of the tangent line to the graph of the following function at x=2. 8 ) 2 ( 4 ) 2 ( 6 ) 2 ( 4 6 ) ( 2 4 3 ) ( 2 = - = - = + - = f f x x f x x x f 2. Find the x values where the following function has a horizontal tangent: ) 1 )( 2 ( 0 2 3 0 2 3 ) ( 2 2 2 3 3 3 1 ) ( 7 2 2 3 3 1 ) ( 2 2 2 2 3 - - = + - = + - = + - = + + - = x x x x x x x f x x x f x x x x f x – 2 = 0 or x – 1 = 0 x = 2 or x=1 3. Find the derivative of the following functions. (a.) x e y 3 - = x e y 3 - = (b.) ) ln( 2 x y - = x y 2 - = (c) ) ln( 4 x y = x y x y 4 ) ln( 4 = = 4. Use basic rules of differentiation to find the derivatives of each function below. a. 2 3 3 4 2 y x x = - + - 2 8 6 y x x x = -
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b. 2 3 y x = First change to exponent form: 2 3 3 6 y x y x - - = = - c. 3 5 y x = Change to 1 3 5 y x = ( 29 ( 29 3 1 3 3 2 3 1 5 3 5 3 y x y x - - = = d. 1 y x = change to: 1 2 y x - = ( 29 ( 29 1 2 2 2 3 2 1 2 1 2 y x y x - - - = - = - 5. Use the quotient rule to find the derivative function for the function below. DO NOT SIMPLIFY
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This note was uploaded on 04/23/2008 for the course MATH 2140 taught by Professor Geving during the Fall '07 term at Belmont.

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2140_Test_4__Solutions_Fall_07 - Review for Test 4: 1....

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