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Unformatted text preview: ASSIGNMENT 3  SOLUTIONS Problem 1. We are told that g (2) = h (2) = 0. The product rule tells us that f (2) = g (2) h (2) + g (2) h (2) hence f (2) = 0. Problem 2. For the derivative of the first function we apply the product rule, and then the generalized power rule: ( x x 2 + 1 ) = x x 2 + 1 + x ( x 2 + 1 ) = x 2 + 1 + x 1 2 ( x 2 + 1) 1 2 (2 x ) = x 2 + 1 + x 2 x 2 + 1 = ( x 2 + 1) + x 2 x 2 + 1 = 2 x 2 + 1 x 2 + 1 The second function is actually (2 x 2 1) 6 , because of the even exponent, so ( (2 x 2 1) 6 ) = 6(2 x 2 1)(2 x 2 1) = 24 x (2 x 2 1) . For the third function, the quotient rule seems like the way to go. Fortunately, we can simplify: x 16 1 x 4 + 1 = ( x 8 1)( x 8 + 1) x 4 + 1 = ( x 4 1)( x 4 + 1)( x 8 + 1) x 4 + 1 = ( x 4 1)( x 8 + 1) = x 12 x 8 + x 4 1 Now x 16 1 x 4 + 1 = 12 x 11 8 x 7 + 4 x 3 ....
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This note was uploaded on 07/31/2011 for the course MATH 102 taught by Professor Maryamnamazi during the Spring '10 term at University of Victoria.
 Spring '10
 MARYAMNAMAZI
 Calculus, Logic, Derivative, Product Rule

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