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HW3 Solutions - ASSIGNMENT 3 SOLUTIONS Problem 1 We are...

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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 . Problem 3. Consider the first function. By the generalized power rule, we have d dx ( 1 - x 2 ) = 1 2 (1 - x 2 ) - 1 2 ( - 2 x ) = - x 1 - x 2 Using the quotient rule, we obtain: d 2 dx 2 ( 1 - x 2 ) = - d dx x 1 - x 2 = - x 1 - x 2 - x ( 1 - x 2 ) 1 - x 2 We have already computed ( 1 - x 2 ) . Therefore d 2 dx 2 ( 1 - x 2 ) = - 1 - x 2 + x 2 1 - x 2 1 - x 2 = - 1 - x 2 + x 2 (1 - x 2 ) 1 - x 2 = - 1 (1 - x 2 ) 1 - x 2 . Now consider the second function. By the quotient rule, we have: d dx 2 x x - 1 = (2 x ) ( x - 1) - 2 x ( x - 1) ( x - 1) 2 = 2( x - 1) - 2 x ( x - 1) 2 = - 2 ( x - 1) 2 Hence, by the generalized power rule, we get
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