FA06Ma.9Hw10solutions

FA06Ma.9Hw10solutions - Math 1a, Section 1 Solutions to...

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Unformatted text preview: Math 1a, Section 1 Solutions to Homework 10 1. 6.17. 21, 22, 31 p249 (a) Given f ( x ) = x x . Express as f ( x ) = exp ( x log x ). Therefore, f ( x ) = exp ( x log x )(log x + 1) (1) (b) Given f ( x ) = (1 + x )(1 + e x 2 ). The derivative is f ( x ) = (1 + e x 2 ) + (1 + x ) e x 2 2 x (2) (c) Given f ( x ) = (sin x ) cos x + (cos x ) sin x . Express f ( x ) = e cos x log sin x + e sin x log cos x (3) Then the derivative is given by f ( x ) = e cos x log sin x (- sin x log sin x + cos 2 x sin x ) + e sin x log cos x (cos x log cos x- sin 2 x cos x ) (4) 2. 6.26.2 p268 By the fundamental theorem of calculus, 2 f ( x ) f ( x ) = f ( x ) sin x 2 + cos x (5) Therefore, when f ( x ) 6 = 0, f ( x ) = sin x 2(2 + cos x ) (6) Therefore, f ( x ) = Z sin xdx 2(2 + cos x ) = Z- d cos x 2(2 + cos x ) =- 1 2 log(2 + cos x ) + C (7) Finally, note that f 2 (0) = 0, hence C = 1 2 log(3) 3. 6.26.13 Instead of the way it is in the book, we write p ( x ) = d + d 1 x + d 2 2 x 2 and f ( x ) = e x p ( x ). First, we differentiate)....
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FA06Ma.9Hw10solutions - Math 1a, Section 1 Solutions to...

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