HW12sol - + 1) = 1 -x 2-x . 2. Evaluate the integral. (...

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M408C, Homework 12 — Solution Instructor: Guoliang Wu 1. For each of the following functions, find the domain and then differentiate it. (a) f ( x ) = ln ln ln x Since the domain of ln is x > 0, we must have ln(ln x ) > 0 ln x > 1 x > e . The domain of f ( x ) is ( e, ). By chain rule f 0 ( x ) = 1 ln ln x 1 ln x 1 x . (b) g ( x ) = sin - 1 (2 x + 1) Since the domain of sin - 1 is [ - 1 , 1], we have - 1 2 x + 1 1 - 2 2 x 0 ⇒ - 1 x 0. The domain of g is [ - 1 , 0]. By chain rule, g 0 ( x ) = 1 p 1 - (2 x + 1) 2 · 2 = 2 p 1 - (4 x 2 + 4 x
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Unformatted text preview: + 1) = 1 -x 2-x . 2. Evaluate the integral. ( Hint: Integration by parts, and then long division. ) Z x ln( x + 1) dx. Solution: Let u = ln( x + 1) ,dv = xdx,du = 1 x +1 dx,v = x 2 2 . Z x ln( x + 1) dx = ln( x + 1) x 2 2-Z x 2 2 1 x + 1 dx = x 2 2 ln( x + 1)-1 2 Z ( x-1) + 1 x + 1 dx = x 2 2 ln( x + 1)-1 2 x 2 2-x + ln( x + 1) + C. 1...
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This note was uploaded on 12/03/2010 for the course M 408c taught by Professor Mcadam during the Fall '06 term at University of Texas at Austin.

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