quiz5sol - Math 221 Summer 2008 C.Gmez o D.Milovich Quiz...

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Unformatted text preview: Math 221 Summer 2008 C.Gmez o D.Milovich Quiz 5 (Solutions) August 4, 2008 1. [5 points] Find the following antiderivative. 1 x 1- Don't get confused by the notation: 1 x 1- Substitute u = log3 x = 1 x 1 - log2 x 3 ln x ln 3 . dx log2 x 3 dx = log2 x 3 dx x ln 3 , 1 x 1 - (log3 x)2 so dx x dx. Then du = = (ln 3)du. Therefore, dx = ln 3 du = (ln 3) sin-1 u + C = (ln 3) sin-1 log3 x + C. 1 - u2 2. [5 points] Find the line tangent to y = (sin x)cos x at ( , 1). 2 By definition ab = eb ln a , so y = (sin x)cos x = e(cos x) ln sin x . Therefore, the quickest method of solution is logarithmic differentiation. ln y = (cos x) ln sin x 1 dy = (cos x) y dx 1 m = cos 1 2 m = (0) 1 1 1 sin x 1 sin 2 cos x + (ln sin x)(- sin x) cos + ln sin 2 2 - sin 2 0 + (ln 1)(-1) = 0 + 0(-1) = 0 So, the tangent line is described by y = 1 + 0(x - /2), or just y = 1. ...
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This note was uploaded on 08/08/2008 for the course MATH 221 taught by Professor Denissou during the Summer '07 term at University of Wisconsin.

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