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Math Help - find the derivative of the function..

  1. #1
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    find the derivative of the function..

    ok so here is the problem http://www.flickr.com/photos/23409594@N02/4277189853/ needless to say my answer is not what the book has and i would like to know why. for starts their answer has a natural log in it? i separated the natural log in the problem and took the derivative using the product rule. so where did i go wrong? well take a look at the problem and tell me what you think. sorry its problem number 55.
    Last edited by slapmaxwell1; January 15th 2010 at 04:12 PM. Reason: wrong link for photo
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  2. #2
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    Quote Originally Posted by slapmaxwell1 View Post
    ok so here is the problem homework problem on Flickr - Photo Sharing! needless to say my answer is not what the book has and i would like to know why. for starts their answer has a natural log in it? i separated the natural log in the problem and took the derivative using the product rule. so where did i go wrong? well take a look at the problem and tell me what you think. sorry its problem number 55.


    show your attempt so maybe someone can spot any mistake(s).
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    ok here is my work, it is broken into two parts, the paper i was using was a little long..anways here it is..my final answer was -4/(stuff) you will see it on the paper. thanks in advance.. problem number 55 001 on Flickr - Photo Sharing! and here is part 2 of the same problem problem number 55 002 on Flickr - Photo Sharing!
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    y = \frac{1}{2}\left(\frac{1}{2} \ln{\frac{x+1}{x-1}} + \arctan{x}\right)

    y = \frac{1}{2}\left[\frac{1}{2} \ln(x+1) - \frac{1}{2}\ln(x-1) + \arctan{x}\right]

    y' = \frac{1}{2}\left[\frac{1}{2(x+1)} - \frac{1}{2(x-1)} + \frac{1}{1+x^2}\right]
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    Quote Originally Posted by skeeter View Post
    y = \frac{1}{2}\left(\frac{1}{2} \ln{\frac{x+1}{x-1}} + \arctan{x}\right)

    y = \frac{1}{2}\left[\frac{1}{2} \ln(x+1) - \frac{1}{2}\ln(x-1) + \arctan{x}\right]

    y' = \frac{1}{2}\left[\frac{1}{2(x+1)} - \frac{1}{2(x-1)} + \frac{1}{1+x^2}\right]

    yes that is essentially what i was doing..i think i got carried away towards the end as i was trying to match what the back of the book's answer. thanks again!
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