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Math Help - Derivative of abs function

  1. #1
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    Derivative of abs function

    What is the derivative of this function:

    abs(\frac{x}{1+x^2})

    This is the shot I took:

    \frac{x}{1+x^2} and \frac{-x}{1+x^2}
    so \frac{(x-1)(x-1)}{(x^2+1)^2} and \frac{(x+1)(x+1)}{(x^2+1)^2} would be the derivative of those functions respectively

    But I really don't understand the fundamentals of this problem - the absolute value is really throwing me and I don't know how to properly deal with it. Thanks so much if you can help!
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  2. #2
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    \left|\frac{x}{x^2+1}\right| = \frac{x}{x^2+1} for x \ge 0

    \left|\frac{x}{x^2+1}\right| = \frac{-x}{x^2+1} for x < 0


    for x \ge 0 ... f'(x) = \frac{(x^2+1) - x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}

    for x < 0 ... f'(x) = -\frac{1-x^2}{(x^2+1)^2} = \frac{x^2-1}{(x^2+1)^2}
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  3. #3
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    Thanks a lot! I guess what I really needed was a little reassurance (and maybe someone to check my algebra!!)
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