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Posted by on Tuesday, January 17, 2012 at 10:35am.

y =ln((x^2+1)/(x^2-1))^1/2

Find the first derivative for this ln square division.

  • calculus - , Tuesday, January 17, 2012 at 3:28pm

    y =ln((x^2+1)/(x^2-1))^1/2
    y = 1/2 ln((x^2+1)/(x^2-1))
    y = 1/2(ln(x^2+1) - ln(x^2-1))

    y' = 1/2 (2x/(x^2+1) - 2x/(x^2-1))
    y' = x/(x^2+1) - x/(x^2-1)
    y' = -2x/(x^4-1)

    Now, if you're a glutton for punishment, just apply the chain rule:

    y =ln((x^2+1)/(x^2-1))^1/2
    y' = (1/2)/[(x^2+1)/(x^2-1)]^(-1/2) * [(2x)(x^2-1) - 2x(x^2+1)]/(x^2-1)^2

    I'm not!

  • calculus - , Tuesday, January 17, 2012 at 3:32pm

    Hmm. I seem to have misread the problem. You have the square root of the log?

    In that case,

    y' = -2x/[(x^4-1)*ln((x^2+1)/(x^2-1))^1/2]

    go to wolframalpha and show steps

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