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December 20, 2014

December 20, 2014

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

Find the first derivative for this ln square division.

- calculus -
**Steve**, Tuesday, January 17, 2012 at 3:28pmy =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 -
**Steve**, Tuesday, January 17, 2012 at 3:32pmHmm. 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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