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