calculus
posted by wimpkid .
y =ln((x^2+1)/(x^21))^1/2
Find the first derivative for this ln square division.

y =ln((x^2+1)/(x^21))^1/2
y = 1/2 ln((x^2+1)/(x^21))
y = 1/2(ln(x^2+1)  ln(x^21))
y' = 1/2 (2x/(x^2+1)  2x/(x^21))
y' = x/(x^2+1)  x/(x^21)
y' = 2x/(x^41)
Now, if you're a glutton for punishment, just apply the chain rule:
y =ln((x^2+1)/(x^21))^1/2
y' = (1/2)/[(x^2+1)/(x^21)]^(1/2) * [(2x)(x^21)  2x(x^2+1)]/(x^21)^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^41)*ln((x^2+1)/(x^21))^1/2]
go to wolframalpha and show steps
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