Posted by Angie on Saturday, September 22, 2012 at 10:45am.
assuming you mean
f(x) = √((x^2+1)/(3x+1))
f = u^1/2 where u = (x^2+1)/(3x+1)
so,
f' = 1/2√u u'
u = f/g where
f = x^2+1 and v = 3x+1
so, u' = (f'g-g'f)/g^2 = (3x^2+2x-3)/(3x+1)^2
so f' = 1/[2√((x^2+1)/(3x+1))] * (3x^2+2x-3)/(3x+1)^2
= (3x^2+2x-3)/[2(3x+1)3/2(x^2+1)1/2]
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