[tex]f'(x)=\dfrac{8x^4-2x^2-\left(28x^4+83x^2-3\right)\ln{\left(x^2+3\right)}}{2x^2\left(x^2+3\right)\left(4x^2-1\right)^4}[/tex]
It can work well to consider the function in parts. Define the following:
... a(x) = (1/2)ln(x^2+3)
... b(x) = x(4x^2-1)^3
Then the derivatives of these are ...
... a'(x) = (1/2)·1/(x^2 +3)·2x = x/(x^2+3)
... b'(x) = (4x^2 -1)^3 + 3x(4x^2 -1)^2·8x = (4x^2 -1)^2·(4x^2 -1 +24x^2)
... = (4x^2 -1)^2·(28x^2 -1)
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Putting the parts together
f(x) = a(x)/b(x)
f'(x) = (b(x)a'(x) -a(x)b'(x))/b(x)^2 . . . . . rule for quotient of functions
Substituting values, we have
... f'(x) = (x(4x^2 -1)^3·x/(x^2 +3) -(1/2)ln(x^2 +3)·(4x^2 -1)^2·(28x^2 -1)) / (x(4x^2 -1)^3)^2
We can cancel (4x^2 -1)^2 from numerator and denominator. We can also eliminate fractions (1/2, 1/(x^2+3)). Then we have ...
... f'(x) = 2x^2(4x^2 -1) -(x^2 +3)ln(x^2 +3)·(28x^2 -1)/(2x^2·(x^2 +3)(4x^2 -1)^4))
Simplifying a bit, this becomes ...
... f'(x) = (8x^4 -2x^2 -ln(x^2 +3)·(28x^4 +83x^2 -3))/(2x^2·(x^2 +3)(4x^2 -1)^4))
