Derivation of f(x):
[tex]f'(x) = \frac{g(x)*8x-4x^2*g'(x)}{(g(x))^2} -\frac{1}{x+2}[/tex]
Since g(3) = 6 and g'(3) = 3
[tex]f'(3) = \frac{g(3)*8*3-4(3)^2*g'(3)}{(g(3))^2} -\frac{1}{3+2} \\f'(3) = \frac{6*8*3-4*9*3}{36} -\frac{1}{5} \\f'(3) = \frac{36}{36} -\frac{1}{5} =1-\frac{1}{5} =\frac{4}{5}\\ f'(3) = 0.8[/tex]
Thus f'(3) = 0.8
Hope that helps!
A couple of identities I used in derivation:
[tex]\frac{d}{dx} (ln x) = \frac{1}{x}[/tex]
