Answer:
[tex]\frac{-1}{3x^2-x}[/tex]
Step-by-step explanation:
using these information:
g(x)=ln2x then g'(x)=[tex]\frac{(2x)'}{2x} =\frac{2}{2x}=\frac{1}{x}[/tex]
h(x)=In(3x - 1) then h'(x)=[tex]\frac{(3x-1)'}{3x-1} =\frac{3}{3x-1}
f'(x)=g'(x) - h'(x) =[tex]\frac{1}{x} - \frac{3}{3x-1} =\frac{-1}{3x^2-x}[/tex]