How to get first derivative? My question is to find the first derivative using product and/or quotient rule
$$f(x) = {(x^2+1)(x^2-2) \over 3x+2}.$$
The solution to the problem is:
 $$f'(x) = {[(2x)(x^2-2)+(x^2+1)(2x)](3x+2)-(3)[(x^2+1)(x^2-2)] \over (3x+2)^2}.$$
I'm having problems getting the same solution would someone be able to help me by showing me the correct way of getting the first derivative of this question

When you have expressions which just contains products, quotients and powers, logarithmic differentiation makes life easier.
For your case
$$f = {(x^2+1)(x^2-2) \over 3x+2}\implies \log(f)=\log(x^2+1)+\log(x^2-2)-\log(3x+2)$$ Differentiate both sides
$$\frac{f'}f=\frac{2x}{x^2+1}+\frac{2x}{x^2-2}-\frac 3{3x+2}=\frac{9 x^4+8 x^3-3 x^2-4 x+6 }{(x^2+1 )(x^2-2 ) (3x+2 ) }$$
Now 
$$f'=f \times \frac{f'}f={(x^2+1)(x^2-2) \over 3x+2}\times \frac{9 x^4+8 x^3-3 x^2-4 x+6 }{(x^2+1 )(x^2-2 ) (3x+2 ) }$$ Now, simplify.
