First, notice that the function is the quotient between two expressions. Then, first we need to use the quotient rule to find the derivative:
[tex]\frac{d}{dx}\frac{g(x)}{h(x)}=\frac{h(x)\cdot\frac{d}{dx}g(x)-g(x)\cdot\frac{d}{dx}h(x)}{h(x)^2}[/tex]
In this case, g(x) = 3x+2 and h(x) = x^3. Then:
[tex]\begin{gathered} \frac{d}{dx}f(x)=\frac{d}{dx}\frac{3x+2}{x^3} \\ =\frac{x^3\cdot\frac{d}{dx}(3x+2)-(3x+2)\cdot\frac{d}{dx}x^3}{(x^3)^2} \end{gathered}[/tex]
Now, notice that the derivatives of 3x+2 and x^3 appear in the numerator. Use the power rule to find the derivative of those expressions:
[tex]\frac{d}{dx}x^n=^{}nx^{n-1}[/tex]
Then:
[tex]\begin{gathered} \frac{d}{dx}(3x+2)=3x^0+0=3\cdot1+0=3+0=3 \\ \\ \frac{d}{dx}x^3=3x^2 \end{gathered}[/tex]
So, the differentiation continues:
[tex]\begin{gathered} \frac{x^3\cdot\frac{d}{dx}(3x+2)-(3x+2)\cdot\frac{d}{dx}x^3}{(x^3)^2} \\ =\frac{x^3\cdot(3)-(3x+2)\cdot(3x^2)}{(x^3)^2} \end{gathered}[/tex]
Finally, simplify the expression:
[tex]\begin{gathered} \frac{x^3\cdot(3)-(3x+2)\cdot(3x^2)}{(x^3)^2} \\ =\frac{3x^3-(3x\cdot3x^2+2\cdot3x^2)}{x^6} \\ =\frac{3x^3-(9x^3+6x^2)}{x^6} \\ =\frac{3x^3-9x^3-6x^2}{x^6} \\ =\frac{-6x^3-6x^2}{x^6} \\ =\frac{-6x-6}{x^4} \\ =\frac{-6(x+1)}{x^4} \end{gathered}[/tex]
Therefore, we used both the power rule and the quotient rule to find the derivative, and the derivative is:
[tex]\frac{d}{dx}\frac{3x+2}{x^3}=-\frac{6(x+1)}{x^4}[/tex]
