Derivative of y=(ln3x)/x^3

1. Apply the quotient rule, which is:

   $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

   $f{\left(x \right)} = \log{\left(3 x \right)}$ and $g{\left(x \right)} = x^{3}$.

   To find $\frac{d}{d x} f{\left(x \right)}$:

   1. Let $u = 3 x$.

   2. The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$.

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 3 x$:

      1. The derivative of a constant times a function is the constant times the derivative of the function.

         1. Apply the power rule: $x$ goes to $1$

         So, the result is: $3$

      The result of the chain rule is:

      $$\frac{1}{x}$$

   To find $\frac{d}{d x} g{\left(x \right)}$:

   1. Apply the power rule: $x^{3}$ goes to $3 x^{2}$

   Now plug in to the quotient rule:

   $\frac{- 3 x^{2} \log{\left(3 x \right)} + x^{2}}{x^{6}}$

2. Now simplify:

   $$\frac{1 - 3 \log{\left(3 x \right)}}{x^{4}}$$

The answer is:

$$\frac{1 - 3 \log{\left(3 x \right)}}{x^{4}}$$