Derivative of log3(3x)/3^2x-1

1. Differentiate $x \frac{\frac{1}{\log{\left(3 \right)}} \log{\left(3 x \right)}}{9} - 1$ term by term:

   1. Apply the quotient rule, which is:

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

      $f{\left(x \right)} = x \log{\left(3 x \right)}$ and $g{\left(x \right)} = 9 \log{\left(3 \right)}$.

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

      1. Apply the product rule:

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

         $f{\left(x \right)} = x$; to find $\frac{d}{d x} f{\left(x \right)}$:

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

         $g{\left(x \right)} = \log{\left(3 x \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

         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:

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

         The result is: $\log{\left(3 x \right)} + 1$

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

      1. The derivative of the constant $9 \log{\left(3 \right)}$ is zero.

      Now plug in to the quotient rule:

      $\frac{\log{\left(3 x \right)} + 1}{9 \log{\left(3 \right)}}$

   2. The derivative of the constant $-1$ is zero.

   The result is: $\frac{\log{\left(3 x \right)} + 1}{9 \log{\left(3 \right)}}$

The answer is:

$$\frac{\log{\left(3 x \right)} + 1}{9 \log{\left(3 \right)}}$$