Derivative of (x/(2x-1))*ln(3x)

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 $g{\left(x \right)} = 2 x - 1$.

   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. Differentiate $2 x - 1$ term by term:

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

      2. 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: $2$

      The result is: $2$

   Now plug in to the quotient rule:

   $\frac{- 2 x \log{\left(3 x \right)} + \left(2 x - 1\right) \left(\log{\left(3 x \right)} + 1\right)}{\left(2 x - 1\right)^{2}}$

2. Now simplify:

   $$\frac{2 x - \log{\left(x \right)} - \log{\left(3 \right)} - 1}{4 x^{2} - 4 x + 1}$$

The answer is: