Derivative of ln(1+3*x)/((3*x))

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = \log{\left(3 x + 1 \right)}$ and $g{\left(x \right)} = 3 x$.

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

   1. Let $u = 3 x + 1$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(3 x + 1\right)$:

      1. Differentiate $3 x + 1$ 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: $3$

         The result is: $3$

      The result of the chain rule is:

      $$\frac{3}{3 x + 1}$$

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

   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: $3$

   Now plug in to the quotient rule:

   $\frac{\frac{9 x}{3 x + 1} - 3 \log{\left(3 x + 1 \right)}}{9 x^{2}}$

2. Now simplify:

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

The answer is:

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