Derivative of y=lg(3x+1)

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 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: $3$

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

      The result is: $3$

   The result of the chain rule is:

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

4. Now simplify:

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

The answer is:

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