Derivative of 8ln*(9x+1)

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

   1. Let $u = 9 x + 1$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(9 x + 1\right)$:

      1. Differentiate $9 x + 1$ 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: $9$

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

         The result is: $9$

      The result of the chain rule is:

      $$\frac{9}{9 x + 1}$$

   So, the result is: $\frac{72}{9 x + 1}$

2. Now simplify:

   $$\frac{72}{9 x + 1}$$

The answer is:

$$\frac{72}{9 x + 1}$$