Derivative of f(x)=1/(8x+1)+6×

1. Differentiate $6 x + \frac{1}{8 x + 1}$ term by term:

   1. Let $u = 8 x + 1$.

   2. Apply the power rule: $\frac{1}{u}$ goes to $- \frac{1}{u^{2}}$

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(8 x + 1\right)$:

      1. Differentiate $8 x + 1$ 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: $8$

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

         The result is: $8$

      The result of the chain rule is:

      $$- \frac{8}{\left(8 x + 1\right)^{2}}$$

   4. 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: $6$

   The result is: $6 - \frac{8}{\left(8 x + 1\right)^{2}}$

2. Now simplify:

   $$6 - \frac{8}{\left(8 x + 1\right)^{2}}$$

The answer is:

$$6 - \frac{8}{\left(8 x + 1\right)^{2}}$$