Derivative of (x^2-1)/(3x+1)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = x^{2} - 1$ and $g{\left(x \right)} = 3 x + 1$.

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

   1. Differentiate $x^{2} - 1$ term by term:

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

      2. Apply the power rule: $x^{2}$ goes to $2 x$

      The result is: $2 x$

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

   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: $3$

   Now plug in to the quotient rule:

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

2. Now simplify:

   $$\frac{3 x^{2} + 2 x + 3}{9 x^{2} + 6 x + 1}$$

The answer is:

$$\frac{3 x^{2} + 2 x + 3}{9 x^{2} + 6 x + 1}$$