Derivative of (x-2)(ln(x-1)-ln(x+1))

1. Apply the product rule:

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

   $f{\left(x \right)} = x - 2$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Differentiate $x - 2$ term by term:

      1. Apply the power rule: $x$ goes to $1$

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

      The result is: $1$

   $g{\left(x \right)} = \log{\left(x - 1 \right)} - \log{\left(x + 1 \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $\log{\left(x - 1 \right)} - \log{\left(x + 1 \right)}$ term by term:

      1. Let $u = x - 1$.

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

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

         1. Differentiate $x - 1$ term by term:

            1. Apply the power rule: $x$ goes to $1$

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

            The result is: $1$

         The result of the chain rule is:

         $$\frac{1}{x - 1}$$

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

         1. Let $u = x + 1$.

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

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

            1. Differentiate $x + 1$ term by term:

               1. Apply the power rule: $x$ goes to $1$

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

               The result is: $1$

            The result of the chain rule is:

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

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

      The result is: $- \frac{1}{x + 1} + \frac{1}{x - 1}$

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

2. Now simplify:

   $$\frac{2 x + \left(x - 1\right) \left(x + 1\right) \left(\log{\left(x - 1 \right)} - \log{\left(x + 1 \right)}\right) - 4}{\left(x - 1\right) \left(x + 1\right)}$$

The answer is: