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

1. Let $u = \frac{1 - x}{x + 1}$.

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

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

   1. Apply the quotient rule, which is:

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

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

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

      1. Differentiate $1 - x$ 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: $-1$

         The result is: $-1$

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

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

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

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

         The result is: $1$

      Now plug in to the quotient rule:

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

   The result of the chain rule is:

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

4. Now simplify:

   $$\frac{2}{x^{2} - 1}$$

The answer is:

$$\frac{2}{x^{2} - 1}$$