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

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

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

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

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

      1. Apply the quotient rule, which is:

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

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

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

         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: $1$

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

         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: $-1$

         Now plug in to the quotient rule:

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

      The result of the chain rule is:

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

   So, the result is: $- \frac{2}{\left(1 - x\right) \left(x + 1\right)}$

2. Now simplify:

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

The answer is: