Derivative of ((x-1)/(x+1))^4

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

2. Apply the power rule: $u^{4}$ goes to $4 u^{3}$

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

   1. Apply the quotient rule, which is:

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

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

      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 $x + 1$ 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:

   $$\frac{8 \left(x - 1\right)^{3}}{\left(x + 1\right)^{2} \left(x + 1\right)^{3}}$$

4. Now simplify:

   $$\frac{8 \left(x - 1\right)^{3}}{\left(x + 1\right)^{5}}$$

The answer is: