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

1. Apply the quotient rule, which is:

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

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

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

   1. Let $u = x + 1$.

   2. Apply the power rule: $\sqrt{u}$ goes to $\frac{1}{2 \sqrt{u}}$

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

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

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

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

         The result is: $1$

      The result of the chain rule is:

      $$\frac{1}{2 \sqrt{x + 1}}$$

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

   1. Let $u = 1 - x$.

   2. Apply the power rule: $\sqrt{u}$ goes to $\frac{1}{2 \sqrt{u}}$

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

      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: $-1$

      The result of the chain rule is:

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

   Now plug in to the quotient rule:

   $\frac{\frac{\sqrt{1 - x}}{2 \sqrt{x + 1}} + \frac{\sqrt{x + 1}}{2 \sqrt{1 - x}}}{1 - x}$

2. Now simplify:

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

The answer is: