Derivative of 2*x/(1-x^2)

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

   1. Apply the quotient rule, which is:

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

      $f{\left(x \right)} = x$ and $g{\left(x \right)} = 1 - x^{2}$.

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

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

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

      1. Differentiate $1 - x^{2}$ 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: $x^{2}$ goes to $2 x$

            So, the result is: $- 2 x$

         The result is: $- 2 x$

      Now plug in to the quotient rule:

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

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

2. Now simplify:

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

The answer is:

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