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

1. Apply the quotient rule, which is:

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

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

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

   1. 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: $-2$

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

   1. Let $u = x^{2} + 1$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x^{2} + 1\right)$:

      1. Differentiate $x^{2} + 1$ term by term:

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

         2. Apply the power rule: $x^{2}$ goes to $2 x$

         The result is: $2 x$

      The result of the chain rule is:

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

   Now plug in to the quotient rule:

   $\frac{4 x^{2} \left(2 x^{2} + 2\right) - 2 \left(x^{2} + 1\right)^{2}}{\left(x^{2} + 1\right)^{4}}$

2. Now simplify:

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

The answer is:

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