Derivative of y=(x²-4)²/(2x²+1)³

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = \left(x^{2} - 4\right)^{2}$ and $g{\left(x \right)} = \left(2 x^{2} + 1\right)^{3}$.

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

   1. Let $u = x^{2} - 4$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x^{2} - 4\right)$:

      1. Differentiate $x^{2} - 4$ term by term:

         1. The derivative of the constant $-4$ 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:

      $$2 x \left(2 x^{2} - 8\right)$$

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

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

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

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

      1. Differentiate $2 x^{2} + 1$ 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: $4 x$

         The result is: $4 x$

      The result of the chain rule is:

      $$12 x \left(2 x^{2} + 1\right)^{2}$$

   Now plug in to the quotient rule:

   $\frac{- 12 x \left(x^{2} - 4\right)^{2} \left(2 x^{2} + 1\right)^{2} + 2 x \left(2 x^{2} - 8\right) \left(2 x^{2} + 1\right)^{3}}{\left(2 x^{2} + 1\right)^{6}}$

2. Now simplify:

   $$\frac{4 x \left(13 - x^{2}\right) \left(x^{2} - 4\right)}{\left(2 x^{2} + 1\right)^{4}}$$

The answer is:

$$\frac{4 x \left(13 - x^{2}\right) \left(x^{2} - 4\right)}{\left(2 x^{2} + 1\right)^{4}}$$