Derivative of (x^3-2*x)/(x+2)

1. Apply the quotient rule, which is:

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

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

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

   1. Differentiate $x^{3} - 2 x$ term by term:

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

      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: $-2$

      The result is: $3 x^{2} - 2$

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

   1. Differentiate $x + 2$ term by term:

      1. The derivative of the constant $2$ is zero.

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

      The result is: $1$

   Now plug in to the quotient rule:

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

2. Now simplify:

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

The answer is: