The problem asks to find the derivative of the function $g(x) = \frac{x^3 - x}{x^2 + 1}$ using the quotient rule. The quotient rule is given by:

$\frac{d}{dx} \left( \frac{f(x)}{h(x)} \right) = \frac{f'(x)h(x) - f(x)h'(x)}{(h(x))^2}$

Here, the numerator is $f(x) = x^3 - x$ and the denominator is $h(x) = x^2 + 1$.

Step-by-step solution:

1. Find $f'(x)$: $f(x) = x^3 - x$ $f'(x) = 3x^2 - 1$

2. Find $h'(x)$: $h(x) = x^2 + 1$ $h'(x) = 2x$

3. Apply the quotient rule: $g'(x) = \frac{(3x^2 - 1)(x^2 + 1) - (x^3 - x)(2x)}{(x^2 + 1)^2}$

4. Simplify the numerator: First, expand $(3x^2 - 1)(x^2 + 1)$: $(3x^2 - 1)(x^2 + 1) = 3x^4 + 3x^2 - x^2 - 1 = 3x^4 + 2x^2 - 1$

   Now, expand $(x^3 - x)(2x)$: $(x^3 - x)(2x) = 2x^4 - 2x^2$

   Subtract the two expressions: $3x^4 + 2x^2 - 1 - (2x^4 - 2x^2) = 3x^4 + 2x^2 - 1 - 2x^4 + 2x^2 = x^4 + 4x^2 - 1$

5. Final expression: $g'(x) = \frac{x^4 + 4x^2 - 1}{(x^2 + 1)^2}$

Thus, the derivative of $g(x)$ is:

$g'(x) = \frac{x^4 + 4x^2 - 1}{(x^2 + 1)^2}$

Would you like further clarification or details?

Here are 5 related questions to extend this concept:

1. How does the quotient rule differ from the product rule?
2. What happens if the function had a higher degree polynomial in both the numerator and the denominator?
3. Can you apply the chain rule to find the derivative of $g(x)$?
4. What would the derivative of the same function look like if calculated without the quotient rule?
5. How would you interpret the derivative $g'(x)$ in terms of the slope of the function?

Tip: When simplifying the results from the quotient rule, expand terms carefully and combine like terms step by step to avoid errors.