To differentiate the given function $y = \sqrt{\frac{x^3 - x}{4 - x^2}}$, we can proceed by applying the chain rule and quotient rule. Let's go through the steps to find $\frac{dy}{dx}$:

Given: $y = \sqrt{\frac{x^3 - x}{4 - x^2}}$

1. Rewrite the square root as a power: $y = \left( \frac{x^3 - x}{4 - x^2} \right)^{\frac{1}{2}}$

2. Apply the chain rule: To differentiate $y = \left( u \right)^{\frac{1}{2}}$, where $u = \frac{x^3 - x}{4 - x^2}$, we get: $\frac{dy}{dx} = \frac{1}{2} \left( \frac{x^3 - x}{4 - x^2} \right)^{-\frac{1}{2}} \cdot \frac{d}{dx}\left( \frac{x^3 - x}{4 - x^2} \right)$

3. Differentiate the inner function using the quotient rule: For $u = \frac{x^3 - x}{4 - x^2}$, we use the quotient rule: $\frac{d}{dx}\left( \frac{f}{g} \right) = \frac{f'g - fg'}{g^2}$.

   • Let $f = x^3 - x$ and $g = 4 - x^2$.
   • Then, $f' = 3x^2 - 1$ and $g' = -2x$.

   Plugging these into the quotient rule: $\frac{d}{dx}\left( \frac{x^3 - x}{4 - x^2} \right) = \frac{(3x^2 - 1)(4 - x^2) - (x^3 - x)(-2x)}{(4 - x^2)^2}$

4. Simplify the derivative: Now we substitute this result back into the chain rule expression: $\frac{dy}{dx} = \frac{1}{2} \left( \frac{x^3 - x}{4 - x^2} \right)^{-\frac{1}{2}} \cdot \frac{(3x^2 - 1)(4 - x^2) + 2x(x^3 - x)}{(4 - x^2)^2}$

5. Express the final result: Putting it all together, we have: $\frac{dy}{dx} = \frac{1}{2} \cdot \frac{(3x^2 - 1)(4 - x^2) + 2x(x^3 - x)}{(4 - x^2)^2} \cdot \left( \frac{x^3 - x}{4 - x^2} \right)^{-\frac{1}{2}}$

This is the derivative of the function. Let me know if you would like further simplification or a step-by-step breakdown of any part.

Would you like additional details or explanations?

Here are five follow-up questions to explore this topic further:

1. How can we simplify the expression further if needed?
2. What would the second derivative of this function look like?
3. Can this function be differentiated using a substitution to simplify the process?
4. How would we approach finding critical points for this function?
5. How does the domain of the function affect the differentiation process?

Tip: Always double-check if there’s a simpler form of the function before differentiating; simplification beforehand can sometimes make the differentiation process easier.