The given function is:

$y = \frac{e^{x^4}}{\sqrt{2 - x^3}}.$

We need to differentiate $y$ with respect to $x$ to find $\frac{dy}{dx}$. We'll use both the quotient rule and chain rule to handle this differentiation.

Step 1: Apply the quotient rule

The quotient rule states:

$\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2},$

where $f(x) = e^{x^4}$ and $g(x) = \sqrt{2 - x^3}$.

Step 2: Differentiate the numerator $f(x) = e^{x^4}$

Using the chain rule, the derivative of $e^{x^4}$ is:

$f'(x) = \frac{d}{dx} e^{x^4} = e^{x^4} \cdot \frac{d}{dx}(x^4) = e^{x^4} \cdot 4x^3.$

Step 3: Differentiate the denominator $g(x) = \sqrt{2 - x^3}$

First, express $g(x) = (2 - x^3)^{1/2}$. Using the chain rule, we differentiate as follows:

$g'(x) = \frac{1}{2} (2 - x^3)^{-1/2} \cdot \frac{d}{dx} (2 - x^3) = \frac{1}{2} (2 - x^3)^{-1/2} \cdot (-3x^2).$

Thus:

$g'(x) = -\frac{3x^2}{2 \sqrt{2 - x^3}}.$

Step 4: Apply the quotient rule

Now substitute into the quotient rule formula:

$\frac{dy}{dx} = \frac{e^{x^4} \cdot 4x^3 \cdot \sqrt{2 - x^3} - e^{x^4} \cdot \left( -\frac{3x^2}{2 \sqrt{2 - x^3}} \right)}{(2 - x^3)}.$

Step 5: Simplify the result

The expression becomes:

$\frac{dy}{dx} = \frac{e^{x^4}}{(2 - x^3)} \left( 4x^3 \sqrt{2 - x^3} + \frac{3x^2}{2 \sqrt{2 - x^3}} \right).$

This is the derivative of $y$ with respect to $x$.

Do you need further details or clarifications on any step?

Here are 5 related questions you might find useful:

1. How does the chain rule apply to exponential functions like $e^{x^4}$?
2. What are some common mistakes made when applying the quotient rule?
3. Can the quotient rule be avoided by simplifying the expression beforehand?
4. What are the limitations of using the quotient rule in more complex functions?
5. How can you check if the derivative is correctly simplified?

Tip: Always double-check your derivatives by confirming each step, especially when using multiple rules like the quotient and chain rule together!